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Neither. The formula makes only sense for average Ic.Is collector current in IGBT static power loss RMS or peak value?
This leads to the following formula:Neither. The formula makes only sense for average Ic.
Vce(sat) * duty% * Ic
(integral of) Vce(t)*Ic(t) dt (and the integral taken over a complete period).
to get power, you must divide by (integral of) dt (with the integral taken over a complete period).
Hi,
This leads to the following formula:
Vce(sat) * duty% * I(average)
(I don´t agree with this)
For a constant voltage drop (where Vce is independent of current, which is not true)
one rather may use:
P(tot) = Vce(sat) * I(average)
Because in I(average) there already should duty cycle be involved.
Klaus
The confusion is from the way the OP wrote the formula.
Already,
Ic_avg = Duty * Ic_pk
Thus
Vcesat * Ic_avg = Vcesat * Duty * Ic_pk
...
1. Consider an ideal diode; it has zero forward drop and zero forward resistance and infinite reverse resistance and zero turn on and turn off delays.
2. Consider an ideal pulse train; 0 to x volts, positive going, 50% duty.
3. when the voltage is zero, the power is zero and this is 50% of the time. When the voltage is x, the power is again zero (because the diode has zero resistance).
4. An ideal diode has zero power dissipation (under all conditions).
5. Now consider the same diode with a forward drop of v but zero forward impedance (theoretically not possible but then...); if the pulse train has amplitude > v, the dissipation will be very large else it will be zero.
6. Better consider the forward biased diode as a exponential graph: something like i(v)=exp(k.v); that is a realistic curve.
7. Now apply a square wave on this: the dissipation will be zero when the voltage is zero (50% of the time) and the dissipation will be v*i(v)=v*exp(k.v) per cycle; to get power you multiply by the frequency.
8. For real diodes, you can substitute i(v) graph from the diode data sheet.
9. If the square pulse train has high frequency, you need to consider the turn on and turn off delays (diode capacitance) and make intelligent guesses.
Are integrated diodes there only for last resort protection from overvolting/reverse volting CE junction or are there applications which completely utilize and rely on the integrated diode?