I need to drive a highside mosfet switch from a GPIO pin. The high voltage supply can be between 42.5V and 90V. I was searching around for the simplest solution I could find (small, low cost, few external components). I found a few options, but then cam up with this on my own:



I didn't find this particular type of example online, but it seems like the simplest way to achieve my desired functionality. Does anyone have feedback or ideas on what I might be missing or what to watch out for? I like the simplicity of this circuit because it is very easy to set the drive parameters:

GPIO = 0 to 3.3V
R2 sets the gate drive current: I_pulldown = (3.3 - Vbe)/R2. In this example I_pulldown = (3.3-0.7)/2.5K roughly 1mA.
Set R1 based on desired Vgs of mosfet:Vgs =(R1/R2) * (3.3-Vbe).
Gate pull up will be comparable but a bit slower than pull down due to RC effects. (when BJT turns off).
Setting R2 is a trade off between on IQ and gate switch speed, but works for low speed applications (~100us to 1ms switch time)

Waveforms in QSPICE look really clean, but still need to test on breadboard:

 

It’s pretty good as long as you don‘t need high frequency switching.I’d put a zener between gate and source just to be on the safe side. Any voltage between 10V and 18V.

 

The circuit with the NPN should work.

The one with the PNP won't, since the collector-base junction is forward-biased by the plus supply voltage.
A PNP needs a negative voltage from collector to base for normal transistor operation.

 

Does anyone have feedback or ideas

I would use this arrangement with a 10 volt zener diode.
I noticed both mosfets shown in your schematic are not rated for 90 volts.

EDIT: Added R4 as required.

 

I would use this arrangement with a 10 volt zener diode.

You still need a resistor across the zener to assure rapid and complete turn-off. I just measured a 12 V zener, and got 4.x Mohm in one direction and open circuit in the other.

The FET's input capacitance and the added resistor determine the turn-off time. With a typical value of 1.2 nF, a 100 K resistor yields a fall time of approx. 360 us (to 3 time constants).

ak

 

I would use this arrangement with a 10 volt zener diode.

You also need a resistor (e.g. 10kΩ) across the Zener to discharge the gate capacitance and turn off the MOSFET.

And R1 needs to be a 1W resistor, since is dissipates about 1/2W when Q1 is on.

 

And R1 needs to be a 1W resistor, since is dissipates about 1/2W when Q1 is on.

To the TS: If you have a desired turn-on time, that can be used to optimize the value of R1 for minimum power dissipation.

ak

 

I would use this arrangement with a 10 volt zener diode.
I noticed both mosfets shown in your schematic are not rated for 90 volts.
View attachment 310265

What turns the MOSFET off?

 

You also need a resistor (e.g. 10kΩ) across the Zener to discharge the gate capacitance and turn off the MOSFET.

And R1 needs to be a 1W resistor, since is dissipates about 1/2W when Q1 is on.

Current through R1 is (3.3V-Vbe)/2.5k = 1mA
So power in R1 is 1mA^2*12.5k = 12.5mW

 

The second circuit is potentially useful with a bit of rearrangement:
Make Q2 NPN, and connect the base to 3.3V
Then you have a common-base circuit which will drive the MOSFET. Its only disadvantage is that it is limited to the amount of current available from your logic output that drives unit.

 

Current through R1 is (3.3V-Vbe)/2.5k = 1mA

If you read my post, it is referring to R1 in post #4.

 

If you read my post, it is referring to R1 in post #4.

Indeed you did - I misread it - my apologies.

 

Current through R1 is (3.3V-Vbe)/2.5k = 1mA
So power in R1 is 1mA^2*12.5k = 12.5mW

He was responding to the schematic in post #4.

P(R1) = 0.49 W.

ak