There's a bunch of topics around this but i haven't found something that fits my requirements. I'd imagine this is a fairly common thing though so any pointers would be appreciated

Basically i want to design a circuit that can use mains power anywhere (85-265v AC) and power a simple system with max power consumption of around 100mA.


  • 85-265v AC input
  • 3.3v +-5% DC output
  • 10-100mA current
  • Size as small as possible
  • ~10mV ripple
  • Cheap (less than ~$10)

This is for a networked device and will be totally enclosed (no buttons etc) so i don't think it has to be isolated although it of course wouldn't hurt.

I've looked at existing PCB mount PSUs but the price is a bit too high. Also they can be fairly big (mostly tall) and provide much more current than i really need. A simple solution with AC -> Transformer -> Diode bridge -> Voltage regulator would probably work but the transformer size becomes very big. Also i'm unclear how this will work with the universal voltage range. Basically i probably want something similar to the 3rd solution here: https://electronics.stackexchange.com/a/41944/6809

The components from Power Integrations seem to be what i should be looking at but i'm slightly unsure where to start. Also very few of the examples give 3.3v out and usually at a much higher wattage than i need (= i'm thinking i can make something smaller/cheaper as i only need .1 Amp)

  • 3
    \$\begingroup\$ I would seriously consider plugging in one of those tiny USB chargers for mobile phones, like the HTC chargers, and an LDO to bring the voltage down to 3.3 Volts, if this is a small run. For larger runs, the same logic applies as used in the miniature mobile phone chargers: Use a switching step-down design, perhaps with a piezoelectric transformer instead of an inductive, if size is paramount. See my answer here for an idea. \$\endgroup\$ Commented Apr 26, 2013 at 16:38
  • \$\begingroup\$ The idea is to do a bigger run so a usb charger won't work. Also size becomes an issue. What i'd like to replicate is how a mobile charger works though \$\endgroup\$
    – Antti
    Commented Apr 27, 2013 at 9:46
  • \$\begingroup\$ Hmm... Then a switching regulator using piezo transformer (if you want isolation) can give you a tiny footprint. Else a capacitor voltage divider for step-down, then a full-bridge and reservoir capacitor, but that might actually end up being of comparable size. \$\endgroup\$ Commented Apr 27, 2013 at 9:54
  • \$\begingroup\$ Try adding a lm7833 ic to some power supply.... \$\endgroup\$ Commented Jan 22, 2018 at 9:01

2 Answers 2


A device like the Power Integrations LinkSwitch-TN allows you to implement a simple buck converter that converts rectified mains voltage to a reasonably-well-regulated non-isolated low voltage with reasonably good efficiency.

If you need low ripple after the fact, you can put a linear regulator on the stepped-down output of the buck and minimize your resistive losses.

I've used these as primary-referenced control supplies and have had good success with them.

enter image description here


If it definitely doesn't need to be isolated from the AC I'd consider trying a capacitor dropper circuit feeding a zener diode. Here's one I found:

enter image description here

It came from this website. This circuit produces 6.2V but there is no reason why a 3.3V zener shouldn't be used to produce 3.3V output.


  • 1
    \$\begingroup\$ It could work, but there are some major issues. The main one is that if you design it to work across the entire Vin range then you can have a lot of power being dissipated by the resistor and zenier at max Vin. It could be several watts in the resistor, and a watt in the zenier. Also, the component values all need to be tweaked to work for 3.3v @ 100 mA. These issues makes this type of supply less attractive at 100+ mA. \$\endgroup\$
    – user3624
    Commented Apr 26, 2013 at 18:28
  • \$\begingroup\$ Also, to elaborate on what someone else was saying: A cap has a certain impedance at 50/60 Hz. Think of this impedance as a resistance at a certain frequency. The two caps in the circuit effectively form voltage divider at 50/60Hz. You can calculate the output of this voltage divider by first calculating the impedance at 50/60 Hz and then doing the normal voltage divider math. This gets a little more complex, because your load is in parallel with the low-side of the V-divider. I am oversimplifying things quite a bit, so I recommend that you simulate this (I did). \$\endgroup\$
    – user3624
    Commented Apr 26, 2013 at 18:35
  • 1
    \$\begingroup\$ @DavidKessner I'm not saying it's perfect or even near perfect. I'm offering it up as an example of getting a low voltage DC from an AC power source without a transformer. For lower voltage main ac supplies, it'll start to flag a bit. For the voltage divider bit you can't neglect the bridge and use a simple voltage divider analogy - I think i'll simulate it next week unless you're prepared to post what you've got David? \$\endgroup\$
    – Andy aka
    Commented Apr 26, 2013 at 18:46

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