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I am reverse engineering a consumer product that I have, and the circuit seems to be mains voltage (85-250VAC) -> resistor (10ohm) -> bridge rectifier -> buck regulator -> linear regulator -> load.

The current is around 0.3A, output voltage is 3.3V. I would expect the resistor would be a very inefficient way of dropping the voltage. The application requires a very small power supply circuit and so a flyback with a transformer can not be used.

Is using a resistor + buck regulator as a power supply okay, or is there a much more efficient way of doing so?

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  • \$\begingroup\$ Buck regulators are non-isolated, so there are less dangerous ways of doing it. \$\endgroup\$
    – Neil_UK
    Commented Aug 22, 2020 at 20:51
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    \$\begingroup\$ Which buck regulator, and which linear regulator? Can you show us the schematic? \$\endgroup\$
    – Bruce Abbott
    Commented Aug 22, 2020 at 21:01
  • \$\begingroup\$ When you say "the current is around 0.3A" do you mean the input current or the output current? \$\endgroup\$
    – Peter Green
    Commented Aug 24, 2020 at 20:07

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$$\color{red}{\boxed{\text{A non-isolated mains AC power supply is a dangerous device - be warned}}}$$

However, if you want some recommendations I'd seek out Power Integrations and consider using this type of buck arrangement: -

enter image description here

Adding a 10 ohm resistor in series with the input is not unheard of as a way of cutting down the inrush current and, if you don't care too much about efficiency it can be many tens of ohms. The one above is a fusible resistor so maybe that is the part you actually saw.

Here's where you can find details on the LinkSwitch-TN2 device.

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  • \$\begingroup\$ Thanks, the schematic you linked is very similar to the one I am looking at. It's for a smart plug application so I would assume it can be non-isolated since likely the device plugging into it will be isolated, or at mains voltage. \$\endgroup\$
    – willieb3
    Commented Aug 22, 2020 at 23:03
  • \$\begingroup\$ @willieb3 has your question been addressed? \$\endgroup\$
    – Andy aka
    Commented Aug 27, 2020 at 7:40
  • \$\begingroup\$ @willieb3 if we are done on this old question you should an answer as "accepted" else raise another query on comment. Take the 2 minute tour to understand why folk give their time for free to help others. \$\endgroup\$
    – Andy aka
    Commented Oct 26, 2021 at 8:11
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The resistor is not there to drop the voltage during normal operation. You don't make it clear whether the 0.3A is input current or output current, but even if the 0.3A is the input current a 10 ohm resistor will still only drop 3 volts.

The resistor is there to limit the current during sudden changes in voltage, either when the mains supply is initially turned on or when there is a sudden spike on the mains.

Non-isolated power supplies based on buck converters or capactive droppers, can be used for mains, but extreme care is required, you always have to be aware that your "circuit ground" may be at a dangerous voltage. So you have to treat the entire circuit like it is operating at live mains.

Personally if I was reverse engineering a device that used such a supply, one of my first steps would probably be to replace the supply with an isolated one so I could work on the device safely.

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