I stumbled across this circuit on page 8 of the Nano Pi Duo 2.

What purpose does it serve?

I recognize the MOSFET arrangement to be a reverse voltage protection circuit and the PNP transistors form a current mirror I believe.

From my Spice simulation, it seems to do reverse voltage protection and suppress some short transients (<30 ns.)

Can anyone else provide more insight into this?

PMOS and PNP protection circuit.

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    \$\begingroup\$ It is an "ideal diode". I.e. a "diode" with minimal forward voltage drop. \$\endgroup\$ – Math Keeps Me Busy Feb 27 at 14:42
  • \$\begingroup\$ @MathKeepsMeBusy why the comment and not an answer? \$\endgroup\$ – Jonas Daverio Feb 28 at 1:17
  • \$\begingroup\$ @Jonas I did answer. Commented first, because I wanted to be first! \$\endgroup\$ – Math Keeps Me Busy Feb 28 at 1:54
  • \$\begingroup\$ @MathKeepsMeBusy fair enough \$\endgroup\$ – Jonas Daverio Feb 28 at 17:39

What is the point / name of this circuit?

The name of the circuit is an "ideal diode".

The point of the circuit is to conduct current in one direction only, like a diode, but to have a minimal forward voltage drop.

It is often used when combining multiple power supplies or sources, where the voltage drop of a conventional diode is undesirable.

It could also be used to protect a circuit from reverse applied voltage.

The principle of operation of this particular "ideal diode" is that if the voltage on emitter of Q1 is greater than the voltage on the emitter of Q2, the gate of M1 will be brought low, and M1 will conduct.

Addendum: This Power Electronoics Tips Article supplied to me by @Grant gives further details.

  • \$\begingroup\$ So the PNP transistors prevent a high voltage on the output from flowing back to the input? As opposed to using just the p-channel mosfet with the gate connection to ground which would conduct in that scenario? \$\endgroup\$ – Grant Feb 27 at 15:44
  • \$\begingroup\$ Yep looks so (powerelectronictips.com/inexpensive-ideal-diode-mosfet-circuit) \$\endgroup\$ – Grant Feb 27 at 15:49
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    \$\begingroup\$ yes. So if you had another power supply connected the same way to the output, the one with the highest voltage would "win", and supply the current. That, of course, assumes that your BJT transistors Q1 and Q2, and your resistors are matched well enough so there isn't some offset which makes a slightly lower volage supply "win". \$\endgroup\$ – Math Keeps Me Busy Feb 27 at 15:50
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    \$\begingroup\$ @Grant with your permission, I will add the link your provided to my answer. I will credit you. \$\endgroup\$ – Math Keeps Me Busy Feb 27 at 15:54
  • \$\begingroup\$ Sure! Go right ahead \$\endgroup\$ – Grant Feb 27 at 16:02

The 2 PNPs and 10k resistors form a comparator. This particular arrangement is called a 'common-base' comparator, and is appropriate for applications like this where the input voltage is the same as the available supply and a small amount of comparator's bias current is acceptable.

If the PNPs are matched, then when there is no or small load, Q2's emitter will be lower V than Q1's emitter. Since the bases are common, Q2 has a lower VBE than Q1, and so will conduct less current. You can calculate by how much, but for example, 0 mV will mean 10x less current. Thus (ignoring PNP base currents), the voltage top across R2 will be a lot less than that across R1, and the gate V of the MOSFET will be quite low -- the MOSFET will be 'ON'.

At zero load current, Q2 and Q1's emitters will be at the same V (if the FET is ON), and so R1 & R2 would have equal currents. This would mean that the gate of M1 would be equal to the base go the PNPs (i.e. about 0.7 V below supply) -- and the MOSFET would be off. This is a contradiction -- what happens is that the current that Q2 consumes pulls its emitter lower, thus increasing the VDS drop across the FET -- the circuit will stabilize with some small drop (10's of mV) across the FET

In a reverse load situation (out > in), Q2's emitter will be higher than Q1 -- Q2 will turn on, and pull its collector nearly as high as its emitter; this will makes M1's VGS nearly 0 and it will be off -- no current will flow back.


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