Is your IGBT gate-driver power supply optimized? – Part 2

In part 1 of this series, we discussed how to correctly select the control voltage for an IGBT.  This time, you’ll learn more about the isolation requirements and how to calculate the correct IGBT drive power.

For any industrial motor drive, potential separation of the input circuit (low-voltage) and the output circuit (high-voltage) has to be ensured. The low-voltage side interfaces with the control electronics, whereas the high-voltage side is connected to the IGBTs. The separation is necessary, because the emitter potential of the upper IGBTs is switched between the DC+ and DC- potential of the DC-bus, which can range in the hundreds or thousands of volts. Depending on the application, the corresponding standards for clearance and creepage distance have to be observed as well as compliance with the test voltages. Some typical standards observed are: IEC60664-1, IEC60664-3, IEC61800-5-1, and EN50124-1.

In the simplest case, it may be sufficient to separate only the upper IGBTs of a half-bridge from the lower IGBTs. This is generally possible if the microcontroller is also referenced to the DC- potential. A subsequent separation of the interconnection to the user interface is advised or required, depending on the application. This is mostly to apply basic isolation from noise and common-mode ground effects. In high-power applications, separation takes place at every IGBT, each driver with its own power supply as shown in Figure 1.

Figure 1. 3-phase inverter with isolated gate-drive (Note: All gate-drivers are powered with individual isolated power supplies)

The complexity for the power supply can be simplified for those switched that have their emitter on DC- potential, as shown in Figure 2.

Figure 2. 3-phase inverter with isolated gate-drive (Note: Lower gate-drivers are powered with a common power supply)

Now, lets’ learn how to calculate how much gate-drive power is needed for an IGBT.  While driving an IGBT, the transition between the two gate voltage levels requires a certain amount of power to be dissipated in the loop among gate driver, gate resistors and IGBT. This figure is typically known as “drive power - PDRV.” This drive power is calculated from the gate charge QGate, the switching frequency fIN and actual driver output voltage swing ΔVGate:

PDRV = QGate * fIN * ΔVGate

If there is an external capacitor CGE present (auxiliary gate capacitor), then the gate driver also needs to charge and discharge this capacitor as shown in Figure 3.

Figure 3. IGBTs with gate drive circuitry for gate power calculation

The value of RGE is not influencing the required drive power as long as CGE is fully charged and discharged during one cycle. The required drive power becomes:

PDRV = (QGate * fIN * ΔVGate) + (CGE * fIN * ΔVGATE2)

It should be noted that the drive power does not depend on the value of the gate resistor or the duty cycle as long as the switching transition goes from fully on to fully off and back. Also, these equations are true in non-resonant gate drives. This is the total drive power required by the IGBT but the gate driver which is driving the IGBT also consumes some power.  This power consumption should be added to get the final value for gate drive power.

PDRV = (QGate * fIN * ΔVGate) + (CGE * fIN * ΔVGATE2) + Pdriver

Did this cover what you need to know for optimizing your IGBT gate driver? What more would you like to learn?