MOSFET Gate Driver Calculator (Power and Gate Resistor)
MOSFET Gate Driver Calculator (Power and Gate Resistor)
Gate-drive power from Qg, Vgs and the switching frequency — and, more usefully, how it splits between the driver’s own output resistance, your external gate resistor and the MOSFET’s internal gate resistance, because only the first of those heats the driver. With the peak source and sink currents, the turn-on time the Miller charge and your Rg actually give, the driver’s junction temperature, and the bootstrap capacitor for a high-side stage.
Gate drive
Qg 100 nC, Qgd 35 nC, 12 V drive with a 5.5 V plateau, 200 kHz, Rg 10 Ω, driver 2.5 Ω up and 1.2 Ω down, Rg,int 1.8 Ω
Gate-drive power, where it goes, and the time it buys
- Qg
- total gate charge at your drive voltage, from the gate-charge curve
- Qgd
- the flat part of that curve — the charge that moves the drain, and so the charge that sets the switching time
- R_HI, R_LO
- the driver’s own source and sink output resistances
- Rg,int
- the MOSFET’s internal gate resistance; its share of the power is dissipated in the die, not in the driver
- V_plateau
- the Miller plateau gate voltage; the drive voltage minus this is what actually pushes current during the transition
- ΔV
- the droop the bootstrap rail may take before the driver’s undervoltage lockout trips
Worked example
Qg 100 nC, Qgd 35 nC, 12 V drive with a 5.5 V plateau, 200 kHz, Rg 10 Ω, driver 2.5 Ω up and 1.2 Ω down, Rg,int 1.8 Ω
Gate-drive power = Qg × Vgs × f = 100 nC × 12 V × 200 kHz = 240 mW
Turn-on loop = 2.5 + 10 + 1.8 = 14.3 Ω, so the driver takes 17.5% of the half-cycle's 120 mW — 20.98 mW
Turn-off loop = 13.0 Ω; adding the two halves, the driver takes 32.06 mW, Rg takes 176.2 mW and the MOSFET's own 1.8 Ω takes 31.72 mW — and 32.06 mW + 176.2 mW + 31.72 mW is the whole 240 mW
Peak current = 12 V ÷ 14.3 Ω = 839.2 mA sourcing and 923.1 mA sinking
During the plateau only 12 − 5.5 = 6.5 V is left to push charge, so the gate current falls to 454.5 mA and the drain takes Qgd ÷ that = 77 ns to move
Driver dissipation 32.06 mW + 24 mW quiescent = 56.06 mW, which on a 60 °C/W package at 40 °C ambient is a junction temperature of 43.4 °C
Bootstrap: 101 nC per cycle ÷ 1.0 V droop = 101 nF minimum, against 83.33 nF from the 10 × Cg rule → 150 nF (E6)
Where a bigger gate resistor takes you
| Rg | Turn-on time | In the driver | In Rg | In the MOSFET die |
|---|---|---|---|---|
| 0 Ω | 23.15 ns | 117.8 mW | 0 W | 122.2 mW |
| 2.2 Ω | 35 ns | 73.85 mW | 91.38 mW | 74.77 mW |
| 4.7 Ω | 48.46 ns | 52.03 mW | 135.9 mW | 52.05 mW |
| 10 Ω | 77 ns | 32.06 mW | 176.2 mW | 31.72 mW |
| 22 Ω | 141.6 ns | 17.17 mW | 206 mW | 16.85 mW |
| 47 Ω | 276.2 ns | 8.728 mW | 222.7 mW | 8.531 mW |
| 100 Ω | 561.6 ns | 4.274 mW | 231.6 mW | 4.168 mW |
Driving a power MOSFET’s gate, and where the heat goes
A MOSFET gate is a capacitor with an awkward, voltage-dependent value, so it is specified by charge instead: Qg is the charge needed to take the gate from zero to the drive voltage. Deliver that charge f times a second from a supply at Vgs and the supply gives up Qg·Vgs·f — 240 mW at the defaults here. None of it is stored: what goes into the gate at turn-on comes back out at turn-off, and every joule ends up in the resistances of the gate loop.
Why the split matters more than the total. 240 mW is nothing spread over a board and everything concentrated in a small-outline driver package. The gate loop is three resistances in series — the driver’s own output resistance, your external Rg, and the MOSFET’s internal gate resistance — and the current through all three is the same, so the power divides in proportion to resistance. At the defaults the driver takes 32.06 mW, Rg takes 176.2 mW and the MOSFET’s own 1.8 Ω takes 31.72 mW inside its die, where it adds to the junction temperature the MOSFET loss calculator computes. Turn-on and turn-off are counted separately because the driver’s pull-up and pull-down resistances are usually different — pull-down is normally the smaller of the two, because a slow turn-off is the dangerous one in a half bridge.
Qgd, not Qg, sets the switching time. This is the part most pages get wrong. The gate-charge curve has a flat section — the Miller plateau — where the gate voltage stops rising because all the incoming current is going into the gate-drain capacitance as the drain voltage swings. Nothing at the drain moves before that plateau or after it, so the transition time is the plateau’s duration: Qgd divided by the gate current available at the plateau voltage, which is (Vgs − Vplateau)/Rtotal and not the peak current. At the defaults that is 35 nC ÷ 454.5 mA = 77 ns, against a peak current of 839.2 mA that only flows for an instant at the start. It also explains why a 5 V drive on a MOSFET with a 4 V plateau is so poor: the numerator collapses.
The trade-off, stated plainly. A bigger Rg slows both edges in direct proportion, which lowers dV/dt and dI/dt and so cuts conducted and radiated EMI, reduces ringing on the drain, and moves heat out of the driver and into a resistor you can choose the size of. It costs switching loss in the MOSFET, because the drain voltage and drain current overlap for longer — and that loss is usually the larger number. The MOSFET loss calculator takes the transition times from this page and turns them into watts in the die, and the heatsink calculator turns those into a temperature. Separate turn-on and turn-off resistors, with a diode across the turn-on one, are the usual way out when you want a slow turn-on and a fast turn-off.
The bootstrap supply. A driver for a high-side switch has its reference pin tied to the switch node, which swings the full rail, so it needs a floating supply. The cheap answer is a capacitor charged through a diode from the low-side rail whenever the switch node is near ground. It has to hold up the driver through the whole on-time, supplying Qg plus the high-side quiescent and leakage currents for that time, and the droop is that charge divided by the capacitance. The familiar rule Cboot ≥ 10 × Cg is exactly the same calculation with the droop set to a tenth of the drive voltage — worth knowing, because it means the rule silently assumes your undervoltage lockout has that much room. Two things the arithmetic will not tell you: the capacitor must be recharged in the off-time, so a high duty cycle at a high frequency can make a bootstrap impossible however large the capacitor, and the bootstrap rail cannot start up at all until the low-side device has switched at least once. Use the figures from your part’s datasheet; typical values vary widely between manufacturers.
Frequently asked questions
How do you calculate gate drive power?
P = Qg × Vgs × fsw. At 100 nC, 12 V and 200 kHz that is 240 mW. It is dissipated entirely in the resistances of the gate loop — none of it is stored in the gate and none of it reaches the load.
How much of the gate drive power heats the driver?
The fraction R_HI/(R_HI + Rg + Rg,int) of the turn-on half, and R_LO/(R_LO + Rg + Rg,int) of the turn-off half. With a 2.5 Ω pull-up, 1.2 Ω pull-down, 10 Ω external resistor and 1.8 Ω internal resistance that is 32.06 mW of the 240 mW — 13.4%. Shrink Rg and that share climbs fast.
What gate resistor should I use?
There is no single right answer: it is a deliberate trade between edge speed and EMI. Start from the switching time you can afford — Qgd × Rtotal ÷ (Vgs − Vplateau) — and check the peak current Vgs ÷ Rtotal against what your driver is rated to supply. Values between 2 Ω and 22 Ω cover most hard-switched designs.
Why does the MOSFET switch slower than the peak gate current suggests?
Because the drain only moves during the Miller plateau, and at the plateau the gate voltage is already partway up, so the driving voltage is Vgs − Vplateau rather than the full Vgs. At the defaults that drops the gate current from 839.2 mA to 454.5 mA.
How do you size a bootstrap capacitor?
Cboot ≥ (Qg + Ileak × D/f) ÷ ΔV, where ΔV is the droop the driver’s undervoltage lockout allows. At 100 nC, 250 µA of leakage, 80% duty, 200 kHz and 1 V of droop that is 101 nF, so 150 nF from the E6 series. Then check that the off-time is long enough to recharge it.
Does the MOSFET’s internal gate resistance matter?
Yes, twice. It is in series with your Rg, so it slows the edge whether you want that or not and puts a floor under the switching time; and its share of the gate-drive power is dissipated inside the die, adding to the junction temperature. On large modules it is the dominant term.
Related calculators
References
- Texas Instruments. Fundamentals of MOSFET and IGBT Gate Driver Circuits, SLUA618A, March 2017 (revised October 2018). PGATE = VDRV · QG · fDRV; the turn-on and turn-off power shared between RHI/RLO, RGATE and RG,I in proportion to resistance; the gate current at the Miller plateau and the drain transition time it gives.
- Texas Instruments. Bootstrap Circuitry Selection for Half-Bridge Configurations, SLUA887A. Cboot ≥ Qtotal ÷ ΔVHB with Qtotal = QG + IHBS·Dmax/fsw + IHB/fsw; the simple rule Cboot ≥ 10 · Cg; ΔVHB = VDD − VDH − VHBL.
- Erickson RW, Maksimović D. Fundamentals of Power Electronics, 3rd ed. Springer, 2020. The switching transition, the gate-charge model and the loss that the overlap of drain voltage and drain current produces.