One of the major problems that is to be solved in an electronic circuit design is the production of low voltage DC power supply from Mains to power the circuit. The conventional method is the use of a step-down transformer to reduce the 230 V AC to a desired level of low voltage AC. The most simple, space saving and low cost method is the use of a Voltage Dropping Capacitor in series with the phase line.

Selection of the dropping capacitor and the circuit design requires some technical knowledge and practical experience to get the desired voltage and current. An ordinary capacitor will not do the job since the device will be destroyed by the rushing current from the mains. Mains spikes will create holes in the dielectric and the capacitor will fail to work. X-rated capacitor specified for the use in AC mains is required for reducing AC voltage.

## Schematic of the Capacitor Power Supply Circuit

**X Rated capacitor 400 Volt**

Before selecting the dropping capacitor, it is necessary to understand the working principle and the operation of the dropping capacitor. The X rated capacitor is designed for 250, 400, 600 VAC. Higher voltage versions are also available. The Effective Impedance (Z), Rectance (X) and the mains frequency (50 – 60 Hz) are the important parameters to be considered while selecting the capacitor. The reactance (X) of the capacitor (C) in the mains frequency (f) can be calculated using the formula:

X = 1 / (2 ¶ fC )

For example the reactance of a 0.22µF capacitor running in the mains frequency 50Hz will be:

X = 1 / {2 ¶ x 50 x 0.22 x( 1 / 1,000,000) } = 14475.976 Ohms 0r 14.4 Kilo ohms.

Rectance of the capacitor 0.22 uF is calculated as **X = 1/2Pi*f*C**

Where **f** is the 50 Hz frequency of mains and **C** is the value of capacitor in Farads. That is 1 microfarad is 1/1,000,000 farads. Hence 0.22 microfarad is 0.22 x 1/1,000,000 farads. Therefore the rectance of the capacitor appears as 14475.97 Ohms or 14.4 K Ohms.To get current I divide mains Volt by the rectance in kilo ohm.That is 230 / 14.4 = 15.9 mA.

Effective impedance (Z) of the capacitor is determined by taking the load resistance (R) as an important parameter. Impedance can be calculated using the formula:

Z = √ R + X

Suppose the current in the circuit is I and Mains voltage is V then the equation appears like:

I = V / X

The final equation thus becomes:

I = 230 V / 14. 4 = 15.9 mA.

Therefore if a 0.22 uF capacitor rated for 230 V is used, it can deliver around 15 mA current to the circuit. But this is not sufficient for many circuits. Therefore it is recommended to use a 470 nF capacitor rated for 400 V for such circuits to give required current.

**X Rated AC capacitors – 250V, 400V, 680V AC**

**Table showing the X rated capacitor types and the output voltage and current without load**

**Rectification**

Diodes used for rectification should have sufficient Peak inverse voltage (PIV). The peak inverse voltage is the maximum voltage a diode can withstand when it is reverse biased. 1N4001 diode can withstand up to 50 Volts and 1N4007 has a toleration of 1000 Volts. The important characteristics of general purpose rectifier diodes are given in the table.

So a suitable option is a rectifier diode 1N4007. Usually a silicon diode has a Forward voltage drop of 0.6 V. The current rating (Forward current) of rectifier diodes also vary. Most of the general purpose rectifier diodes in the 1N series have 1 ampere current rating.

**DC Smoothing**

A Smoothing Capacitor is used to generate ripple free DC. Smoothing capacitor is also called Filter capacitor and its function is to convert half wave / full wave output of the rectifier into smooth DC. The power rating and the capacitance are two important aspects to be considered while selecting the smoothing capacitor. The power rating must be greater than the off load output voltage of the power supply.

The capacitance value determines the amount of ripples that appear in the DC output when the load takes current. For example, a full wave rectified DC output obtained from 50Hz AC mains operating a circuit that is drawing 100 mA current will have a ripple of 700 mV peak-to-peak in the filter capacitor rated 1000 uF.

The ripple that appears in the capacitor is directly proportional to the load current and is inversely proportional to the capacitance value. It is better to keep the ripple below 1.5 V peak-to-peaks under full load condition. So a high value capacitor (1000 uF or 2200 uF) rated 25 volts or more must be used to get a ripple free DC output. If ripple is excess it will affect the functioning of the circuit especially RF and IR circuits.

**Voltage Regulation**

Zener diode is used to generate a regulated DC output. A Zener diode is designed to operate in the reverse breakdown region. If a silicon diode is reverse biased, a point reached where its reverse current suddenly increases. The voltage at which this occurs is known as “Avalanche or Zener“ value of the diode. Zener diodes are specially made to exploit the avalanche effect for use in ‘Reference voltage ‘regulators.

A Zener diode can be used to generate a fixed voltage by passing a limited current through it using the series resistor (R). The Zener output voltage is not seriously affected by R and the output remains as a stable reference voltage. But the limiting resistor R is important, without which the Zener diode will be destroyed. Even if the supply voltage varies, R will take up any excess voltage. The value of R can be calculated using the formula:

R = Vin – Vz / Iz

Where **Vin **is the input voltage,** Vz** output voltage and **Iz **current through the Zener

In most circuits, Iz is kept as low as 5mA. If the supply voltage is 18V, the voltage that is to be dropped across R to get 12V output is 6volts. If the maximum Zener current allowed is 100 mA, then R will pass the maximum desired output current plus 5 mA .

So the value of R appears as:

R = 18 – 12 / 105 mA = 6 / 105 x 1000 = 57 ohms

Power rating of the Zener is also an important factor to be considered while selecting the Zener diode. According to the formula **P = IV**. P is the power in watts, I current in Amps and V, the voltage. So the maximum power dissipation that can be allowed in a Zener is the Zener voltage multiplied by the current flowing through it. For example, if a 12V Zener passes 12 V DC and 100 mA current, its power dissipation will be 1.2 Watts. So a Zener diode rated 1.3W should be used.

**LED Indicator**

LED indicator is used as power on indicator. A significant voltage drop (about 2 volts) occurs across the LED when it passes forward current. The forward voltage drops of various LEDs are shown in Table.

A typical LED can pass 30 –40 mA current without destroying the device. Normal current that gives sufficient brightness to a standard Red LED is 20 mA. But this may be 40 mA for Blue and White LEDs. A current limiting resistor is necessary to protect LED from excess current that is flowing through it. The value of this series resistor should be carefully selected to prevent damage to LED and also to get sufficient brightness at 20 mA current. The current limiting resistor can be selected using the formula:

R = V / I

Where **R** is the value of resistor in ohms, **V** is the supply voltage and **I** is the allowable current in Amps. For a typical Red LED, the voltage drop is 1.8 volts. So if the supply voltage is 12 V (Vs), voltage drop across the LED is 1.8 V (Vf) and the allowable current is 20 mA (If) then the value of the series resistor will be

**Vs – Vf / If = 12 – 1.8 / 20 mA = 10.2 / 0.02 A = 510 Ohms.**

A suitable available value of resistor is 470 Ohms. But is advisable to use 1 K resistor to increase the life of the LED even though there will be a slight reduction in the brightness. Since the LED takes 1.8 volts, the output voltage will be 2 volts less than the value of Zener. So if the circuit requires 12 volts, it is necessary to increase the value of Zener to 15 volts. Table given below is a ready reckoner for selecting limiting resistor for various versions of LEDs at different voltages.

**Circuit Diagram**

The diagram shown below is a simple transformer less power supply. Here 225 K(2.2uF) 400 volts X rated capacitor is used to drop 230 volt AC. Resistor R2 is the bleeder resistor that remove the stored current from the capacitor when the circuit is unplugged. Without R2, there is chance for fatal shock if the circuit is touched. Resistor R1 protects the circuit from inrush current at power on. A full wave rectifier comprising D1 through D4 is used to rectify the low voltage AC from the capacitor C1 and C2 removes ripples from the DC. With this design, around 24 volts at 100 mA current will be available at the output.This 24 volt DC can be regulated to required output voltage using a suitable 1 watt Zener. It is better to add a **safety fuse** in the phase line and an MOV across the phase and neutral lines as safety measure if there is voltage spike or short circuit in the mains.

Caution:Construction of this form of power supply is recommended only to those persons experienced or competent in handling AC mains. So do not try this circuit if you are not experienced in handling High voltages.

**The drawback of the Capacitor power supply includes**

**No galvanic isolation**from Mains.So if the power supply section fails, it can harm the gadget.**Low current output**. With a Capacitor power supply. Maximum output current available will be 100 mA or less.So it is not ideal to run heavy current inductive loads.- Output voltage and current will not be stable if the AC input varies.

**Caution**

Great care must be taken while testing the power supply using a dropping resistor. Do not touch at any points in the PCB since some points are at mains potential. Even after switching off the circuit, avoid touching the points around the dropping capacitor to prevent electric shock. Extreme care should be taken to construct the circuit to avoid short circuits and fire. Sufficient spacing must be given between the components.

The high value smoothing capacitor will explode, if is connected in the reverse polarity. The dropping capacitor is non-polarized so that it can be connected either way round. The power supply unit must be isolated from the remaining part of the circuit using insulators. The circuit should be housed in metal case without touching any part of the PCB in the metal case. The metal case should be properly earthed.

Was just surfing the net for circuits and got on to this post..

I must say i have never understood any transformerless supply circuits ever…but in this post the whole concept got clear and now i may make one by the guidance of this post…

Thank you for such a deep and proper explanation which is in simple language

hi ,i just want to know when in circuit AC wire connects to capacitor and other to diode 1N5408 in a series or bridge of 3 which transistor we can attach at the end and then it connects to loop of wire and small magnet place over loop of wire.

Thanks for the feedback.I will post more articles in easy to understand way so that even a new comer can make use of it

Capacitor Power Supply:

is it use in the circuites with variable current(200ma-10ma)?

I have Transformer 12volt 40 ampire

How i can make a circuit for this transformer.

How much volt capacitor & how much ampire diode should i use for this.

Please help me…

How can I correct the power factor of such a circuit easily?

Your artical is very good for power supply circuit.

Please guide me which value of AC capacitor is to be use in my circuit as, I want, from 230 Volt,A.C., 50 C/S to 50 volt A/C, for 30 ma current circuit and after connect bridge rectifier to convert DC. Please also guide capacitor calculation, selection guide, bidder resistor ect.

Sir with all due respects can i have an circuit of 12 vdc to 600vac 5ma inverter circuit sir.

DC at this voltage is not that difficult–can be accomplished via a fly-back or boost type converter.

Generation of AC is a difficult job due to the requirement of a very special transformer.

i like your article

thanks for this……….

sir, I have 105j capasitor, Can I reduce the current outpput to 20ma..I want use it to circuit them for led’s lamp

member

dear Mr Mohankumar

I have circuited led lamp with 60 leds, I made it in 3 string. each string has 20 leds , I use A 105j . as you said, if I make the circuit to 3 string it will deliver for about 23ma each string,

Sir., How much watt (power) my circuit is? and how do I calculate for some circuit if I make with different number of leds?

Dear sir i assemble the same circuit, but the bleeder resistor and neutral resistor is burned out. so please help me to define how to choose the bleeder resistor value.

Hello Sir,

I’m using 24 VDC 150 Watt DC motor. I have 24 VDC power supply with 240 Watt capacity. When I turn ON this motor, power supply get trip due to inrush current.

I get to know to avoid this should use one capacitor in parallel to motor and it will avoid this tripping.

Is it right ? How to decide capacitor rating for this application ?

I need converter 24 v dc to 12 v dc 60 w who can help me

Anup panchal: the neutral resistor is rated 1W, that mean Imax = 100mA. The circuit is dessigned to supply 140mA max. When the load draws 140mA, power dissipated by R1 will be (0.14)(0.14)(100) = 1.96W. Of course, it will burn out.

If the load will draw a higher current, say 200mA, the output voltage will drop to compromise as the circuit has been designed to supply a maximum current of 140mA only. When the current is so high that the power dissipated by the bleeder resistor is higher than 1W, the bleeder resistor will burn out as well.

If the connected load draws 140mA or less, all the components should be fine.

Dear sir,

I am a non technical person but verymuch interested in electronics. I have not understood how the reactance for 0.22 uF capacitor

under 50 hz.A.C. mains is coming to 14.4 ohms. I am getting

by the formula X = 14462.98 ohms. Kindly explain. Thanks in advance. V.G.Krishna, Hyderabad.

Rectance of the capacitor 0.22 uF is calculated as X=1/2Pi.f.C

Where f is the 50 Hz frequency of mains and C is the value of capacitor in Farads. That is 1 microfarad is 1 / 1,000,000 farads.Hence 0.22 microfarad is 0.22 x 1 / 1,000,000 farads. Therefore the rectance of the capacitor appears as 14475.97 Ohms or 14.4 K Ohms.

To get current I divide mains Volt by the rectance in kilo ohm

That is 230 / 14.4 = 15.9 mA.

The detailed explanation is added in the writeup

Thank you Sir, for your kind prompt reply.

Your explaination read as,

Rectance of the capacitor 0.22 uF is calculated as X=1/2Pi.f.C

Where f is the 50 Hz frequency of mains and C is the value of capacitor in Farads. That is 1 microfarad is 1 / 1,000,000 farads.Hence 0.22 microfarad is 0.22 x 1 / 1,000,000 farads. Therefore the rectance of the capacitor appears as 14475.97 Ohms or 14.4 K Ohms.

To get current I divide mains Volt by the rectance in kilo ohm

That is 230 / 14.4 = 15.9 mA.

My question is what is voltage drop by this capacitor if connected to 230 volt, 50 c/s A/C supply?

hi sir, how to select a varistor for 230v and if 120v mains are available and how to read a varistor. any table or easy explaination.

this circuit is OK.But how to increase the current for up to 400mA.because in this circuit is giving only 100mA. please reply how to change the circuit to get the 400mA current

Hi, here is a transformerless power supply with 30 volts and 1000 mA output.

thanks…guys…please send the details of the circuit ..details of components…… ..is there any other circuit……..

Dear Sir,

Could you please let me know the value of 5k capacitor in uf?

I’ll be very thankful to you.

Regards

Atir

I think you are asking for 5 PF capacitor. 5 PF is 0.000005 UF.Letter K is usually used in High voltage capacitors and it means + / – 10% tolerance

CAP TOLERANCE

C = +/- 0.25PF

D = +/- 0.5PF

F = +/- 1%

G +/- 2%

J +/- 5%

K +/- 10%

M +/-20%

Z +80% / -20%

sir,

I have some confusion. The function of capacitor C1 is to only provide series reactance to step down AC mains, it is nothing to do with energy storage. if i simply put one high watt resistor, will it work for the same purpose.

Please elaborate the significance of capacitor C1.

If we use a high watt resistor in series with AC, current will be reduced but huge quantity of heat will be generated and energy will be lost due to heat. Capacitor power supply eliminates the chance of heat generation and it gives some isolation from mains, until the dielectric of the capacitor breaks due to short circuit or high voltage.

hi there, what a great article! covers just about everything. well written and very clear. thanks a lot. dee.

Thanks for the feedback

SIR,THANK YOU VERY MUCH FOR YOUR SIMPLE YET VERY USEFUL CIRCUIT.I HAVE TESTED IT AND WORKING GREAT.I HAVE LEARNT MANY THINGS STUDYING YOUR POSTS.YOU ARE DOING GREAT SIR.I HAVE A CONFUSION REGARDING POLYESTER FILM CAPACITORS WHICH SAY 2G104J AND 2A104J.BOTH ARE GREEN COLORED.HOW TO ASCERTAIN THE VOLTAGE RATING OF SUCH CAPACITORS?WHAT ABOUT A BROWN COLORED ONE?

What is the device of the MOV?

MOV is Metal Oxide Varister, a kind of component used to protect circuits from AC spikes. If the input AC increases above the limit, it will short circuit and break the fuse

Thank you…

Here you can read more about the MOV http://electroschematics.com/5224/metal-oxide-varistor/

HELLO SIR,CAN I GET YOUR EMAIL ADDRESS?

THANKS N BEST REGARDS.

excuse me sir, can i ask you more about MOV?

What is different between circuit have fuse and fuse + MOV?

1. What affects the size of the output voltage

2. What affects the size of the current in this circuit.

size of the curent should be given reactance of capacitor, I think. Is it true? But I don’t know what affects the size of the output voltage.

Output voltage and current depends on the Capacitive Rectance of the capacitor used.Higher value capacitor like 2.2 uF will give around 100 milli ampere current and 50 volts AC.This depends on the input AC voltage

OK. Thanks. I think it.

sir,

in one of my experiments with capacitive power supply, to have 4V DC output with DC input instead of AC, i found that bleeder resistor R2 across capacitor C1 is very useful.

It provides path to DC to be injected.

Please suggest will it (R2) have any adverse effect on performance of power supply with AC input or high voltage DC input if its value and power rating properly selected?

In capacitor power supply, the bleeder resistor is included to remove the stored current quickly when the circuit is unplugged. Usually a 220-470 K 1 watt resistor is used for the purpose. If high volt DC is used as input, increase the value of the bleeder resistor to 1 M or 2.2 M so that it will not affect much to the output

I have 2 Amps Load in DC Supply, and how much Capacitor range i want to connect for protection purpose?

nice idea

can i use 1M;1/4W bleeder resistor for .22uF ? plz suggest me……..

Use 470K to 1M bleeder resistor for the AC capacitor to remove stored current quickly when the unit is unplugged. Low value resistor may heat up.Ideal value is 470K or 560K 1W

A capacitor is in series with resistance of 30 ohms and connected to a 220 volt ac line. The reactance of the capacitor is 40 ohms. What is the current in the circuit ? Help me with the answer

What is the value of the AC capacitor. See the table showing output current and voltage of common capacitors given in the article. Convert the rectance into Kilo ohms by dividing it with 1000. Then the current is supply voltage / rectance in Kilo ohms. Resulting current is in Milli amps

i got very helpful knowledge from this discussion

sir,

is it essencial to use bleeder resistor except for safety purpose….what r the basic difference whenever we use bleeder resistor or not for .22uF ? is any variation in voltage and current.please advice me….

Dear Raghvban

The bleeder resistor parallel to the AC capacitor is a must in capacitor power supply. The AC capacitor is rated 400V,600V or up to 2 KV. The capacitor charges fully up to this high voltage, and keep this high voltage for many days even if the unit is not connected to AC lines. This can give a lethal shock, if the pins of the AC cord is touched. If you add a bleeder resistor, this stored current will be removed immediately when the cord is un plugged.Adding a high value bleeder resistor like 470K or IM will not affect the properties of the capacitor, voltage, current etc in practical conditions.

Need to have a conversation with you sir, your explaninig things very easyly understandable,great stuff.

Dear

Please let me know about life time. its good and long life tine?

Thank

Thanks Mr. D, Mohankumar,

I like Your tutorials, the simple man ore woman can understand now how to calculate if its needed.

I hope to see and find more of Your tut’s.

Greets from Estonia,

Daniel.

Thank you for posting the circuits (this and others) on the internet.I found them with google, searching for key words like “led circuit capacitor bridge rectifier 120 220 volts AC” and “how to calculate voltage output of AC series capacitor afterr bridge rectifier”.

Until recently, I was under the assumption that C1 only regulates current, which would vary somewhat with the always changing AC input voltage (0 volts to V*sqrt(2) volts to 0 volts again), 50 or 60 times a second. If C1 drops the voltage, is that output voltage constant, or does it vary ? Also, for a given V, C1, XC1, f, etc what are the steps/formulas needed to calclate the output voltage after the bridge rectifier ?

Thank you !

In this circuit, where and how can we connect a TVS (transient voltage suppressor) diode ? Same way as the MOV (metal oxide varistor) in the circuit ? I connected a TVS across the input wires of another led circuit (led string) and it kept frying the fuses. I am thinking it was because I had no resistor before the TVS, but I can’t tell for sure why and I am trying to figure out what I did wrong and how to incorporate TVS in my circuits and use them with MOV, together.

Hi,

Good article, very well explained,

I have few queries

1- How do you caclulate R1, R2 and MOV from your example circuit.

Thanks in advance.

Mr. Mohan Kumar,

Your above article is excellent, but we readers are bombarded with doubts, which if answered could help us in many (understanding of electronics and the various functions each components tend to play in a circuit)ways. but you have not responded since october 2010, Sir are you on vacation!

pls respond

How to calculate output voltage ? can I get 150volt 50mA from capacitive transformer-less power-supply. Please do reply. As this article have no info about selecting the capacitor for desired output voltage.

sir i thank for your great contribution. i tried this circuit but the R1,100 ohm 1 W in the neutral line is burned off and another problem is the R3, 1 W is becoming hotter. kindly reply.

sir can i use the R1 in the phase line before the dropping capacitor.

do i want to increase the wattage of resisters(R1 & R3)

Abhin raj, I think you are describing what happens when C1 fails, so make sure C1 and the connections/wiring going to C1 are in working order. R1 acts as a fuse and should burn out when C1 fails and that is what did for me, when I did not have a R2, which keeps R1 from burning and protects the circuit when C1 fails.

Sir, please let me know about the life time of this circuit? Is it 24/7 or not?

What about power factor of this circuit? From my (extremely limited) understanding the reactive power if this power supply is used, say, to light a LED, will be much higher than the true one. Is this at all something to be concerned about in circuits with such low currents?

hi

dear sir i read Your whole article and I like it very much, sir you write this article very clearly and simply. Even a first year student of Electronics or Electrical can understand this article.I personally admire your work.

Thank you very much.

hi

Dear sir ,I read Your whole article and I like it very much, sir you wrote this article very clearly and simply. Every one can be banifitted from this artical.

Dear Sir Can we take 1A current from this circuit to change the capacitor value.

Thank you very much.

Hi Sir

Thank you very much for the explanation. This circuit was puzzle to me and you made it so clear.

Regards

gud day sir,,,

im nterested of how kvar can make my electrical consumption can be lessen

can u gave me the circuit diagram of capacitor bank and their sample computation

for my 100 amp residential building,,,

by the way we are using a line to ground power supply

It’s the best time to make some plans for the future and it’s time to be happy. I have read this post and if I could I desire to suggest you some interesting things or tips. Perhaps you can write next articles referring to this article. I want to read more things about it!

Dear Sir You Article is great to the greast.

Sir Pls Drop a mail to me regarding is power supply

This article is very understandable. Thank you very much for that. And we are grateful if you can answer us. Hope you will be able to do that soon because we are waiting.

And I will add another to the question list. Can we calculate the output DC voltage by seeing this circuit as a voltage divider?, where the resistance of the capacitor works as the 1st resistor and the resistance of the load works as the 2nd resistor.

In other words if R1 = 1st resistor and R2 = 2nd resistor,

the output voltage = [Input Voltage]*[R2/(R1+R2)]

Is it work that way?

thanks again.

Anyone not experienced with power electronics should understand that this circuit is not isolated from the power line and risk of electrocution is possible. These types of power supplies are only used where there is no possibility of touching any electrically conducting part of the power supply or conducting part of any component connected to the power supply or its DC output.

Sir,

Your article is great and i really appreciate it.

Regards

hi sir,

hope u are fine ,,,sir i have to ask u a question

i hve (center tapped)transfomer whose specification is:

input=220V

oupt=12×2 V

power=8W

i want to give (operate) 12V dc motor,and i am using LM7812 regulator for 12V but when i connect the motor ,after some time the regulator comes soo much hot,,,,,what should i do ???please hlp me ,,i shall be very grateful to you .waiting for ur nice reply

regards

bilal

hi this is vinodh i’m looking for the thing that how to suppress an interference in the o/p of power supply.

Some refeers to connect the capacitor in the o/p, but the thing is i’m working in high freq like 640Khz. pls suggest some thing and how to connect

Sir,

I want to design I/p is 230V A.C 5A

and I want to get out put is 12V DC and 3A is it possible for capacitive power supply?

Hello Sir,

greeting for you.

I want to design 45VDC and 5VDC power supply for DC motor.

i/p 230VAC, Transformer output 45VAC

need to get 45VDC (for DC motor) and 5VDC for microcontroller

Please send me the circuit for my requirement.

Thanks and regards

Sekhar

Hello Sir,

I have 42vAC transformer (i.e. 230VAC to 42VAC stepdown transformer), I want to get 42VDC (from AC to DC) to drive DC motor. Please send me circuit diagram for converting 42VAC to 42v DC. and also i need to get 5 V DC from same 42 V DC supply to connect microcontroller.

Transformer is not center tapped. I need to drive 5 amps dc motor.

I request you to send me circuit diagram for this.

Sekhar

Hello Sir,

greeting for you.

I want to design 45VDC and 5VDC power supply for DC motor.

i/p 230VAC, Transformer output 45VAC

need to get 45VDC (for DC motor) and 5VDC for microcontroller

Please send me the circuit for my requirement.

Emil ID: bapanapalle@gmail.com

Thanks and regards

Sekhar

What is the necessity of R1=100R,1W in capacitor (transformer less)power supply circuit? It’s connected series in capacitor

dear sir, i hav a doubt in power supply circuit.

i need to charge a 12V battery, i used 3 resistors & LM7818 regulator to drop the voltage from 22 to 14.5V.

the voltage gets dropped, but there is no output current from this connection..

i need to charge a 12V battery,so what type of components should i use to get desired 14.5V & required current.

and how much current is required to charge a 12V battery? pls reply sir……..

thank you so much. your article helped a lot.

Hello Mohan, i first came across a transformer-less power supply while repairing a LED sign flasher an i was amazed. i did not understand at-all what the heck that was but my eyebrows were really raised. when i Google i was refered to your notes which have turned out to be very useful. thanks for the good work man.

Jack from Kenya (East Africa)

Hello Sir,

I have 42vAC transformer (i.e. 230VAC to 42VAC stepdown transformer), I want to get 42VDC (from AC to DC) to drive DC motor. Please send me circuit diagram for converting 42VAC to 42v DC. and also i need to get 5 V DC from same 42 V DC supply to connect microcontroller.

I request you to send me circuit diagram for this.

Sekhar

Hi Sekher,

You have’nt mentioned the current required.

Hi Shekhar, It is very simple. But before you get a diagram please let me know the Transformer you have is Center tapped or what?

hello sir i want the circuit for 7watt high power LED to connect the main pls send to me the circuit to my mail

ID :(pkn0808@gmail.com)

Dear sir currently I am working on capacitive power supply of 3.3V .

I am using 1uF Capacitor with 330K bleeder resistor.

my voltage get drop down slowly

Dear sir currently I am working on capacitive power supply of 3.3V .

I am using 1uF Capacitor with 330K bleeder resistor.my voltage get drop down slowly.

please tell me what will the effect of running this capacitive power supply on ups instead of mains power supply

Dear Sir,

Very good artical and nice question answering.

Sir I want to know how smoothing capacitor value and voltage calculated ?

What are the parameters to be considered ?

sir i want to known how output voltage of x-rated capacitor calculated in above circuit ?

Dear Sir,

thax 4 sharing ur knowledge with us…

sir,

what is the capacitor importance here?

what happens if the capacitor is not present in the design?

explain me…………….

Dear Sir,

thanks for a good article. It might be good to stress even more clearly that the minus of the DC side is not GND or not floating, but possibly tied to a high potential instead.

I’m constructing an op-amp sense to the speaker terminals of an amplifier for switching on a separate sub-woofer amplifier. The relay/op-amp board would be powered with this kind of a power supply. I’m happy I understood the potential issue before entering the realization stage of the project. I think I would have destroyed the amplifier.

The DC side must not be connected to the outside world.

Is this correct?

Dear sir please tell me, how will we determine the value of capacitor in microfarade & volt after bridge rectifier.

pls tell me how to kwon the values of a capacitor voltage in a circuit diagram.for example am working on a subwoofer amplifier as a project but the circuit diagrams i have does not indicate the values of the capacitor voltage values.pleas help me out

splendid work . thnx sir

D.Mohankumar,

May I add to the many comments of appreciation.

This is possibly the best tutorial of it’s type available today. Very well presented & obviously aimed at aspiring electronics newcomers.

I would suggest including a more detailed BOM specifically using value terms to describe components & not the shortcut descriptions because these are not used in the typical parts cataloges we mostly use to source components.

It’s a small point but you can see from several comments that this has caused some confusion.

Again sir, thankyou for an exceptional contribution.

Thank you for such a detailed and well thought out explanation – a rare gift indeed.

Cheers

Gerry

dear sir

i am fahad,from calicut.now i am in dubai working in GEEPAS electronics as a service engineer.

sir this circuit is very useful in my life.

sir i have a doubt,what is the problem when replace the capacitor with a high value resistor?

Dear sir,

How do i get 230 volts DC output so as to supply a load of 50 series connected LEDs?

Hai,

I would like to know about any shock hazard on the LV(12V) side of the circuit.

I have constructed a similar x,mer less power supply for miniature circuits but i always had a doubt of shock hazard if the same is used as battery charger or as a power adapter for any utility.

could you please clarity.

The entire circuit is live, as is EVERYTHING connected to it. Although the voltage across the output terminals is low, the terminals will be at mains potential relative to ground for the negative portion of the incoming AC waveform. Also if phase and neutral are reversed (if plugged into an unpolarised outlet for example) the current limiting will be in the neutral line, not the phase line, increasing the shock risk further, a 100 ohm resistor and 1N4007 diode being all that stand between the negative output terminal and the incoming mains supply.

@ani: Assuming that you’re using 230V AC as your input, you could switch it to DC by skipping the initial capacitor, and resistor, going straight from the fuses to the diodes. Then it’s into a capacitor to make it more flat, as what’s before the capacitor is a really bumpy signal, rising from 0 to 230V (Well, not exactly 230V. The Diodes steal a volt or two.)and back down to 0V. The capacitor smooths this out a great deal. You could then use some sort of voltage regulator IC, which would cost you another few volts, or you could try to find a zener diode for 230V, again losing a few volts.

This method uses some of the same pieces as the circuit posted up above, but in a different order. I hope this helped?

Don’t forget that once the 230Vac is rectified the output DC peak is 1.414 times as large. i.e 230 x 1.414 = 325Vdc so components must be selected rated above this and you also need to regulate this down to 230 dc with additional components.

Steve

There are a lot of mistakes in the article above and a lot of poor descriptions.

Selection of the dropping capacitor and the circuit design requires some technical knowledge and practical experience to get the desired voltage and current. An ordinary capacitor will not do the job since the device will be destroyed by the rushing current from the mains. Mains spikes will create holes in the dielectric and the capacitor will fail to work. X-rated capacitor specified for the use in AC mains is required for reducing AC voltage.

The above statement is entirely incorrect.

Any capacitor rated at 400 volts will work in the circuit.

An X-rated capacitor is simply better-made and guaranteed to withstand the peaks for the life-time of the capacitor.

Selection of the dropping capacitor

The capacitor is not a “dropping-capacitor”

If you connect the capacitor to the active line and touch it and an earth wire, you will get not only a 240v shock but a 336v shock.

The way the capacitor works is this:

Connect the capacitor to a 100R resistor and place the two components across the “mains.”

The mains voltage will rise and this will put an electrostatic charge on the left-plate. This electrostatic charge will influence the right-hand plate of the capacitor and a charge (called current) will flow in the 100R resistor.

When the mains reverses direction, the “current” will flow in the opposite direction.

Due to the size of the capacitor, this current happens to be 16mA.

When 16mA flows in a 100R resistor, a voltage of 1.6v develops across it.

This is how the voltage develops.

Most of the general purpose rectifier diodes in the 1N series have 1 ampere current rating.

This is entirely untrue.

The “1Nxxxx” means it has one junction.

A 1N5404 is a 3 amp diode.

The explanation of the output voltage is entirely incorrect.

The output voltage depends on the zener voltage. Nothing else.

Suppose we select a 24v zener diode.

The output voltage will be 24v.

You do not need R3 (100R) as the circuit contains a current-limiting resistor R1 (100R).

The current available will be about 8mA for each 100n capacitance of the X-rated capacitor.

It’s that simple.

The way the zener diode works is this:

When no-load is connected, the 8mA flows through the zener diode.

You can take up to 8mA and the output voltage will remain at exactly 24v. As soon as you take more than 8mA, the output voltage will drop. It’s as simple as that.

Exactly the same reasoning applies to an 18v zener or 12v zener.

Dear sir,thanks fr a good job,i would lik to see more of yr posts $ yr personal websit.

Thanks for valuable information.

Editor of this article

thank u for the xplanations about capacitive voltage lowering….sir will it be okey f i ask about your nationality,

greatfully randy

Hi everyone

i guess mr mohan has left the building !!!!!

Its upto us to sort out things

firstly great article, although misleading explanations as pointed by mr Mitchell

i tried out the circuit without the zener, i measured about 180 v dc at the bridge output

then i tried to put1k resistor in series with a blue diode, the the 1k res in series with the load started smoking

anybody can explain why this happened

Remember this:

A 100n cap (104) will deliver 7mA RMS or 10mA peak in full wave. (when only 1 LED is in each string).

Thus 225 will deliver 22 times 7mA or 154mA.

154mA will pass though the 1k resistor.

Wattage (or heat) being dissipated = 0.154 x 0.154 x 1,000 = 23 watts.

Now you can see how dangerous and how stupid the circuit is.

I think you are wrong, first of all, the current thru R4 1K resistor is at maximum 24V/1000Ω + LED resistance => 24V/1470Ω = 16mA.

If you are talking about R3 with the value of 100Ω then the power dissipation will vary depending on power consumption.

He said he tried the circuit without the zener.

Where do you get 24v from??????

I didn’t read what Jay said. I thought you are talking about the complete circuit.

Thanks for the prompt reply

let me come to the point straight i want to power a 1 watt led from this circuit

do i need to put a zener

The led can go upto350 mA . Can i connect the led

across the bridge output without frying it

Yes you can put a 350mA LED on the output. The circuit is basically a CONSTANT CURRENT CIRCUIT.

That’s where everyone is making the mistake.

that’s true . capacitive power supply is niether constant voltage or constant current supply. s input current varies o/p volt and current varies.

But if you want to add 1watt powerled then you can use two 2.2uf x-rated capacitors with 1% tolerance rating.

If you are not using MOV then led life will be reduced .

A rectifing capacitor is must after the bridge. 16volt/1000uf works like charm.

Sorry to bother you again just that i am weak in electrical

as i mentioned earlier i measured 180 volt dc at the bridge output

the led data sheet says vf max is around 4volts

i also read some where that led are current driven

so from what you are saying i should be able to connect the 1 watt led without any problem although the voltage is much higher

The circuit is called a CONSTANT CURRENT SUPPLY.

It will deliver a maximum of 154mA to anything connected to the output.

This includes a short-circuit, a LED or even a 1k resistor.

When no load is applied, the output voltage will be about 315v. If the voltage developed across the load is 100v, the maximum current will be about 100mA. If the voltage is 200v, the current will be 50mA. If the voltage is 300v, the current will be 0mA.

This is something that has never been covered before. ANYWHERE.

Or should i put in a zener

What is the zener going to do???????

I guess i will connect the 1watt led directly to the bridge output.

i have a doubt . i opened a led light in my house that runs off 220mains. inside i saw the same circuit with .22 micro farad 250volt capacitor a bridge rectifier the bridge out is connected to about 15 led go series i measured the ac volt between cap out and neutral i got about 20volts

does this mean that the voltage drops once a load is applied

please clarify

The voltage generated across a load (especially a LED load) is a characteristic of the LED device and cannot be altered when the prescribed current is flowing.

For a high power LED such as 1watt to 10watts, this voltage can be about 4.5v. For a white LED it can be about 3.5v

Hi again

i connected the 1 watt led without R3 the led flashed very brightly and burnt off

i connected another led and this time i connected R3 com measured the current it was around 120mA . Voltage across led was 3.2 volt. voltage across led and R3 together was 15 volts

R1 and R3 were getting very hot but every thing was steady. so again i removed R3 and tried it was working. but somehow the led hot burnt again . i have run there leds at 200mA without any trouble

Any theories on this

ps i am not using the zener and mov

I have to add something

i was doing all the testing with a multimeter

so i was quite frequently making and breaking circuit while measuring current would this have caused the led to burn due to transient high currents . both the times the leds burnt there was no R3. isn’t R3 required to limit any transient currents

You obviously did not discharge the electro before connecting the LED.

So it is not advisable to frequently switch on off this circuit

The 1,000u gives you plenty of protection against spikes but you connected it directly across the LED when it was charged and damaged the LED.

I am using 2.2u not 1000u

C1 is 2.2u but out put voltage extremely high why

I have the 470k across 225k cap shouldn’t that discharge the cap

in the worst case what will be the max current

is it a mistake to connect the led after switching on the circuit: like i did when testing the current

Isn’t the value of R1 and R3 high

As per your calculations current through r3 is 150ma so I sqr R equals 2.25watt

is it ok to send 2.25watt through 1watt res

another doubt : i read in another forum that by using another cap in parallel with the load the sudden inrush of current can be avoided

is this correct

is it a mistake to connect the led after switching on the circuit: like i did when testing the current

ABSOLUTELY

I told you the circuit has lots of mistakes.

The 100R should be 2 watt or higher

i read in another forum that by using another cap in parallel with the load the sudden inrush of current can be avoided

is this correct

That’s what the 1,000u is for.

The 1,000u is to protect the LED

Thankss a lot i think that should do it

i will post once i have tried the circuit

Hi

i tried the circuit without R3 zener and mov

i connected the 1 watt led as load

it worked out good the current was around 120mA

all components were fine except R1 which is getting very hot

R1 is 100R 1 watt

any ideas

Jay can you put the final circuit diagram ?

Make it 2watt

Is it ok if i don’t use R1 and R3

You need one of the resistors to buffer the huge inrush of current when the circuit is connected and the mains happens to be at a high part of the voltage-curve.

What is the minimum value of R3

can i use 56ohm

You can use any value at all but remember that a high value provides more protection and reduces the inrush current to the electrolytic.

Thank you for your patience in answering all my questions . i have learnt a lot . The circuit i built is working fine except for R3 which is getting heated up i will have to get a 2 watt res

what if i use two 56 ohm 1 watt res in series

Yes.

That’s 112R 2watt

Colin Mitchell: Your response to jay basically wrong:

- Two 56 ohm 1 watt res in series have: The double 112R but the (same 1 watt power).

May be you can go again to the Australia’s beginners school to learn serial and parallel adding.

To: JR Joserra Zierbena

You don’t know what you are talking about.

Learn about Capacitor-fed Power Supplies before you make any more comments.

Hi Mr Mitchell thanks

Are there any other postings by you

i certainly would like to read them

we having problem with voltage. sometimes we are having about 35 volts ac current, so how can i get 35 volts ac to 230 volts ac current

Come to Australia. We have 230v AC all day, every day, rain, hail or shine.

Colin Mitchell: Your response to jay basically wrong:

- Two 56 ohm 1 watt res in series have: The double 112R but the (same 1 watt power).

May be you can go again to the Australia’s beginners school to learn serial and parallel adding.

You can go to… even every day, rain, hail or shine!

Hi

can i use an lm317 in series to get regulated current

i know this is already a constant current source but the current varies with input voltage

but the voltage out from bridge is around 180 lm317 rated max volt in is 60 volt i think

so the question can i use lm317 as current regulator in this circuit

is there any other ic that can handle higher input volts

No.

No.

NO.

NO.

if i use 220 volt DC as input then what we have to change in the cicuit to get only 9 volt DC for LED @ 20 mA.

where will u get 220dc from

u will get it if u use a bridge across your mains ac

instead of putting the bridge there first it is better you put the capacitor so you get current control to run your led

for the values of capacitors look up the article

if you still insist on dc i guess you have to go with resistor you can hook up a resistor in series

you can get res value from R=V/I

R=220/20X1000000=11M

NEAREST STD VALUE IS 10M OR 22M SO USE TWO 22M IN PARALLEL

but still the above ckt is ok if you want it off 220v mains

sorry a blunder 1mA is divided by 1000 so

R=220/20×1000=11000ohm or 11kohm nearest std value is 12 k ohm

I am getting x2 275 v ac rated capacitor from a nearby shop can i use it or should i use ones with rating above 400 v

doesn’t 275 v ac rating mean rms value

275v AC is ok

Thanks

I have these ceramic caps rated at 2kv

can i use them they look very small and feeble

please clarify

What is the value printed on the ceramic cap?

224K

2KV

224K

2KV

Each ceramic capacitor of 220n will deliver a constant current of 15mA on 240v supply when using a bridge.

Dear Sir ,

I am using capacitive power supply. designed for 50mA Iout current. I have doubt in selecting capacitor voltage rating.I have supply of 230V AC lines. So can I use 250v rating? and X2 rated is compulsary?

Dear Sir ,

I am using capacitive power supply. designed for 50mA Iout current. I have doubt in selecting capacitor voltage rating.I have supply of 230V AC lines. So can I use 250v rating? and X2 rated is compulsary? 1

X2 capacitors are needed for commercial products. You can use 250v ceramic or poly for your own supply.

Thank you Colin sir …

1. Still I have one doubt according to IEC664 standard its recommended that X2 rated capacitors should be used for such applications as these capacitors has to pass the test of few kV voltage test.

2. the application I am using having a tiny micro controller and a 12V relay and a tact switch well I have provided a acrylic cover to it to protect from shock. whether its a safe design without using X2 rated capacitors? or what exactly it differs from normal ceramic or poly capacitors?

Please focus more light if I am going wrong.

Your suggestions are valuables

Hi,

can anybody help me as I have doubt regarding the capacitor selection. what is difference between normal ceramic cap and X2 rated capacitors.please help

I have already explained:

X2 capacitors are needed for commercial products. You can use 250v ceramic or poly for your own supply.

There is no difference between any of the capacitors.

230 Volt,A.C., 50 C/S to 50 volt A/C, for 30 ma current

A capacitor-fed power supply delivers 7mA for every 100n capacitance.

Pls i’m trying to design a 70v power supply and am confiuse of how to select my components like diode, capacitors resistors can u help me?

Just use the same components as shown in the circuit above.

Hi, Colin Mitchell

it’s possible to appear spike current at the start up condition. I’m measure this circuit with an Osciloscope.

Pls, help me how to reduce this ,

Or in the other way, it must be transferred to resistor or something??

Put a 1k resistor in series with the capacitor

Colin mitchell

Thank you for your help.

you mean to drawn the current passing 1kOhm..

hmm..any other solution?

how about adding some material like inductor,,can it work??

Pls

You can use an inductor with a DC resistance of about 100 ohms but it is more expensive than a resistor.

Yes..its more expensive if adding a inductor..

so,,adding an resistor is cheaper solution..

okay,,i will try..

thank you for you help

hey,,,mitchel,,,

i asking you again…

the x rated capacitor dimension is too big..

can we use a DC capacitor?

pls give me suggestion about that?

sir i have 4440 double ic audio amplifire bord it requires 12 volt 3 amps so how can get it please tell me and also how much capacity doide na dcapacitor can i use please tell me

Use a transformer for the 3 amp power supply.

is this circuit safe or dangerous due to all voltage drop on capacitor not voltage is step down.

A capacitor-fed power supply is very dangerous.

They are illegal in Australia and I don’t recommend them AT ALL.

You can get a power supply from an old computer for $1.00, so why risk it?????

Mr. Colin,,

I don’t know about capacitor-fed,,

why it’s very dangerous?

A capacitor-fed power supply is dangerous because if you touch one of the wires and a toaster you will get a 345v shock.

this topic is an excellent guide how to calculate the right resistance and capacitance values for LED lighting fixture,i am not an expert but i learned a very useful tool in my next electronic project…keep up the good work

thanks

sir, i want to do the project on our issues for power supply. there was a lot of confusion to select a theme.. but my project should be useful to society .. please guide me.

sir i want to a power supply transformerless how we calculate the capistor and resistor veleo for difrent voltage and current

thanks very much

hello,

I have done and attained the voltage I want but the reisistor R3 100ohms creates heat and increasing around 60 degee C, Is this normal or there’s something wrong on the circuit. what can I do to decrease the temperature of R3…

Note: I use this to supply my 3 watts LED downlight. Please can you give me any recommendation and I very appreciate a lot on your help

thanks

Use a large wattage 100R or put 4 x 100R in the circuit by placing two in series and the other two in series then connecting the first two across the other two and soldering them to the circuit.

This will give you 100R with four times the originl wattage dissipation.

Very nice article! One thing that i don’t understand at all – and i’ve been searching for the solution for quite some time: how do I calculate how much voltage the main capacitor will drop? I can see that the current is calculated with “I = V / X” but how about the voltage?

I’d really appreciate an answer. Thanks!

View this as a simple series circuit:

Peak capacitor V = Peak Mains V – DC output voltage.

Generally, the capacitor voltage will be only slightly lower than the mains voltage.

That is very interesting. So the output is close to max (mains) if I have no load – it basically supplies as much voltage as the circuit draws from it, right?

I asked this because I am building a led bulb just to see if I can. I used an online calculator to get the exact values for the components in relation to my leds. But the problem is that the calculator outputs raw values for the capacitor (microfarads) – in my case they needed approximation. I thought that using a slightly different value for the capacity would “give” more volts to the circuit, thus burning my leds. Seems that I was wrong though.

The calculator is found here (I hope I can post this here):

http://danyk.wz.cz/ledzar_en.html (bottom of page)

Many thanks for your time!

Hello

Greeting for you

I find this is very useful for me.

But I think there are an error between your schematic and your calculation for C1,

In schematic the value of C1 is 225 that equal 2.2 uF but you use in formula 0.22 uF.

please take care about this.

Best & Thanks.

Note to our readers: check our work and alert others when things do not add up–we sometimes make mistakes, but electrons never make mistakes.

Thanks for alerting us on this one. However, this may not really be a discrepancy as 0.22uF can be used as well–the table supplied lists the results for several capacitor values including 2.2uF, but 0.22uF just happens to be missing–duh?

Another issue not mentioned is the power rating of the zener is not indicated –for the higher current versions, it should be 5W–and be careful to derate the 5W zener to about 1.5W–I know from experience that they run extremely hot. The 1W zener must be derated to 0.5W as well. To determine zener power, assume that all current goes through the zener when the load is completely disconnected, and of course, P = E * I

In calculating average current the 0.9 Average/RMS factor (0.637/0.707) comes into effect when dealing with sine waves–keep this in mind to make sure that you have sufficient current for your application–allow at least about 20% additional to make sure that the zener remains in conduction at all times.

Also the table should have the minimum filter capacitor size indicated as 1000uF is grossly large in most cases.

The 0.22u is just an example.

The table in the discussion is correct.

0.22u provides about 10mA

2u2 provides about 100mA

@Jim, I see. Is there a formula to calculate the output voltage after the AC passes through C1? This is what bothers me actually.

I have been using an online calculator for the values of my led bulb components, but for the capacitor I needed to approximate its capacity to the nearest one I can buy. This must also affect the amount of output voltage which might damage the leds I suppose?

Thanks for your time!

@Alex, Because the input voltage is much greater than the output voltage (Ein >> Eo), the voltage difference and the capacitive reactance form a limited current source that varies little with changes in the output voltage. In short, the output voltage is set by the zener voltage.

@Alex, Regarding your LED current, it is set via the LED series limiting resistor.

R_series = (Zener V – LED V drop)/ desired LED current

Otherwise, you must have adequate current available to keep the zener in conduction under low line voltage conditions.

e.g. if LED current = 100mA, total available current should at least = 100mA + 20% or 120mA.

Otherwise, you must have adequate current available to keep the zener in conduction under low line voltage conditions.

This is not true.

You can use up ALL the current from the capacitor-fed power supply. The zener does not have to be kept in conduction!!!!!!!

Yes, you are correct, but what happens if the load then increases slightly? Due to the existence of the current source feed, as soon as the zener drops out of conduction, the voltage will drop like a rock. Good design practice adds safety factor.

Good, healthy discussion going on here…

How do I calculate how much voltage the main capacitor will drop?

Answer from Keith:

Generally, the capacitor voltage will be only slightly lower than the mains voltage.

This is not true.

As the delivered voltage increases, the current-capability of the capacitor-fed power supply decreases.

For instance, if the input voltage is 240v and the output is 5v, the delivered current will be 5mA for 100n.

But if the delivered voltage is 100v, the current will be 2.5mA.

It all boils down as to semantics or what “much greater” really means. I say that much greater indicates greater than a factor of 10 or so–if so, this starts to fall apart when the output voltage exceeds about 24V with a 240V mains. For most practical purposes, your voltage requirement will not exceed 24V.

If the output is 100V, then the current source is definitely reduced and yes, the output current will be less–the capacitor size calculation will be more complex in this case.

“as soon as the zener drops out of conduction, the voltage will drop like a rock. ”

This is not true.

You don’t understand the concept of adding a zener.

You have got the whole concept around the wrong way.

The zener voltage should be higher than the voltage required by the LEDs (or any device) on a capacitor-fed power supply.

This means the zener will be taking NO CURRENT and all the available current will be delivered to the LOAD.

The zener is just a safety device.

It is included for the following purpose:

Suppose the load takes less current due to one of the items in the load either failing or being turned off.

The voltage across the other devices will increase and to prevent the voltage increasing too much and damaging them, a zener is included. It is also included if you have say a 100u 25v electrolytic on the output. If the load is removed, the output voltage will rise to 300v and the electrolytic will be destroyed. The zener prevents the voltage rising above the zener voltage.

As I said before, the load can take all the current from the capacitor-fed power supply. If it requires less current than the supply can deliver, the voltage will rise and the zener will come into conduction.

This can be set up either way–firstly, as you indicate as a current source to drive LEDs with a safety zener clamp, or as a voltage regulated supply in which the load current is always less than the current source with the zener picking up the difference.

Application dependent.

I need to read more about the principles behind this power source.

My whole problem arised from the fact that in the other schematic I found (which looks like the one on this page except it doesn’t have the following parts: the zener, the red led and R4) I couldn’t find the way the voltage was lowered. I can’t post links here but you can find the schematic by googling “LED lamp online calculator”.

So I guess that my best shot is using this power source – the one that includes the Zener for regulation. My bulb (~ 400 lumens) would use 14 leds rated 2.9 to 3.2 volts and maximum 100 mA. I just need to calculate the components and it should work.

Many thanks for taking the time to answer my questions.

14 x 3.1v = 43v

2u2 capacitor will deliver 100mA

You can use the circuit above. The zener will need to be 47v at 5 watt rating and the 100R will need to be 1 watt.

You don’t need the zener or the 100R

You can put the electro directly across the LED string. The 100R 1watt on the input is needed to prevent the LEDs being damaged.

The only problem with removing the zener is this:

If one LED dies, the elecro will blow up.

It’s all clear now. Thank you for the help, it’s much appreciated. When it’s done, i’ll make a comment about the results. Cheers :-).

Hi Sir

Thanks for your response regarding my last comments..

Sorry I try to increase the voltage output of the circuit and decide it around 12 volts+ but whenI try to load it with 3 WATTS LED lights the voltage drop to 9.5 volt.

Is there anything else I can do to increase the voltage to 10.5 volts with load?

thank you very much

Ariel

If you are talking about a single 3-watt LED, you will need to build a 1-amp capacitor-fed power supply.

For 1Amp you will need a 10u 400v X2 capacitor.

hi Sir

Thank you very much, the LED is on Series connection (3 led’s). Do I only change the X rated capacitor to 10uf 400v and the rest are still the same value? capacitor resistor etc?

Ariel

The problem is this:

x2 capacitors are very expensive and very hard to get.

2u2 400v will cost a few dollars.

Hi

Sorry if Iam going to change from 2.2uf to 10 uf.400v do I have to change also the Diode, resistor, capacitor and zener? Please give what do I do and what the value on each

thanks

Very large capacitor…

What is your mains voltage?

What is your required output voltage?

Required load current?

Hi

230V 50hz output 10.5 volts at 3 watts LED’s

Bad choice–available current far too high and will cause many problems–select proper value via the matrix in the article.

If you are talking about 1watt LEDs, the current will be 300mA and you will need 2 x 2u2 in parallel (x2 400v) capacitors.

The 100R’s will need to be 47R at 15watt and you can see how the power supply is getting quite “out of hand.”

Sir,

can u tell me what value of capacitor and resistors to be used in a case were we have to step down 230v ac supply means to get dc 5v, 20ma output.

330n X2 capacitor and 5v1 400mW zener

Hi [mr] colin mitchell plz can u just tell the maximum current/voltage for charging each battery and how charger drop current and voltage

I don’t want to help you too much because a capacitor-fed power supply is very dangerous and I don’t recommend it AT ALL.

i want to know that what can i change out put voltage of a capacitor.

Actually:-*what am refering to is that …i want you to help me in some cases apart from this project {capacitor%power%supply}..i am not experience enough to correct circuits from any authors just little..sentimentality..as@now i, m extremely confused plz/plz MR={COLIN-MITCHELL}…you the only person to through more light in my problem..as a matter of..that i am good in electronics despite that i found it difficuits..corrects little error …all what i need from you is..give more details of the listed below {1}why rough d.c is suitable for charging..if its true how to select {capacitance}uf for d.c ripples {2}can you just tell the maximum current/maximum voltage for charging 3.7v860mah= 6v4.5AH=12v7ah=12v200ah=12v180ah=..and others batteries {3}sk100=sl100=TIP127..bd140=bd139 datasheet.. {4}and what are the main electronic component that drop voltage and current in a circuits…explain how it happen {5}fomular to calculate primary/secondary currents of a transformer for me to figure out the AWG is suitable for a typical xtran..also how to calculate the efficiency of a transformer and why xtrans can=ever be 100%.. .and some relevant example for better understanding.. {6}the difference between half wave/full wave retification in a circuits..in anyway which one is the best voted for electronics project {7}how to regulate output voltage without droping voltge {8}how to regulate output current without droping current {9}capacitor is not well understood@all…what are the other uses of capacitor except.from.smoothing d.c time delay….. {9}How to calculate current limiting resistor for zener diode and their watt {10}fomular to calculate a sensible value resistor for LDR {11}fomular to calculate darlinton pair of transistor current……..

colin mitcell help ooooooooo..:

colin mitcell help me plz responed to me :

what are the common type of failures in capacitive drop power supply ? and what corrective actions should we take to improve efficiency or durability or life of the circuit?

Use only X2 capacitors

What is the formula to determine the No load output voltage And Load Current?

dear how i get 220v ac to 14v 2amp and 10amps with out transformer only capistor suply pleas help me for this circuit

I don’t want to help you too much because a capacitor-fed power supply is very dangerous and I don’t recommend it AT ALL.

Anything over 500mA is getting dangerous and expensive.

Sir

thank u very much….pls help me about the different capacitors available and there areas of applications..

Sir,

Please tell me about different capacitors and there uses and applications

..

For power supply applications like this the capacitor of choice is the metalized polypropylene X or Y type with X for Line to Line applications and Y line to neutral.

The only other acceptable type that I know of are oil impregnated paper that are used for permanent split capacitor motor applications. Generally they are greater than 1uF and are rated for 440VAC.

The old Sprague Vitamin Q series is also oil impregnated paper and can hold up under the AC voltage stress, but unfortunately they are available only as old stock or used.

sir,

how to calculate voltage across capacitor,in table it was given 22ma,10v for 334k.. pls explain sir

A capacitor in a capacitor-fed power supply only does one thing. It allows a certain current to pass or flow though it.

The voltage on the input side of the capacitor is the same on the output side of the capacitor.

The capacitor has no effect on the voltage. It only passes or limits the current.

This is the first point to understand.

Now, why is the voltage on the output of the capacitor only say 10v when the capacitor is connected to a 240v supply?

The voltage on the output of the capacitor is determined by the LOAD.

This is how the output voltage becomes 10v:

The capacitor can only supply 10mA.

When the load receives 10mA, a voltage develops across the load and in this case the voltage is 10v.

That is how the voltage is developed. It has nothing to do with the capacitor. If you put another load on the circuit, the 10mA may develop 35v across it. Another type of load may only develop 5v or 1v. This is why I say the capacitor has no control over the voltage on the output. It is the load that determines (or generates) the voltage.

sir,

thanking you for clear explanation sir,what will happen(necessity of) without 100R resistor..

sir ,

suppose if we using for 1 led or series of led bulbs for X rated capacitor ,then how should we determine the value & power of THAT 100R series resistor.waiting for ur reply

Hi

In this circuit R1 and R3 is getting very very hot in 10-15 seconds i tried replaced with 2 watt and 5 watt but the problem is same, only getting hot time increased to 15-25 seconds. Please help me where i did mistake.

What is the value of the capacitor?

Dear Colin Mitchell

Capacitor is Metallized Polyester Film Capacitor 450V 225K marking on capacitor is (MM225K 450Vob)

Dear Colin Mitchell

Capacitor is Metallized Polyester Film Capacitor 450V 225K marking on capacitor is (MM225K 450Vob)

Thanks for reply.

A 225 capacitor will allow 140mA to flow.

140mA through a 100R resistor will produce a wattage (loss) of 2 watts.

That’s why the resistor is getting hot.

Thanks for your reply. Please suggest what to change resistor or capacitor(what rating has to put) i only require to run one led and 12v relay(approx 30ma). Please suggest that this circuit will work 24hours continues(24×7) or not.

How many mA for the relay?

Sorry for wrong typing, relay coil is 12v and consumes 30mA.

What type of relay is it? 30mA is very small.

How many mA for the LED?

yes it is very small PCB mounted Relay(JQC -3F(T73))Coil Ratings: Nominal Operating Power : 0.36W

So i converted watt to mA (I =V/P, I = 12/0.36 = 33mA)and LED is 5MM 20mA.

So i think i need total of 53mA(33mA+20mA). I am new to electronics and confused also. Thank you Mr. Colin Mitchell for helping me.

You need 53mA. Use 105 capacitor or put two 225 capacitors in series.

This will give 70mA.

53mA plus 10mA for the zener and 7mA for the indicator LED.

Thanks Mr.Colin Mitchell

I relapsed 225 to 105 and output getting is 60mA. But the problem remain same the R1 and R3 is getting very very hot in 60-70 seconds i tried replaced with 2 watt resistors but result is same. I am not able to understand where i am doing mistake Please help!

R1 and R3 will dissipate 1watt with 105 capacitor instead of 2watts when using 225 capacitor.

This is the best the circuit can do.

This type of circuit is very wasteful.

Thanks Mr.Colin Mitchell

for giving your valuable time for replying my queries. Is there any other circuit can build transformer less which give output of 50mA 12V DC from input 220V AC which will work 24×7 without any problem.

You don’t need R3. Remove R3.

Replace R1 with 33R.

The circuit will not be more efficient and the resistor will not get as hot.

Mr.Colin Mitchell

awesome now the circuit is working fine no heating. thanks for your help.

Mr Colin,

I want to know one thing, if you can recommend some load disconnect mechanism, and that should be of Latching relay type.

e.g. As a load connect disconnect mechanism for Home Appliances and that should be electronic operated not a manual one.

Purpose is not to have a continuous current consumption that’s why latching type relay is required.

hi mr. can i touch the power supply after R4?

If your mains is not connected properly and your body completes the circuit to earth, you will receive a severe electrical shock. It is especially dangerous to curious children.

While you may wire the mains correctly, you are now depending upon how well the building outlets are wired wherever this power supply circuit may be plugged in…

This circuit is intended for “double insulated” applications where all circuitry (power supply and load) is fully enclosed within an insulating plastic enclosure.

colin sir ,

suppose if we using for 1 led or more series of led bulbs for X rated capacitor ,then how should we determine the value & power(WATTAGE)of series resistor.pls ans me

The series resistor does not change.

thank you colin sir, ihave through your website and seen youtube interviews

thanks all contributors..especially colin. this is a worth read thread.. got a really gud idea about capacitor power supplies..

Sir,i have a doubt.I am a final year Electronics and Communication Engineering student.Can we use optocoupler at tha output of this circuit to provide safety for load if supply fails.Because i am going to use it in water level indicator

While opto-couplers are great at isolating signals, they cannot isolate power. This is a bad idea since the water is probably at ground potential and there is extreme risk of electrical shock hazard. Check out my recent submission on Wall Warts & Wall Transformers.

Sir,please suggest me a safe power supply for water level indicator

If using this circuit:

electroschematics.com/5764/simple-water-level-indicator-2/

the power supply need not be regulated. It can accommodate up to 7 level probes if all sections of the ULN2003 or ULN2004 are utilized. I would perhaps make most of the LEDs the same color except for the top ones. Brightness is adjusted via the series resistors. Then use a transformer isolated 12VDC wall wart like one of the ones in this article:

electroschematics.com/6938/of-wall-warts-wall-transformers/

…and use 1K (or greater) series resistors for the LEDs because 100Ω is altogether too low.

Thank you sir.If i have any doubt,i will ask u

Sir, transformer makes my cicuit big.I need some other alternative to make my water level indicator compact.Please help me

Please reply

Hi,

This was great explanation for transformerless power supply

how to calculate the value of R1 & bleeder resistor R2 connected parallel to capacitor

Always use 100R and 470k

please can you help me how to calculate the ratings and value should be use in power supply circuit we are using these are all components bridge rectifier 4 capacitor and lm-7815 & lm-7915 and again we are connecting inductor to the out of the 7815 and 7915 and parallel we are connecting2 capacitor output of lm7815,7915 and we connecting other two capacitorto output of inductors we should get out put voltage +15v and-15v please explain briefely

i dont know the value of capacitor

A capacitor power supply is not suitable for your application.

colin mitchell sir

yestarday i saw its not good application but to practice how to find the value and ratings resistor capacitor we should known the details of design i am in entry level so please what are the steps to taken to calculate the value please send me the details

colin mitchell i did not get ur point as per coustmer required the hardware design want to do . please send ur email.id i will send the circuit and disscus

Dear colin/Mohankumar,

im new to electronics ,i have a small question in the above figure ,they mention phase and neutral.if by mistake someone inter change the phase wire and neutral wire.this will damage the circuit or not

The mains can be connected either way around. After all, the voltage is changing from one direction to the other about 50 times a second, so the wires can be connected either way.

thankyou Mr collin,i have a small cofusion ,which is written in the beginning of the post(The most simple, space saving and low cost method is the use of a Voltage Dropping Capacitor in series with the phase line.)

That means if dropping capacitor is series to the neutral line(when we plug),then also the circuit works fine,without a problem .please confirm

It does not matter if the capacitor is in the “active” line or “neutral” line because the mains is changing 50 times a second and the result is the same.

Dear Sir,

Please tell me the design for 350 mA Current

Put 3 x 2.2 microfarad X2 capacitors in parallel.

thanks sir i will try and give you reply

please tell me about led in series and led in parallel

I want to give supply to my computar speaker 9volt 0.1A

supply is it possible not to have trsformar to convert 240V 50Hz to DC 9volt .1amp can you give me the circuit

for such to convert 240V AC 59HZ to 9volt DC 9.1amp please let me know as early as possible and oblige

I want to give supply to my computar speaker 9volt 0.1A

supply is it possible not to have trsformar to convert 240V 50Hz to DC 9volt .1amp

The answer is NO

“sir, I have 105j capacitor, Can I reduce the current output to 20ma..I want use it to circuit them for led’s lamp”

A 105 capacitor in full wave will deliver about 70mA.

Use 3 strings of LEDs so each string will see about 23mA

member

“sir, I have 105j capacitor, Can I reduce the current output to 20ma..I want use it to circuit them for led’s lamp”

A 105 capacitor in full wave will deliver about 70mA.

Use 3 strings of LEDs so each string will see about 23ma

oke thaks sir,,your explanation is very help

full for me

“sir, I have 105j capacitor, Can I reduce the current output to 20ma..I want use it to circuit them for led’s lamp”

A 105 capacitor in full wave will deliver about 70mA.

Use 3 strings of LEDs so each string will see about 23mA”

Sir.,how many leds should be set each string in series, in order I have a long life led circuit lamp?

I looking forward to hearing from you.

A 105 capacitor in full wave will deliver about 70mA.

Use 3 strings of LEDs so each string will see about 23mA”

“Sir.,how many leds should be set each string in series, in order I have a long life led circuit lamp?”

Use 1 LED to 50 LEDs in EACH string. The same number in each string. You need 470 ohms in each string to reduce the current when turning the circuit ON and the resistor will balance the currents.

wow..you are very informative..Thank I like your articles

dear Mr collins

I have circuited led lamp with 60 leds, I made it in 3 string. each string has 20 leds , I use A 105j . as you said, if I make the circuit to 3 string it will deliver for about 23ma each string,

Sir., How much watt (power) my circuit is? and how do I calculate for some circuit if I make with different number of leds?

as you know,,I’m Realy beginer to electronics

Each LED is 50 milliwatts to 70 milliwatts

sir, i am working out on a circuit sir, my doubt is i have to run around 100 amperes current through the capacitor so please suggest me which type of capacitor i need to use sir.

you cant run 100 amperes current through the capacitor

Hello, and what a great article you have written.

My business Distributes a capacitor system that works both domestically and commercially, IE: bars, restaurants, and Hotels.

We have been achieving savings of between 30 to 48% when the system is wired in line to the 3 phases via the differential in a loop.

My question is can you send me a wiring diagram of how to, for example, wire on a 5 story hotel floor by floor so as not to change the installation of the building, as this would be illegal in Spain without the energy suppliers permission.

We currently install our plug in capacitors on each floor at the main Distribution panel on each floor.

I have a new Distributor and he would like a wiring diagram, can you help please.

Hello sir,

This article really helped me a lot.

My question is regarding drawback.

Low current output. With a Capacitor power supply. Maximum output current available will be 100 mA or less.So it is not ideal to run heavy current inductive loads. why?

But If I can design required Xc with proper selection of capacitance so that i can have current more than 100ma, so why this is limited?

please confirm my answer

current more than this will destroy the capacitor.

I appreciate what you are doing here. I would really need your help but would prefer email communication. i have a project am working on. Its complete. i need analytic judgement. further details will be through email.

Dear Sir,

I have a circuit which input is 230 Ac and out put is 120 v DC but suddenly it has started giving 350V DC output, its a transformer less circuit which consist two 22 micro farad 450v capacitor and much more

” it has started giving 350V DC output,”

It will always be 350V DC output

how to calculate out put volt this circuit ? please help me sir.

The OUTPUT VOLTAGE of all transformerless power supplies will be about 50% HIGHER than the mains voltage if a LOAD is not connected. That’s RIGHT: The output of a 120v CAPACITOR POWER SUPPLY (transformerless power supply) will be about 180v and a 240v mains transformerless power supply will be about 345v.

How do you get a 12v or 24v supply????

It works like this: The transformerless power supply is a CURRENT-DELIVERED power supply. In other words we have to talk about CURRENT-VALUES and not voltages.

For a bridge circuit (called a full-wave design) it will deliver 7mA for each 100n. Suppose we have 220n. We have 15mA available.

We take the 15mA and say: How many volts will develop across a 100R load? The answer = 0.015 x 100 = 15v. I f we use 82R the voltage will be about 12v. If we use 220R the voltage will be 33v. That’s how the output voltage is developed.

If you add another 220n across the 220n, the voltages will be DOUBLE. It’s as simple as that.

so,volt is load dependent.

pls sir, what is problem with a 230v and 400v transformer and rectifier that keeps tripping the DCCB (Breaker)

Put a fuse in the circuit in place of the breaker and see if it blows.

this have to do with a 230V and 400V transformer and rectifier plane.

how many banks of a 1.6v,90ah batteries can get 148vDC

Dear Sir,

I am currently working on capacitor charging circut. The capacitor rating is 2200uf 250v. Can u suggest me schematic to charge this capacitor.

Thank you.

Mr.harish

you can charge this capacitor upper power supply circuit jest open Zener diode and don’t touch it when you give power .

Dear Mr Faridul Islam,

My input is 120Vac on the secondary of the transformer, and the DC ouput required is 120Vdc to charge these capacitors. How do I design a regulator for 120Vdc? I intend to use linear regulator using transistors. Is it possible?

Thanks.

The capacitor rating is 2200uf 250v. Don’t you mean The capacitor rating is 2200uf 25v.

Dear Mr Colin Mitchel

Yes the capacitor rating is 2200uf 250v and 2200uf 160v.

the capacitor rating is 2200uf 250v

It must be very BIG

Get a flash camera and charge the electro from the output.

Dear Mr Colin,

Yes! Its a big capacitor only. Now i need to charge them @ 120vdc for which i am working on. I need some idea on regulator circuit for obtaining this constant voltage of 120v dc

How fast do you want to charge them

Dear Mr Colin

Have not got a thought on the charging time. Few ms may be. Its just a vague answer. No sure!! Will get back. Please propose a idea if it were to be milli seconds range.

mS is too quick. It has to be seconds.

Dear Mr Colin

Sure. Will get back once i study it again.Any idea on regulator circuit for 120v for seconds charging.

Thanks.

What is your input voltage

Hi.

My Input to bridge rectifier is 120vac. i.e the secondary of the transformer.

120vac will give you 170v DC to charge the 2,200u to 170v via a 1k 2watt resistor

Colin,

Right. But i shall need a 120v as my dc voltage. If i’m not wrong i need to regulate the rectifier o/p to obtain it. How do i do that!!!

Hi.. I am getting about 149vdc at the o/p of capacitor filter 200uf/400v and 1k 2w resistor. Got to bring down to 120v constant DC voltage. pls Help.

Hi.. I am getting about 149vdc at the o/p of capacitor filter 200uf/400v and 1k 2w resistor. need a 120vdc constant at my o/p.

A 220u 400v electro is placed across the output of the bridge.

The voltage across the 220u will be about 170v. DC

Why do you want 120v DC?

How much current do you want?

How long do you want to draw the current?

Sir I connect 225k capacitor series with my multimeter and check the output voltage, but I saw something against my expectation. That is the voltage varies from max 230 to 0. In my expectation it must be nearly 25v. But when I put 1M resistor parallel with capacitor the output willbe nearlly 118 volt. I get the same range for some other values of capacitor. Is it correct, pls rply.

Sir i want to connect 5 blue leds in series I have 24v 1amps transformer …. my question is how much power need for tis cicuit…

How much current do the LEDs take

25ma ….

If each blue LED has a characteristic voltage-drop of 3.5v, the total voltage-drop will be 17.5v for 5 LEDs. This means you have 24v – 17.5v = 6.5v to be dropped across the current limiting resistor.

But a 24v transformer specifies the voltage as an AC RATING and when this AC voltage is converted to DC via a bridge and electrolytic, the output will be 24 x 1.4 = 34v. There will be 2v lost across the bridge, making the output voltage 32v. This means the voltage to be dropped across the current-limiting resistor will be 14.5v.

If the current is 25mA, the resistance of the current-limiting resistor must be 580 ohms. Use 560 ohms

Tq Colin sir I understand from u …… so now i want to now about CA Blue color 4′ inch 7 segment . I not get data sheet .. Now how much current & voltage need for each segment…. Inside each segment they connect 5 leds in series . pls HELP ME SIR..

For power supply we can use IN4007 diode and Electrolytic how much uF to use….???

IN4004 diode and Electrolytic 470uF

Tq Colin sir I understand from u …… so now i want to now about CA Blue color 4′ inch 7 segment . I not get data sheet .. Now how much current & voltage need for each segment…. Inside each segment they connect 5 leds in series . pls HELP ME SIR..

so now i want to now about CA Blue color 4′ inch 7 segment . I not get data sheet .. Now how much current & voltage need for each segment…. Inside each segment they connect 5 leds in series

I cannot help you

why ,,,,,???? for five blue leds in series we can give –560ohms /32v means …. Then we can give same voltage with —560ohms resistor to each segment ….

Min and max input voltage for 7805 Ic…help me

Sir, please explain

how to design half wave power supply of 10 volt 100ma using 684k capacitor. Actually I want to make 4 min timer circuit using cd4060. Triac is use to switch AC to external bulb.

hello sir ,

i need ur help , for making a led message display , http://www.electronicsforu.com/electronicsforu/lab/ad.asp?url=www.electronicsforu.com/electronicsforu/circuitarchives/view_article.asp?sno=354&title=LED-Based%20Message%20Display………plz check this website and if i wantt to use thousands of led instead odf 172 led what change in this circuit plz help.

Hiii,

my Problem is, Power factor of my circuit is in (-ve) side, can you help me..

sir,i I have one doubt .How to choose the capacitor rating for single phase motor .THE motor rating is VOLTAGE=230v,CURRENT=6.7A,FR=50HZ,HP=1,SPEED=1430.PLEASE SENT THE VALUE.

hi

can we add mor then 1 led for thise circuit

can we put 10 or morr leds

thanks

Remove the zener diode and increase the voltage of the electro and you can add up to 50 LEDs in series

thanks Mr colin

Mr Colin

can i find mobile solar circuit charger about 1350 MA

i put solar cell and diod 4007 and 4 v batary 2 AM

but dosnt charge the mobile phon and thy battry is full ?

how i can solve the problem

if there is any circuit solar charger for mobile (cill phone )

thanks

Lethal! Never use a transformerless supply of this type. Anything connected to it can easily be raised to full mains potential even without any faults. As an alternative, there are hundreds of wall-wart psu’s to be found in trash: they either have a conventional safe transformer inside, or a pulse-mode circuit. Both give the necessary isolation from mains. Some have a built in rectifier, and some also a regulator. It is easy anyway to add those, and they provide a much more stable supply for your equipment.

Could save a life. Yours – or someone else’s! Take care.

Thanks Howard, I have been warning everyone about the dangers here for some time. While some think that they are essentially getting something cheap, they are sacrificing safety (both human and equipment). Consider an oscilloscope that is attached to such…and the scope is plugged into a USB port on your PC (that is grounded)…and then you throw the switch (turn on the fault current). Something just may blow up in your face and take your valuable equipment with it –or you just happen to bridge your body across the hot side of the line to ground. Get a wall wart power supply.

http://www.electroschematics.com/6938/of-wall-warts-wall-transformers/

Hello Jim, glad to make your acquaintance. The great electronic experimenters are young people: least concerned with safety, and the most likely to try any schematic they come across. The mains is no joke. Fifty years on, I still remember what it felt like to touch the 240 volt live chassis of a cheap radio and get a jolt right through to my feet standing on a damp stone floor. Momentarily thrown unconscious, but fortunately still around to tell the story!

Hi there. Can I use this capacitive circuit on a modified sine-wave inverter system? I used sensor lights and they both failed after a few minutes. Thanks for a great article.

No ,its not possible.

Dear Sir,

I want Cap drop power supply with 3.3V /.250A is it possible with below calculation please let me know.

X = 1 / {2 ¶ x 50 x 2.7 x( 1 / 1,000,000) } = 1150 Ohms

&

I = 230 V / 1150 = 0.2A

Yes, i like it

each 100n provides 7.5mA

That’s ALL you have to know.

Hey,Hey! hold on guys…This is a simple circuit 4simple applications requiring a few milliamps only. If u want2 draw more current over what Mr Collins & other well meaning people state then be prepared 4the consequences.The circuit can suddenly become unstable esp if the cap,s are stressed out2much as they usually are.One other point is that exploding caps esp of ratings >225or more can also unsettle lots of stable folks around the said pwr unit.For Lo pwr or 4curiousity ok…more than that use the transformer approach.Its mush safer n stable.(Its not live also!)Jus 4 the rec I had observed many Chinese LED powered flashlight batteries getting spoilt using this ckt.Rewire the same by using a Zener of 12Vand then 2a 7806 with a small clip on heatsink(Heats up when batt is Lo/dead.)try2use a 1000/12V and a 470/12 at the input&@the output.Ur batt will last a very,very long time.U dont have2 worry abt getting overcharged also(The reg limits the I flow2the Batt after it reaches full charge…)(:-)Take care Bye..

hell sir, i want a 75 v dc output voltage. so which capacitor i can can use in series with mains. thanks in advance

It doesn’t work like that. ALL capacitor-fed power supplies produce 350v output and it is the load that determines the final voltage along with the value of the capacitor.

Good day mr. Colin.

I will try 3 strings of up to 50 leds, at 20ma each string i guess 70-80 ma will be enough. My question is about durability, (having read all the coments about failing capacitors) i want to make a signal and use this at least 9 hours a day, an equivalent wall wart is more durable, or both options have about the same performance?

Also, if i want to add a ne555 (in a 4th string) can i use the same power supply (having modified the values, assuming 12v for the ne555, reducing the number of leds in this string), or should i use another power supply for the 4th string?

Thanks in advance.

The way to work out the current in a capacitor-fed power supply is to take the capacitor on the front-end and find its reactance. For a 2u2 capacitor, it is 1,450 ohms. This will allow a current to flow of 160mA. If the power supply is 20v, we take off 10% of 160mA= 140mA.

This is the current that will flow when the zener in the power supply is about 12v.

The wattage of the 100R on the front-end = .14 x.14 x 100 = 1.96watts and it will burn out at 1watt.

The voltage across the 100R in the power supply = .14 x 100 = 14v Power lost = 14 x 14 / 100 = 1.96 watts and a 0.5watt resistor will burn out.

I used this circuit (removing R3) with a ne555, and 12 strings of 4 green leds (6 strings sink + 6 source) and it works very well, leds are bright enough. But how is this possible, as I am using a 224K. I’m on 120 V 60 HZ. According to the figures here X=12K, so I=10 mA. So where does the mA are coming? Zener? Ne555? Or the calculationes in this circuit are wrong?

Hi Alberto,

You are right that the maximum current is about 10mA. You can measure it with an ammeter in series with the LED.

Since you mentioned about sink and source, the two 6-strings of 4 green LEDs possibly lighten up alternatively. Each of the 6 strings will pass through roughly 1.5mA, that’s not much current for one LED. However, when the 6 x 4 = 24 LEDs all lighten up at the same time, you will have the “bright enough” perception.

Both the zener and the 555 timer will consume a little current, leaving less than 10mA for the LED to consume.

The use of transformerless power supply dated back at least 15-20 years ago if not earlier. It was a historic innovation at that time when transformer power supply that is heavy, bulky and cost-ineffective is the mainstream power supply in small electrical and electronic gadgets.

Transformerless power supply are normally for loads that consume a small current ranging from a few mA to a few tenths mA. In recent years, they are also designed to supply larger current like those inside a LED light bulb. Every transformerless power supply is in fact custom design specific to the current load of the application, and is put inside an insulated enclosure. The risk of electric shock is minimal. In comparison to its transformer counterpart, it is more cost-effective in manufacturing and more compact. They are not supposed to be used as a general power supply.

For anyone who are interested in this kind of circuits, treat them for educational purposes, learn how and why the circuits work. One can surely try to build one to suit a specific application, but certainly not using it as a general-purpose power supply.

The use of a transformerless power supply dates back at least 80 years. It was one of the earliest ways to get HT and filament voltages for a radio.

The filament voltage for the valves was taken from a very large wire wound resistor and some of the European valves had a very high filament voltage so that 2 of them could be connected directly across the mains.

Any form of transformerless power supply is banned in Australia and yet electronics suppliers are still selling auto-transformers.

Mr John Lam: You are right, this circuit isn’t for everyone and shouldn’t be taken lightly insulation-wise. Danger is still present, if you’re not careful. As you mention it, this circuit should be used only for the tenths of mA, which is my idea. Still I was amazed to see how well it worked within that range. Tomorrow I’ll use the ammeter and maybe later post some other thoughts.

Thank you for your swift response.

Saludos.

please see attached file

Hi follow me,

Let’s talk about Q.5 first. When the circuit is connected to mains, C1 starts charging. At that instant moment, the current flows through the circuit is at its maximum. If the mains is at its peak value at that instant moment, the high current may trip the mains circuit breaker, and that’s the reason for the inclusion of R2 to limit the in-rush current when plug-in.

The value of R2 is chosen so as to limit the in-rush current but at the same time won’t dissipate too much heat. When mains is 200V, Vpeak = 1.414 x 200V = 283V; with R2 = 1kΩ, Imax (inrush) = 283 / 1k = 283mA.

Because of the inclusion of D2, which is necessary for the sequential charging and discharging of C1 in the circuit, R2 is conducting all the time (both in positive and negative half cycles), the power dissipated by R2(assuming 60mA current) = I^2 x R = (0.06)(0.06)(1000) = 3.6W. For safety, add roughly 50% margin, R2 should be rated for 5W. That explains why R2 was very hot when you ran the circuit.

Q.3 and Q.4, both are based on the specifications of the LEDs. Try reading this link http://en.wikipedia.org/wiki/Light-emitting_diode.

Q.1 Refer to other comments in this post. Take time to read through everything in this post before asking the question again.

Q.6 Theoretically, you can use a resistor in place of C1. The impedance of C1 at 200V 50Hz is 3.183kΩ (Z = 1 / 2πfC). At 60mA, power dissipated by an equivalent resistor = (0.06)(0.06)(3183) = 11.5W. Realistically, such resistor will be very big and not economical and practical to build. An ideal capacitor will only go through the charge / discharge cycles and will not dissipate heat. Practically, resistor will be used only when the circuit is designed to supply current of a few mA.

Q.7 Calculations are a bit complicated to show here as they involve vector calculations. Up to 3 decimal places, the equivalent impedance of C1 and R1, which are in parallel, is in fact the same as the impedance of C1 which is 3.183kΩ. Since R2 is in series, the total impedance of these three components is 3.336kΩ. (Using Ztotal^2 = Zc^2 + Zr^2)

Assume the total voltage drop of the four LEDs is 10.3V, and the voltage drop of D1 is 0.7V. Total voltage drop is 11V. The average current supplied by the half-wave rectified circuit (passing through the LEDs) = (Vmax – total voltage drop) / (Z x π) = (1.414 x 200 – 11) / (3.336 x 3.1416) = 25.94mA.

When calculating current passing through R2, use full-wave rectified formula, Irms = (Vmax – total voltage drop) / (Z x 1.414) = 57.61mA. So voltage drop across R2 = I x R2 = 57.61V.

The voltage drop across R1 and C1 in parallel = 57.61 x 3.183 = 183.37.

You may see that the total voltage drop does not add up to 200V. It is because individual voltage drops are out of phase to each other. When using vector calculations to add up for the total, the answer will be exactly 200V.

Google “how to calculate voltage drop in RC circuit” for more information. Hope these will answer most of your questions.

This is a half-wave circuit and the 105 will deliver 35mA through ALL the LEDs at the same time.

The current will actually be 70mA for half-cycle and 0mA for the other half-cycle.

The 1k resistor will dissipate about 1 watt

sir, i have connected 70 no.of 5mm multicolour led in series with 1 uf poly capacitor in ckt. it draws 10.68 ma current. please give me justification of that,

The 1u only delivers an average of 35mA when it is driving one to 5 LEDs. When you connect more LEDs to the circuit, the voltage on the top of the string is 70 x 2.5v = 175v approximately and the difference between the incoming voltage and the voltage being delivered to the LEDs is 240v – 175 = 65v. The 1u will only deliver 30% of 35mA or about 10mA.

what is that funda of 30%

voltage at output of bridge showing 260volt dc, and incoming of bridge supply voltage is 208 volt ac. across 1 uf it’s showing 56 volt ac. kindly explain my doubts……..

Hi kalamkar,

You mentioned bridge. Did you use a four-diode full-wave bridge rectifier OR use the exact circuit posted by “follow me”? If applicable, provide me with the value of R1 and R2 used. In addition, measure and give me the values of the AC mains voltage and the total DC voltage drop of the 70 multicolor LED in series. Moreover, tell me whether the measurements are from an analog or digital multimeter.

I will then justify all your measurements.

There is no DC in the circuit.

” the total DC voltage drop of the 70 multicolor LED in series. ”

Both of you have made the same mistake.

There is NO DC in the circuit.

The voltage across the LEDs is PULSATING DC and cannot be measured by either type of multimeter.

Hi Mr. Mitchell,

DC voltage may be steady, pulsating, or even has AC superimposed on it. When a voltage does not change its sign, the voltage is termed DC voltage. Even rectified AC with a smoothing capacitor is in nature a pulsating DC with minimal ripple voltage.

Cheap digital multimeter may not be able to measure DC voltage accurately when the DC voltage has been coupled with AC voltage. However, when the DC voltage is steady or just pulsating, either an analog or cheap digital multimeter will measure the average value of DC waveform, especially here we are dealing with either a full-wave or half-wave sinewave pulsating DC voltage.

Hi kalamkar,

You are connecting 70 multicolor LEDs in series, of which the total DC voltage drop is based on the specification of individual LED. The current that the circuit can supply is dtermined partly by the voltage drop of the LED string. On the other hand, the voltage drop of the LED string is based on the amount of current flows through. That means they are mutually dependant on each other.

To justify your measurements, the DC voltage drop of the LED string must be measured.

“However, when the DC voltage is steady or just pulsating, either an analog or cheap digital multimeter will measure the average value of DC waveform, especially here we are dealing with either a full-wave or half-wave sinewave pulsating DC voltage.”

So? What value is the multimeter going to read? How are you going to interpret the result?

Any calculations are worthless.

How does a cheap multimeter measure AVERAGE DC ?????

The whole circuit is much more complex than we are covering here.

For a start, LEDs are not like an ordinary LOAD. They are OPEN-CIRCUIT until the characteristic voltage of the whole string is reached.

This is not important when you have a few LEDs but when you have 70 LEDs in series, the operation of the circuit is completely different.

The characteristic voltage of a LED changes very little from a few mA to about 20mA, but LEDs from different batches can be quite noticeable. 70 LEDs in series is going to have a minimum characteristic drop of 120v and can be as high as 210v for white LEDs.

This means the voltage across the 1u is going to be between 150v and 225v and this will reduced its ability to deliver current. The LEDs will only turned on for less than 30% of the cycle and they will not be very bright.

Sir, Please check your above calculation. the X must be equal to 1.44k and not to 14.4k, then I=159mA. Please check it.

Varuna.

i made 25 led series connection in which i found 1 big problem that when dc side disconnected accidently & reconnect it damaged my all led in seconds, what will be the reason & how i control this, please guide me……………….

What value zener diode did you use?

no i didn’t used any zener diode

Without a zener diode, the electrolytic charges to 345v and when you connect the LEDs they BLOW UP.

ok sir, i will try with zener .if possible kindly clarify your answer posted on 6th nov.

still i had a doubts

for 70 no led it will require 3*70=210 volts dc. i used 1 mf capacitor which having xc=3184.71 for 50hz. on 15 ma current it,s drop is 47.770volt. whwn we reduce this from 230v rms it remains 183 volt rms on dc side it becomes 164dc then how my series run……

What colour LEDs are you using

multicolour led==having vf=3.2

sir, i have also 1 doubt that for filtering i have used 10uf,400v electroly. capacitor, if i creased that value can i get high dc voltage.

For 70 LEDs you just need a bridge and one current limiting resistor.

For 70 LEDs you need a bridge and then two 1k2 resistors in series. Put the electrolytic on the end of the resistor to 0v. Now add two more 1k2 resistors and on the end of the 4th 1k2, put the 70 LEDs.

what is the formula for selection of filter capacitor please tell me?

Is there is increase in dc current or higher brightness by increasing filter capacitor value?

Are you talking about a filter capacitor or a supply capacitor

Respected sir, I am talking about filter( electrolytic capacitor)……

Let’s start from scratch.

I have 25 white LEDs. Specifications of the white LED are, Voltage: 3.2-3.4VDC, Forward Current: 20mADC. The LED series string will be driven by a capacitive power supply composed of R1 and C1 connected to 120VAC @60Hz and a bridge rectifier (1N4004)and a smoothing capacitor C2(10uF 250V).

Based on specification, the LED string will drop (3.2 x 25)VDC = 80VDC, and roughly 20mADC current is required, we first have to calculate the value of the limiting capacitor.

1) Roughly calculate the peak voltage available on the input:

Vpeak = Vrms – Vload = [120 x SQRT(2)]V – 80V = 90V

Here we ignore the voltage drop of the bridge rectifier.

2) Calculate the impedance on the input that will give DC 20mA:

Z = [(90 x 2 ÷ π) / 0.02]Ω = 2.865kΩ

Here we ignore the resistance of the resistor in series.

3) Calculate the value of the limiting capacitor:

Z = 1 / 2πfC ==> C = 1 / 2πfZ = 0.9259uF; closest is 1uF.

Therefore, a 1uF capacitor is required to build the circuit.

After building the circuit, we measure the volatge and current at different points of the circuit, and the value of the components connected using a cheap multimeter (cost C$10) and compared with the theoretical values. By doing so, we will have a better understanding of how and why the circuit works as well as learn some basic calculations.

Actual figures: AC mains = 122.8VAC,

Actual resistance of R1 = 219.5Ω

Actual capacitance of C1 = 1.043uF, ==> reactance = 2.543kΩ;

Total impedance of R1 + C1 = SQRT ( R1^2 + C1^2) = 2.55246kΩ

Measured voltage drop of LED string = 80.2VDC

Measured average current passing through LED = 21.80mADC

Now calculate the theoectical current:

[122.8 x SQRT(2) - 2x0.7 - 80.2} x2 ÷ π] / 2.55246 = 22.96mA; the 2×0.7 is the voltage drop of the bridge rectifier in each half cycle when two diodes are conducting.

Compare the calculated value of 22.96mA with the measured value of 21.80mA, the difference is about 5% which is very satisfactory taken into consideration measurements are taken by a cheap meter and the actual components are not ideal components.

Other measurements worth of interest are:

Measured current on mains input = 31.85mAAC

Measured Voltage of C1 = 73.2VAC

Measured Voltage of R1 = 6.98VAC

Total voltage of R1+C1 should be 73.53VAC based on these figures. However,

Measured Total Voltage of R1+C1 = 73.8VAC, a 0.37% error.

In my circuit, a smoothing capacitor with an appropriate higher value will theoretically smooth the current flow and reduce the flicker of the LED that is in fact invisible to human eyes, but certainly has no effect on voltage.

The image file is the circuit for the related discussion.

There is one type error in my last explanation.

When I mentioned that we want the capacitor to at least maintain 50% of its initial charge, the equation used is “time required = ln(1/2) * RC = 0.693RC”; it should be “time required = ln(2) * RC = 0.693RC”. 0.693 is from ln(2) which is correct in the original equation.

Although that type error doesn’t affect subsequent mathematical deductions, it’s better to provide you with the right concept.

sir if i used half wave rectiier i.e. 1 single diode then what should be the values, as i tried & lost my all leds….waiting for your reply

thanks for your valuable answer sir, but my basic query still unanswered….what i want to know that if i choose correct filter capacitor is there any increse in brightness or not?

Hi sunil,

Before I provided you with the above explanation, I read all your questions posted since Nov. 5. You questions ranged from why a certain value of the limiting capacitor will deliver different current for different load voltages to the justification of various voltage drop at different points in the circuit. My explanations probably answer all your curiosities if not helping you to design a capacitive circuit from scratch.

My circuit didn’t include a 80V zener diode because the supply current is only about 22mA that probably will be all consumed by the LED string leaving the zener diode unconductive. A zener diode requires a few mA current to function probably. The only benefit to include a zener diode in my circuit is the protection of the LED string in case the LED string is accidently diconnected and re-connected again or there is a loose connection. However, the rating of the zener should be rated for 2W or higher (P = V * I = 80 * 0.021 = 1.68W). For your 75 LED string with 10mA current, P = 225 * 0.01 = 2.25W or higher.

As mentioned, increasing the value of the electrolytic smoothing capacitive would smooth the current flow but the effect was insignificant to human eyes, so the calculation for choosing it is skipped.

Since you are eager to know, here is a simple way to calculate.

Discharge rate of a capacitor is an exponential function (related to e = 2.71828). However the equation introduced here is a linear approximation that will work sufficiently well as ripple time is a small percentage of the stable time in the analysis.

The term Time Constant, T = RC, is the time taken for the capacitor to drop its charge or current to 1/e (0.3679) of its initial value. For practical use in smoothing, we want the capacitor to at least maintain 50% of its initial value, and the time required = ln(1/2) * RC = 0.693RC. Remember this is the time the supplied voltage (currednt) is in a more stable condition. We now call it Time(stable).

For a full-wave rectified output, the time of each half sine-wave, which is actually the ripple, is 1/2f. For 60Hz mains, Time(ripple) = 1/120 = 0.008333. For 50Hz, Time(ripple) = 0.01.

Ripple Ratio = Time(ripple) / Time (stable), so a ratio of 0.05 means 5% ripple.

That means: Ripple Ratio = (1/2f) / (0.693RC)

Rearranging, C = 1 / [(2f)(0.693R)(Ripple Ratio)]

For circuit with V and I available, (like our circuits), we can substitute R by V and I. So C = I / [1.386fV (Ripple Ratio)]

Example, based on my circuit, I = 0.0218A, Voltage = 80.2V, f = 60Hz, Ripple = 0.05 (5%) ==> C = 65.37uF.

For 1% ripple, C = 326.84uF.

For your circuit, assume full-wave rectified, I = 10mA, V = 75V, f = 50Hz, and you think 5% ripple is acceptable, C = 38.48uF, for 1% ripple, C = 153.92uF.

For half-wave rectified circuit, multiple the answer by 2 as T(ripple) = 1/f instead of 1/2f.

Let me know if you notice any difference in brightness by increase the value.

There is one type error in my last explanation.

When I mentioned that we want the capacitor to at least maintain 50% of its initial charge, the equation used is “time required = ln(1/2) * RC = 0.693RC”; it should be “time required = ln(2) * RC = 0.693RC”. 0.693 is from ln(2) which is correct in the original equation.

Although that type error doesn’t affect subsequent mathematical deductions, it’s better to provide you with the right concept.

Adding a reservoir capacitor (it is not a filter capacitor) will increase the brightness considerably. Changing from 1u to 10u to 100u will not make much difference.

” RC = 0.693RC”

One “Physics” point to note is this:

It is silly providing an accuracy to 3 decimal places when a capacitor is +50% -20% tolerance.

thank you very much sir, i’ll definietly try this phanda

i am having 1 problem in full wave rectifier with capacitor.

problem is:

A fwr with capacitor filter having load resistance 150 ohm and caspacitance is 100 microfarad. calculate ripple factor. so in this what frequency should i took for calculation????

Hi pradnya,

Equations for calculating FWR ripple voltage or ripple percentage in any electrical or electronic textbooks alreday taken into consideration that ripples are twice the mains frequency, and the denominator includes the factor 2f where f is the mains frequency. Therefore, the frequency used should be just the mains frequency in such equations.

Unless the equation you used is a modified equation found on a website that specifies the “f” in the equation is already ripple frequency, you should always use the mains frequency for “f”.

If your mains frequency is 50Hz, using the general equation found in most textbook, Ripple% = 1 / (2fRC) = 66.67%, meaning the capacitor value is too small.

100Hz

sir i’m have connected a same circuit but i’m getting 218v across the bridge rectifier so could tell me the solution for bcoz i’m stucked over here

i have made 25 led series by using 5 parrallel path. 5 led in each sstring, can any one provide me running light circuit ,

or can i use this circuit in any running light circuit . if any one having ,, then plz provide. so that it help me to complete my project.

What is your supply voltage?????

sir,after so much try i got one running light belt circuit for 300 led i.e. 50 led in one path.

i want to run my 25 led earlier mentioned circuit on it, kindly suggest me requisite changes in the attached circuit. i tried so much but can’t succeed. plz help

i want to connect our circuit output ( un filtered )to attached circuit, kindly provide me the connection guidance accordingly

waiting for reply sir

where r all guys ? no solution

hi john lam suppose i want to run 25 series led series by using 1 diode what will be the values of capacitor and resistor , my input volttage is 230 volt ac …..waiting for your valuable reply

250v

is anybody hear me ? don’t worry i will not ask any new question……….

25 or 50 LEDs will work on the PCB shown above

no sir i have connected 25 led , all are blown within 10 sec. ,

If 50 LEDs work, then 25 LEDs will work. It is a constant current circuit

No sir, they have used half wave I.e. Used 1 diode and supply given to 6 paths….. 6 scrs gate fired sequentially.

25 led come across half rectified 230 volt circuit,peak 325 volt. So my basic question is suppose I want our capacitor switching circit to use full wave,is that possible?

Hi sunil,

Just looking at the PCB with different components on it, I was unable to figure out what kind of circuit it is. So I gave no comment until I now better understand the circuit by summing up all the information you gave in your last few postings.

You said the LEDs are driven by half-wave rectified DC, meaning it would automatically turn off during the negative half cycle. The original circuit is designed to connect 50 LEDs (50 x 3V = 150V) and can safely connected to HWR that has an effective of Vrms = Vpeak / 2 = 162V. However, when you only connect 25 LEDs (25 x 3V = 75V), they will blown off right away with an applied voltage of 162V.

Hi sunil,

I have to admit my last explanation is TOTALLY wrong. Mr. Colin Michell is right that with a peak voltage of 325V, the LED string would blown off instantaneously.

The part that I agree to “the LEDs are driven by half-wave rectified DC, meaning it would automatically turn off during the negative half cycle” remains valid. Of course, it’s only a possibility until the PCB is flipped over for the connections.

The board clearly shows fullwave AND SMOOTHED DC to the LEDs

The board clearly shows fullwave AND SMOOTHED DC to the IC and halfwave to the SCR’s to turn them off.

DC will not turn them off.

“The original circuit is designed to connect 50 LEDs (50 x 3V = 150V) and can safely connected to HWR that has an effective of Vrms = Vpeak / 2 = 162V. ”

This is entirely incorrect.

The half-wave rectifier still delivers 345v and any LEDs less than this will blow up immediately.

50 LEDs are equal to a zener diode and must have a limiting resistor.

Stop putting false information on the web. It just makes my job harder.

sir you are correct 50 leds are with limiting resisitors nearly 2400 ohm in each path.]

but still my basic question is not answered regarding how to use capacitive power supply circuit in this circuit to run leds without connectingg series resistor to them?

before that i want to share some info. with you that i disconnected diode and disconnected the return path of main supply next to ic and inserted our capacitive power supply circuit but it only blinks at starting and not run as i desired ..please find remedy on it so that i can save resisitive power loss in led series……

I have absolutely no idea what you are doing but you have to remember to discharge the capacitor.

ok just tell me how we can series resisitor in led path?

ok how we can eliminate series resisitors in led series path ?

what i did for that i have attached pcb,s reverse side image. which didn,t succeed

my basic goal is use capacitor power supply circuit for running light and i hope genious persons like you would help me to achieve this ……

You need a series resistor to absorb spikes and when the circuit is turned ON. It can be about 1k. The capacitor limits the current.

Provide a photo of the underside of the PCB shown in the photo above.

i can’t understand which photo u want, i already posted image of underside of pcb

if possible kindly provide me circuit diagram of what u want to say

“i already posted image of underside of pcb”

Where????

underside of pcb

Hi sunil,

The PCB bottom is NOT from the same PCB that shows components. Though I am sure the LED strings are driven by FWR DC, can you post a photo of the component side of this exact board?

Provide the component no. of the IC if you can read from it. I want to figure out the exact reason for burning your 25-LED string. The zener probably is for regulating voltage for the IC, and should be of comparatively low voltage. It doesn’t sound like the 25-LED string was burnt by the high voltage of the smoothing capacitor.

Annway, I will appreciate if you can take a photo of the component side of the exact PCB that shows the bottom side.

Another typo, FWR should read as HWR.

I completed the circuit up to bridge and then I kept both ends open. did not connect zenor or any thing (no load) and then I measured the output across bridge outputs. It was 218V. After connecting 550K direct between bridge out no zenor no smoothing capacitor; still it shows 218V across bridge output. Is it right or I thought It should show 15V.

hi john i have to use our caopacitor FWR supply instead of this diode , is it possible to achive ? for ref i have attache dthe back side of the respective pcb

Hi sunil,

Attached is the circuit of your PCB. The 2.4kΩ resistor may be locate just directly in front of the HWR diode. For simplicity, I have only drawn one of the six outputs of the IC.

The PCB has both the capacitive and resistive power supply. The capacitive supplies power to the IC, whereas the resistive part feeds the LED strings.

When you only connect a 25-LED string, the current is higher than connecting 50-LED string. That might be the reason of the blown out.

Cut the PCB at the two locations marked with “x” and connect your capacitive power supply output to the circled + and -. Without cutting the lower “x”, the current from the CPS flows to the ground and will not light up your LED string.

Give me the number of the IC if you can see it.

Sunil,

I don’t know why you want to do it this way. Just remember the LED string will stay on forever because the power supplied now is FWR instead of the original HWR. You might consider using the original PCB and increase the value of the current-limiting resistor.

hallo john sir, you r saying correctly by doing this scr will gone in on stage only, is there any option instead of scr….

Sunil, you still haven’t provided me with the part no. of the IC. Even if you couldn’t read from it because of masking, at least let me know.

I can understand the satisfaction of using something built by oneself and your insistence of connecting your own capacitive power supply. Replace one SCR with a small NPN transistor, one example is 2N3904. Viewing from the bottom of the PCB, the SCR has KGA from left to right, connect the NPN in EBC order.

Try cutting only the HWR connection, it should work. You may need to lower the value of the 10kΩ resistor to increase the base current. It all depends on the specification of the NPN used. Let me know if it works in addition to providing the part no. of the IC.

Hi sunil, I forgot to mention one thing that’s very important.

For your capacitive power supply, DO NOT connect any smoothing capacitor unless a zener around 75V has also been connected. Otherwise, you have to MAKE SURE you have discharge the capacitor using a 1kΩ resistor before completing all the connections. Or your LED string will blow out again.

how to calculate voltage as per following table.

Listen,Listen

It Is Very , Very Important4me2stress that this circuit is for study &evaluation Purposes BY VERY WELL EXPERIENCED SENIOR PLAYERS FAMILIAR WITH THE TRADE AND SHOULD BE STRICTLY OFF LIMITS TO ALL BEGINNERS MANY OF WHOM, SADLY ARE HARDLY THE KIND WHO SHOULD HAVE ANYTHING2DO WITH THESE CIRCUITS.Why? 2 imp reasons, Its live,second no matter how well its planned, these circuits are technically “Wonky” as most of the components are leading a rather stressful life out there and its unreliable at its worst especially the main cap which is at the core of the whole setup. Its very easy2goof Up & screw up or worse get screwed up in the whole seemingly but essentially a technically complicated process.Beginners STAY OUT OF THESE CIRCUITS ESP THE NERDS AND ROOKIES OF WHOM THERE ARE PLENTY..Bye Take Care…Apna Khyal Rakhna(:-(

Much agreed! Applications for such should be limited to circuits that can be easily isolated via electrically insulated enclosures to prevent electrocution and damage to eyes from flying debris. Though simple and cheap, you are not getting something for nothing –essentially trading safety for a little money.

My recommendation: Use wall wart power supplies wherever possible.

no , it’s quite good n reliable ciruit. if properly all parts are soldered n keep in insulated cover/box.

Dear Mohankumar

Based formula X = 1/(2Pi*f*C),

1 = 2Pi*f*C*X

Your article has C value, But how to fin C or X value? if :

I have Vin = 220v, Iin = 2A

Need V = 3v and I = 50A

How to find R or C value without guesting 1 by 1 value??

I out cannot be larger than I in

Hello Sir,

Sir which type of capacitor is this.?

is it Metalized polypropylene film capacitor ya

metalised polyester capacitor.?

It is called x2

Hello Sir,

I want to connect 18 LED having voltage rating 3.2 to 3.4V and current rating is 150mA . I have selected the X rated capacitor value is 177J 400V to that i have connected R2= 470K (1/4 W) Resistance . Diodes are are 1N4007 (full wave bridge rectifier ) after that i have connected C2= 47uF 100V and R3= 470k Resistor parallel to that Capacitor. when i have switch on the 230V 50Hz supply without connecting the LED i.e without load it is showing the output voltage across R3=470k resistor is 105V but after 30sec to 40sec the C2= 47uf 100V Capacitor get burnt. Then i have removed that capacitor and measured the voltage across the Diode it is showing 285V …. Now please tell mi how to get 58V DC at 150mA supply from this 285V what is the rating of C2=capacitor that should i connect so that i will not blown..

There is not such capacitor as 177

The LEDs produce the voltage. They MUST be connected.

Sorry Sir its 175J 400V

For half wave, each 100n delivers 3.5mA For full wave it is 7mA

100n = 104 1u = 105 175 = 1u7

1u7 with half wave = 3.5 x 17 = 60mA

1u7 with full wave = 7 x 17 = 120mA

TOO MUCH CURRENT

Use half wave and put 2 x 175 in SERIES.

ok sir , what should be the value of smoothing capacitor ..

I’m using this circuit .. is it correct one ??

10u to 47u @100v

You must add the LEDs or the electro will bow up.

You must put 4 x 1u7 in series with the bridge or the LEDs will blow up

Thank you sir,

I have connected 2.2uF 400V C1 capacitor along with 470K (1/4 W) remaining circuit is same as previous and i have connected 18 led having rating 3.4V 150mA .

The circuit is working fine but the current at AC is 124mA @ 258V AC . and Across LED the voltage is 2.998V & Current is 134mA .

I want to reduce the Input Current of AC 124mA to 60 to 80mA but at output side the values should not be change i.e voltage across led should be 3V and current is 134mA .

How to do this

put 2 2u2 in series or use 1u

when i ll put the 2 2u2 400V the voltage will be 400V or 200V i.e will it remain same or get half. and at out put side the across led will get 134mA 3V ??

The voltage will be 345v and 60mA

Thank you sir Input current become 70mA but the current across the led array is decreased i.e 55mA and intensity of led decreased

Change the 1u

All I want to decrease the input wattage by reducing the input current to below 40mA . but at same time the output should have current 130mA or grater than that. How to do this please help mi sir

Changed to 1uF 400V Input AC current 76mA and at across LED array it is 72mA

Excellent! Very useful and informative!

Hello Sir What will be the power factor of above circuit if the X rated capacitor is 205J 400V . What will the power consumption for this circuit for AC Voltage 230V

“What will be the power factor of above circuit if the X rated capacitor is 205J 400V”

The power factor is not a value that is needed for any type of capacitor-fed power supply.

“What will the power consumption for this circuit for AC Voltage 230V”

The power consumption for a capacitor-fed power supply is almost zero, that’s why it is used.

It means AC Power Consumption doesn’t matter here all we need to calculate DC power . Am i Correct Sir ??

It means there is almost no heat generated in the capacitor.

Suppose i want make 12W led array then i have to used 0.5W 24 LED so that Wattage will be 12W but at same time what is AC power consumption will be P= 230 X .150 = 34.5W (without PF) so how the array is consuming the only 12W in AC n DC both side ??

Of course the AC wattage is 12watts and the DC wattage is 12watts because nothing is lost in the capacitor.

In other words, there is nothing lost in the conversion.

When i have measured the current and Voltage across the 10W Led bulb at

AC side it is V= 230V and current is 40mA the P= 230X.043 = 9.996W at same time at DC the array of 0.2W 45LED having again same consumption P= 0.2X 45 = 9W so total power is approximately same at both side but that led bulb contains Transformer based driver.

While in this Drive the Above drive the AC Consumption for 205J 400V is

P= 230 X .144= 33.4W so if want to use this circuit for 12W led buld i need to connect 0.5W X 24 LED so P= 0.5 X 24 = 12W

but in actual the it is consuming 33.4W so total loss will be 33.4 – 12 = 11.4W

all these things i have said is that correct or Wrong

Can you construct ENGLISH sentences . . .

Sorry sir I’m weak in English.

When i have measured the current and Voltage across the 10W Led bulb at

AC side it is V= 230V and current is 43mA so the power is P= 230X.043 = 9.996W. at same time when i have measured power at DC side it is P= 0.2X 45 = 9W for 0.5W Led and 45 led conncted in series array . so total power consumption for AC & DC side is approximately same. The Led bulb contains Transformer based driver.

While in capacitor based Driver which is given above circuit the power consumption at AC side is

P= 230 X .144= 33.4W for 205J 400v X rated capacitor. so if i want to use this circuit for 12W led bulb i need to connect 0.5W X 24 LED so power consumption at DC side will be P= 0.5 X 24 = 12W

but at AC Side power consumption is 33.4W. And At DC side power consumption is 12W .

so how we can say that bulb is 12W if it is consuming 33.4W.

I have already told you that there are no looses with a capacitor-fed power supply

Nothing is getting HOT, so where are the losses ?????