To get precise voltage to a circuit, the simplest and easy way is to use a Zener diode.It uses the reverse breakdown voltage characteristics to maintain a fixed voltage across it

Zener diode ZD 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. The zener output voltage is not seriously affected by the resistor and the output remains as a stable reference voltage. But the limiting resistor is important, without which the zener diode will be destroyed. Even if the supply voltage varies, the series resistor will take up any excess voltage. The value of the series resistor 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 R1 to get 12V output is 6volts. If the maximum zener current allowed is 100 mA, then R1 will pass the maximum desired output current plus 5 mA. So the value of R1 appears as

R1 = 18 – 12 / 105 mA = 6 / 105 x 1000 = 57 Ohms

Power dissipation

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.

As a rule of thumb, a minimum of 10% of the total current should flow the zener leaving the rest available to the load. By increasing the value of the series resistor, it is easy to reduce the current flowing through the zener but allow a minimum of 5 to 10 mA current through the zener. Then only it will reach the avalanche state to give output voltage.

You can use a variable resistor in the place of R1so that accurate breakdown voltage can be obtained. But use a dummy load (LED with 1K series resistor) to fix the output voltage as desired. Adjust variable resistor till the Vout is reaching the required output voltage.

I made capacitive power supply (680K400V || 1M ohm) of rating 12V 60mA.but I got 230V at the rectifier output is it right or wrong..?

My ckt has following steps

1.68Moh0V,1Mohm

2.Rectifier

3.10K ,5 W(zener current limiting resistor)

4.zener diode 27V

5.7812 Vregulatorgulator

@neon.

To avoid that problem, use a zener of 6.8V and a resistor of 62ohms.

Hello Sir,

i am trying to use 12v battery to my load. i need 12v to get the correct light intensity of the load. but as time progress the battery power diminishes and brightness reduces. so my question is either by using any voltage regulator will increase the time of brightness. if yes then which one will be appropriate zener or an IC.

waiting for your reply.

Sir,

I need to get a 3.3v reference. I am using this zener after 7805. But the output of 7805 varies from 4.2-5.7. The voltage across the zener also varies as the output of 7805 varies. I adjusted to 3.29 volt by giving 470ohm resistor. If I use 680 ohm, the voltage across zenar is near about 3.0v. Why is this variation? Can i make a precise reference using this zenar(3v3) for two decimal points?

Low voltage zeners (below 6.2V) are sloppy and become progressively worse as voltage rating decreases –voltage is highly dependent upon current.

Instead of a zener and 7805, I recommend a single LM317 with proper voltage setting resistors without zener.

It is difficult to increase the power rating of Zener by connecting them in parallel.If you want a high power supply, some circuits are in electroschematics

Sir, Thanks for your reply . I have solved this problem . Today morning I went to the market and bought a 8.2E 5W Micron Resistor . Now the heat is in totally under normal condition . And Zener diode also is not getting any heat at all .

But still one problem is left there . That is current limit of Zener . This 1/2W Zener cann’t pass 600mA Current . So, I need 2W Zener to pass 600mA current . But sir, I have searched the whole market for 2W Zener diode . Zener is very rear in my market . Few of these shops keep Zener diode and they have not more than 1/2W Zener . So, I cann’t buy the 2W Zener .

Sir, If I add four 1/2W Zener in parallel then will it be work like as 2W . I mean – 1/2W + 1/2W + 1/2W + 1/2W = 2W ( In parallel ) . can I make 2W in this way and will it pass 600mA current ? Sir , please reply .

– Thanks

– Neon

We cannot pass this high current through the zener. It will heat up.Use a 7806 regulator IC or use the circuit,High current voltage regulator in the site