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23rd April 2019, 07:34 #1
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igbt power losses collector current
Is collector current in IGBT static power loss RMS or peak value?
Vce(sat) * duty% * Ic

23rd April 2019, 08:45 #2
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Re: igbt power losses collector current
Hi,
P = V x I
this means at any time the power P(t) = V(t) x I(t)... this is true in any case.
For ideal resistors: P(tot) = R x I(RMS)^2
for constant (I mean pure square shaped) current: I(RMS) = sqrt(duty cycle) x I(peak)
But an IGBT neither is an ideal resistor nor you may expect pure square shaped current waveform.
So you may use the formula, but I strongly recommend to add some headroom.
***
Additionally there is swtiching loss: in time where the IGBT is neither fully ON nor fully OFF. > at every swtiching edge. Time is the important parameter, thus it mainly depends on gate drive strength and frequency.
***
Best is if you can provide worst case current waveform and voltage waveform.
KlausPlease donīt contact me via PM, because there is no time to respond to them. No friend requests. Thank you.
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23rd April 2019, 08:53 #3
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Re: igbt power losses collector current
Is collector current in IGBT static power loss RMS or peak value?
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23rd April 2019, 09:19 #4
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Re: igbt power losses collector current
Hi,
Neither. The formula makes only sense for average Ic.
Vce(sat) * duty% * I(average)
(I donīt 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.
KlausPlease donīt contact me via PM, because there is no time to respond to them. No friend requests. Thank you.
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23rd April 2019, 09:30 #5
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Re: igbt power losses collector current
Would be nice to have a working igbt simulation.
igbt.zip

23rd April 2019, 12:44 #6
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Re: igbt power losses collector current
Vce(sat) * duty% * Ic
The correct formula will be (integral of) Vce*Ic dt (and the integral taken over a complete period).
The mistake is:
If you use Vce(sat) you under estimate the losses during the switch transition (turn on and turn off)
The inclusion of duty% in the formula is wrong; it will be automatically corrected for if the current or voltage becomes zero,
The values should be instantaneous values; neither RMS nor peak. To make this clearer, I should have written the above formula as:
(integral of) Vce(t)*Ic(t) dt (and the integral taken over a complete period).
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23rd April 2019, 13:01 #7
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23rd April 2019, 14:13 #8
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Re: igbt power losses collector current
Whats the issue with the LTSPICE IGBT simulation?

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23rd April 2019, 15:08 #9
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Re: igbt power losses collector current
I guess, not particularly an IGBT simulation problem, but a problem with the specific model (too complex, weird equations, whatsoever). LTspice seems to stop acquiring waveforms in similar cases. Would be a good question at the Yahoo LTspice forum, the unofficial LTspice support forum.
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23rd April 2019, 15:51 #10
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Re: igbt power losses collector current
to get power, you must divide by (integral of) dt (with the integral taken over a complete period).
Actually I wanted to focus on a graphical determination of the power from the waveforms but that will take more effort (on my part).
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23rd April 2019, 20:15 #11
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Re: igbt power losses collector current
dividing by period works fine
it is, after all, the same thing
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24th April 2019, 00:34 #12
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Re: igbt power losses collector current
Whats the derivation for average diode power dissipation from a 50% duty square input?
Last edited by Zak28; 24th April 2019 at 00:39.

24th April 2019, 02:01 #13
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Re: igbt power losses collector current
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.

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24th April 2019, 06:12 #14
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Re: igbt power losses collector current
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
The OP should have specified whether his Ic in his formula is average or peak.
If Ic in the formula is peak value, then the formula is correct.
C_Mitra, the OP refers to static power loss which is conduction loss. Switching loss is dynamic.Last edited by Akanimo; 24th April 2019 at 06:23.

Akanimo.

24th April 2019, 21:19 #15
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27th April 2019, 01:02 #16
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Re: igbt power losses collector current
Last edited by Zak28; 27th April 2019 at 01:11.

27th April 2019, 03:28 #17
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Re: igbt power losses collector current
yes
take 2 IGBTs
tie the emitters together, tie the gates together, break one leg of an AC line, connect each new end to a collector
apply 5V to the gate emitter, and AC current will flow through CE 1 and the diode 2
when current reverses, current moves through CE 2 and diode 1
very nice AC on off switch for line
add a transformer to isolate the GE
diode has same current capability as CE

27th April 2019, 04:28 #18
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Re: igbt power losses collector current
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?
In real life, most loads have some capacitance and inductance. The diode takes care of the induced voltage at turn off. This is clearly obvious.
Then you consider the current capacity and the heat dissipation of the built in diode.
The diode is designed based on the max voltage, max turn off time (speed; the smaller, the faster it is) and the max current capacity.
The load inductance is not under control of the designer; you must use your design principles to use the correct load suitable for the application.
One thing you can do easily: use a slower turn on and turnoff. But you need to consider the device dissipation in mind.
Contrary to popular belief, hard turn on/ off are not always the best option. At the cost of some efficiency, you can get greater reliability.
The integrated diodes have been optimized for performance and it is not wrong to rely on the integrated diodes for general applications.
So the answer to your question is "Yes"!
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