What the MOSFET Gate Resistor Calculator does
This calculator sizes a MOSFET gate resistor from the device's gate charge. It gives the peak gate current the driver has to supply, the turn-on and turn-off transition times, the power the gate drive itself costs, and the switching and conduction losses in the device - then shows the same figures for a range of gate resistors so the trade-off is visible rather than argued about.
The key quantity is Qgd, the gate-drain or Miller charge, not the total Qg. During the transition the gate voltage sits on a plateau while that charge moves, and the time it takes is Qgd divided by the current the driver can push through the gate resistor.
How to use it
- Take Qg and Qgd from the data sheet's gate-charge table, at a drain voltage and current close to yours - both figures change with operating conditions.
- Enter the driver voltage and the plateau voltage. The plateau is read from the gate-charge curve, and is usually a volt or two above the threshold.
- Enter the driver's own output resistance; it is in series with your gate resistor and, on a strong driver, may be a small part of the total.
- Enter the power stage details to get switching and conduction losses at your frequency and duty cycle.
- Read the table: it is the same device with different gate resistors, so you can see exactly what a slower turn-on costs in watts.
Reading the results
Turn-off is slower than turn-on for the same resistor, because the driver only has the plateau voltage to work with rather than the full drive voltage. A separate turn-off path - a diode bypassing the gate resistor - is the standard fix where that matters.
Gate drive power is Qg x Vdrive x frequency, and it is shared between the driver and the gate resistor in proportion to their resistances. It is small in absolute terms, but at high frequency it can be the thing that overheats a small driver package.
The switching loss shown assumes the classic triangular voltage-current overlap. It is a reasonable estimate for a clamped inductive load and it ignores reverse recovery, Coss and layout inductance, all of which add real loss.
Worked example: a 48 V half-bridge at 100 kHz
A device with Qg = 30 nC and Qgd = 10 nC, driven from 12 V with a 4.5 V plateau, through a 2 ohm driver and a 10 ohm gate resistor. The total is 12 ohms, so the initial peak gate current is 12 / 12 = 1 A, and during the plateau it is (12 - 4.5) / 12 = 0.625 A.
The turn-on transition is therefore 10 nC / 0.625 A = 16 ns. Turn-off is slower: the driver pulls down from the plateau with only 4.5 / 12 = 0.375 A, so it takes 26.7 ns.
At 100 kHz switching 5 A at 48 V, the switching loss is 0.5 x 48 x 5 x (16 + 26.7) ns x 100 kHz = 0.512 W, conduction at 20 milliohms and 50% duty is 0.25 W, and the gate drive costs 30 nC x 12 V x 100 kHz = 36 mW. Raise the gate resistor to 22 ohms and the switching loss roughly doubles - which is the price of the slower, quieter edge.
Formulas and scoring rules
- Total gate resistance
Rtotal = Rdriver + Rgate- Peak gate current
Ipeak = Vdrive / RtotalAt the very start of turn-on, before the gate has charged.- Plateau current, turn-on
I = (Vdrive - Vplateau) / Rtotal- Plateau current, turn-off
I = Vplateau / RtotalLower, which is why turn-off is slower.- Transition time
t = Qgd / I(plateau)The Miller charge divided by the current available to move it.- Gate drive power
P = Qg x Vdrive x fswSplit between driver and gate resistor in proportion to their resistances.- Switching loss
P = 0.5 x Vds x Id x (t_on + t_off) x fswThe triangular-overlap approximation for a clamped inductive load.- Conduction loss
P = Id^2 x Rds(on) x dutyUse the hot Rds(on): it roughly doubles from 25 C to 125 C.
What the gate resistor is actually for
With no gate resistor at all, the gate loop - driver output, trace, gate and source inductance - rings, and the dV/dt at the drain can be high enough to cross-conduct a half-bridge through the Miller capacitance. The gate resistor damps that loop and slows the edge in a controlled way.
So the choice is not "as small as possible". It is the smallest value that keeps the gate waveform clean, the drain dV/dt within what the rest of the circuit tolerates, and radiated emissions under control. Start from the loss numbers here, then look at the gate and switching-node waveforms on a scope with a short ground spring - a long ground lead shows ringing that is not there and hides ringing that is.
Separate turn-on and turn-off paths
Because turn-off has less voltage available to drive the gate, it is slower for a given resistor - exactly when you often want it faster, to avoid shoot-through and to keep the device out of its linear region. The standard arrangement is a diode in parallel with the gate resistor, oriented so that turn-off current bypasses it, sometimes with a smaller resistor in series with the diode.
The other common addition is a resistor from gate to source, typically 10 kOhm. It does nothing during switching, but it holds the gate off when the driver is not powered or is in high impedance - which is what stops a half-bridge from turning itself on during power-up.
Limitations: what the result does not prove
- This is a charge-based estimate, not a simulation. Reverse recovery in the opposing diode, the output capacitance Coss, and parasitic inductance in the power and gate loops all add real loss and ringing that are not modelled.
- Qg and Qgd depend on drain voltage and current, so the data sheet figure only applies near its stated test conditions.
- Rds(on) roughly doubles between 25 C and 125 C; using the cold figure understates conduction loss substantially.
- The result is indicative and must be verified by measurement. A qualified engineer should check the design against the device data sheet, the driver's peak current rating and the thermal design before it is built.
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Standards and sources
- Texas Instruments - Op amps for everyone design reference - checked 19 Sep 2026
- Texas Instruments - Fundamentals of MOSFET and IGBT gate driver circuits
- Infineon - MOSFET power losses calculation using the data sheet parameters
Frequently asked questions
How do I choose a MOSFET gate resistor?
Work out the transition time from Qgd and the current the driver can supply, then pick the smallest resistor that still gives a clean gate waveform and an acceptable drain dV/dt. Typical values are 2 to 20 ohms for a hard-switched converter; the table above shows what each choice costs in switching loss.
Why is Qgd more important than Qg?
Because the drain voltage transition happens entirely during the Miller plateau, while Qgd is being moved. Qg sets the average drive current and the gate drive power; Qgd sets how long the device spends with both voltage and current on it, which is where switching loss comes from.
What happens if the gate resistor is too small?
The gate loop rings, the drain dV/dt rises, and in a half-bridge the Miller capacitance can turn the opposite device partly on - cross-conduction, which shows as a current spike and heat with no obvious cause. Radiated emissions get worse too.
What happens if it is too large?
The transitions lengthen, so switching loss rises in proportion, and the device spends longer in its linear region. At high frequency this quickly becomes the dominant loss. The delay also eats into the dead time of a bridge, which can cause shoot-through at the other end of the problem.
Why is turn-off slower than turn-on?
The driver pulls the gate down from the plateau voltage, typically 4 or 5 volts, whereas turn-on pushes with the full drive voltage minus the plateau. Less voltage means less current through the same resistor. A diode that bypasses the gate resistor on turn-off restores the speed.
How much power does gate drive consume?
P = Qg x Vdrive x fsw. A 30 nC device driven at 12 V and 100 kHz costs 36 mW, shared between the driver and the gate resistor. It is small, but at 500 kHz with a large device it becomes hundreds of milliwatts inside a small driver package.
Do I need a resistor from gate to source as well?
Yes, as a matter of practice. Around 10 kOhm from gate to source holds the device off when the driver is unpowered or high-impedance, which prevents a half-bridge from turning itself on at power-up. It has no effect on switching speed.
Can I use this for an IGBT?
The gate drive arithmetic is the same, because an IGBT's gate is also a capacitance charged by a driver. The loss model is not: an IGBT has a current tail at turn-off that dominates its switching loss and is not captured here. Use the device's data-sheet switching energies instead.
Last reviewed by the A2Z.Tools team against the sources listed above.
