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  • This 12 Volts transformerless power supply take advantage of the fact that a Zener diode is also a normal diode that conducts current in the forward direction. During one half wave, the current flows via D1 through the load and back via D4, while during the other half wave it flows via D3 and D2. Bear in mind that with this circuit (and with the bridge rectifier version), the zero voltage reference of the DC voltage is not directly connected to the neutral line of the 230-V circuit.

    Transformerless Supply Circuit Schematic

    12 volts transformerless power supply diagram

    This means that it is usually not possible to use this sort of supply to drive a triac, which normally needs such a connection. However, circuits that employ relays can benefit from full-wave rectification. The value of the supply voltage depends on the specifications of the Zener diodes that are used, which can be freely chosen. C2 must be able to handle at least this voltage.

    The amount of current that can be delivered depends on the capacitance of C1. With the given value of 220nF, the current is approximately 15mA. A final warning: this sort of circuit is directly connected to mains voltage, which can be lethal. You must never come in contact with this circuit! It is essential to house this circuit safely in a suitable enclosure.

    
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    11 Responses to "12 Volts Transformerless Power Supply"

    1. Pentaconto says: on August 23, 2010 at 1:31 pm

      In the Circuit Diagram are not labeled the parts (C1, C2, D1, D2, D3, D4), the explanation of how work the circuit is not good because this problem. Please correct this in the circuit diagram.

      Greetings from Barcelona (Spain)

    2. dear sir

      can i use this circuit for your Automatic Gate Lamp circuit

      thank you…

    3. I am not sure how this circuit might benefit the relatively inexperienced readers seen here.
      The output current is minute, barely capable of driving a bright LED, & that requires further voltage reduction, details of which are not shown.
      Nor do you suggest an appropriate fuse size, or include an MOV. Surely bearing in mind your warnings these are almost mandatory, that is if you wish to be part of the solution not the problem.

    4. O.C.Bessong says: on July 18, 2012 at 6:24 pm

      Pls sir, i need a complete writeup about this circuit. Thank you!

    5. thank you i need a complete writeup about this circuit…..

    6. pls send me simple circuit to produce 12v dc without using transformer

      • kbdprasadrao@gmail.com says: on October 23, 2012 at 8:05 am

        it is really simple but failed for many times

      • John Lam says: on August 25, 2014 at 6:08 am

        This circuit is quite innovative in the sense that it uses two rectifer diodes instead of four in a bridge rectifier to get a full-wave rectified output. Of course, it has to use two zener diode instead of one generally used to accomplish this.

        Let’s do some calculations:

        1) 220nF capacitor is for limiting the current of the whole circuit, the impedance, Zc = 1 / (2 * pi * f * c), where f = 50Hz; Zc = 14.469 kilo-ohms.

        2) 100 ohm / 1W resistor is to limit the inrush current when first plug-in. Since the value of this resistor is significantly low compared to Zc, the total impedance of the resistor and the capacitor in series (Ztotal) is the same as Zc to 3 decimal places. Formula of total impedance for RC in series = SQRT (Zc^2 + R^2). Do the calculations, identical…

        3) Current generated = V / Ztotal
        = [(Vmax - Vzener) / SQRT(2)] / Zc
        V = (Vmax -Vzener) / SQRT(2) = 211.52V

        Current generated = 211.52 / 14.469k = 14.62mA

        4) Power dissipated by 100 ohms resistor = I^2 * R = 0.0214W.
        Therefore, any 1/8W and higher resistor is good.

        5) Vmax = 220 * SQRT(2) = 312V, so the 220nF capacitor selected should be at least 350V. The 250V suggested is guaranttee to burn out very soon.

        6) Since the current generated from this circuit is only 14.62mA, the voltage drop on the rectifier diode should be 0.6V. Therefore the output voltage = Vzener – Vdiode
        Voutput = 12V – 0.6V = 11.4V

        Power dissipated by the diode = VI = minimal

        The 1N4004 that can handled a RMS reverse voltage of 280V selected is a good selection.

        7) Maximum power dissipated by zener diode without load
        = VI = 12 * 0.0146 W = 0.1752W. So 1/2W or up is good.

        8) The specifications selected for the smoothing electrolytic capacitor is fine. For substitution, 22uF or up is good for the circuit current.

        In case you want to connect the output to LED, just change the value of the two zener diode to match the volatage of the LED plus 0.6V, or if necessary, with a resistor in series to drop the voltage.

        That should cover everything for those who want to understand the circuit and math.

      • John Lam says: on August 25, 2014 at 7:26 am

        My mistake in one of the calculations.

        6) Since the zener diode connects directly to AC instead of the usual rectified DC output, the potential difference across the zener diode should be 12V (reverse bias) + 0.6/0.7V (forward bias) for a total of 12.6/12.7V.

        The output voltage should be 12.6/12.7V minus 0.6/0.7V (forward voltage drop for the rectified diode) = 12V.

        That means the output voltage is the same as the voltage of the zener diodes selected. No subtraction is necessary.

    7. dear sir, pls send compelete circuit diagram. and sown component value.

    8. sir,
      this is reliable circuit that can use for industry purpose

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