I tried to make a MOSFET reverse polarity protection for a 24V power supply. The MOSFET I had available has a Vgs max of +/-20V so I came up with the design below (I only had a 7V5 zener). The OK LED serves as power indicator, the other LEDs are test load.
When the power supply is connected correctly it works as expected (measured Vgs of -7V4). But when I reverse the polarity the red LED goes on and I measure a Vgs of -3V6. How is this possible?
Thanks, Jenalee K
IRF6215 .------o------+^+------. | | ||| | | z === | | 7V5 A | | | | | | | o--------' .--o--. | | | | | OK V -> green V -> - red /+\\ - led - ^ -> led 24V ( ) | | | \\-/ | '--o--' | | | | .-. .-. | | |1k5 | |2k2 | | | | | | '-' '-' | | | '------o---------------' (created by AACircuit v1.28.6 beta 04/19/05
Take a look at that MOSFET symbol. Notice the reverse diode poking out from the substrate connection.
A power MOSFET has a parasitic diode (that is, a diode that results from the semiconductor arrangement necessary to make the MOSFET, not one that they put in on purpose). So, it will conduct in reverse even if the gate is biased off.
Does the power supply itself have crowbar overcurrent protection? If so, you might just be able to put a beefy reverse-biased diode across the terminals. Reverse the polarity, lots of current flows, power supply voltage-limits or shuts off.
See "OverAndReverseVoltageProtection.pdf" and "PerfectDiodeForChargerIsolation.pdf" on the S.E.D/Schematics page of my website.
...Jim Thompson
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Old Latin teachers never die...they just decline
I know about the parasitic diode, that's why I put the MOSFET in with its source connected to the load. The behaviour is the same when I swap the source and the drain. I've tried several MOSFETs of the same type and they all do the same thing. Maybe they are all faulty?
In fact I've traced the problem to the LED in the gate path. If I connect the zener to the source instead of the drain (as suggested by Slavek) *and* if I short the OK LED it works fine. With OK LED it doesn't.
Every time anybody posts the FET reverse polarity protection circuit here, at least one person fails to notice that the FET is "backwards," and that the body diode will therefore be under reverse bias when the power supply polarity is reversed. This time it was Walter Harley's turn, I guess. ;-)
The problem with your circuit is that the PFET gate must be connected to a voltage which is negative (with respect to the source) when polarity is correct, and positive or zero (with respect to the source) when polarity is backwards. If you look at your original circuit, there is no reason to expect the gate to be positive or zero with respect to the source when the supply polarity is reversed. In fact, as the OK LED goes into reverse breakdown (another problem with your original circuit) the Zener will be under forward bias, thus guaranteeing that the gate will be only one diode drop higher than the lowest Voltage node in the whole circuit.
So, the gate is at, let's say, 0.6 V. If the FET were off, then the source would be at 24V. But a Vgs of -23.4 would turn the FET on strongly, so the FET must not be off. If the FET were completely on, then Vds would be 0, and Vgs would be positive, so that isn't possible, either. So the FET must be somewhere in between on and off. Which means that the source will be at around 0.6 + Vth. Which means significant current will flow backwards through the load. Oops.
In the classic version of the circuit, where the power voltage is less than Vgs max of the FET, you just connect the gate to the negative rail, the drain to the positive rail, and the source to the load.
In your case, the 24 Volt supply is too much for the gate, so you need the Zener. The problem is that you put the Zener in the wrong place. Put the Zener from the gate to the negative rail. The Zener current-limiting resistor should probably go to the source, not the drain, as you have it.
Let's see if I can draw it:
IRF6215 .-------------+^+--+---. | ||| | | | === \\ | | | /1k5| | | \\ | | | / | | +-+ | | | | | o--------' .--o--. | z | | | 7V5 A green V -> - red /+\\ - led - ^ -> led 24V ( ) | | | \\-/ | '--o--' | | | | | .-. | | | |2k2 | | | | | | '-' | | | '------o---------------' When polarity is correct, the body diode conducts, causing the 1k5 resistor to drop around 15 or so volts, which will turn on the FET before the body diode can fry.
When the polarity is reversed, the drain will be at zero volts, the DC resistance of the load will pull the source up to 24 Volts, and the Zener, under forward bias, will pull the gate up to 24 Volts. With Vgs=0, the FET will be off.
You don't need the OK diode, and if you put it in series with the Zener, it may cause the FET to turn on slightly when polarity is reversed. This is because it could keep the gate from getting pulled up all the way to 24 V. Although, even then, I think the circuit will eventually turn off under reverse polarity.
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