Thanks for your valuable reply, i think you have the right, the power consumed by a MOSFET is (A x RDSon), but my case is to use 4 MOSFETS type IRFP460 to drain 30V, 20A not 20V 30A

Here we have 30A to drain from the 8 Mosfet

I still don't quite follow the circuit. If the FET is not on hard then Rds will rise and the FET itself becomes a noticeable power wasting device. And when Rds goes up then that reduces load amps.

Do the FET's need to handle 20A or 30A? And when they turn on will they be at spec Rds or or they in some non-linear region allowing them to have higher Rds?

Actually i received the Heatsink already and i mount 4 IRFP460 on it, well for 30V, 2A without FAN, the heatsink start to heat until you can not touch it

30V is where exactly, measured between FET Drain and Source? What's the supply voltage?

 

I still don't quite follow the circuit. If the FET is not on hard then Rds will rise and the FET itself becomes a noticeable power wasting device. And when Rds goes up then that reduces load amps.

The circuit (in post #1) is a constant current sink. The MOSFET(s) plus their source resistor(s) - R1 - act as a variable resistor where the voltage drop (30v) is partly across the MOSFET and partly over R1. For example, if R1 was 0.1ohm, at 20A it would drop 2v, so there would be 28v across the MOSFET. With multiple MOSFETs each would carry a portion of the current - in this case 20/8A = 2.5A so a 1ohm resistor might typically be used (Vr1= 2.5v, so Vds = 27.5v.) Each MOSFET has its own opamp feedback loop.

The opamp sets the gate voltage to maintain the desired voltage across the source resistor (opamp -input) to be equal to the input voltage at the opamp +input. Rds is indeterminate (but of course is equivalent to Vds/Id) and its variability with dissipation is not important as the opamp maintains the necessary gate voltage to keep Id constant irrespective of Vds. Dissipation is, of course, Id * Vds (2.5 * 27.5 = 68.75W, 6.25W dissipated in each R1).

Hope that helps.