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I will use MCU that has 5V output that should connect to ISL9V5036S3ST IGBT. IGBT will drive ignition coil that has 0.8 Ohm resistance connected to 12V (+/- 3V) power supply. IGBT will be on the low side. Spark will occur after every 15 milliseconds and longer (around 200 milliseconds). Coil will be charged for couple ms.

What I'm trying to figure out is how to calculate resistor value that is needed (if any) between MCU output pin and IGBT gate and how much current will flow from and to gate.

Also I'm not sure if gate driver has any additional benefits for current application.

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Also I'm not sure if gate driver has any additional benefits for current application.

Given the speeds you are operating at you shouldn't need a gate driver but, watch out for \$C_{IES}\$: -

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You should consider \$C_{IES}\$ because it is approximately 2 nF and, the gate resistor (if you use one) will have a slight effect. However, if the gate resistor is 1 kΩ then the activate time (5x CR) is still going to be only around 10 μs so maybe not a showstopper given your operating speeds.

Coil will be charged for couple ms

I don't know what the primary inductance is but it might be typically 5 mH for a vehicle ignition coil and you have to ask yourself is 2 ms long enough to get energy into the coil. At 2 ms, with a 12 volt supply and 5 mH, the primary current will rise up to 4.8 amps. That's a total energy stored of 57.6 mJ.

I mention this because I'm not overly confident that this is sufficient for your needs. Anyway, that's for you to decide. Also be aware that there will be a flyback voltage that superimposes itself on the primary and this cannot be clamped. There will also be an extra back-emf due to leakage primary inductance that usually needs to be quenched. So, double check that against the voltage rating of the IGBT (390 volts peak).

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According to this post it has a relative high gate resistance in order of 100 Ohm to 1 kOhm.

Perhaps you do need this way:

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT:

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enter image description here

The datasheet values are obtained with RG=1k, so there might be a reason for that.

If you look the internal structure of the this IGBT hybrid, you can see that it has a kind of over voltage protection circuit. When the voltage rises at certain level the prot. circuit will put the transistor in conduction mode again, so that excessive energy will be dumped by the transistor itself.

If you put a gate resistor with too low value, the protection won't work, since you will drain the gate.

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  • \$\begingroup\$ What effect this circuit will have regards to high gate resistance? It seems that it will change charge and discharge time of the gate. \$\endgroup\$ Commented Sep 27, 2021 at 19:48
  • \$\begingroup\$ @featherbits Yes, it's the standard way of IGBT drivers, but see edit. In this specific case it shall not be used. \$\endgroup\$ Commented Sep 28, 2021 at 9:32

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