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Topic: Gate drive requirements of IGBTs (Read 1 time) previous topic - next topic

jtw11

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Well the collector-emitter breakdown voltage is 400V - should be a clue...  You probably don't want to clamp the spike anyway


The devices are advertised as self clamping, is the breakdown voltage where the voltage at which the device clamps? "should be a clue" certainly seems to imply so, but then you say you probably don't want to clamp, so I think maybe not?

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Again depends on how fast you want to switch (faster reduces dissipation in the main circuit, but increases noise and RFI).


So say my gate driver is capable of delivering 5A peak gate drives, but I want only to deliver peak 1A to switch the gate at a speed I want (I've got a good app note from Texas outlining very well how to calculate gate currents, speeds etc), then I just size the series gate resistor to limit current to 1A? Given the switching speeds, would it be neccesary to size up this resistor physically? All my resistors on my board are currently 0603 SMD parts...

Cheers for your help so far!

MarkT

Well the device has an avalanche rating - that's how much energy it can take in a single breakdown event...

To calculate the dissipation in the gate resistor I think the formula is   f x C x V^2  where f is switching frequency,
C is the effective capacitance of the gate, V is the gate drive voltage.   Or you can substitute total gate charge for
C x V, giving  f x V x Q  (if you think about it that's just I(average) x V).  R is really the total resistive loss in the gate circuit
of course, but if you add a gate resistor it is usually dominant.
[ I won't respond to messages, use the forum please ]

danavsbelly

yeah,  device has an avalanche rating - that's how much energy it can take in a single breakdown event.thank

jtw11

#8
Jan 10, 2013, 11:04 am Last Edit: Jan 10, 2013, 11:36 am by jtw11 Reason: 1
I see - thanks for the gate resistor calculation info!

I'm still a little confused about whether or not I need to / don't need to clamp. All the performance curves talk about being valid only for V clamp < 430V. That seems to imply that I need to design a voltage clamp into the circuitry? ...but then Figure 1 and Figure 2 are on about 'Self Clamped Inductive Switching'.

But given there's no clamp voltage given, only self clamped inductive switching energy, instead of worrying about actual clamping, just calculate the energy delivered by the inductive spike, and always keep that within the avalanche energy spec.

I feel very lost having never used IGBTs, or really BJTs for that matter...

Next question, current limiting - how do I go about limiting the output current to a certain value - say I only want 8A out, I don't want to use an external ballast resistor. So I assume the easiest way to do this is use a ballast resistor on my board, in a TO220 package or something to limit the current, but can I not limit current by only supplying the gate with a certain voltage?

The resistor approach dosen't seem feasible, even at my max duty of 33% - I'd be dissipating nearly 20W in the ballast resistor!

Or actually, at dwell times of only around 3ms - given the inductance of the primary itself, I'd imagine the current dosen't rise all that fast and it may be safe to drive the primary directly without a ballast resistor? EDIT - Or of course, use only coils that do not require a ballast resistor......

MrOnion

This topic is close to my question so i'll just post here if you don't mind ;)

I scoped the signal coming from an existing ignition system to a IGBT/etc driver. It stays at 12V when the system is turned on and drops to 7V (delta 5V) when an ignition event occurs (triggers on falling edge). Heres the screenshot:



It's a steady looking 3.5ms square signal and I just want to ask out of curiosity how to replicate this kind of signal? I was planning to use an mosfet driver with 12V, but the fact that it does not drop to 0V when triggered got me curious.

For a n00b like me this is weird but for you this might be something obvious ;). Thanks!

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