# IGBT gate drive calculator

Published on: Feb 17, 2016. Updated: Dec 06, 2017.

To insure that we switch our IGBT fast enough to lessen losses, slow enough to avoid ringing and the drive circuitry is stable we have to calculate the power, current and peak currents in our drive circuit. Detailed explanation further down.

A much more detailed description will be available in the GDT / driver chapter of the DRSSTC design guide, here is only presented limited information needed to use the calculator.

The results are only valid and absolute minimum for a single gate, so if you have 4 IGBTs in a fullbridge, you need atleast 4 times the driving power available.

Adjust the gate resistor so that the gate drive peak current is lower or equal to the capabilities of your IGBT driver IC or circuit.

Switch between the input fields to automatically calculate the values.

 Gate charge of IGBT(IXGN60N60 example) nCoulomb Switching frequency Hz Vgate (on) Volt Vgate (off) Volt Gate resistor (ext) Ohm Gate resistor (int) Ohm Gate drive power needed Watt Gate drive current needed mA Gate drive peak current(*) mA DRSSTC specific See description below Optional: BPS Hz Optional: on-time in uS uS Gate drive power needed Watt Gate drive current needed mA

DRSSTC specific: optional BPS and on-time are DRSSTC specific for calculating a reduced power need at lower duty cycles. BPS is how many times a second the IGBT is turned on for the duration of the on-time, these two figures gives a average duty cycle.

### Mathematic used

When developing the power supply for the driver section it is important to know how much it has to deliver. The power needed to drive a IGBT gate is given by the energy needed to drive the gate and the frequency at which it should be driven. The energy needed is the gate charge(QC) times the difference between turn-on and turn-off voltages from the driver or GDT.

$P_{\mbox{gate drive}}=Q_{\mbox{Gate}}\cdot\left ( V_{\mbox{Gate(on)}}-V_{\mbox{Gate(off)}}\right )\cdot\mbox{f}_{switching}$

Enough current also has to be supplied to charge and discharge the input capacitances of the IGBT in order to switch the IGBT on and off. This calculated current is the minimum average output current of the driver output stage per channel:

$I_{gate}=I_{GE}+I_{GC}=Q_{Gate}\cdot\mbox{f}_{switching}$

If the gate peak current is increased, the turn-on and turn-off time will be shorter and the switching losses lessened. Switching faster has disadvantages too like ringing voltage on the gate that could result in overvoltage spikes from switching too fast across the internal stray inductance of the IGBT module. (*) A theoretical peak current can be easily calculated, the IGBTs internal gate resistor must be taken into account and the result is in practice lower as the internal stray inductance limits this theoretical value.

$I_{\mbox{gate peak}}=\dfrac{\left (V_{\mbox{Gate(on)}}-V_{\mbox{Gate(off)}}\right )}{\left (R_{\mbox{gate resistor external}}+R_{\mbox{gate resistor internal}}\right )}$

In the data sheet of an IGBT driver, a maximum peak current is given, as are the minimum values for the gate resistors. If both these maximum and minimum ratings are exceeded, the driver output may not be able to properly drive the IGBT.