Electronic – Gate capacitance vs. Gate charge in n-ch FETs, and how to calculate power dissipation during charging/discharging of the gate


I am using a MOSFET driver (TC4427A), which can charge a 1nF gate capacitance in about 30ns.

The dual N-ch MOSFET I am using (Si4946EY) has a gate charge of 30nC (max) per fet. I am only considering one for now as both on the die are identical. I am driving the gate to 5V. (It is a logic level fet.)

Does this mean I can apply Q = CV to work out the capacitance? C = 30nC / 5V = 6nF. So my driver can turn the gate fully on in about 180ns.

Is my logic correct?

Gate resistance of the MOSFET is specified at a max. of 3.6 ohms. Will this have any effect on the calculations above? The driver has a 9 ohm resistance.

Is there any significant difference for when the gate is discharged instead of charged? (turning off the fet.)

As a side question, during the 180ns the fet is not fully on. So Rds(not-quite-ON) is quite high. How can I calculate how much power dissipation will occur during this time?

Best Answer

Like endolith says you have to look at the conditions for parameters. the 30nC are a maximum value for \$V_{GS}\$ = 10V. The graph on page 3 of the datasheet says typically 10nC @ 5V, then C = \$\frac{10nC}{5V}\$ = 2nF. Another graph also on page 3 gives a value of 1nF for \$C_{ISS}\$. The discrepancy is because capacitance isn't constant (that's why they give a charge value).

The gate resistance will indeed have an influence. The gate's time constant will be (9\$\Omega\$ + 3.6\$\Omega\$) \$\times\$ 2nF = 25ns, instead of 9\$\Omega \times\$ 2nF = 18ns.

In theory there will be a slight difference between switching on and off, because when switching off you start from a higher temperature. But if the time between on and off is small (lots of margin here, we talk about tens of seconds) temperature is constant, and the characteristic will be more or less symmetrical.

About your side question. This isn't usually given in datasheets, because the current will depend on \$V_{GS}\$, \$V_{DS}\$ and temperature, and 4-dimensional graphs don't work well in two dimensions. The only solution is to measure it. One way is to record \$I_D\$ and \$V_{DS}\$ graphs between off and on and, multiply both and integrate. This transition normally will happen fast, so you'll probably can measure only over a few points, but that should give you a good approximation. Doing the transition more slowly will yield more points, but the temperature will be different, and hence the result will be less accurate.