Simple Mosfet driver

Question

I need to switch 8 IRF9540 (P-Channel Mosfets) with about 32kHz (so 1 MOSFET about 4kHz) so that only one is On at one time. I thought about using the decade counter 4017 for this, and the 555 Timer for driving the 4017. Because of the low output current of the 4017 the Mosfet would take a long time to switch completly or am I thinking wrong? Is this a possible layout for a simple Mosfet driver for switching the mosfet faster than without the npn transistor?

EDIT: Maybe it's helping I have to switch 4A at maximum. It should switch with about 32kHz (so 1 MOSFET about 4kHz)

^{simulate this circuit – Schematic created using CircuitLab}

respectively what is the differnce between these two schematics in switching speeds?

^{simulate this circuit}

Answer 1

Yes, the transistor-resistor driver is helping.

A quick look at the datasheet of the MOSFET show Qin = 38nC and Cin = 860pF.

With I=2mA direct drive, switch time is t=Q/I=19us. With your driver, pulling up, the RC time constant is t=R*C=40ns (way faster), and pulling down, that BJT must have some current gain. So yes, it appears to help quite a bit.

That is the "back of the envelope" approximation. You can simulate it very easily to check for other "2nd order" effects.

Answer 2

I will leave it to others to question if you need to do this at all...

The big problem with your circuit is that the R is in the source, and you do not have enough VGS to turn it on. You need gate voltage to go well above 5V (eg 10-15V) if you expect to get 5V at the source. You will only get a couple of volts across R in your circuit.

Dual NPN/PNP pairs make an easy, cheap, single component driver e.g. BC847BPDW1T3G

Also it does not draw idle current, nor require a very low value of R for fast operation

^{simulate this circuit – Schematic created using CircuitLab}

BTW 74HC4017 has a pretty low output R anyway (prob 50 ohm) - it is able to deliver far more than 1-2mA. You can run it on up to 6V to get better drive to the fets. Still won't let turn the fets on enough to supply 5V - you will need 8-15V depending on fet.

Answer 3

The speed limiting thing in your circuit is the relatively weak pull up resistor R1 that has to pull the gate of the IRF9530 up towards 5V when Q1 is turned OFF.

Instead of Q1 and R1, you can use a digital logic IC with a push-pull output configuration like for example the SN74AHC04 capable of driving from 8mA up to 50mA (depending on the type).

But you could also build your own push-pull driver with a PNP transistor instead of R1. There might be a lot more options...

Answer 4

Here is a question/answer on MOSFET gate driving. Summary: knowing the required gate-charge to go fully on or fully off, and the equivalent Cgate, besure you can switch that gate voltage is less than 1 microsecond (300nS tau is about right), to avoid Safe Operating Area failure.

Big loads turns MOSFET's on despite pull-down resistors

Answer 5

Have you considered using high speed high-side MOSFET driver instead of transistor? They specifically designed to keep MOSFETs reliably Off and to switch them faster. Something like IX4427

Edit: oops... I see @Dean Franks already suggested this