_________________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
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._________________
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.
That's the least of your problems.The DCP would be used for controling the gate of the Mosfet.
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.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.
You've passed the test.God I love beautiful gadgets like that. Does that mean I'm an engineer?
Cool! No I mean really cool.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.
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