# How does this transformerless 5V power supply work?

I recently took apart a crock-pot cooker that stopped working correctly.

• One that dealt with the high-voltage for controlling output to the heating element
• The other had a 4-bit microcontroller and a few buttons/LEDs.

I was surprised to see no transformer anywhere, and very few parts, considering there was a signal labeled +5V for powering the microcontroller.

I reverse-engineered the schematic:

simulate this circuit – Schematic created using CircuitLab

Is it me or is this a really crappy design?

I was very surprised to see that the DC GND was connected directly to the 120VAC L!

Is the 5V DC basically just a rectifier with a voltage divider?

• No this is not a "crappy design", it just uses a capacitive dropper circuit to make 5 V for the control board. You didn't reverse engineer this properly because in your schematic the heating element can barely get warm as it is in series with C22 so the current will be severely limited. I expect the heating element to be connected directly to the node marked "N". – Bimpelrekkie Jan 7 '18 at 20:26
• I hope that you do realize that touching the 5V point, when the crock-pot is plugged in, can very easily kill you. – jsotola Jan 7 '18 at 20:29
• You may also have swapped L and N (it would make more sense for N to be the "0V" rail) and reversed the 1N4001s. – Brian Drummond Jan 7 '18 at 20:52
• See Microchip's AN954: Transformerless Power Supplies: Resistive and Capacitive. – CL. Jan 7 '18 at 21:09
• this L/N swapping thing is quite confusing as long as you're using an AC power source symbol with polarity. Technically it does not matter if L and N are swapped (many types of mains plugs are not directional anyway) but I'd agree with @Brian that readybility would be improved if you flipped the schematic vertically so that N/"GND" is bottom and +5V on top. – dlatikay Jan 8 '18 at 9:50

1. It's you.

2. The 1.5uF film capacitor acts as a reactive (lossless) dropping element. D21 shunts negative current away and D22 conducts positive current to the filter capacitor C23 and it is limited by the 5.1V Zener diode D23. The 1M bleeder resistor (across the 1.5uF) prevents a shock from touching the plug pins after it's pulled. R21 limits the peak current to about 7A if plugged in at an AC peak.

This kind of circuit requires all elements that could come in contact with a human to be isolated from the mains for safety, including any kind of display, switches etc. Any breach in that insulation (including you opening up) could lead to a potentially fatal shock.

As in the comments, your heater should go directly to the N side of the mains. There's probably an overtemperature fuse cutout in there somewhere too, maybe buried near the heater, and probably an overcurrent fuse somewhere.

• I fixed my schematic: The heater does indeed go right to N. And D2 -> D22. – Jonathon Reinhart Jan 7 '18 at 22:17
• Also, the +5V label is right on the PCB. I was mostly interested to understand why the microcontroller might have been acting strangely; I would usually suspect a flaky power supply. – Jonathon Reinhart Jan 7 '18 at 22:19
• Maybe C23 has shuffled off its mortal coil. Just a guess. Heat + e-cap is not a good combination. Either your diodes are all shown wrong or the markings on the PCB are wrong. – Spehro Pefhany Jan 7 '18 at 22:27
• Yeah (upon seeing photos), replace C23 before anything else. It's way too close to heat sources. Try to get a higher temperature rated part than whatever POS capacitor is in there, even if it ends up being a lot bigger (assuming you can fit it safely). – Spehro Pefhany Jan 7 '18 at 22:50
• This kind of circuit is often referred to as a "capacitive dropper". The capacitor C22 acts like a series resistor to drop most of the AC voltage, but without the heat dissipation that a resistor would produce. – Simon B Jan 7 '18 at 23:29

The triac needs a control signal relative to the power line it is controlling. So if the triac is going to switch the L side of the load it needs a control signal relative to mains L.

As such it greatly simplifies the design to use mains L as the "ground" voltage for the electronics. Obviously this means that the electronics are live but that can be dealt with by insulating the control buttons and knobs.

The 22 ohm resistor limits the effect of spikes on the power supply. Without it a fast spike (including a badly-timed connection of the mains power) could cause extreme currents to flow.

The capacitor has a reactance of about 1.8K so it limits the current to about 67mA.

The 1 Megohm resistor ensures that the capacitor is discharged when the mains is disconnected.

The Diodes convert the AC current to DC. We need both diodes as there must be a path for current to flow in the capacitor during both half-cycles of it's AC waveform.

Finally the Zener diode and capacitor control and smooth the output voltage.

This type of power supply does have some drawbacks

• The draw from the supply is basically constant and has to be designed around the maximum load the power supply will run.
• The power factor is terrible. The DC output current available is less than half the RMS AC input current.

But overall for providing a little bit of current at 5V for a control circuit that needs to be mains referenced it's not a terrible option.

I've seen this approach used in PowerLine Metering. In that case, 0.68uF was used. The circuit came to my attention because of small errors in metering, especially at the ZERO load point.

Turned out the magnetic field of the 80mA thru the capacitor was magnetically coupling into the +- current sense wires across the 100 microOhm shunt. The error voltage was approximately 100 nanoVolts and caused big errors when monitoring the power consumption of minimal loads, such as 10 watt bulbs.