Gate drive

Intent

Read the Miller plateau and its charge from a switch’s gate-charge curve.
Estimate the gate current and the time the drain voltage takes to switch from the driver’s voltage and the gate loop’s resistance.
Size a bootstrap capacitor for a high-side driver.

Theory

The Miller plateau

A MOSFET’s gate charges through its gate loop: the driver’s output, an external resistor and the device’s internal gate resistance, together $R_G$ . At turn-on there is a stretch where the gate does not rise: while the drain voltage falls, the gate-source voltage stays constant, on the Miller plateau, and all the gate current available from the driver is diverted to discharge the gate-drain capacitance [Balogh 2018]. The charge the gate takes meanwhile is the gate-drain charge $Q_\mathrm{GD}$ of the data sheet’s gate-charge curve [Vishay Siliconix, Power MOSFET Basics]. The gate itself sits at the plateau voltage $V_\mathrm{pl}$ , so the driver’s on voltage $V_\mathrm{GS}$ (the drive voltage, as in the gate-drive loss) less the plateau drives the gate current:

I_G = \frac{V_\mathrm{GS} - V_\mathrm{pl}}{R_G}

$I_G$: gate current on the plateau
$R_G$: gate-loop resistance
$V_\mathrm{GS}$: gate-drive voltage
$V_\mathrm{pl}$: Miller plateau voltage

Gate current on the Miller plateau at turn-on✓ verified by 7 tests

Assumptions, convention and source

Assumptions

the gate voltage stays at the plateau voltage while the drain voltage makes its transition; the driver switches between V_GS and 0 behind the gate-loop resistance

Notes

with the gate held at the plateau, what the driver's voltage exceeds it by drives the current through the gate loop's resistance; a plateau close to the drive voltage leaves little current, and a slow turn-on

Source

Balogh 2018, MOSFET turn-on (the Miller plateau); Alexander & Sadiku 2017, Ch. 2 (Basic Laws: Ohm's law); derived here

Derivation

step-by-step sympy derivation

Equation id

gate.I_on · equations.yaml entry

At turn-off the driver pulls the gate to zero. On the plateau the drain voltage now rises, and the gate current is the current that charges the gate-drain capacitance [Balogh 2018, MOSFET turn-off]; the plateau voltage itself drives it:

I_G = \frac{V_\mathrm{pl}}{R_G}

$I_G$: gate current on the plateau
$R_G$: gate-loop resistance
$V_\mathrm{pl}$: Miller plateau voltage

Gate current on the Miller plateau at turn-off✓ verified by 7 tests

Assumptions, convention and source

Assumptions

the gate voltage stays at the plateau voltage while the drain voltage makes its transition; the driver switches between V_GS and 0 behind the gate-loop resistance

Notes

the driver pulls the gate to zero, so the plateau voltage itself drives the current out; with the same gate-loop resistance at both edges and a plateau below half the drive voltage, turn-off draws less current than turn-on and takes longer (a driver that sinks more strongly than it sources, or a turn-off diode across the gate resistor, changes that)

Source

Balogh 2018, MOSFET turn-off (the Miller plateau); Alexander & Sadiku 2017, Ch. 2 (Basic Laws: Ohm's law); derived here

Derivation

step-by-step sympy derivation

Equation id

gate.I_off · equations.yaml entry

The plateau lasts as long as the gate current takes to deliver $Q_\mathrm{GD}$ , and the drain voltage makes its transition meanwhile:

t_\mathrm{pl} = \frac{Q_\mathrm{GD}}{I_G}

$t_\mathrm{pl}$: time on the plateau
$I_G$: gate current on the plateau
$Q_\mathrm{GD}$: gate-drain (Miller) charge

Time on the Miller plateau✓ verified by 7 tests

Assumptions, convention and source

Assumptions

the gate voltage stays at the plateau voltage while the drain voltage makes its transition; the driver switches between V_GS and 0 behind the gate-loop resistance

Notes

the plateau lasts while the gate delivers the gate-drain charge at a constant current, and the drain voltage makes its transition meanwhile; a larger gate current shortens it in proportion

Source

Vishay Siliconix, Power MOSFET Basics, Q_GD, the charge taken on the Miller plateau; Alexander & Sadiku 2017, Ch. 1 (Basic Concepts: charge and current); derived here

Derivation

step-by-step sympy derivation

Equation id

gate.t_pl · equations.yaml entry

With the same gate-loop resistance at both edges, a plateau below half the drive voltage leaves less current at turn-off than at turn-on, so the switch turns off more slowly than it turns on; a smaller resistance at turn-off (a driver that sinks more strongly than it sources) raises the turn-off current in the same equation. Each cycle the driver delivers the whole gate charge, and the gate loop’s resistances dissipate the gate-drive power between them (gate-drive loss).

The bootstrap capacitor

A high-side switch’s gate is driven from a supply that floats on the switch node. A bootstrap capacitor, charged while the switch node is low, is that supply [Texas Instruments 2023]: while the high side is on, it alone supplies the gate charge and whatever the driver draws, and its voltage droops by that charge over its capacitance. For an allowed droop:

C_\mathrm{boot} = \frac{Q_\mathrm{boot}}{\Delta V_\mathrm{boot}}

$C_\mathrm{boot}$: bootstrap capacitance
$\Delta V_\mathrm{boot}$: allowed bootstrap droop
$Q_\mathrm{boot}$: charge drawn from the bootstrap capacitor

Bootstrap capacitance for an allowed droop✓ verified by 7 tests

Assumptions, convention and source

Assumptions

the bootstrap capacitor alone supplies the charge while the high-side switch is on; it recharges only while the low side conducts

Notes

the capacitor's voltage falls by the charge it supplies divided by its capacitance; halving the allowed droop doubles the capacitance needed

Source

Alexander & Sadiku 2017, Ch. 6 (Capacitors and Inductors); derived here

Derivation

step-by-step sympy derivation

Equation id

boot.C · equations.yaml entry

A capacitor below the minimum value can let the driver’s undervoltage lockout turn the high-side switch off [Texas Instruments 2023].

GaN switches

The maximum gate-source voltage of an eGaN FET (EPC’s enhancement-mode GaN switch) is low compared with silicon MOSFETs [Efficient Power Conversion (EPC), eGaN FET Drivers and Layout Considerations]. The gate loop is a loop too: its inductance develops a voltage over every edge of the gate current (loop voltage), and it rings with the gate’s capacitance, so the gate’s voltage can overshoot the drive voltage toward that lower limit.

Worked example

A driver, a gate loop and a switch’s gate-charge curve:

Gate-drive example

Symbol	Value	How it is obtained
$V_\mathrm{GS}$ gate-drive voltage	10 V	given
$V_\mathrm{pl}$ Miller plateau voltage	4.5 V	given
$R_G$ gate-loop resistance	5 Ω	given
$Q_\mathrm{GD}$ gate-drain (Miller) charge	10 nC	given
${I_G}_{\mathrm{on}}$ gate current on the plateau	1.1 A	Gate current on the Miller plateau at turn-on
${t_\mathrm{pl}}_{\mathrm{on}}$ time on the plateau	9.09 ns	Time on the Miller plateau
${I_G}_{\mathrm{off}}$ gate current on the plateau	900 mA	Gate current on the Miller plateau at turn-off
${t_\mathrm{pl}}_{\mathrm{off}}$ time on the plateau	11.1 ns	Time on the Miller plateau

The gate draws 1.1 A on the plateau at turn-on, and the drain voltage switches in 9.09 ns. At turn-off the plateau drives only 900 mA, and the transition takes 11.1 ns.

A bootstrap capacitor for that switch’s driver:

Bootstrap example

Symbol	Value	How it is obtained
$Q_\mathrm{boot}$ charge drawn from the bootstrap capacitor	50 nC	given
$\Delta V_\mathrm{boot}$ allowed bootstrap droop	500 mV	given
$C_\mathrm{boot}$ bootstrap capacitance	100 nF	Bootstrap capacitance for an allowed droop

Try it

▶ Open in the equation explorer: Time on the Miller plateau
Preset from the example bench-gate

Sweep the gate current: the transition’s time falls in inverse proportion. A smaller gate resistor raises the current, and a driver with a higher voltage does too, as long as the gate stays within its rating.

Bench exercise

With a voltage probe on the gate and another on the drain of a low-side switch, both referred to its source at circuit ground, find the plateau during turn-on: the gate’s rise stops while the drain voltage falls. A switch whose source is not at ground needs differential probes: a probe’s ground lead is tied to the oscilloscope’s earth. Measure the plateau’s voltage and length. The probe sees the gate pin, outside the internal gate resistance, so the plateau it shows sits above the die’s at turn-on and below it at turn-off, by the gate current times that resistance. Compare the plateau’s length with the gate-drain charge over the gate current the equations give; the data sheet gives that charge at its own test drain voltage, and a larger swing takes more. Repeat at turn-off, and with a larger gate resistor.

Gotchas

A plateau close to the drive voltage leaves little voltage across the gate loop, so little current and a slow transition (gate current at turn-on).
Turn-off slower than turn-on: with the same gate-loop resistance at both edges and the plateau below half the drive voltage, the gate’s current at turn-off is the smaller of the two (gate current at turn-off).
A bootstrap capacitor that is too small droops until the driver’s undervoltage lockout turns the high side off [Texas Instruments 2023].
An eGaN gate’s lower voltage limit [Efficient Power Conversion (EPC), eGaN FET Drivers and Layout Considerations] leaves less room for the ringing the gate loop’s own inductance adds.

Go deeper

Application note Fundamentals of MOSFET and IGBT Gate Driver Circuits (SLUA618A) — L. Balogh, Texas Instruments · retrieved 2026-09-23
MOSFET switching behaviour, gate charge and gate-drive power, and practical gate-drive circuits.
Application note Power MOSFET Basics: Understanding Gate Charge and Using it to Assess Switching Performance (AN608A) — Vishay Siliconix · retrieved 2026-09-24
The gate-charge curve read interval by interval, and what each charge says about the switching times.
Application note Bootstrap Circuitry Selection for Half-Bridge Configurations (SLUA887A) — Texas Instruments · retrieved 2026-09-24
Choosing the bootstrap capacitor, diode and resistor of a half-bridge driver, and what goes wrong when they are too small.

Quiz

What does the gate-source voltage do while the drain voltage falls at turn-on?
1. It rises at its fastest
2. It falls to zero
3. It oscillates
4. It stays constant: the Miller plateau
Answer and explanation
Correct answer: D All the gate current the driver can supply discharges the gate-drain capacitance as the drain voltage falls, so the gate voltage stays on the plateau.
A 12 V driver drives a gate through 4 ohm in all; the plateau is at 4 V. What is the gate current on the plateau at turn-on?
1. 3 A
2. 2 A
3. 1 A
4. 4 A
Answer and explanation
Correct answer: B The driver's voltage less the plateau, 8 V, across the 4 ohm gate loop.
The gate-drain charge is 20 nC and the gate current on the plateau is 2 A. How long does the drain voltage take to switch?
1. 40 ns
2. 1 ns
3. 10 ns
4. 100 ns
Answer and explanation
Correct answer: C The plateau lasts the gate-drain charge over the gate current: 20 nC over 2 A.
The gate loop has the same resistance at both edges; the plateau is at 3 V and the drive voltage is 10 V. Which transition is slower?
1. Turn-on
2. They take the same time
3. It depends only on the gate-drain charge
4. Turn-off, because the plateau voltage drives less current than the 7 V left at turn-on
Answer and explanation
Correct answer: D At turn-on 7 V drives the gate current; at turn-off only the 3 V plateau does, through the same resistance.
A bootstrap capacitor supplies 40 nC while the high side is on, and may droop by 0.4 V. What capacitance does it need at least?
1. 100 nF
2. 16 nF
3. 10 nF
4. 1 uF
Answer and explanation
Correct answer: A The charge over the allowed droop: 40 nC over 0.4 V.
What can happen when a bootstrap capacitor is too small?
1. Its droop can let the driver's undervoltage lockout turn the high-side switch off
2. The low-side switch cannot turn on
3. The switching frequency doubles
4. Nothing: the driver regulates its own supply
Answer and explanation
Correct answer: A The capacitor alone supplies the high side while it is on; too little capacitance droops below the driver's lockout threshold.

The numbers in the examples, presets and exercises on this site are synthetic: round teaching values, not data from any real design.