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Make sure the peak current is higher than the peak current in your circuit. Assuming that the load is switched on long enough for the current to be limited by the resistance, then Imax=14V/3.2Ω≈4.4A.
Choose a diode that can withstand a peak current higher than 4.4A.
Next, power dissipation. This will depend on how often you turn on and off the load. Each time you turn it off, the diode will dissipate power. The current through the diode decays exponentially from 4.4A, with a time constant τ=5.5mH/3.2Ω=1.71ms
The power dissipated in the diode in this case can be aproximated by the product of its voltage at 4.4A (say 0.7V) and the peak current, considered as a rectangle of duration τ/2=0.855ms, averaged over the pweriod T.
Thus, the average power dissipated in the diode is:
Pdave=(Vd*Imax*τ/2)/T, where T is the period of the load switching action.
For example, assume you switch the load on and off every 50ms. Then:
Pdave=0.7*4.4*0.855/50=0.052W
Choose a diode that can handle that average power. If you switch every 10ms, then you get Pdave=0.263W
This is the average power, which you need to use to calculate the junction temperature of the diode and possible heatsink requirements.
(Note that the peak power is 0.7*4.4=3.08W. The diode must be able to withstand that, too, which is really equivalent to the diode being able to withstand the 4.4A current, so do not neglect the peak current just because the average power is low).
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