AN-944 (v.Int)

Figure 4. Basic Gate Charge Test Circuit

The required gate drive current is derived by simply dividing the gate charge, 15 X 10

10

applies 14 volts to the gate, for instance, then a drive impedance of about 50 ohms would be required. Note that throughout the

“flat” part of the switching period (Figure 3), the gate voltage is constant at about 7 volts. The difference between the applied 14

volts and 7 volts is what is available to drive the required current through the drive circuit resistance.

0.038 Watts. Even though the 150 mA drive current which flows during the switching interval may appear to be relatively high.

the average power is minuscule (0.004%) in relation to the power being switched in the drain current. This is because the drive

current flows for such a short period that the average power is negligible. Thus actual drive power for MOSFETs is minute

compared to bipolar requirements, which must sustain switching current during the entire ON condition. Average drive power, of

course, increases at higher frequencies, but even at 5 MHz it would be only 1.9W.

3.

The Gate Charge Curve

The oscillograms of the gate-to-source voltage in Figure 2 neatly delineate between the charge required for the gate-to-source

capacitance, and the charge required for the gate-to-drain, or “Miller” capacitance. The accompanying simplified test circuit and

to-source capacitance starts to charge, and the gate-to-source voltage increases. No current flows in the drain until the gate

continues to rise and the drain current rises proportionally. So long as the actual drain current is still building up towards the

shuts off; the potential of the drain now is no longer tied to the supply

enforced by the circuit, whilst the drain voltage starts to fall. Since the

gate voltage is inextricably related to the drain current by the intrinsic

transfer characteristic of the DUT (so long as operation remains in the

“active” region), the gate voltage now stays constant because the

“enforced” drain current is constant. For the time being therefore, no

further charge is consumed by the gate-to-source capacitance, because

the gate voltage remains constant. Thus the drive current now diverts,

charge now contributes exclusively to discharging the “Miller”

capacitance.

and hence the total drive charge is typically higher for the “Miller”

comes out of the “active” region of operation. (In bipolar transistor

terms, it has reached “saturation.”

The gate voltage is now no longer constrained by the transfer characteristic of the device to relate to the drain current, and is free

source. The time scale on the oscillogram of the gate-to-source voltage is directly proportional to the charge delivered by the

drive circuit, because charge is equal to the product of current and time, and the current remains constant throughout the whole

not represent “switching” charge; it is simply the excess charge which will be delivered by the drive circuit because the amplitude

of the applied gate drive voltage normally will be higher (as a matter of good design practice) than the bare minimum required to

accomplish switching.

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