Why does a high PWM frequency heat the driver more?
At every turn-on and turn-off transition of a MOSFET the voltage and the current are both high for a short moment, and at that moment energy turns into heat. Since the number of transitions per second is directly proportional to the frequency, the switching loss also grows in direct proportion to the frequency. Conduction loss, on the other hand, is independent of frequency and is determined by P = I² × RDS(on).
The loss in a semiconductor switch consists of two components:
- Conduction loss: The power the current flowing through the switch dissipates in its resistance while it is on, P = I² × RDS(on). It is not affected by frequency but grows with the square of the current: 10 mΩ at 15 A means about 2.25 W, and doubling the current makes this loss four times larger.
- Switching loss: The loss occurring at the transition moments. A fixed amount of energy is spent at every transition; since the frequency determines how many transitions occur per second, the total loss is proportional to the frequency.
That is why a high frequency brings quietness and low current ripple while consuming the thermal budget. If cooling is limited the board temperature rises, and because the RDS(on) of the MOSFET increases with temperature the conduction loss grows and the process feeds itself. For this reason the thermal calculation must be done not with the 25 °C datasheet value but with the normalised RDS(on) versus junction temperature curve.
The variable the user can actually control is the current: a correctly sized motor, adequate cable cross-section and a current selection with a safety margin are the three items that reduce heating the most.
The full topic: Locked rotor and thermal protection