# How to estimate IGBT Junction Temperature and understand the datasheet

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#### jamesh77

##### New Member
Hi All,

I'm trying to understand how you would estimate the junction temperature of an IGBT and i am getting myself all confused.

Im trying to use the attached IGBT datasheet tutorial provided by ST.

It is based on the STGFW40V60D datasheet. http://www.st.com/en/power-transistors/stgw40v60df.html

It mentions that PToT (the maximum power dissipation for a given case temperature) is as follows :. PTOT = (TJmax - TC )/ Rth J-C
Where TJmax = 175°C, TC = measured case temperature, Rth J-C igbt = 0.53°C. Where TC = 25°C .:. PToT = 283W

PToT can also be worked out by PToT = Vce * Ic

Question A) ( I am unsure whether Vce is the average voltage accross the collector and emitter or this is the saturation voltage of the IGBT, if so how do i measure this?)

This further leads on to looking at section "2.4.9 Maximum transient thermal impedance (ZthJC)" where TJstart == Tc

Tj = TJstart + ΔTj = TJstart + ZthJc (tp,δ) * Ptot = TJstart + k(tp,δ) * RthJc * Ptot

Looking at the provided graph assuming
δ = 0.5
tp = 0.000025
fs = 0.00005
TC = 25°C
Rth J-C igbt = 0.53°C

i would say k(tp,δ) doesn't show on the graph. So wrongly or rightly i would assume K = 0.5 as the response looks fairly flat for 10^-5. so for Tj = TJstart + k(tp,δ) * RthJc * Ptot = 25°C + 0.5 * 0.53°C * Ptot.

Question B) : For Ptot can i use the 283w i calculated earlier or do i need to measure Vce and Ic?

If Ptot == 283w then Tj == 99.995°C is this correct? and is this telling me the peak junction temperature or the steady state temperature?

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#### Diver300

##### Well-Known Member
Ptot is normally the average power. What are you switching with the IGBT?

#### kubeek

##### Well-Known Member
I am unsure whether Vce is the average voltage accross the collector and emitter or this is the saturation voltage of the IGBT, if so how do i measure this?)
that is the actual momentary Vce times the current at that instant. This all averaged out. The equation is simply only for a steady DC situation, in all other cases you have to measure and calculate the power dissipated, or at least simlate it.
The equation you use assumes the case is steadily held at 25°C (with active refrigeration, liquid nitrogen or whatever means necessary), which in practice is very much not achievable.

As for the thermal impedance... Where do you assume that k=0.5 and 1e-5s from? How is the device actually being used?

#### jamesh77

##### New Member
Ptot is normally the average power. What are you switching with the IGBT?
The IGBT is switching a motor.

that is the actual momentary Vce times the current at that instant. This all averaged out. The equation is simply only for a steady DC situation, in all other cases you have to measure and calculate the power dissipated, or at least simlate it.
The equation you use assumes the case is steadily held at 25°C (with active refrigeration, liquid nitrogen or whatever means necessary), which in practice is very much not achievable.

As for the thermal impedance... Where do you assume that k=0.5 and 1e-5s from? How is the device actually being used?
I assume k= 0.5 as tp = 0.000025 as tp = (1/fs)/2 (where fs equals 20khz) and the duty cycle is 50%. ( I assume k as that value of tp is not shown on the log graph.

The IGBT is certainly not held at a constant temperature but I assume that for a given situation the IGBT case temperature will stabalise. It will therefore be this value that i will be using to estimate the junction temperature as the junction temperature should've reached a maximum.

The device is effectively PWM a high voltage motor. The duty cycle could also go above the 50% shown on the graph but again I assume this is when you use the Dc thermal impedance value of 0.53 as K will equal 1?

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#### dr pepper

##### Well-Known Member
If the motor is either on or off then the calcs for dissipation are much simpler, if however you are driving the motor with Pwm then things get more complex, depending on switching freq the device will spend some time in its linear region where dissipation is high, so you need to calc switching time & work out linear dissipation as well as that caused by vsat, this is why driver ic's are used to keep speeds up & dissipation down.
Are you just switching on/off are are you controlling speed with Pwm?

#### jamesh77

##### New Member
If only that was case. Yes I'm regulating speed with PWM.

#### dr pepper

##### Well-Known Member
Is this a college/uni question or is it practical.
I wonder in these situations how many actually bother doing all the calcs, it may well be quicker just to bolt the tranny on a known heatsink and measure the temp and back calc from there.

#### jamesh77

##### New Member
This is practical too, how would i work out the junction temperature from a known heatsink then?

It would've been nice to work it out theoretically based on the values in the datasheet and the given use case but it seems pretty hard.

#### dr pepper

##### Well-Known Member
I often cant be bothered with theory and do a practical.
If you use a 1 degree c per watt reference 'sink and the 'sink rises 10 degrees above ambient then you are chucking approx 10 watts into it.
I say approx as there is some thermal resistance inside the tranny and from the tranny to the 'sink, but your taking this into account doing this experiment.
Just multiply the number of watts your throwing into the 'sink by its degrees per watt for the final application heatsink and add the ambient temp to get its running temp.
To work out the junction temp you look at the thermal resistance of the washer your using or thermal heatsink compound so you have the tranny to 'sink resistance, then add that to the junction to case resistance in the datasheet, multiply this by the wattage dissipated and add to the 'sink running temp.
This is a ballpark way of working it out, so long as junc to case or tranny to 'sink resistance isnt silly high which for a to220 or to247 is unlikely.

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#### kubeek

##### Well-Known Member
This is a ballpark way of working it out, so long as junc to case or tranny to 'sink resistance isnt silly high which for a to220 or to247 is unlikely.
I just recently looked up an IGBT in TO-247, and Rthjc is 0.53 K/W. Add to that a few tenths for the thermal interface of case to heatsink, then some 0.5K/W for the insulating pad and you can easily overheat the tranny with just a hunderd or so watts, even if the heatsink stays perfectly cool at 25°C, even though the datsheet talks about 283W.

I am still not sure how to wrap my head around that thermal impedance graph (mainly what is the Ptot used there), but since the OP is using 50% duty cycle at quite high frequency the difference in peak junction temperature between the "proper" approach and using just the average power will be negligible (as the k=0.5 probably means, but it just doesn´t fit with my calculation). At very high peak power and low duty cycle the thermal impedance will become important, but not in this case.

#### dknguyen

##### Well-Known Member
but since the OP is using 50% duty cycle at quite high frequency the difference in peak junction temperature between the "proper" approach and using just the average power will be negligible (as the k=0.5 probably means, but it just doesn´t fit with my calculation).
EDIT: Nvm, I miss-assumed the context of what you were talking about. I agree. If your is dissipating power continuously, you can just use average power in your calculations as long as you calculate switching losses and conduction losses independently before summing them together. No need for the thermal impedance stuff as per kubek.

Trying to take into account thermal impedance when your system is under continuous duty cycle is kind of like trying to use your sprint times to estimate your marathon times.

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#### dr pepper

##### Well-Known Member
Looks like you've had some good advice there jamesh77, how you getting on?

#### jamesh77

##### New Member
Thanks for everyone's help.

Would i be correct in saying that the equation you provided dr pepper where .:. Rthjc(IGBT) + Rthjc(Sink) * Ptot + Tsink = Tj-average .:. Ptot = Tsink-Tambient / Rthjc(sink) shows me the average approx junction temperature.

Where as the Thermal impedance graph is used to show the peak junction temperature for a given point in time?

As the figures i provided for Tc etc above where purely figurative and used so you could help show me how to estimate it.

The challenge i have is that the duty cycle is forever changing for a given load, as the motor breaks through the load the duty cycle reduces to maintain a set speed. However the temperature of the sink and igbt is still obviously increasing.

(different igbt to the example from the application note) As an example of what i have measured at a given point in time: Tc = 63 , Tsink = 57. Vce p-p = 307v. Duty cycle = 65%. IC = 1.65A. Rthjc(IGBT) = 1.56K/W Fs = 20kHz.

Using the equation Ptot = Tjmax - Tc / Rthjc ................with a Tc of 63 and Tjmax = 150. For a case temperature of 63 the maximum power dissipation is only 55 degrees c. Yet calculating the Ptot = (Vce p-p * 65%) * (IC * 35%) = 93.53w....i must be going about measuring Vce wrong.....What do you think? All i can think is maybe it is VCEsat that needs to be measured here and because of the scale on the scope as the p-p voltage is 307v as the IGBT switches on I can't see that the VCE is higher than 0v...

#### kubeek

##### Well-Known Member
The transistor is not conducting Ic at the same time as it is blocking Vce, so you cannot multiply those. The power loss of the IGBT consists of switching losses and conduction losses. Switching losses you can estimate from the datasheet to be approximately 200uJ for turn on and maybe 50uJ for turn off, times switching frequency. Conduction losses are Ic times Vcesat times duty cycle.
From this you should get 0.65*1.65A*1.2V + 250uJ*20000/s = 1.287W + 5W = 6.3W. Nothing to worry about, at six watts you are close to not needing a heatsink at all. (BTW a mosfet would most likely be a better choice in such application)

#### ronsimpson

##### Well-Known Member
at six watts you are close to not needing a heatsink at all.
I agree with most of what you said but;
Junction to air resistance (no heat sink) is 50C/watt. At 6 watts that 300C + room temp = 325C. Which is twice the max junction temp.

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#### kubeek

##### Well-Known Member
Yes, that is why is said "close"  