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? :)

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. :(

  • \$\begingroup\$ Q1's body diode will be reverse-biased in normal operation. \$\endgroup\$ – The Photon May 25 '12 at 20:37

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.

  • \$\begingroup\$ What is the CAD program used to draw the schematic? Great answer btw. \$\endgroup\$ – abdullah kahraman May 28 '12 at 17:07
  • \$\begingroup\$ @abdullahkahraman Not CAD. Microsoft Visio. Thanks. \$\endgroup\$ – Telaclavo May 28 '12 at 17:59
  • \$\begingroup\$ Wow it does a great looking job! \$\endgroup\$ – abdullah kahraman May 29 '12 at 18:30

This is designed for ultra low dropout, load dump surge protected current limited and voltage clamped with auto recycle shutdown and warning.

So GATE is an open collector NPN ( with internal charge pump from V+) Pin 4 via 10KΩ to turn on Q1 and then later Q2 via 1MΩ to turn on slower.

Q1 and Q2 are OFF at Vgs=0 and have a threshold 1~2.5V where they start to turn ON and fully ON at 12V

Q3 serves as a protection to Q2 ( oops my neighbour reversed the jumper cables) so when Vbat is negative Q3 and R7 turn ON to shut OFF Q2 (open).. if optional D3 is shorted, there is a standby drain of 1mA through D1,which protects Q3 and R7 . Thus with the optional D3 , it is reverse biased, Q3 is off and Q2 and Q1 are controlled by GATE on the Active surge suppressor device.

Pretty complicated... Cheap design is a 10A/60V Schottky diode that drops 0.3V at 3A and then you drop to 12V with standard LDO clamp... You only need this ultra low switches for linear Lamps and stuff that are very voltage sensitive on brightness. If you designing a protected 12.0V port... no need for all this stuff.

  • \$\begingroup\$ Right. I don't necessarily want an extra FET on the board. A single diode is far simpler. Just trying to understand the circuit operation. It may sound weird, but figuring out all the stuff mentioned in my edit on the original post felt very rewarding. :) \$\endgroup\$ – Toby Lawrence May 25 '12 at 20:40
  • \$\begingroup\$ So, with D3 in place... it stops D1 from conducting any current during normal operation hence the reduced standby current? \$\endgroup\$ – Toby Lawrence May 25 '12 at 20:43
  • \$\begingroup\$ If no D3 and a zero ohm resistor is used in its place ( ie jumper) then the R7 is always conducting in forward and reverse modes. "D1 is not a body diode, it is for reverse polarity protection of Q3 when Vbat is normal polarity." then in reverse Batt. D1 is OFF and D3 & Vbe is ON . to shut off Q2. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 25 '12 at 21:01
  • \$\begingroup\$ Gotcha. Makes sense now. \$\endgroup\$ – Toby Lawrence May 25 '12 at 21:03

What's not shown in your diagram is the body diode of the FET at Q2. Here's the "full" diagram of Q2, according to its datasheet:

enter image description here

In order for the circuit to start, the diode between drain and source will carry current from VIN to the main regulator circuit. As Telaclavo points out in another answer, after the regulator has begun operation, it will use a charge pump to bias Q2's gate above VIN, turning it on and reducing the voltage drop through Q2.

In a reverse-polarity situation (VIN = -12 V), Q3 will be turned on, and it will pull the gate of Q2 low, shutting it off and preventing current being pulled out of the rest of the circuit.

  • \$\begingroup\$ But given the orientation of the body diode on Q2, should it implicitly keep itself off, assuming no outside voltage at the gate, when Vin is negative? \$\endgroup\$ – Toby Lawrence May 25 '12 at 20:37
  • \$\begingroup\$ Without something to pull the gate low, it could pick up a charge from any small leakage path, which would turn Q2 on. It needs to be actively pulled towards VIN in the event of reverse connection. \$\endgroup\$ – The Photon May 25 '12 at 20:39
  • \$\begingroup\$ Aha! That makes sense now. \$\endgroup\$ – Toby Lawrence May 25 '12 at 20:42
  • \$\begingroup\$ @Telaclavo, does my edit resolve your concern? \$\endgroup\$ – The Photon May 26 '12 at 0:14

It is kind of standard reverse polarity protection circuit (having a high side driver available of course). A simple series diode is replaced with mosfet's body diode, and mosfet is normally on in order to minimize voltage drop. For example it is found (with minor variations) in bq40z60EVM from TI.

I find D1 completely redundant, R7 protects emitter-base junction from excessive reverse current. My design works fine without it, please correct me if I'm missing something.


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