FET Switch Not Shutting Off

Question

I am looking at a circuit that seems to be having a problem. This circuit is being used to switch a supply from an automotive battery. It delivers voltage to a quite dynamic load, normally at 6A but can handle 24A pulsed loads. Here is the schematic.

When switched on, the lower node on R2, is grounded. The zener diode D2 clamps the gate-source voltage to 10V and the FET is fully on. Everything works great. OK...

When switched off, the lower node on R2 is, in an ideal world, open. I am seeing quite a number of units coming back with the FET, Q1, damaged. In the field, I found that the FET is not turning off, Vgs above the threshold voltage, and the FET in the linear region. I'm seeing a Vgs peaking at 4V, keeping the FET on. I saw numerous problems with the circuit.

First, I found there can on the order of 1.4 uA of leakage current through a protection diode, not shown but on the lower node of R2. Across 100 kohm, this would be on the order of 1.4V caused by this leakage current. Meanwhile, 100 kohm of biasing is insanely large. I am reducing this to 10k, R2 to 2k.

I am concerned about the 7000 pf of gate capacitance on the FET. Any noise on the drain and source will surely couple on to the gate. I'm seeing around 2 Vpp of ripple at nearly 40 kHz due to the dynamic load. I am seeing this signal on the gate and believe the zener, D2, forward conducting is causing the Vgs voltage to be further increased.

Here are the questions:

1: I've been reading up about Miller capacitance. Would a solution be to add something like a 47 ohm resistor in line to the gate, attenuating the high frequency noise?

2: This does not need to be fast switching. Could I simply put something like 0.1 uF from the gate to ground to drive the noise coupled through the gate capacitance into the ground? At 40 kHz, the gate capacitance is nearly 650 ohm and the 0.1 uF to ground would be nearly 40 ohm.

Answer 1

It's the 100kΩ resistor, R1, that would develop a \$V_{GS}\$ due to 1.4μA leakage current through it - R2 is not a problem where leakage current is concerned. Reducing R1 to 10kΩ is very likely to solve any issues in that respect.

Using a zener diode to clamp maximum \$V_{GS}\$ mandates the use of R2 to limit current through that diode, and R2 necessarily causes gate potential to vary more as a result of miller current. R2 at 20kΩ will obviously develop a larger voltage due to miller current than, say, 100Ω. To mitigate miller current, you require a much tighter hold on gate potential, a much lower impedance source of whatever potential \$V_G\$ is supposed to be there.

Reducing R2 commensurately with the reduction in R1, might be sufficient to limit \$\Delta V_G\$. I don't know where you intend to put your hypothetical 47Ω gate resistor where it isn't in series with R2 worsening the problem. R2 is the main culprit where miller is concerned, and of course any additional impedance leading up to R2, which you (inconveniently) don't show us.

It's always a good idea to limit \$V_{GS}\$ with a zener diode, so I won't propose changing that. MMBZ5240 seems to be a 10V device, so presumably you're aiming for \$V_{GS}\$ slightly under 10V. That's fine, but since this MOSFET would tolerate up to 20V \$V_{GS}\$, and if the battery supply is 12V, why not use a 15V zener diode and dispense with R2 altogether, or make it 100Ω, to limit diode current during short supply spikes? If battery voltage is 24V then the story is different, because the diode would have to be well under 20V \$V_{GS(MAX)}\$, and well under the supply voltage, again mandating a diode-current-limting R2. Battery voltage (and whether this is in an actual vehicle) would be good information to include in your question, since it's very relevant to any solution.

Assuming you still want your 10V \$V_{GS}\$ clamp, how small a resistance can R2 be before current and power dissipation issues arise? Due to zener diode D2, the largest voltage across R1 is 10V, and the largest voltage across R2 is the remaining 2V (assuming a 12V supply). To avoid having to use high power resistors, keep power 100mW or less:

$$ \frac{{V_{R2}}^2}{R_2} = \frac{(2V)^2}{R_2} = 100mW \rightarrow R_2 = 40\Omega $$

DC "on" current in R2 would be

$$ I_{R2} = \frac{2V}{40\Omega} = 50mA $$

We should also reduce R1, to help suppress \$V_{GS}\$ changes due to miller current while the MOSFET is off. Assuming the same 100mW cap:

$$ R_1 = \frac{(10V)^2}{100mW} = 1k\Omega $$

If you are willing to use a beefier R1, to dissipate more power, then we can do even better. The smallest appropriate R1 is when it develops 10V while passing all 50mA:

$$ R_1 = \frac{10V}{50mA} = 200\Omega $$

It would dissipate power while the MOSFET is on:

$$ P_{R1} = I^2 R_1 = (50mA)^2 \times 200\Omega = 0.5W $$

All that looks reasonable, and already you've solved the impedance problem, for both conditions MOSFET on (\$R_1\parallel R_2=33\Omega\$ to DC) or off (\$R_1=200\Omega\$ to DC):

^{simulate this circuit – Schematic created using CircuitLab}

I remind you that since we don't see what's below R2, there may be something down there with much higher impedance to ground than the direct connection I made here, which will necessarily add to R2.

This is 50mA of continuous load on the battery, while the MOSFET is on, though. Can your "driver" circuitry (that you omitted) even sink that current? If that's a problem, another solution may be to employ an emitter follower to provide the low impedance to ground while permitting precise gate potential control:

^{simulate this circuit}

You control \$V_{CTL}\$ somehow, making it +12V or 0V, which would result in +12V or +1.5V at Q1's base. Q1's \$V_{BE}\$ will translate +1.5V to \$V_X=+2.2V\$, leaving \$V_{GS}=12V-2.2V=9.8V\$ across D1, insufficient to cause it to break down and conduct. Q1 becomes a very low impedance source of +2.2V, able to sink away most of the miller current coupled from M1's drain. Strictly speaking, you don't need R2 here, but I've left it there as a safety precaution to limit current to 200mA if for some reason \$V_X\$ falls below +2V. There is only 3mA of total quiescent current when the MOSFET is on.

If you are fine with 50mA quiescent current, but your "driver" can't sink that, then enlist the help of a common-emitter transistor to offload that responsibility:

^{simulate this circuit}

Unfortunately any "correct" or useful answer will need to account for what's below R2 in your schematic, what's controlling things from down there.

Lastly, 2V peak-to-peak "noise" at the drain is excessive, and can be easily mitigated with proper decoupling close to the MOSFET's drain:

^{simulate this circuit}

Not only does this reduce the amplitude of noise there, it will reduce the rate of change of drain potential \$\frac{dV_D}{dt}\$, the phenomenon responsible for miller current in the first place. Don't forget to also provide adequate decoupling right at the source of that noise, close to the load.

Answer 2

There are a few ways to damage a MOSFET:

1. too high gate voltage- that would seem to be covered by the 10V Zener
2. Too high Vds- a possibility if the user can connect the battery to an automotive electrical system
3. Excessive dissipation or exceeding SOA limits

I think we should concentrate on 3. While the 100k resistor is quite high and will probably result in switching times of milliseconds, the SOA is fairly generous in this region.

SOA limits.

Still, I would definitely decrease those resitors by at least 10:1, though I don’t think it’s the actual cause.

One thing you may not have considered is that batteries discharge and if you don’t equip your circuit with a 100% reliable Undervoltage Lockout (UVLO) the MOSFET will eventually find itself in linear mode and respond in undesirable manners such as the die damaging itself, or even unsoldering itself and falling off the board. As you probably know, “on” or partially “on” is what generally happens when the die is damaged, but not badly damaged enough to blow the bonding wires off (“off”).