You need an active cooler sink & fan, the size of a CPU heatsink (est.) from an old PC tower.
Active Load Design Specs
- I = 5A max
- R = 0.5 min
- P = I^2*R = 12.5W load
- Load = V(R)=IR=2.5V
- Vdd= 5V
- FET Vds also = 2.5V = Vdd-V(R) must deliver 5A*5V=25W
- Case Temp rise = 25’C max.(pref.)
- Load thermal Rth = 2 ‘C/W = 25/12.5 ‘C/W .
Notes
If dummy load R is a 25W 0.5 Ohm resistor with sink mount flange then both FET and fixed R load can share heatsink with a rating of 1’C/W which is like a standard modern tower CPU heatsink & fan 5V with silver oxide heat grease & FET drain TO-220 insulator.
This way the desired 25W max. power is shared by 2 parts to increase surface area to the sink.
To compute final thermal resistance, and temp rise, the equivalent R network adds all series and parallel parts to the shared heating Rca (case to ambient).
Added
Also if you want 5A always choose a 50 mV max drop or less.
Even if you use a 2W resistor in this case for R3 that is excessive.
R3 temp rise
I^2R = 450 mW for 3A and 1.25W for 5A. To estimate R temp rise using Pd/Max *125’C. So for a 2W part dumping only 1.25W heat rises 80’C which will sizzle water.