# igbt power losses collector current

1. ## igbt power losses collector current

Is collector current in IGBT static power loss RMS or peak value?

Vce(sat) * duty% * Ic  Reply With Quote

2. ## 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.

Klaus

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3. ## Re: igbt power losses collector current

Is collector current in IGBT static power loss RMS or peak value?
Neither. The formula makes only sense for average Ic.

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4. ## Re: igbt power losses collector current

Hi,

Neither. The formula makes only sense for average Ic.
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

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5. ## Re: igbt power losses collector current

Would be nice to have a working igbt simulation.

igbt.zip  Reply With Quote

6. ## Re: igbt power losses collector current

Vce(sat) * duty% * Ic
You are almost correct but:

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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7. ## Re: igbt power losses collector current Originally Posted by c_mitra (integral of) Vce(t)*Ic(t) dt (and the integral taken over a complete period).
gives energy - Volts Amps time = joules
to get power, you must divide by (integral of) dt (with the integral taken over a complete period).

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8. ## Re: igbt power losses collector current

Whats the issue with the LTSPICE IGBT simulation?  Reply With Quote

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9. ## 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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10. ## Re: igbt power losses collector current

to get power, you must divide by (integral of) dt (with the integral taken over a complete period).
Sorry, I goofed. In fact, I wanted to include the term (1/T) before the integral but ...

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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11. ## Re: igbt power losses collector current

dividing by period works fine
it is, after all, the same thing

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12. ## Re: igbt power losses collector current

Whats the derivation for average diode power dissipation from a 50% duty square input?  Reply With Quote

13. ## 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.  Reply With Quote

14. ## Re: igbt power losses collector current Originally Posted by KlausST 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.
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.  Reply With Quote

15. ## Re: igbt power losses collector current Originally Posted by Akanimo The confusion is from the way the OP wrote the formula.
Ic_avg = Duty * Ic_pk
Thus
Vcesat * Ic_avg = Vcesat * Duty * Ic_pk
...
I generalized a little bit too much here. This is applicable only to Continuous Conduction Mode with small ripple.  Reply With Quote

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16. ## Re: igbt power losses collector current Originally Posted by c_mitra 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?  Reply With Quote

17. ## 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  Reply With Quote

18. ## 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?
If you are trying to control a purely resistive load (say a heater), then they are non-functional; they have no role to play.

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 turn-off. 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.  Reply With Quote