Miller Capacitance causing IGBT to turn on in half-bridge

Question

Some time ago I built a H-bridge with 5 paralleled IRFP250N and after a couple of blown-up IR2104s I came up with this driver design that was more about protecting the IC than anything else:

Where:

• HO is upper MOSFET gate
• LO is lower MOSFET gate
• Vs is half bridge output

(R2 is not connected. It was designed so I can put another chip in there. R1, R4, R5 are shorted if I recall. R17, R18, R19, R21, R22, R23 are there so I could experiment with different values. Gate resistors are directly soldered to the transistors).

It worked well, and I never had a problem with it. Switching frequency was 400 Hz with about 700 nS turn on and off times. Then I bought two 2MBI200U2A-060 modules for cheap and hooked up these drivers. When no power is applied to the half-bridge this is the waveform on the gate of the lower IGBT:

It looks good to me. Then after applying 30 V to the bridge this is what happened:

The yellow waveform is Vge and blue Vce of lower IGBT. This is without the load. There are snubbers added but there is still some ringing on the output, but this is not what I am concerned about.

The voltage on the gate suddenly rises well above the turn-on threshold and the IGBT actually starts conducting because I can see increased power consumption when this spike goes above 7 V. I tried different gate resistors and diodes as well as adding a PNP gate discharging circuit (using I think it was TIP42C paired with the second transistor connected in Darlington) and the max spike voltage was around 5 V.

When I connect the load it gets a bit better but still not pretty. With a 100 Ω resistor and diode on the gate the turn on time was about 7 μs and turn off about 1 μs but at this point having these times so large is just a loss of power.

I tried different combinations of gate resistors, diodes, and even a second totem pole driver but I can't get rid of this spike. I know I can use negative voltage at the gate but this makes the circuit so much more complicated. How can I prevent this?

I think that this has something to do with Miller capacitance but I'm not sure. The datasheet says that this module has 14 nF input capacitance, so quite high. In the end I plan on using this bridge with 150 V, 180 A running at 400 Hz and I fear that with 150 V both transistors will turn on and blow up.

Any help is appreciated.

Answer 1

A bootstrap driver is wholly unsuitable for this type of IGBT. Notice Cres vs. Coes curve in the datasheet (also see edit below): this means for a voltage step on the collector, half that voltage appears on the gate. They're pretty awful as transistors go these days; but what do you expect from a 20-ish year old design, I guess. (Plenty fine for say motor drives, which was probably the main application they were used in.)

Devices like this, must operate with low enough gate drive impedance, and slow enough dV/dt on the switching node, that the differentiator thus formed (between the C-G-E capacitor divider, and total equivalent gate resistance), does not drop enough voltage to turn on the transistor, thus causing destructive switching (shoot-through). V_ge(off) could be as high as a couple volts below V_ge(th), but it is for this reason that it must be considerably less: the datasheet recommends -15 V.

Turn-on dV/dt is likewise controlled by R_G and the same equivalent capacitance, so you simply wire up whatever gate drive you have, and adjust R_G until switching is just safe enough under all operating conditions.

Your circuit may still work, but R_G must be much higher to get dV/dt low enough to support it. This is a direct consequence of V_GE(off) being only a few volts below V_GE(th), rather than more than 15 below.

Typically an isolated gate driver is used, with 30 V total supply (or 15 V supply and an H-bridge output, but this makes desat protection difficult to add -- another highly recommended feature!). This solves both the issue of isolation between control and, usually mains voltage (600 V devices are typically used on 240 V industrial equipment), and the difficulty (or impossibility!) of using bootstrap type drivers.

Do you really need IGBTs for such low voltages? -- is this a convenience thing, this module is cheaper than alternatives? Is it really cheaper when the lost efficiency (V_CE(sat) ~ 2 V, max. ~97% efficiency) is accounted for? Switching loss at 400 Hz at least isn't a problem, either way. (Indeed with MOSFETs, your application might benefit from the reduction of harmonics, or component size, by using PWM at higher frequency; this will require more care in the inverter layout, and more design of the controls, of course, so I understand if this isn't an option at this time.)

I guess you were originally working with MOSFETs, but blew the driver, which is most likely be due to poor layout. This will not be affected by going to IGBTs. The switching frequency doesn't matter: it's speed that kills, how fast it goes between on/off. Layout is very much a part of the circuit and cannot be ignored!

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Edit: Manufacturer explains the parameters here:
Technical Terms and Characteristics
Particularly Fig. 2-13: the capacitances are separated, using a different definition from most MOSFETs. This explains why C_oes can be lower than C_res. This also means you must use the sums of these values, as appropriate, in calculations.

Also note the typo on page 3, they forgot to edit the "Output capacitance" description to say C-E capacitance. Presumably, they are using the latter definition.