Gate drivers — the critical link in power-device performance: part 2

Thus far, we have examined the general issue of power devices and their characteristics. This part goes into the details of the power-device gate driver.

Gate-drive details

To drive current into the gate, the positive rail’s voltage should be high enough to ensure full saturation/enhancement of the power switch, but without exceeding the absolute maximum voltage for its gate. While this voltage value is a function of the specific device type and model, IGBTs and standard MOSFETs will generally be fully on with a 15-volt drive, whereas typical SiC MOSFETs may require closer to 20 volts for a full on-state.

The negative gate-drive voltage situation is a little more complicated. In principle, for the off-state, 0 volts on the gate is adequate. However, a negative voltage, typically between -5 and -10 volts, enables rapid switching controlled by a gate resistor. An appropriate negative drive ensures that the gate-emitter off-voltage is always actually zero or less.

Figure 1. Even a small emitter inductance at point ‘x’ between a switch and the driver reference, due to layout considerations, can induce an opposing gate-emitter voltage when the switch is turning off, causing turn-on/off “jitter.” (Image: Murata Power Solutions)

This is critical because any emitter inductance (L), shown at point ‘x’ between a switch and the driver reference in Figure 1, causes an opposing gate-emitter voltage when the switch is turning off. While the inductance may be small, even a very small inductance of 5 nanohenries (nH) (a few millimeters of wired connection) will produce 5 volts at a di/dt slew rate of 1000 A per microsecond (A/μsec).

A negative gate-drive voltage also helps to overcome the effect of collector/drain-to-gate Miller-effect capacitance C_m, which injects current into the gate drive circuit during device turn-off. When the device is driven off, the collector-gate voltage rises, and a current of value Cm × dVce/dt flows through the Miller capacitance, into the gate-to-emitter/source capacitance C_ge, and through the gate resistor to the driver circuit. The resulting voltage V_ge on the gate can be sufficient to turn the device on again, potentially causing shoot-through and damage, as shown in Figure 2.

Figure 2. Using a negative gate-drive voltage can overcome the shortcomings that occur due to the presence of the Miller-effect capacitance within a MOSFET or IGBT. (Image: Murata Power Solutions)

By driving the gate negative, this effect is minimized. For this reason, an effective driver design requires both positive and negative voltage rails for the gate-drive function, however, unlike most bipolar dc-dc converters which have symmetrical outputs (such as +5 V and -5V), the supply rails for the gate driver are usually asymmetrical, with a positive voltage that is greater than the negative voltage.

Sizing the converter’s power rating

A critical factor is the amount of current the gate-driver converter must provide, and thus its power rating. The basic calculation is fairly straightforward. In each switching cycle, the gate must be charged and discharged through the gate resistor R_g.

The device’s datasheet provides a curve for the gate charge Q_g value, where Q_g is the amount of charge that needs to be injected into the gate electrode to turn the MOSFET on (drive) at specific gate voltages. The power which must be provided by the dc-dc converter is derived using the formula:

P = Q_g × F × V_s

Where Q_g is the gate charge for a chosen gate voltage swing (positive to negative), of value V_s and at frequency F. This power is dissipated in the internal gate resistance (R_int) of the device and external series resistance, R_g. Most gate drivers need a power supply of less than one to two watts.

Another consideration is the peak current (I_pk) required to charge and discharge the gate. This is a function of V_s, R_int, and Rg. It is calculated using the formula:

I_pk = V_s/(R_int + R_g)

In many cases, this peak current is more than the dc-dc converter can provide. Rather than relying on a larger, more costly supply (operating at a low duty cycle), most designs instead supply the current using “bulk” capacitors on the driver supply rails, which are charged by the converter during the low-current portions of the cycle.

Basic calculations determine how large these bulk capacitors should be. However, it is also important that they have low equivalent series resistance (ESR) and inductance (ESL) so as not to impede the transient current they are delivering.

The final part looks at additional issues associated with the gate driver and its power source.