# Is this the correct way to calculate IGBT power loss?

I am having trouble obtaining the power dissipation of IGBTs. It does not have a buit in diode and it isn't driving an inductive load. So I read in an application note that IGBT power losses are actually the sum of 2 kinds of power losses - 1. Static power loss 2. Switching loss

Datasheet

I figured this is the proper way to obtain overall power losses:

For my particular IGBT

• Desired Switch Frequency = 2Hz
• Duty Cycle = 2%
• Vce(sat) = 2.10
• Ic =130A
• Eon = 2.25mJ
• Eoff = 0.95mJ
• Ets = 3.20mJ

Static power loss = Vce(sat) * Ic * duty cycle

2.1 * 130 * 0.02 = 5.46w

Switching loss = Ets * Switching Freq

3.2e-3 * 2 = 0.0064w

lastly overall power dissipation is = Static power loss + Switching Loss

5.46 + 0.0064 = 5.4664w

please let me know if this is correcct or please show the correct derivation.

As noted by a member the correct values must be referenced in the Eon/off curves.

The Vce(sat) is also probably varied a bit but insubstantially so it will be kept identical to the previous derivation.

The correct values for Id=130a are: Eoff = 3mJ Eon = 7mJ Ets= 10mJ

Static Power Loss = 2.1 * 130 * 0.02 = 5.46w

Switching Loss = 10e-3 * 2 = 0.02w

The proper overall power dissipation turns out to be 5.46 + 0.02 = 5.48

the difference is ~2mW and it seems to make a bigger difference for higher frequency switching operation.

• "Static Losses" are called "Conduction Losses" and they depend on the type of the transistor (majority or minority carriers). Your calculations are a bit crude, but this depends on the function of the igbt. Does it have an anti-parallel diode? Why only 2Hz of switching freq and 2% of DC? also check this out mitsubishielectric.com/semiconductors/files/manuals/… Commented Aug 22, 2018 at 8:39
• What is crude about the derivations I used? Commented Aug 22, 2018 at 8:52
• What is the circuit. this device might be ok at 2% duty, but then the freewheel path will be seeing 98% duty. Likewise are you sure 2% is enough? The forcing voltage isn't stated
– user16222
Commented Aug 22, 2018 at 10:32
• The IGBT is discharging a 85v cap which peaks its current to ~130a for ~10ms and that peak is not quite flat rather triangular so its not a long lived current peak. There is no inductive load facing the IGBT and it also does not have a built in diode Im not sure what you meant by freewheel path, Im quite certain you are referring to a diode. Perhaps the diode you mean is the actual IGBT emitter and collector which is just a PN junction diode? Commented Aug 22, 2018 at 11:03
• I mention "freewheel diode" as the circuit was unknown (hence the query in earlier comment). If the load was inductive there must be a freewheel path and at 2% duty, the diode would need to take the full current for 98% of the time. This doesn't appear to be the case here
– user16222
Commented Aug 22, 2018 at 14:10

Fundamentally yes but not quite

You have captured the basic concept

$P_{total} = P_{cond} + P_{sw} + P_{leak} \approx P_{cond} + P_{sw}$

I have added in leakage losses for completeness but this only becomes of significance one you get to series stacked devices.

Conduction loss
This is the subtle difference. Conceptually it is $P_{cond} = V_{ce} \cdot I_c$ however, $V_{ce}$ might not be good enough. This needs to be the saturation voltage at your operating point. Now you may have read the datasheet and chosen $V_{ce} = 2.1V$ when $I_c$ equals 130A but there is no named part to cross-reference.

Assuming you have done this correctly then yes $P_{cond} = V_{ce} \cdot I_c$ scaled by duty.

However, there is a more flexible method and that is approximating the IGBT operating point as a series connected DC source $V_{ce0}$ and a collector-emitter resistance $r_c$

In this example I an interested in an operating point around 20A.

1V = 18.942mm
zero @ 17.071mm
therefore Vce0 = 0.901V
dI = 20A
dv = 12.527mm = 0.661V
therefore rc = 0.00331R

The instantanious power eqates to

$P_{cont}(t) = V_{ce0}(t)\cdot I_c(t) + r_c\cdot i_c^2(t)$

The average loss is therefore

$P_{cont}(t) = \frac{1}{T_{sw}}\int_{0}^{T_{sw}} V_{ce0}(t)\cdot I_c(t) + r_c\cdot i_c^2(t) dt = V_{ce0}\cdot I_{c,avg} + r_c\cdot I_{c,rms}^2$

You can then derive the average and rms currents based upon your waveform.

A similar process can be done with the diode IF appropriate. NOTE: thermally the co-packaged diode typically becomes the limiting factor in a half-bridge

Switching loss
AS you have captured... Switching loss is the accumulation of switching energy with respect to the switching frequency

$P_{sw} = (E_{on} + E_{off})\cdot f_{sw}$

However... the Eon and Eoff are typically stated at a given operating point

This means you need to scale the switching energy with respect to your operating point AND the test point. a linear rescaling is usually good enough BUT please check the Eon,off curves to see if a linear interpolation is good enough

$P_{sw} = (E_{on,t} + E_{off,t})\cdot \frac{V_{DC}}{V_{DC,test}}\cdot \frac{I_{peak}}{I_{I,test}}\cdot f_{sw}$

Likewise there is is something comparable with the diode as you accumulate the reverse recovery charge.

The real fun is deriving the average and rms current waveforms and there are given equations for the profile all related to modulation depth

• This is why I like this method. The 1st pass at least helps down select. This 2nd pass refines the losses. The problem you will face is the DIODE... I don't know the load nor the circuit, but 2% duty implies 98% duty on the freewheel path
– user16222
Commented Aug 22, 2018 at 10:31
• While Vge does influence Vce, you wouldn't be driving an IGBT with anything more or less than 15V typically. Above that and you risk causing a latchup, lower and the losses are too painful. The Vce_stat you tool of 2.1V is at Vge of 15V but more importantly... Ic of 75A, ie its too low
– user16222
Commented Aug 22, 2018 at 10:34
• Added datasheet. Also, I used the absolute maximum Vce(sat) voltage in the derivation since it would be a bit more safe. After looking over figure 4 in the datasheet it turns out Vge varies the Vce(sat) quite substantially. Far surpassing the initial table stating 2.1v max. Thanks to your description I now know the proper values for Id=130a are: Eoff = 3mJ Eon = 7mJ Ets= 10mJ Commented Aug 22, 2018 at 10:39