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understanding IGBTs

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IamSmooth

Thu Aug 20 2009, 07:25PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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I just wanted a separate thread to make sure I understand IGBTs properly. This thread does not involve any circuit designs.

If I understand correctly (I hope), as long as Vce and current (I) operate within the SOA with sufficient heat sinking, the chip should be safe.

One can conduct higher currents and use higher Vce if the IGBT is
1. fully open (full conduction)
2. the conduction is in the order of msecs or usecs

If the above statements are reasonably correct, what I don't fully understand is what is the difference to the chip when it is switching ON/OFF compared to when it is just ON.

Finally, let's say a current conduction of 10A and 100v is outside of the SOA. I would destroy the IGBT if I did this. If the chart says I can have these values for 100usec (10khz), does this mean that (with sufficient heat sinking) I can turn the gate ON/OFF at a frequency of 10khz or higher? What is my limitation on the duty cycle? Is it that the on period can not exceed the 100usec?

Dr. Slack

Fri Aug 21 2009, 07:12AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
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Posts: 1659

what I don't fully understand is what is the difference to the chip when it is switching ON/OFF compared to when it is just ON.

When it's ON, it's taking 10A with a Vds of the order of a volt or two, so dissipation of the order of 10s of watts.

When it's switching ON/OFF, it's taking 0 to 10A, with a Vds of 100 to 0v, with a power dissipation of the order of 100s of watts, higher for a rail higher than 100v.

The difference is that it's disspiating an order of magnitude or two more power in the transition.

Is it that the on period can not exceed the 100usec?

No, it measn that the *transition* period when it's dissipating high power must not be longer than 100uS.

A power device must meet several limitations simultaneuosly. Amongst these are peak power (as defined by the SOA graph) and average power (as defined by the Tjmax, Rjamb etc).

Consider the average power dissipation of a switching device to be made up of the sum of two components.

a) The first is easy to predict and control, it is the ON state rresistive loss. Take the ON current, multiply by the Vds in the ON state (relatively constant for a BJT or IGBT, based on resistance for a FET) to get the ON power. Multiply by the duty cycle if it can never vary up to 100%.

b) The second confuses many people. While the device is in the transition between ON and OFF, the instantaneous disipation can be very high, 100s or even 1000s of watts. However, this period lasts only for uS. It is best to integrate the power loss over the switching time, and call it a switching ENERGY loss in JOULES. This loss occurs every time the device switches. The switching POWER is then the switching ENERGY per second, = switching loss * the number of switch events per second, or 2 * switching loss* switching frequency.

You can reduce the switching POWER by reducing either the switching frequency, or the switching ENERGY, or both.

You can reduce the switching ENERGY by driving the device faster through the linear region, so it is dissipating the kW pulse for less time.

Many data sheets, especially for IGBTs, specify the switching loss in Joules, under particular drive and load conditions. It is usually somewhat different for the turn on and the turn off directions.

The total dissipation, a+b, must not exceed the data sheet average disspiation. The peak dissipation, b by itself, must not exceed the SOA graph.

This goes for all power devices, including IGBTs, FETs, BJTs. It sort of includes tubes as well, but they tend to be much more robust to large peak powers, so don't have the same critical short pulse SOA graph, but it's still the same principle for average power. I somehow can't see too many people building a SMPS using tubes, but I'm afraid that I can't resist generalising results from one area of technology to another where the basic principles are the same. Actually, a relay when it's arcing is in the high dissipation transition region, many relays specify a maximum energy that the contacts can absorb before they weld, equaivalent to a SOA, but maybe that's a generalisation too far?

IamSmooth

Fri Aug 21 2009, 12:32PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

Thanks Neil. There is still one point that got missed, and one you brought up that I did not appreciate. It has to do with the confusing terminology of ON and ON/OFF.

I see from your explaination that there is significant energy dissipated during the transition from the OFF to the ON state, and less energy when the device is fully on as the voltage drop across the device is small.

So the question is what is the difference when on leaves the device fully ON (frequency at the gate = 0) and when is using the device in a pulsed mode, where the gate opens the device hard (takes it into full saturation, is that correct terminology?)?

You see, to me the heating of the device would seem the same if I leave the device ON for 1sec, or pulse the gate at 10khz for 4 seconds with a duty of 25%.

HV Enthusiast

Fri Aug 21 2009, 03:14PM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
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MODERATOR NOTE:

Just a note, all these threads about "how" components work, are bordering on the rule about how you should be doing your own homework. There are more than enough resources on the internet that describe how transistors work and so forth . . .

Lets please try to keep these to a minimum . . .

And I guarantee, if someone posts a thread about how a resistor works, it will be locked! :P

Dr. Slack

Fri Aug 21 2009, 04:05PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

You see, to me the heating of the device would seem the same if I leave the device ON for 1sec, or pulse the gate at 10khz for 4 seconds with a duty of 25%.

Quite why you should want to do this comparison is beyond me. However, there is enough information in any particular device data sheet, and my previous post, for you to work it out yourself.

HV Enthusiast

Fri Aug 21 2009, 06:06PM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
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Posts: 3068

IamSmooth wrote ...

You see, to me the heating of the device would seem the same if I leave the device ON for 1sec, or pulse the gate at 10khz for 4 seconds with a duty of 25%.

Thats not correct. The conduction losses would be equivalent, but the switching losses would be much higher in the 10kHz example.

Total Diss = Conduction Losses + Switching Losses

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