Electronic – How does this reverse polarity circuit work

mosfetpower supplyprotection

The circuit in question uses the LT4356:

example circuit

So, I'm trying to get a grip on how this circuit works precisely. I've prototyped the pictured schematic minus everything to the left of D2. So, essentially, no reverse polarity protection… it's just the main IC controlling Q1 to allow power through. That part I understand just fine.

When adding the reverse polarity protection via all that extra stuff before D2, everything get super fuzzy for me. I know Q1 and Q2, based on their datasheet, have body diodes. Thus, I know that when everything is hooked up correctly, Q2 will let power flow and everything past it should work normally. Conversely, I know when polarity is reversed, the body diode will block power flow.

After that, I'm pretty confused. I can roughly reason the operation of D1, Q3, R7 and D3 as a group, but together with the rest of the circuit…. I'm confused on how they work / help the circuit in either a normal or reverse polarity situation.

Anybody able to shed a little light on this for me? ๐Ÿ™‚

EDIT: I realized, after some more graph paper and highlighter action, that when there is normal polarity that Vbase will be the same as Vemitter, so Q3 should effectively be off.

I'm still sort of unclear, though… when Q3 starts conducting during a reverse polarity situation… is all it's doing is pulling the gate for Q1/Q2 to ground to stop them from turning on? If so, why even bother with Q3? Q1 would have its body diode to allow enough current to pass to turn things on normally, and otherwise, it would block reverse polarity. Confused on the benefit of Q3 still. ๐Ÿ™

Best Answer

This is a simplified view to the problem:


In normal operation (Vin is positive, and Vout has reached its target value), the (high) current flows through the channels of Q2 and Q1. No current flows through the body diode of Q2, in normal operation (it does flow, during startup, as The Photon says). The reason to have two MOSFETs (instead of one MOSFET (Q1) and one Schottky diode (in place of Q2)) is exactly this one. To avoid the voltage drop that otherwise we would have across that Schottky diode.

Why does the current flow through both channels, in normal operation? Because both are on. The LT4356 uses an internal charge pump to generate a voltage Vg higher than Vgs_th for those MOSFETs (which is 2.5 V max). Imagine Vg is around 10 V (actually, between 4.5 V and 14 V). Vg1 is 10 V above Vout. So, Vgs(Q1)=10 V > 2.5 V=Vgs_th, and Q1 is on.

Q3 is on only for negative voltages below -1.4 V. So, in normal operation, Q3 is off. No current flows through the 1 Mohm resistor, and Q2 sees exactly the same Vg as Q1. So, Vg1=Vg2. How about Vgs for Q2? How much is it? Well, if Vout is (for instance) designed to be +12 V, and Vg is 10 V above it, then Vg1=Vg2=22 V. Q2 is on if its Vgs is higher than 2.5 V. For Q2 not to be on, Vin should be higher than Vg2-Vgs_th=22-2.5=19.5 V (!), which will never happen, in normal conditions. In normal conditions, Vin will be only slightly above Vout. So, Q2 is on in normal operation, and its body diode is just short circuited, contributing to zero voltage drop (which was the reason to put there a second MOSFET).

When Vin is reversed, and below -1.4 V, Q3 is on, that makes Vgs(Q2)=0, and there is no way that Q2 may conduct. Also, its body diode will be reversed biased, so it won't conduct, either. Since Q2 is in series with Q1, it does not matter what Q1 does, because no current will flow through any of them, and the load will be safe.

More: the reason for this complexity is that a silicon MOSFET is a device that can carry current in both directions, but can block only in one direction (due to the unavoidable body diode). If that body diode wasn't there, a MOSFET would be an ideal switch (able to carry and block in both directions), and a single MOSFET would be enough. Given that the diode is there, the only way to build a bidirectional-carrying bidirectional-blocking switch with them is by placing two of them in anti-series. With their gates tied together and also either a) (ideally) their sources tied together, or b) their drains tied together (as is the case, here).

GaAs MOSFETs don't have the body diode, and therefore a single device works as an ideal switch.