Transformerless Capacitor Dropper Power Supply

Transformerless Capacitor Dropper Power Supply

I already posted about transformerless power supplies in the article, Transformerless DC Power Supply. Here we will see how to design a capacitor dropper power supply. Capacitor power supplies are simple, low cost and light weight solution for providing dc supplies to circuits which require low currents. It is low cost and light weight since there is no bulky transformers.

Circuit Diagram

Transformerless Capacitor Power Supply 12V 40mA

Transformerless Capacitor Power Supply 12V 40mA

The 200mA fuse will protect the circuit from mains during shot circuit or component failures. The 275V MOV (Metal Oxide Varistor) will protect from power supply spikes or surges. The X Rated Capacitor C1 is the core part of this power supply as it will drop the excess mains voltage across it. The excess energy will not dissipated as heat as we are using capacitor dropper instead of resistor. The resistor R1 is the bleeder resistor for capacitor C1. Which will discharge the capacitor when the supply is switch off, so it will prevent any shocks due to capacitor charge. Resistor R2 is provided to prevent excess transient current that can flow when the power supply is switch on.

Diodes D1 ~ D4 constitutes a bridge rectifier which will rectify the input ac power. Among these D1 & D2 are zener diodes. So the rectified output will gets clipped at its zener voltage. Capacitor C2 is the filtering capacitor which will filter the rectified ac voltage.


Capacitor Dropper Power Supply - Working Animation

Capacitor Dropper Power Supply – Working Animation

Working is self explanatory in the above animation. In the positive half cycle diodes D1 & D2 gets forward biased and current flows through the load. Output voltage will be clipped by the zener effect for diode D2. In the negative half cycle diodes D2 & D3 gets forward biased and the output voltage will get clipped by the zener effect of diode D1.


Maximum Current

Current I = V/Z, where V is the voltage and Z is the impedance.
Capacitive Reactance XC1 = 1/(2πfC), where f is the frequency and C is the capacitance.

  • XC1 = 1/(2 x 3.14 x 50 x 680 x 10-9) = 4683Ω
  • X1 = XC1 // R1 = (XC1 x  R1)/ (XC1 +  R1) =  (4683 x  470 x 103)/ (4683 + 470 x 103) = 4637Ω (Parallel Resistances)
  • Zener Voltage, Vz = 12V
  • Vin = 230V
  • Diode Drop, Vd = 0.7V
  • I = (Vin – Vd – Vz)/(X1 + R1) = (230 – 0.7 – 12)/(4637 + 100) = 0.046A = 46mA.

Component Ratings for 12V, 40mA Supply

  • As per the above calculations, C1 = 680nF, 400V
  • VX1 = X1 x I = 4637 x 0.046 = 213.3V
  • PR1 = I2R = V2/R =(213.3)2/470,000 = 0.1W
  • R1 = 470KΩ, 0.25W
  • PR2 = I2R = (0.046)2 x 100 = 0.2116W
  • R2 = 100Ω, 0.5W
  • Zener Diode Power, Pz = Vz x Imax = 12 x 0.046 = 0.552W
  • D1, D2 = 12V, 1W Zener
  • D3, D4 = 1N4007, since 1000V PIV

Note : It is better to choose power ratings of resistors greater than the double of the dissipated power.


In our experiment we used resistors with higher rating than we got in calculation. You don’t need to use this much big resistors. Here we used a load resistor of 300Ω to test the current driving capability.

12V Capacitor Dropper Power Supply on Breadboard

12V Capacitor Dropper Power Supply on Breadboard

Output Voltage = Vz – Vd = 12 – 0.7 = 11.3V

12V Capacitor Dropper Power Supply Output

12V Capacitor Dropper Power Supply Output


Don’t try this circuit if you don’t have much experience with electronics. Care should be taken while testing or using this circuit. Don’t touch at any points of the circuit since some points of this circuit is at Mains Potential. After constructing and testing, enclose the circuit in a metal casing without touching PCB and metal case. The metal case should be properly earthed to avoid shock hazards.


Share this post

Comments (11)

  • swapneel sukhdani

    how do we increase the o/p current

    January 6, 2017 at 12:57 pm
  • John Philip Apulog

    how to know what fuse is needed in the circuit.

    February 20, 2017 at 11:44 am
  • manoj

    it is depends on your load.that how much it consume

    September 8, 2017 at 6:28 pm
  • Sunbeam systems

    Dear sir , kindly advise a design for 12v with 35amp for led module .

    October 6, 2017 at 11:24 pm
  • George

    the maximum current that can be achieved with this type of circuit is 300ma, to get this current you have to use higher value capacitor which is 1uf(105).

    November 3, 2017 at 8:20 pm
  • george

    just increase the value of x-rated capacitor to its maximum value of 1uf

    November 3, 2017 at 8:21 pm
  • Sunbeam systems

    Many Thx for ur reply.
    Can I multiply this circuit and connect it Parallely to obtain max current/ Amp .pl advice

    November 4, 2017 at 9:27 am
  • Hussaindeen Mohamed Azeem

    what will the output if we use 1.5uf 450VAC AC motor capacitor

    November 5, 2017 at 6:13 pm
  • snow

    i m using 684pf , 400v capaacitor… can it provide 12 v also? and can it on 12v relay / 6v relay circuit . please reply asap.. 🙂

    November 18, 2017 at 11:26 pm
  • matthayichen

    I don’t think the warning goes far enough.


    Use a transformer. It could save your life.

    March 13, 2019 at 3:55 pm
  • Chand Sayyad

    I have used 2.2 uF 400V poly. capacitor for dropping down the AC voltage to 24 volt 100 mA.
    But output after capacitor and bridge rectifier is DC 220 V 100 mA.
    Input voltage is 240 V AC.
    why it is 220V instead of 24V?

    August 6, 2019 at 5:22 pm

Leave a Reply

Your email address will not be published. Required fields are marked *