# Rapid Reverse polarity

This is the circuit for reverse voltage protection, R2 is the load of a motor. The problem is that there is a test where the changing of polarity is very fast ( <5ms) and the MOS burns.

Power supply is 32V and Vds in MOS is up to 200V, also gate voltage is fixed to 15V by Zener for not reaching max Vgs voltage that is +-20V, so MOS should withstand.

Any idea of what is happening here? Thanks a lot.

Here is the oscilloscope trace, when blue (Vds) goes down is when it dies:

Legend:
Yellow trace: Vgs
Blue trace: Vds

• Where does the 12 uH come from? Cable inductance? – winny Jul 1 '16 at 11:16
• You have no snubber / freewheel diode on your motor. You need to add one. Why are you using an unswitched FET? This circuit just stays on (until the FET fails). – Transistor Jul 1 '16 at 12:45
• @transistor, you can see that the unswitched FET was intended as the reverse protection that the OP describes. – TonyM Jul 1 '16 at 13:08
• @transistor See my comment on Tony's answer. The FET IS switched as the DS connections are swapped compared to normal use, relying on the FET being able to operate in two quadrants. – Russell McMahon Jul 3 '16 at 22:33
• @Russell: Thanks, but I gave up trying to figure out what this was about. – Transistor Jul 3 '16 at 22:49

Your FET will have an internal source-drain diode as a consequence of its semiconductor construction. According to your FET's datasheet, it can carry up to 34 A continuously with a 0.9..1.2 V drop across it.

The motor won't act like a resistor so I'm assuming that its normal running current is 32/12 = 2.66 A. Please can you verify that value and provide its stall (start-up) current, too.

If the FET in your circuit is the way up you've drawn it... (CORRECTED) When powered normally, the FET's Gate-Source will be forward biased and the FET will be on. The FET's internal diode will not be forward biased enough to conduct and all FET current will flow Source-Drain. When the supply is reversed, the FET will be off as its Gate will be a D1 diode drop lower than its Drain. The FET's internal diode will be reverse-biased and so you will have your protection.

If the FET in your circuit is the other way up to what you've drawn... When powered normally, the FET will conduct the motor current. When the supply is reversed, the internal diode will conduct and your motor will run backwards. So no reverse protection that way up.

Note that if your motor is a brushed DC motor, which your circuit suggests it is, it will generate EMI, and were you to run and power this through a diode, it would make the EMI much worse. That's something to bear in mind if EMC is important to an application. Putting diodes in supplies for higher-frequency circuits is a bad idea for EMI, despite the odd designer regarding them as 'magic one-way wires'. I've done a fair amount of EMC testing and this was showed this time and again :-(

ORIGINALLY, THIS REPLY WAS BASED AROUND AN INCORRECT AND WRONG STATEMENT ABOUT THE FET OPERATION AND HAS BEEN CORRECTED ABOVE...WHICH LEAVES IT WITHOUT A POINT TO MAKE. BUT I'VE LEFT IT HERE SO THE COMMENTS ON IT REMAIN. THE ORIGINAL STATEMENT WAS: If the FET in your circuit is the way up you've drawn it... When powered normally, the FET's internal diode will carry the full motor current and the FET will be reverse-biased. When the supply is reversed, the FET will be off as its Gate will be a D1 diode drop lower than its Drain. This circuit is therefore no different to putting a big diode in the supply and doing away with M1/R1/D1/C1.

• Thanks Tony, the MOS is placed the way I've drawn, it's the common circuit for reverse polarity. The current consumption of the motor is about 3A in normal operation and in starting it is up to 7A. Rds is 32mohm so voltage drop in MOS is very small. If I connect reverse polarity in static there is no problem, but if I change the polarity very fast, then the fireworks happend. – Pau Jul 1 '16 at 15:00
• @Pau, thanks and that leads to: do you agree with my conclusions on the FET/internal diode? – TonyM Jul 1 '16 at 15:05
• Tony - Your diode comment is actually incorrect due to the unusual way that the FET is used. This cct relies on the FET being a 2 quadrant device which is turned on when Vgs is positive (for an N Channel device) regardless of Vds polarity. So in this mode the DS polarity is reverse to usual in normal operation. This means that the DS diode is forward biased in normal operation. The MOSFET is also turned on in normal operation so the diode is a 'bonus'. In reverse polarity operation the FET Vds is normal but the gate is now reverse biased so the FET is off. – Russell McMahon Jul 3 '16 at 22:31
• @RussellMcMahon, yes of course, silly mistake on my part - I'll correct my reply, thanks :-) – TonyM Jul 4 '16 at 5:10

I'd try

• No C2 or lower C2

• Feed C2 from input side of L1 if possible

It is possible that the C2 / 1n cap is holding Vgs positive as the supply reverses so that the FET still conducts with reverse polarity applied. You have not said what the motor does if reverse polarity is applied but it may draw excessive current. This may not be the problem.

Removing C2 should prevent the above but reduces noise immunity. You could try a lower value of C2 or R1.

The L/R time constant seems too short to be relevant and the LC time constant is also very short. Sometimes you can get resonant conditions with LC or RLC circuits that may cause and effect some time after switching, but this seems unlikely here.

Can R1 be supplied from the input side of L1?

What is the load current if reverse voltage is applied?

• Thanks rusell, I also have a 2.2uF capacitor in parallel with power supply. For the gate charge, there is one thing I can not explain. In measurements, when polarity changes, Vgs goes from 15V to almost 0V and then to 2V. I have the oscilloscope graph but I don't know how to add it here. – Pau Jul 4 '16 at 6:37
• (Now I'm actually looking at the FET operation properly :-) ) I agree with @RussellMcMahon, it looks like the gate capacitance is holding it on. This consists of C2 and the FET's internal gate-source capacitance, which I think is 2..3 nF (not got datasheet now), meaning C2 is redundant anyway and can go. You want to discharge this sharply so look at replacing C2 with an R2 of 2 K/250 mW and an R1 of 10 K. I haven't calculated the timings but a true reversal would suck the gate capacitance current out through R1 to - 32 V so the smaller, the faster. R2 leaks it away in switch-off. – TonyM Jul 4 '16 at 17:47
• I tried by removing the C2 capacitor but still the same, I'll try with the resistor of 2K. Here is the graph, yellow is Vgs and blue is Vds, when blue goes down is when it dies. imgur.com/XvbALAj – Pau Jul 5 '16 at 5:28

Some suggestions.

Have you simulated this circuit to get a feel for M1's gate current at the ±64V rising and falling edges during the voltage reversals? M1's gate-source capacitance $C_{gs}$ might be capacitively coupling one or both those voltage transitions through M1's gate with enough current to cause a blow through in M1's gate oxide layer.

How necessary is it to have $V_{gs}=15\; V$ via Zener diode D1? Note that M1's Miller plateau is around 4.5 Volts according to the the "Typ. gate charge" graph in M1's data sheet. So clamping $V_{gs}$ to 6 Volts (via a 6 Volt Zener diode) rather than 15 Volts, for example, might suffice for this application, and the reduced $V_{gs}$ would create less electric field stress across M1's gate oxide layer.

• Thanks Jim for the answer, and yes, I've simulated the circuit but there is a spike in the gate during switching but I didn't see it in measurements. I placed a 15V zener for the channel to be formed as far as possible. I could give a try with a lower voltage zener. – Pau Jul 4 '16 at 6:32

The problem is that a DC motor is not just a resistor. When spinning it acts as a generator, producing a voltage proportional to its speed. The current drawn by the motor creates a voltage drop across its internal resistance. The voltage dropped across this resistance is also the difference between the terminal voltage and the generator voltage, so the motor will speed up or slow down depending on the load and current drawn.

However the motor's armature has inertia, so when power is suddenly removed the motor will slow down but still generate a voltage proportional to its speed.

Let's say your motor has an internal resistance of 3Ω and is drawing 3A. That means its resistance is dropping 9V, so it must be generating 32V-9V = 23V. Now reverse polarity. The motor gets -32V on its positive terminal, but continues to generate +23V internally. The two voltages combine to put 55V across the motor's internal resistance, so it tries to draw 18A at the same time that the FET is trying to turn off.

Now consider what happens at this point. The FET is turning off, allowing the motor's negative terminal to go below ground (as intended). But the motor is still generating voltage internally, so as the current drops its internal resistance soaks up less voltage and the negative motor terminal actually goes below -32V, maintaining a positive voltage between the FET's Gate and Source.

The more the FET tries to turn off, the less current passes through the motor so the more voltage it can put out, keeping the FET partially turned on. this will stabilize at around 6A and -36V - the FET operating in its linear region with ~4V on the Gate.

Result - one burned FET!

• Thank you Bruce, so any suggestion to solve it? – Pau Jul 7 '16 at 8:44
• If you only want the motor to go in one direction then don't use an H-bridge. – Bruce Abbott Jul 7 '16 at 16:52