Simple Transformer less Power Supply

This is the simplest version of a transformer less power supply to power low current devices. It can give 30 to 40 milli ampere current and 12 to 15 volt DC depending on the input voltage.

CX is the most important part of the circuit to drop high volt AC to low volt AC through Capacitive reactance property. Bleeder resistor R1 discharge the stored current from the capacitor when the circuit is unplugged from AC outlet. This prevents shock hazard since the capacitor stores more than 400 volts for many days. The low volt AC is rectified by the full wave diode bridge and made ripple free by C1. So the output gives minimum 12 volt DC @ 30 mA current which can be used to drive 6 LEDs in series.

Caution: There is no galvanic isolation between the power supply and the gadget. So not use this circuit to power costly equipments. If strong spike develops in the AC lines, it may destroy the power supply circuit.

The circuit board is at mains lethal potential. So take adequate safety measures during operation. Enclose the circuit is a shock proof case. Do not try to test or trouble shoot when the circuit is connected to mains. If you are not experienced in handling high voltages, do not construct this circuit to avoid shock hazards.

Circuit in PDF

Simple Transformer less Power Supply

More Information

Dealing with AC Mains

Capacitor Power supply design note

4 responses to “Simple Transformer less Power Supply

  1. Dear SIr i want power supply with 230v ac input and 48v @ 120mA as output .Pls tell me what changes i have to make in the circuit .Waiting for reply.Thanks in advance

  2. Thanks for your feed back and valuable comments. The capacitor used is X rated. If half wave rectifier is used , the brightness of LED may reduce if it is used to drive the LED. In my experience, Zener may sometime heat up and open. So just selected the capacitor that gives 12-15 volts AC.MOV is not included for simplicity and it can be included for safety purpose. Output capacitor can be between 10 uF to 100 uF. The bleeder resistor can be between 470 k and I Meg Ohm. A higher value will take more time to discharge the capacitor. Optimum value is 820 K.

  3. PIV rating of each diode should be stated as greater than Vp +0.7V (bridge rectifier), based on a input of 230V rms @50Hz, minimum PIV would be 325V where Vp = 230 * sqrt 2, you have specified diodes with a PIV of 1000V so is good. If someone used 1N4001 instead this might be a problem as its PIV is 50V.

    Capacitor class of CX should be specified as X2, for safety.

    It could made simpler by replacing the full wave bridge rectifier with a half wave recifier and incorporating a zener diode, everything else staying the same. Drawback being less effcient.

    Might be good to state that there are no safety features like your other circuits where an MOV and fuse has been added.

    would be nice to know what the calculated output ripple voltage would be for a given load RL, and how you sized the output filter capacitor based on 10mS (100Hz full wave) @ 30mA, assume you used Q=C*V, I = Q/T

    How did you calculate the 820KOhm parallel resistor on CX? I just select a high resistance like 1MOhm from practice, haven’t used Vc = Vi *e^(-t/RC).