Calculating base resistor on mosfet AO3400A

[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.

[Ioannis]

You can drive the MosFet directly from PIC as your frequency of activations is really slow.

Gate driver is needed if your frequency is high. Also resistors or ferrite beads are also needed in higher frequency where ringing may interfere with normal operation.

So, you are good to go without them.

Ioannis