# IGBT rating

Discussion in 'General Electronics Chat' started by robert.guttormson, Feb 27, 2010.

1. ### robert.guttormson Thread Starter New Member

Oct 3, 2009
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0
HI,

I need to design an inverter in a half bridge configuration that needs to drive a 2KW motor. For this purpose i am on the lookout for an IGBT that satisfies the power requirements. In this context i should add i am not very comfortable with the concept of IGBTs. I was considering the IRGP50B60PD chip and i would like clarification on some terms i found. The data sheet says Id for this chip is 50A, this means that the amount of current that can pass through the collector-emitter is 50A(in case if this is not right please let me know).

My problem starts with the Vces which it says is 600V. What does this quantity signify? does it mean it can push 50A of current @ 600v ? If that is so and if i am operating at 25C would this IGBT be good enough to power my motor, since P=VI = 50*600 = 30000w o/p ?

Here is the link for the datasheet
http://www.irf.com/product-info/datasheets/data/irgp50b60pd.pdf
Also what does the Vces(br) signify?

TIA,
Bob

2. ### SgtWookie Expert

Jul 17, 2007
22,194
1,763
The Vces is the absolute maximum collector to emitter voltage; if that is exceeded, the IGBT will break down and be destroyed.

What is the voltage powering the motor?

The Id is the equivalent MOSFET rating.

If you have a continuous collector current of 33A, with Vge=15v, Vce(on) will be around 2V, so you'll have to dissipate 66W in the heat sink.

3. ### robert.guttormson Thread Starter New Member

Oct 3, 2009
26
0
Thanks for your prompt reply, the motor is powered using a 110v rating. So the Vce(on) signifies the voltage at which the collector-emitter current is pushed through ? If that is the case then what IGBT rating should i use if i need continuous power output of 2Kw

So the Vge if it is 15v it means the IGBT needs 15v to move into conducting state from an off state ?

4. ### SgtWookie Expert

Jul 17, 2007
22,194
1,763
If the motor is powered by 110v, there will be an 18.2A current flowing through the collector-emitter junction of the IGBT, and Vce will still be around 2v. It will be dissipating somewhere in the vicinity of 35 Watts of power.

I don't know why you're considering using an IGBT for a fairly low-voltage application rather than a power MOSFET? You could easily use a power MOSFET with a low Rds(on). Just for instance, an IRF3415 has a rating of Vdss=150v, Rds(on)=42m Ohms, Id=43A; you would have under 14 Watts of power dissipation. Gate charge would be lower as well. That isn't a recommendation to use that particular MOSFET; there are undoubtedly more modern versions with better specifications available.

Anyway, I think that IGBT's are better used in higher voltage applications - say, 300+V. If you're <200v, you're probably going to be much better off using power MOSFETs.

5. ### robert.guttormson Thread Starter New Member

Oct 3, 2009
26
0
"It will be dissipating somewhere in the vicinity of 35 Watts of power." power dissipated is reflected as heat at the junction of the IGBT?

Preference for IGBTs seem to be driven not out of technical consideration but something else that I am not aware of.

TIA

6. ### SgtWookie Expert

Jul 17, 2007
22,194
1,763
Yes.
You have current flowing through the emitter and collector. There is a voltage drop across the collector and emitter; somewhere in the vicinity of 2v. Since P=EI, and E= about 2v, and I=18.2A, you have around (ballpark) 35 Watts dissipated in the IGBT.

The datasheet doesn't get real specific about Vce at different Ic's, but there's a plot that shows the equivalent of a saturation curve; Vce vs Ic vs Vbe. There doesn't appear to be a lot of difference between the three plots; all Vce's are around 2v as long as Vbe is 15v.

You might look at an IGBT rated for a lower Vces. The Vbe(sat) curves would probably be more attractive.

Sep 30, 2009
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