I am working on a project that involves an IoT device, with a 3.3V microcontroller, powered from a doorbell transformer. In an earlier question, I was attempting to solve an issue with a bridge rectifier failing. One of the answers to that question involved a circuit similar to the one below. However, this circuit is not working correctly.


The doorbell transformer provides about 21V AC. My application requires that the doorbell still functions. The switch allows the transformer outputs to be shorted. This triggers a solenoid that rings the audible bell. To prevent the inductive kick from the solenoid from killing any of the other components, a bidirectional TVS (1) is used to suppress the voltage spike.

Because the MCU does not require much current, a half-wave rectifier (3) is used to convert the 21V AC from the doorbell transformer to about 13V DC. A capacitor is used to smooth the rectifier output (4). Finally, a switching voltage regulator (5) converts the rectified DC voltage to 3.3V to power the microcontroller.

In order to prevent a loss of power to the MCU when the doorbell button is pressed, a 6V zener diode (2) is used to allow power to be supplied to the switching regulator (or at least it is supposed to) while the button is pressed.

I've included links to all the datasheets below:

(1) Bidirectional TVS Diode

(2) Zener Diode

(3) Half-Wave Rectifier

(4) Capacitor

(5) Switching Regulator

I would greatly appreciate any insight into why this circuit fails to power the MCU. Thank you in advance for your help.

  • 1
    \$\begingroup\$ "convert the 21V AC from the doorbell transformer to about 13V DC". You should check your math again. \$\endgroup\$
    – Long Pham
    Aug 24, 2018 at 13:22
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    \$\begingroup\$ I don't understand the purpose of putting a zener diode in series with doorbell (according to what you described, not the schematic). \$\endgroup\$
    – Long Pham
    Aug 24, 2018 at 13:24
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    \$\begingroup\$ Include the transformer and bell in your schematic, and explain if the MCU is never powered or if it loses power when the button is pushed. \$\endgroup\$ Aug 24, 2018 at 13:48
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    \$\begingroup\$ What are the measured voltages open, and bell pressed. BTW 47uF is really not very much, and unlikely to be enough (470uF would be my starting place). And where does GND connect to? Does the 78e33 work from 5V? \$\endgroup\$
    – Henry Crun
    Aug 24, 2018 at 14:10
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    \$\begingroup\$ You need to connect GND to the circuitry on the left, most likely to the negative end of the capacitor. \$\endgroup\$ Aug 24, 2018 at 14:10

1 Answer 1


A doorbell transformer does not produce 21VAC. Or any other voltage. It has an air gap and/or bypass that makes it "soft".

It supplies a mostly constant current, not a constant voltage. That's by purpose so it doesn't make a difference how long and thin and rusty the wires, and wire nuts from the multiple doorbell knobs to the doorbell are. Also, how rusty the most times poorly made doorbell knob contacts are. The bell has always the same volume (selectable by the "voltage" at the transformer), regardless which knob is pressed.

If you want to supply something else in addition ot the doorbell, you better put it in series with it, not parallel. If your device has high impedance (almost any electronics, e.g. your DC converter), you need a local impedance transformer.

  • 1
    \$\begingroup\$ Can you clarify about the "air gap and/or bypass"? It would be great if you could also link to a datasheet for one of these transformers. \$\endgroup\$ Aug 24, 2018 at 14:11
  • \$\begingroup\$ Unfortunately, this is usually already stated in the name "bell transformer". Some manufacturers print the doorbell symbol on the transformer case to state this one is special. I found doepke.de/uploads/tx_doepkeproducts/datenblatt/… which says with small loads or idling, the output voltage may rise. That's understatement. \$\endgroup\$
    – Janka
    Aug 24, 2018 at 14:52
  • \$\begingroup\$ Expect the idle voltage being much higher than the operating voltage. Of those 21VAC idle, there may be a few volts left when the doorbell is ringing. \$\endgroup\$
    – Janka
    Aug 24, 2018 at 14:54

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