I'm trying to develop a switchable 5v/3.3v power supply. I've got a switching 5V regulator that will drive my 3.3v rail. I've looked into using an LDO, and I think I'm going to go that route after investigating this option, but I thought I'd see if my concept is completely impractical, or I've missed something fundamental. I'm kind of expecting both.

So the idea for my 3.3v regulator is to switch my zener (2.4v@5mA) from the low side with a resistor in series with the anode such that it drops the remaining voltage from the desired rail voltage.

The zener resistance is 100 ohm, +80(r1) = 180 ohm. Rds is a few hundred milliohms, which I'm ignoring for now.

Zener drop:3.3v - 2.4v = .9v
Resistor drop: 5mA * 180 = .9v

The circuit: 5v switch on the high side, and 3.3v diode regulator circuit on the low side.

enter image description here

As you can see, I'm hoping to draw 1A out of this to power basically anything; dev boards, sensors, maybe DC motors. What would it take to make this work? Is the premise completely wrong?

  • 3
    \$\begingroup\$ 1) the 3.3 V output will only deliver 3.3 V when you pull 1 A from it 2) this is very power inefficient 3) Why a 2.4 V Zener if you want to make 3.3 V ? 4) the voltage drop of 5 V - 3.3 V = 1.7 V will fall across M2 yet you use that as a switch. This circuit makes no sense and will not work as you expect it will An LDO is really much better than this. \$\endgroup\$ Commented Oct 21, 2016 at 6:58

1 Answer 1


Your circuit is completely confused, sorry, and well beyond repair.

Here is an approach that will work:

enter image description here

You can (say) use 1.68K for R2 and parallel R1 = 1K with a MOSFET and 1.21K resistor. When the MOSFET is on, Vout = 5V, otherwise Vout = 3.3V.

You can modify a very inexpensive buck regulator module to get this to work (remove the pot and resistor and replace with the R1/R2/R3/R4/MOSFET), and I highly recommend this approach.,

Buy a bunch of them because they are cheap and I suspect you might experience the odd problem given from where you are starting.


simulate this circuit – Schematic created using CircuitLab


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.