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Gate drive power and supply

What the gates cost the driver: the charge each one takes over the swing your drive actually makes, the power for the whole bridge, the peak current at the edge, the heat in each gate resistor, and on a QCW the bulk capacitor that has to carry a whole ramp. The datasheet's gate charge is measured over a smaller swing than a GDT gives, and carrying it over to your swing is the step people skip.

partgate on · Vgate off · VR_G(ext), each gate · Ωrunning frequency · kHzdevices in the bridgeburst length · µsbursts a second · BPSdriver supply · Vsag allowed through a burst · V

Gate figures from the list, so there is nothing to look up; choose “your own figures” to change them. Infineon: Q_g 167 nC from 0 to 10 V, C_iss 5030 pF, R_G(int) 0.7 Ω; V_DD 480 V, I_D 26.3 A, V_GS 0 to 10 V; Q_gs 32, Q_gd 87 nC, plateau 6.4 V; ±20 V static, ±30 V AC. From the datasheet (Rev 2.4, 2011-09-27).

whole bridge36.4 Win a burst, 5.09 W averaged

charge a gate309 nC-15 to 15 V; the datasheet's 167 nC is 0 to 10 V

bulk capacitor4.8 mFholds a 1 V sag on 15 V through 2000 µs

charge, the whole swing 309 nC (-15 to 15 V) the datasheet's 167 nC from 0 to 10 V below 0 V +75 nC (C_iss, the switch off) above 10 V +67 nC (curve slope, 13.3 nF) energy a cycle 9.3 µJ a gatea gate, in a burst 4.54 Wa gate, averaged 0.64 W (14.0 % duty)WHOLE BRIDGE (8) 36.4 W in a burst, 5.09 W averaged gate current, average 151 mA a gate in a burstgate current, peak 2.62 A a gate; 21.0 A summedin each R_G(ext) 4.27 W in a burst, 0.60 W averaged driver supply 2.42 A through a burst at 15 V one burst draws 4.85 mC bulk capacitor 4.8 mF for a 1 V sag Inside IPW65R080CFD's gate rating, -20 to +20 V.0.60 W averaged in each gate resistor is over half aquarter watt part's rating: size it for twice that, 1.2 Wor more.21.0 A summed is more than a 9 A driver gives. Nothing ona GDT primary actually reaches it (the leakage inductanceand the driver's own resistance hold it down), so expectslower edges than R_G alone sets.

formula

Q = Q_g + C_iss·(V_from - V_off) + slope·(V_on - V_to)
    slope = (Q_g - Q_gs - Q_gd)/(V_to - V_plateau), or C_iss if not given
E a cycle = Q·(V_on - V_off)      P a gate = E·f      bridge × devices
averaged × duty      average gate current = Q·f      peak = ΔV/(R_ext + R_int)
in R_ext: P·R_ext/(R_ext + R_int)
supply: I = P_bridge/V_s    one burst Q = I·t_on    bulk C = Q/sag

limits

Get the swing right. A datasheet quotes Q_g over a stated swing, usually 0 to 15 V for an IGBT and 0 to 10 V for a silicon MOSFET, and a GDT drives the gate from -15 to +15. On the reference switch that is 309 nC, not the table's 167, and the power goes with it. The gate supply article covers the forum thread where this, the second gate of a leg and the second edge of a cycle were all missed together, a factor of four. This page reproduces davekni's corrected figures from that thread exactly.

Why zero voltage switching does not change it. Gate charge is a state: at a given gate and drain voltage the gate holds a given charge, whichever way it got there. A cycle costs the charge between its two ends. Below the datasheet's swing the switch is off with the bus across it, and the gate is the C_iss the table quotes. Above it the switch is on, the drain is near zero and the Miller capacitance is large, so the curve climbs faster than C_iss says: the table's Q_gs, Q_gd and plateau voltage give that slope. Without them C_iss stands in, and the page marks that line because it undercounts.

The peak is an upper bound. It is the whole swing over the gate resistance, the instant the driver flips with the gate still at the other rail. The driver's own output resistance and, on a GDT, the leakage inductance hold it lower, and summed over a bridge it is usually more than the driver could give anyway (a UCC37321 is a 9 A part). The gate resistor's share of the power is overstated by the same small amount. A bigger gate resistor also slows the turn-off, which the phase lead has to cover.

The bulk capacitor matters on a QCW. A DRSSTC burst of a few hundred microseconds draws little. A ramp of milliseconds draws amps the whole time, and the supply's capacitor carries that, not its rating. The supply field is the rail feeding the GDT's drivers: a full bridge of drivers on 15 V puts ±15 V on a 1:1 transformer.

Published calculators for this take Q_g as given for the drive's swing. This one adds the carrying over and the supply.