power dissipation during turn-on and turn-off
You might think that the transistor getting hotter during those transitions has something to do with the internal voltages and currents and capacitances of the transistor.
In practice, as long as you turn a switch on or off sufficiently quickly, the internal details of the switch are irrelevant.
If you pull the switch completely out of the circuit, the other stuff in the circuit inevitably has some parasitic capacitance C between the two nodes that the switch turns on and off.
When you insert a switch of any kind into that circuit,
with the switch off, that capacitance charges up to some voltage V, storing
CV^2/2 watts of energy.
No matter what kind of switch it is,
when you turn the switch on,
all CV^2/2 watts of energy are dissipated in that switch.
(If it switches really slowly, then perhaps even more energy is dissipated in that switch).
To calculate the energy dissipated in your mosfet switch,
find the total external capacitance C it is attached to (probably mostly parasitic), and the voltage V that the terminals of the switch charge up to just before the switch turns on.
The energy dissipated in any kind of switch is
at each turn-on.
The energy dissipated in the resistances driving the gate your FET is
- V = the gate voltage swing (from your description, it's 5 V)
- Q_g = the amount of charge you push through the gate pin to turn on or off the transistor (from the FET data sheet, it's about 10 nC at 5 V)
The same E_gate energy is dissipated during turn-on, and again during turn-off.
Some of that E_gate energy is dissipated in the transistor, and some of it is dissipated in the FET driver chip -- I usually use a pessimistic analysis that assumes all of that energy is dissipated in the transistor, and also all of that energy is dissipated in the FET driver.
If your switch turns off sufficiently rapidly,
the energy dissipated during turn-off is typically insignificant compared to energy dissipated during turn-on.
You could place a worst-case bound (for highly inductive loads) of
- E_turn_off = IVt (worst case)
- I is the current through the switch just before turn-off,
- V is the voltage across the switch just after turn-off, and
- t is the switching time from on to off.
Then the power dissipated in the fet is
- P_switching = (E_turn_on + E_turn_off + 2 E_gate) * switching_frequency
- switching_frequency is the number of times per second that you cycle the switch
- P_on = IRd = the power dissipated while the switch is on
- I is the average current when the switch is on,
- R is the on-state resistance of the FET, and
- d is the fraction of the time that the switch is on (use d=0.999 for worst-case estimates).
Many H bridges take advantage of the (usually unwanted) body diode as a flyback diode to catch the inductive flyback current.
If you do that (rather than using external Schottky catch diodes)
you'll also need to add in the power dissipated in that diode.