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driving an IGBT

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IamSmooth

Tue Aug 25 2009, 10:16PM

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

I have made a circuit that gives me a high/low signal of 17v/0v to drive an IGBT. The output is from a schmitt trigger using a comparator with the output driving a signal transitor to invert the signal. Every time I connect the IGBT gate to this output the gate voltage never gets above a few volts. Does this have to do with the gate charge? Does the gate require a certain amount of current in order to achieve the voltage to properly drive it?

EDIT:

I found out the problem. I forgot that I was using a microwave transformer to get up to test voltages of 240vdc. Under the full load I drew more current than the transformer was capable of putting out so I never saw any current flow through the IGBT. I will have to rework my test equipment to get up to 240vdc.

HV Enthusiast

Wed Aug 26 2009, 11:35AM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
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The gate for all intensive purposes is simply a capacitor. You drive it with volts and current just like you would charge any other capacitor up.

Just that, the amount of current capability you have determines how quickly you will charge up that gate and discharge it.

Not sure what the application is, but if its one where noise is expected, a negative bias on the gate would be beneficial.

IamSmooth

Thu Aug 27 2009, 02:32AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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I'll keep this on this thread since it involves IGBTs. Maybe someone can clear up some confusion I have about this IGBT's SOA.

On page 3, figure 2 the SOA voltage goes up to 600v with very high currents.

On this link for a more robust IGBT

You can also view the above file here, but the first link might be easier

on page 5 the SOA is broken down into regions defined by the rate of switching. One can see that if the rate of change is too slow one can not allow more than Vce=30v.

So, how is one to determine on the first IGBT how slow one can change before there are problems? Or does one increase Vge as quickly as possible for almost all applications?

I also noted that under the test conditions Rgate was 2.2ohms. I am using 1k @18v to drive the gate. Should this be ok for 240v/8A?

Dr. Slack

Thu Aug 27 2009, 03:17AM

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

The killer is the Miller (I'm a poet and don't knowit)

An IGBT (or FET for that matter) is a high gain inverting amplifier, with a modest Collector (drain) to gate feedback capacitance. To see what a high gain inverting amplifier with capacitive feedback does, put an opamp and a feedback C into a simulator and play with them.

As you trickle (which is the correct word for 1kohm, for 2.2ohm the word wouild be slam) charge into the gate, first of all you raise the voltage from 0 to Vth by charging the static capacitance. As the device begins to conduct, then it becomes an amplifier and for all intents and purposes, the capacitance seen at the gate becomes infinite as the gate current charges the gate-collector capacitance as its voltage falls from rail (600v or 240v) down to saturation. This can take a long time at low current, and it's a long time spent in the high dissipiation SOA-busting region. Finally the device saturates, stops being an amplifier, and you're back to charging the static capacitance. If you look on a data sheet, this dangerous phase is the flat portion of the gate voltage versus time graph.

A very precise analogy is for the temperature of melting ice. As you trickle heat into ice at -20, its temperature rises quickly to freezing point (Vth), then sticks at freezing as it absorbs heat (charge) without changing temperature (voltage), then finally when it's fully melted (bottomed) the temperature rises again.

This is why you should take Conner's advice and go for a gate driver that can deliver an amp or so, get it out of that Miller region as fast as you'd get out of a spacesuit having f@rted.

IamSmooth

Thu Aug 27 2009, 04:49AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Very funny.

I read about the driver chip Steve was talking about, but it says it is for driving MOSFETS. Will it work for an IGBT? Also, it talks about having 9A to deliver to the gate, but it does not mention voltage.

How about this: could I use a PNP transistor for the gate drive? I can connect the emiiter to the 18v rail. I can connect a 2-5ohm resistor from the collector to the gate of the IGBT. I can even put a capacitor from ground to the emitter to increase the burst of current when the PNP is turned on by my base signal going low.

Dr. Slack

Thu Aug 27 2009, 05:16AM

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

As far as all power devices that you would consider for this duty are concerned, IGBTs and FETs look exactly the same. There's a bit of difference on speed which matters for TCs, and on saturation voltage which affects efficiency depending on load voltage and current, but otherwise they're pretty much interchangeable.

You can rest assured that Steve's driver does specify the volts somewhere somehow, and that it will be suitable for both FETs and IGBTs.

Now it's time to get quantative and do some sums, rather than all this hand-waving. Once you nailed all the details, it all becomes very easy.

The maximum dissipation in the linear/Miller region, as there's a resistive load and assuming a 240v rail, will be at half rail on the IGBT, so 120v across 30ohm = 4A, * 120v = 480 watts.

From the SOA graph, how long can you allow 480 watts to be dissipated? The disspiation is lower at both higher and lower Vce, but is a large fraction of this for most of the time, so use 480 watts for the whole slew time as a conservative worst case to keep things simple and safe.

From the data sheet, what is the effect capacitance from collector to gate (the max to be safe)?

If you assume 15v between your driver output and the gate terminal at Vth, with a 1k sourcing resistor, then there's 15mA flowing into this capacitance when you switch on. How fast is it going to slew in volts/second? A worse case than this is switching off, as you now only have 0v from your driver, and *still* Vth at the other end of the 1k resistor, so you may only be pulling 2 or 3mA from this capacitance. How fast is it going to slew now? (Hint, this is why gate drivers specify sink as well as source current, and many people use -ve gate drives to accelerate the off transition).

The collector/gate capacitor has to be slewed through 240v. How long is that going to take? Going on? Going off?

If both times are less than the time for 480 watt SOA, then result happiness, job done. If either is greater, then there's more work to do.

What current do you need to supply to the gate to spend less than the maximum permitted time in the nasty region?
Now you can then ask "will driver configuration x be adequate?" Will it provide at least that current when either sourcing or sinking to a voltage of around Vth?

Steve Conner

Thu Aug 27 2009, 08:53AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Trust me, just use the gate driver chip. It'll save you a lot of hassle.

The PNP transistor thing might get your IGBT turned on quick, but you need to turn it off quickly too.

If you search the forum and archives, you'll see lots of discrete gate drive circuits, using NPN and PNP pairs, or little complementary pairs of MOSFETs. I don't believe that any of them are worth the hassle when you can get it all on a chip for a few dollars.

You asked about voltage: The voltage is just whatever you feed to the gate driver chip. If you supply it with 12V, it'll switch the gate between 12V and 0V.

There's also the TC4420/4422. I list both because I can't remember which one is inverting and which non-inverting.

Finally, re Dr. Slack's discussion, you can switch too fast as well as too slow. Search for the "Goldilocks Gate Drive" thread where we discuss gate drive that's not too hot or too cold...

IamSmooth

Thu Aug 27 2009, 12:09PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Neil and Steve, thanks again. I was thinking about just what you said when I went to bed. I would be able to turn the device on, but I would have to work out a means with another transistor to sink the base to ground to turn the chip off.

I'm going to order a few of the chips today. Right now my NPN driver hits the gate at 18v through a 600ohm resistor. I have an 18v regulator so I will read up on the TC4420. It says the input voltage is typically 1.8v (then it says min 2.4v??). There is no max value. Would it be acceptable to use my comparator output of 17.8v to drive the input? With no max value I would think that anything up to my Vs should be fine.

Also, I looked at the graph of Vge vs the current and I see that Vge=18v is higher than Vge=15v. Does it make a big difference using a gate driver that operates at 15v vs. 18v? Current will charge the gate capacitor, but it is the voltage that opens the CE junction, right?

IamSmooth

Mon Sept 07 2009, 06:17AM

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

I destroyed an IGBT...

Here is a basic schematic of the load my IGBT sees.

I used setup #1. When I did this the IGBT died in a permanent shorted state.
I removed the capacitor and it worked fine. My shunt controller cycled back and forth, and the IGBT lived.

I was thinking of using topology #2, feeling that the sudden inrush of current through the capacitor was the problem. If it is, how are IGBT drivers for say, induction heaters, able to handle the current through the LC tank? Is it that the capacitor values are way smaller in the order of nf vs uf?

Proud Mary

Mon Sept 07 2009, 06:33AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Sorry if it's stating the obvious, but why not study the data sheet and application notes for this device?

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