The attached schematic is taken from the Raspberry Pi drawing. I am trying to understand how it works and I am pretty much stuck with it.
I guess the purpose of the resettable PTC fuse and the TVS diode is clear, being as overvoltage protection. It looks quite obvious that the transistors are providing reverse voltage protection, but I do not understand how. I even put the whole circuit to LTspice and it works.

What I got to is that the BJT transistors are forming a current mirror. So if I am correct the current in R2 and R3 should be the same. This can only happen if the voltage on them is about the 1:5 ratio, so when R3 has near 5V on it, then on R1 should be about 1V.

But having 5V on R3 should switch off the FET, since Vgs is near 0V, right? So how will it be switched on and providing a very low voltage drop?

Can someone shed some light on it for me?

 

I even put the whole circuit to LTspice and it works.

Play around, then, and you could be able to answer your own questions. Put a sine wave on the input voltage (like from 3-6V) and probe all over the place. You're right that the MOSFET will shut off if/when the voltage on R3 approaches 5V.

 

OK, seems reasonable. At first I only checked that no reversed voltage is on the output. So now I checked the voltages and put also a sine wave on the voltage source. This is the first time I am using LTspice, so hope I really did what you meant, see the attached screenshots.

So it looks to me that the FET switches on at about 2.3V, when gate voltage gets lower enough. I also see that the current mirror is not a current mirror here, at least currents are not the same on the resistors. But in fact, this is all I got to.

I made another circuit where I simply put a resistor on the gate and I see basically the same behavior. The only difference that makes a little sense, that in case of the original circuit the gate voltage follows the source voltage, so it cannot exceed the Vgs limit. But in case of 5V this will never be a problem. So what are those transistors about?

Any little help would be appreciated.

 

Hi,

I think the goal is to manage the power supply in case you connect 5V on USB and GPIO. So just add another power supply after the MOSFET and check what happens.

 

I think the goal is to manage the power supply in case you connect 5V on USB and GPIO. So just add another power supply after the MOSFET and check what happens.

Good point! This circuit is not only to protect reverse voltage, but also reverse current, which is something I did not expect before.

So when we have a higher voltage at the output than on the input, the BJT at the input side starts closing, because the base voltage is maintained by the output side transistor. The base current here increases and also the collector current. As a result the gate voltage increases effectively switching off the FET.

Thanks for your response.

(In my simulations I put the wrong label letters for Source and Drain, but I do not think anybody cares.)

 

Hi kralg,

It was interesting for me too, I was not aware of this method to perform this kind of protection.