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

efficiencymosfet

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?

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.
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.