@tasty You are
conceptually incorrect in your application of the 10N60 MOSFET as a "variable resistance" electronic load. This device is intended to be used as a high-speed electronic switch, being either turned ON with low resistance from source-to-drain, or turned OFF with high resistance from source-to-drain. Either of these two states will dissipate minimal power.
It will be difficult to operate your MOSFET as a linear device, with source-to-drain resistance being controlled by the gate-to-source voltage, without electronic feedback, of which you have indicated none. So get with the program and begin posting schematics. For reasons I won't discuss here, it will be difficult to operate power switching MOSFETs without "blowing them up" even at modest power dissipation. For an excellent discussion and a sample circuit of "how to do it" please read
Kerry Wong's web page here.
I was NOT going to jump into this thread because
@Alec_t is competent to give you good advice. Problem is, you don't know WTF you are doing, so it's like trying to explain quantum mechanics to barnyard chickens. The chickens will nod their heads and pretend understanding, but they are just dumb chickens and will never understand quantum mechanics. You OTOH have some hope of eventually "getting it" although you are not there yet. Read on.
An electronic load is a nice thing to have handy during design, but realize that its function is to consume energy and dissipate power. Dissipating power means heat will be generated,
no matter what device you use to dissipate power. Your job, should you choose to quit blowing up components, is to figure out a way to safely dissipate the power to the environment.
I need to pull 5 amps on this 55V supply. So I need 3V on this [0.6 ohm] shunt.
Okay, that means you need to dissipate 5 A x 55 V = 275 W. That's P = V x I as the "formula" for power dissipated in a resistive load with V volts across the load and I amperes flowing through it. That's almost as much as the heat generated by SIX 60 watt incandescent light bulbs (360 watts for six light bulbs versus 275 watts for your application). That's a lot of heat there! Common sense should tell you that the TO-220 package is not capable of this much power dissipation without a significant increase in semiconductor junction temperature, even with a "perfect" heat sink maintained (somehow) at 25C.
Since the current-measuring shunt is in series with the 55 V supply, that 5 A current will produce 5 A x 5 A x 0.6 Ω = 15 W power dissipation in the shunt. That's P = I
2R as the "formula" for power dissipated in a resistive load. Fifteen watts is insignificant compared to two hundred seventy-five watts, but why do you need such a large-valued shunt resistor to measure five amperes of current? How did you decide a 0.6 ohm shunt was necessary?
But let's get back to power dissipating, electronically controlled, loads. For small amounts of power, up to about ten kilowatts or so, and low voltages, up to about a kilovolt or so, bi-polar junction transistors (BJTs) are almost ideal candidates. If higher current than a single device can safely pass is required, it is fairly simple to parallel two or more devices with "current sharing" resistors in series with their emitter leads.
I will try to connect 6 MOSFET in parallel then. I will see.
Well, good luck with that then. Back to
@Alec_t for further help...