Replacing a variable resistor (rheostat) with MOSFET

wayneh

Joined Sep 9, 2010
17,496
That chart is a great illustration of why you get thermal runaway if you try to parallel MOSFETs without designing for it.
 

Thread Starter

Denis De Bremme

Joined Dec 29, 2015
10
You need to close loop control it like I showed above because as the FET heats up everything changes. For example threshold voltage.

Here is a link I found about short duty cycle heat calculations. I've been looking for this myself.
http://www.irf.com/technical-info/appnotes/an-949.pdf
_________________

Thank for the chart Temp vs Vgs and the an-949 pdf.
In your second schematic (analog version) what would be the watt value of R5?

Denis
 

ronv

Joined Nov 12, 2008
3,770
_________________

Thank for the chart Temp vs Vgs and the an-949 pdf.
In your second schematic (analog version) what would be the watt value of R5?

Denis
In that circuit there are 2 FETs so each sees about 3 amps max. So 3^2 X 0.5 is 4.5 watts. So a 10 watt resistor.
 

dannyf

Joined Sep 13, 2015
2,197
The DCP would be used for controling the gate of the Mosfet.
That's the least of your problems.

1) you will need a drive voltage substantially close to 96v to deliver the desired voltage on the load, as configured. It is very difficult to conceive a digital pot that can work at that kind of voltage / dissipation.
2) you will need to figure out how much power will dissipate over the mosfet - do the math and figure out how much money you will need to spend on heatsink.

You can solve some of those problems by moving the load to the drain, but there are reasons, fundamentally sound reasons, that linear controls are NOT used in high power applications.
 

GopherT

Joined Nov 23, 2012
8,009
That's the least of your problems.

1) you will need a drive voltage substantially close to 96v to deliver the desired voltage on the load, as configured. It is very difficult to conceive a digital pot that can work at that kind of voltage / dissipation.
2) you will need to figure out how much power will dissipate over the mosfet - do the math and figure out how much money you will need to spend on heatsink.

You can solve some of those problems by moving the load to the drain, but there are reasons, fundamentally sound reasons, that linear controls are NOT used in high power applications.
To handle the power drop, this poor guy will need a full copper heat sink with high velocity glycol liquid cooling to -30 or so.

image.jpg
 

ronv

Joined Nov 12, 2008
3,770
To handle the power drop, this poor guy will need a full copper heat sink with high velocity glycol liquid cooling to -30 or so.

View attachment 98314
Cool! No I mean really cool.
@Denis De Bremme
Here is another chart for you.
It shows how much current and how much voltage can be controlled by your really big FET.
So in your case there will be 90 volts across the FET at 5 amps while the motor is running.
The fine print is the arrow. It can only do that if the temperature of the case is 25C.
A pretty big heat sink will let the temperature go up say 1 degree C per watt, so in your case the transistor would want to go to 475C.
It can't do that. :(



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