# calculation of the Fet or igbt gate current with pwm signal

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hello.
Some of my lab lessons, we needed to know how to calculate fet or igbt gate current.
For exmpl;
1)driver circuit is a totem poled optoisolator.
2)100ohm gate resistor connected between gate and opto.
3)12Volt opto output.
4)10kHz switching freq by pwm signal.
5)Total Gate Charge (Qg) is 146 nC
6)Gate-Source Charge (Qgs ) is 35 nC
7)Gate-to-Drain ("Miller") Charge is 54nC
8) There is no snubber circuit.
NOTE: 5-6-7th infos are from irf3205's datasheet
Can someone explain how to calculate that current value will flow from opto through the gate resistor ? Or is there any excel file to make that calculation ?
Many thx. HANS.

Edit:
I know gate current is not constant; it varies exponentially as the capacitance charges. Im asking of the max current will flow through the gate when fet is fully on and gate capacitance is fully discharged. Also for short discharging time, a fast recovery diode paralel connected to gate resistor should be better right ?

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There is too much information I do not know:
-Gate peak current? Average current?
-What optoisolator: I will assume the rise and fall time is very fast and the output current is very large and it is a perfect isolator.
-What is the MOSFET Drain current? Assume 40 to 50A Driving a inductive load.

IRF3205 Gate turn on voltage "2 to 4V" but that is for 250uA. We need to look at gate voltage for 50A which is about 5.5V.

MOSFET is off and the gate voltage is zero.
Isolator turns on and the voltage goes from 0 to 12V in 0nS (perfect part).
There is a 100 ohm resistor from 12V to the gate at 0V. 120mA!
The 3450pF Gate capacitor charges up. (actual value is not important) Gate voltage increases up to 5.5V where the current is only 65mA!
Mow the MOSFET starts to turn on and drives 65mA into the gate through Miller effect. Gate voltage holds at 5.5V.
When the Drain drops to near 0V the Miller effect goes away and the Gate voltage starts up again. (simple 12V 100 ohm capacitor charge up)

MOSFET turn off is the reverse of that. (0V source, 100 ohms to gate at 12V) Miller effect and 5.5V. You can work it out.

So the max current is 12V 100 ohms 120mA. During turn on/off time the Miller effect and the Gate turn on voltage of 5.5 volts gets you a current of about 1/2 of the 120mA. That only tells you the peak current and the current during 'Miller time'.

If you want average currant:
During off or on time current =0.
One cycle = 10khz.
You will have to find the time to charge up from 0V to 5.5V + Time for turn on + 5.5V to 12V and the discharge time.
-You will be charging the Miller cap with only 65mA. Last edited:
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