# High-Side NMOS Short Circuit Protection: How to Calculate Maximal Allowed Turn-off Time?

I'm about to build a high-side MOSFET switch that can switch voltages up to 150V to an external load (0.5A nominal). For this application I would like to use the N-Channel MOSFET Vishay SiHH240N6E.

Right now, I'm searching for an appropriate gate driver IC that features some short-circuit protection mechanism. I found that some gate drivers integrate a fast comparator input that can turn-off the MOSFET within some 100's of nanoseconds when an overcurrent occurs. The overcurrent condition is set by an internal reference voltage and an external shunt resistor, as sketched below.

simulate this circuit – Schematic created using CircuitLab

## Question

How can I determine (calculate) the maximal allowed MOSFET turn-off time of a gate-driver IC in case of a short circuit? In a worst-case scenario, I will have to assume that the MOSFET junction temperature is already 95°C when the short occurs.

## My Attempts

I checked the Safe Operating Area graph in the MOSFET's datasheet. During a dead-short, there will flow the maximal drain current $$\I_{DM}=30A\$$ (specified in the datasheet) and the drain-source voltage will be basically $$\V_{DS}=150V\$$. I'm not really sure if the MOSFET will really limit the current to 30A or if it could be even higher, at least for some short time...
However, the SOA-Graph unfortunately does not specify a timing for my scenario:

I then tried to estimate the maximal turn-off time based on the maximal junction temperature and the thermal transient impedance.
Assuming the drain current is really limited by the MOSFET to 30A, the power dissipation will be: $$P_d = I_{DM} \cdot V_{DS} = 30A \cdot 150V = 4.5kW$$ Since the junction temperature is already 95°C, and the maximal junction temperature is 150°C, I will only have a budget of 55°C temperature rise. Therefore, my desired transient thermal impedance (junction-to-ambient) would be: $$r_{thJA} = \frac{\Delta Tj}{Pd} = \frac{55°C}{4.5kW} = 12.22 \cdot 10^{-3} \frac{°C}{W}$$ The transient thermal impedance graph in the MOSFET's datasheet is normalized to the nominal thermal impedance $$\R_{thJA}=42\frac{°C}{W}\$$, so I have to search for a value of ~0.0003. Unfortunately the graph does not cover this range. I could extrapolate the single pulse curve a little bit, but I fear that the curve will not continue linear in real-world:

I don't know whether my calculations and assumptions above are correct. I've never done something similar to this. Any help would be greatly appreciated.

• Have you considered to use an IGBT instead? The main difference is that IGBT exhibits certain characteristic that is used for saturation detection - DESAT, so you can have a gate driver with DESAT, you can place a small inductor and a freewheel diode to limit di/dt, then you get a SC protection. Commented May 5 at 15:02
• Hi @MarkoBuršič. No I haven't considered IGBTs yet but I'll take a look at them. Thank you
– Mau5
Commented May 6 at 5:15

I'm assuming there are enough capacitors on the power rail, and the event is short enough to not significantly discharge them. This is a reasonable hypothesis, because the amount of capacitance you need on a 150V power supply will pretty much always store enough energy to blow the MOSFET into the ceiling, so the situation will be resolved in a couple tens of microseconds anyway.

During a dead-short, there will flow the maximal drain current IDM=30A (specified in the datasheet)

No. The "max pulsed drain current" datasheet spec is the maximum pulsed current it can withstand before the most vulnerable part of the component is vaporized. It does not indicate how much current will flow, that depends only on your circuit. It simply says you should ensure current stays below this limit at all times.

During a dead short, current starts at whatever it was before the short, then rises according to wiring inductance : $$\ \frac{di}{dt} = \frac{V}{L} \$$.

It will then reach a maximum current which is determined by Ohm's law: divide the supply voltage (150V) by all the ESR in series (capacitors ESR, MOSFET RdsON, wires, etc).

So you can assume something like that:

You need to adjust the simulation according to your own circuit parameters. Since the event is very short, no thermal transfer will take place, so the important thing for MOSFET survival is total dissipated energy which determines temperature rise in the semiconductor, right on the surface where the MOSFET actually is.

In the beginning, current (blue plot) rises almost linear due to inductance, power (red plot) rises as the square of time, and energy (green plot) as the cube. After this, current stabilizes to the maximum value, so power is constant and energy increases linearly, but by then it's already too late.

This means a protection that is fast enough to act before the current begins to rise too much ensures much lower total dissipated energy. Speed is absolutely essential. Adding a bit of inductance also helps.

Note that when the MOSFET turns off, any energy in the wiring inductance also must be dissipated. If there is no freewheeling diode it will be dissipated in the MOSFET, and this energy is in I^2, so again faster is better. Purple plot shows energy dissipated in the MOSFET including energy stored in the inductance.

You can use the "Maximum avalanche energy" datasheet spec to estimate how much energy it can absorb safely. In this case, 81mJ. This is given on a MOSFET starting at 25°C so it needs to be derated.

In your case the maximum pulsed current will be the limit, so you should use that instead. It is pretty low, so a little bit of inductance will probably be necessary to prevent current from rising too fast.

• Hi @bobflux. Thank you for the very clear answer. These calculations will then also help me to determine the current trip point for a given turn-off delay.
– Mau5
Commented May 6 at 5:26