# Switchable regulator with a Zener diode

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.

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?

• 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. Oct 21, 2016 at 6:58