Help with MOSFET switching circuit

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
I'm basically going to be doing these calculations.

I'm going to see what you can expect if you drive the gate from the arduino output.

I'll then repeat that using one of the gate drivers I mentioned.

In both cases you'll find that power dissipation will rise with frequency, and there will be some practical maximum frequency. In the case of the direct drive from the arduino the dissipation at a given frequency will be much higher, and subsequently the maximum practical frequency will be much lower.

Having said that, I can't see that you'd want to drive the motor at 60kHz, but you may well want to drive it at much faster than 60Hz too.

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Qg - 25 nC

Vgs(th) - 1 V
Vdd - 5 V
Vss - 0
Vgg -5 V *
Vgg0 - 0V
IL = 3A

Rg - for an arduino max(Vgg - Vgs(th), Vgs(th) - Vgg0) / 0.03 = 133 ohms

Ig(on) = (Vgg - Vgs(th))/Rg = 0.03 A
Ig(off) = (Vgs(th) - Vgg0)/Rg = 0.0075 A

T(on) = Qg / Ig(on) = 833 ns
T(off) = Qg / Ig(off) = 3333 ns

Maximum switching frequency (only applies for 50% duty cycle) = 1 / (T(on) + T(off)) = 240 kHz

(that all sounds dandy doesn't it?)

The power lost in switching is:

Ps = f * (T(on)/2 + T(off)/2) * IL * (Vdd - Vss)
= f * 62.5E-6

static dissipation (at duty cycle d):
P = (IL)2.Rds.d
= 0.315.d

If we want to keep the dissipation under 1W (presumeably to remove the need for a heatsink) at close to 100% duty cycle (the worst case) then we need to solve for:

1 = 0.315 + f * 62.5E-6
so f = 10 KHz.

This means that a direct driving of the mosfet is likely to be acceptable with a 133 ohm gate resistor at a frequency of less than 10kHz.

10kHz is audible, so you either need a gate driver (or a heatsink) if you want to push the frequency higher.

Oct 2, 2014
11
Excellent. Thank you so much for all your help mate.
So i'll just add an external heatsink to the mosfet to stop it combusting?

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