Hey All,

I am trying to drive IRF7749L2PbF MOSFETs with DRV8323 Gate driver, and I thought it would be nice idea to calculate switching time to estimate switching loses

....

Voltage is Aprox 48V from a battery so max 55V, Current consumption is about 60A RMS. I am designing a ESC for BLDC so yes it is an inductive load.

A lot of the specifications you have quoted are typical under one set of conditions. They can be useful if they happen to exactly match your circuit, but it is always worthwhile to get back to some basics.

First, there is the ON channel resistance which depends on the gate voltage. For 60A switching, you will need at least 10V gate drive. It is typically 1.1 milliohms for this device and it increases as the device gets hotter. Doubling it would probably be a good initial design approximation. The Rds power loss is just simple I

^{2}R calculations.

The next loss is due to the charge in the drain to gate and drain to source capacitance. Every time the mosfet switches on, this charge is dissipated in the mosfet. Don't over-complicate things - just add the two capacitances together and don't worry about the fact that the gate is not at 0 volts.

Looking at the curves, at 55v, the total drain capacitance is a bit under 1800pf - that is close enough.

The energy in this capacitor = 1/2CV

^{2} = 2.7uJ. So if you were running at 100KHz, that would be 0.27W of power.

The gate charge from the graphs with a 10V gate drive and 48V on the drain is about 210nC. The gate resistance is 1.1ohms. This charge has to be added and removed every cycle, so at 100KHz, the average current is

100KHz x 210nC x 2 = 42mA (The times 2 factor is for the charge and discharge)

So the power lost by the gate resistance at 100KHz is 42mA x 1.1 = 46.2mW

Now we get to the hardest factor - the losses due to the mosfet still conducting while the drain voltage is rising or falling. Also the rise time changes with the current. If you really wanted to maximize mosfet efficiency, you would drive the gate with +/- 10V with a 10A capable driver. Practically, a driver like the DRV8323 with a 1A source and 2A sink and a 0 - 10V? voltage is more common. This is an interesting driver - it has programmable drive currents and it actually acts as a constant current source/sink.

As we mentioned, at 55V, the drain capacitance is about 1800pF. That effects the rise time and it is dependant on the inductor current. At 1A in the inductor, rise time is = C.dV/I = 100nS. At 60A, this calculation drops to below 2nS. So at the peak current, you can forget it. At low currents, it is relevant.

Now the gate charge is 210nC and the driver IC can sink this at 2A. The gate resistance can discharge the gate at almost 5A, so the 2A current limiting of the driver dominates here. To discharge 210nC at 2A takes 210nC/2 = 105nS. This number is misleading since the mosfet is still on at 5V, so during the 10V to 5V discharge time, the mosfet has not started to turn off significantly.

The mosfet only starts to turn off at 60A when the gate voltage is about 4.5V and it is essentially off by about 3.5V. These numbers change with temperature, so I would say on = 5V, off = 2V. The 0V gate capacitance is about 12000Pf, but this falls with voltage, I am just going to use 10000pF, at 2A. The time taken to discharge the gate by 3V (between 5 and 2V) at a current of 2A is 10,000pf x 3V/2A is 15ns. That is a much better figure.

However, the driver also has to cope with the current from the drain-gate capacitance which is something like 800pF. If we calculate the risetime at 2A, you get dT = 800pF x 55V/2A = 22ns.

The way the maths works out, you have to add these together to get the final mosfet turn off time of 37ns.

The risetime will depend on other issues, like the internal capacitance of the inductor, the properties of the ferrite, etc, but as long as the drain risetime is significantly over 37ns, then the mosfet will not be on significantly while the drain voltage is rising. If at 60A, the inductor can force the drain to rise in 20ns, then the mosfet will still be partially on while the voltage is rising. This can really start to add to the power losses, and it can also be a reliability issue for the mosfet.

This is probably the point at which you would build the circuit and check the risetime. If it is 50ns or more, then the mosfet will probably run extremely efficiently. If it is 25nS at full current, the mosfet is probably getting very hot and you may have to consider a more powerful gate driver. Perhaps also using a negative voltage for turning off the mosfet.

I hope that gets you started. I haven't used any equations other then basic resistance and capacitance ones. If you start this way, then when you see a design guide equations, you can actually work out how it is derived which means you can understand it.

Richard.

Edit: I ignored the output rise and fall times for the driver IC, but it is important. If the gate is charged to 10V and the mosfet doesn't not start to turn off till 5V, the critical thing is that the gate driver output is fully on by the time the gate reaches 5V. I haven't gone through the turn on losses. Hopefully, you can work through that yourself.

Also, I forgot to calculate the gate resistive losses due to the gate-drain capacitance. At a guess, it could be 70mW at 100KHz.

Finally, there can be other significant losses of power. If there is ringing on the drain when the mosfet turns off, then if the drain voltage exceeds 60V, the mosfet will start conducting like a zener diode and it will absorb this energy from the inductor. The mosfets can be used this way, but it is not efficient design. With a good layout and good Schottky diodes, hopefully this never happens.