Calculating base resistor on mosfet AO3400A

[Demon]

Using datasheet:
https://www.aosmd.com/res/datasheets/AO3400A.pdf

My work so far:

I added tD(ON) and tr to get total t

Google AI says:
Ig = Qg / t

Ig = 6 nC * (3 ns + 2.5 ns)
Ig = 6 * 5.5
Ig = 33

And this is my first hurdle; what unit is Ig? mA? uA? nA? I'm seeing contradicting results on forums.

Google AI says "typically nA", but that's not a definitive answer to me.

[amgen]

.
usually, all the gate drive circuits mostly have to do with the frequency the mosfet is engineered to operate. That is where high frequencies matter and the rise and fall times of the mosfet switching are critical..... but with your circuit that doesn't matter much at all...... you could put a 100 ohm, a 1k ohm, a 10k or 100k and all would be ok...... even no resistor would be ok IMO... nice project you have !!!

[richard]

1 columb = 1 amp for 1 second

To apply 6nC of charge in 5.5 nanoseconds takes
6e-9/5.5e-9 = 1.09 amps

It needs a gate driver to get that speed, a pic with 20mA of drive can provide a 6nC charge in 30uS

[amgen]

no gate driver needed for occasional intermittent switch... the rise and fall times are where there is power dissipated in the mosfet...... do a calculation of the power dissipated in that 30 micro second turnon time.

[Demon]

Richard brings up an important point about the PIC's drive current capacity. I'll use 10mA in my calculations.

R = V / I
= 5 / 0.010
= 500

P = V * I
= 5 * 0.010
= 0.050 W (just to confirm resistor package at JLCPCB)

Qg = 6 nC
Ig = 10 mA
t = 0.000 000 006 / 0.010
= 0.000 000 6

0.6 uSec is not a problem for my application. I just want to activate a solenoid long enough to properly activate a limit switch, then let if fall back using the built-in return spring.

The specs for Qg on the AO3400A use:
- Vgs = 4.5V
- Vds = 15V
- Id = 5.7A

I'm nowhere near that with only Vds 8.5V at 1A max.