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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.

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  • \$\begingroup\$ "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/… \$\endgroup\$ – thece Aug 22 '18 at 8:39
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    \$\begingroup\$ What is crude about the derivations I used? \$\endgroup\$ – user33915 Aug 22 '18 at 8:52
  • \$\begingroup\$ You're off by a factor 1000 as far as switching losses are concerned. Ets is 3.2mJ and hence Pts is 6.4mW \$\endgroup\$ – carloc Aug 22 '18 at 9:04
  • \$\begingroup\$ I noticed and fixed it. \$\endgroup\$ – user33915 Aug 22 '18 at 9:14
  • \$\begingroup\$ 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 \$\endgroup\$ – JonRB Aug 22 '18 at 10:32
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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\$

enter image description here

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

enter image description here

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

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  • \$\begingroup\$ 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 \$\endgroup\$ – JonRB Aug 22 '18 at 10:31
  • \$\begingroup\$ 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 \$\endgroup\$ – JonRB Aug 22 '18 at 10:34
  • \$\begingroup\$ 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 \$\endgroup\$ – user33915 Aug 22 '18 at 10:39

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