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Author Topic: Gate drive requirements of IGBTs  (Read 1296 times)
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Hi all,

Quick question - that I feel I know the answer to already. Is it neccesary to drive the gate of an IGBT with a gate driver, given that there's a MOSFET before the BJT?

I'd imagine the answer is the same as for a MOSFET on its own "not essential, no - but for higher switching frequencies yes to reduce switching losses", correct?

Thanks in advance!
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The root of the answer is in the datasheet for the specific IGBT device you are using. It will state it's input capacitance and also it's output current curve Vs input gate voltage at a given collector voltage. From that you can determine what kind of drive capacity you must supply. As most IGBT devices are used in very high voltage and/or current circuits the use of special gate drivers is commonly called for.

Lefty

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IGBTs are typically driven to +15V and -5V on the gate (since IGBTs are designed for higher voltages the
gate voltages are typically higher and the negative gate drive helps reduce the slow turn-off time.  If the
application isn't demanding you probably can just use 12V and a standard MOSFET driver...
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IGBTs are typically driven to +15V and -5V on the gate (since IGBTs are designed for higher voltages the
gate voltages are typically higher and the negative gate drive helps reduce the slow turn-off time.  If the
application isn't demanding you probably can just use 12V and a standard MOSFET driver...

The device I'm using is a dedicated ignition IGBT - a Fairchild FGD3440G2. http://www.fairchildsemi.com/ds/FG/FGD3440G2.pdf - I chose this device as it has the lowest saturation voltage of all the easily available ignition IGBTs.

The voltage I'm switching is only small, ~13V - the inductive spike of course is significantly higher - which leads me to my first question;

1) There is no clamping voltage given, only clamping energy capability. How would one go about determining at what voltage this device clamps the inductive voltage spike?

2) Total gate charge seems to be somewhat higher than most MOSFETs I've worked with, at 24nC - however, the max gate emitter voltage is +- 10V, so I'll drive the gate using a MOSFET driver with 5V, given all the test conditions in the datasheet are quoted at 5V. Does this sound right?

3) All the examples, and test data use a series gate resistor - however, the symbol on page 1 shows a gate resistor, R1 - so why the external gate resistor too?

On the subject on gate resistors, with a 'normal' MOSFET on its own - should one use a gate resistor even when driving the gate with a gate driver?
« Last Edit: January 08, 2013, 08:21:14 am by jtw11 » Logged

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IGBTs are typically driven to +15V and -5V on the gate (since IGBTs are designed for higher voltages the
gate voltages are typically higher and the negative gate drive helps reduce the slow turn-off time.  If the
application isn't demanding you probably can just use 12V and a standard MOSFET driver...

The device I'm using is a dedicated ignition IGBT - a Fairchild FGD3440G2. http://www.fairchildsemi.com/ds/FG/FGD3440G2.pdf - I chose this device as it has the lowest saturation voltage of all the easily available ignition IGBTs.

The voltage I'm switching is only small, ~13V - the inductive spike of course is significantly higher - which leads me to my first question;

1) There is no clamping voltage given, only clamping energy capability. How would one go about determining at what voltage this device clamps the inductive voltage spike?
Well the collector-emitter breakdown voltage is 400V - should be a clue...  You probably don't want to clamp the spike anyway - for an ignition coil the spark plugs conduct before the primary voltage goes too high.
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2) Total gate charge seems to be somewhat higher than most MOSFETs I've worked with, at 24nC - however, the max gate emitter voltage is +- 10V, so I'll drive the gate using a MOSFET driver with 5V, given all the test conditions in the datasheet are quoted at 5V. Does this sound right?
Yup, +/-5V is probably a reasonable range to restrict to (the device has internal zeners I note).  Driving with +5 and 0V will be OK, just
a bit slower turning off.  BTW a logic-level IGBT is not something I've seen before - will see how cheap these are.
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3) All the examples, and test data use a series gate resistor - however, the symbol on page 1 shows a gate resistor, R1 - so why the external gate resistor too?
To limit rise/fall times and reduce gate currents - reducing RFI, to reduce dissipation in the gate and the gate driver.  Also can reducing
ringing on the gate circuit.
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On the subject on gate resistors, with a 'normal' MOSFET on its own - should one use a gate resistor even when driving the gate with a gate driver?
Again depends on how fast you want to switch (faster reduces dissipation in the main circuit, but increases noise and RFI).
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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!
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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.
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yeah,  device has an avalanche rating - that's how much energy it can take in a single breakdown event.thank
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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......
« Last Edit: January 10, 2013, 05:36:28 am by jtw11 » Logged

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This topic is close to my question so i'll just post here if you don't mind smiley-wink

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 smiley-wink. Thanks!
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