I'm switching a 12V 500mA resistive load (LED strip) with a IRF540 MOSFET and it's getting very hot when it's on and conducting.
This confuses me as MOSFETs are supposed to have very low on-resistance and be great for power-switching applications. I'm driving the gate at 12V, and the data sheet says that the IRF540's RDS is 0.077Ω at a VGS of 10V, so at 500mA current it should be dissipating a tiny amount of energy:
P = I2 ⋅ R
P (W) = (0.5)2 (A) ⋅ 0.077 (Ω)
P (W) = 0.01925
... but what I see must be dramatically greater than that, because the TO-220 package gets too hot to touch comfortably after a minute or so. CircuitLab says ≈ 1.8W which seems more reasonable from the heating observed.
What's my beginner mistake?
Edit: If you're reading this and interested in driving LED strips, see also Design considerations for LED strip PWM dimming .
Answer
The only mistake is using the NFET as a high side switch when it should be a low side switch with Vs=0V then with Vgs>=10V you pull down the load cathode and series R from the supply with the drain.
So transistors used as switches (FETs and BJT’s) are always inverting. Vgs is chosen from the specs or as a rule of thumb Vgs>2.5 x Vt(max) the threshold of conduction, also known as Vgs(th).
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