Semiconductor switches and their losses

Intent

• Model a MOSFET that is on as a resistance, and a conducting diode as a forward voltage plus a small resistance.
• Estimate the capacitive turn-on loss and the gate-drive power, and see how both scale with the switching frequency.
• Tell the losses that grow with the switching frequency from those that grow with the load.

Theory

The switches of a converter

An ideal switch has no voltage across it when on and no current through it when off. A converter realizes its switches with transistors, such as MOSFETs, and diodes [Erickson & Maksimović 2020, Ch. 4]. A transistor is turned on and off by its gate; a diode switches by itself, conducting when it is forward-biased and blocking when it is reverse-biased [Erickson & Maksimović 2020, Ch. 4].

Conduction loss

A MOSFET that is fully on behaves as a resistance, its on-resistance, so its conduction loss follows from its rms current (conduction loss). A conducting diode is modelled by a constant forward voltage in series with a small resistance [Erickson & Maksimović 2020, Ch. 3]:

$$P = V_F I_\mathrm{avg} + r_d I_\mathrm{rms}^{2}$$

$P$

power

$I_\mathrm{avg}$

average current

$I_\mathrm{rms}$

rms current

$V_F$

diode forward voltage

$r_d$

diode dynamic resistance

Diode conduction loss✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• periodic steady state

Source

Erickson & Maksimović 2020, Ch. 3 (Steady-State Equivalent Circuit Modeling, Losses, and Efficiency)

Derivation

step-by-step sympy derivation

Equation id

loss.diode · equations.yaml entry

The forward voltage takes a fixed voltage from the circuit whatever the current, so it costs the most efficiency when the output voltage is low [Erickson & Maksimović 2020, Ch. 4].

Switching loss from the node capacitance

When the transistor turns on, it discharges the capacitance at the switch node, which includes its own output capacitance. The energy stored in that capacitance is lost in every period [Erickson & Maksimović 2020, Ch. 4]:

$$P = \frac{C_\mathrm{node} V_\mathrm{sw}^{2} f_s}{2}$$

$P$

power

$C_\mathrm{node}$

switch-node capacitance

$V_\mathrm{sw}$

switch-node voltage at turn-on

$f_s$

switching frequency

Capacitive turn-on loss✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• periodic steady state
• the capacitance is fully discharged at every turn-on
• all of the stored energy is dissipated every cycle

Source

Erickson & Maksimović 2020, Ch. 4 (Switch Realization)

Derivation

step-by-step sympy derivation

Equation id

loss.sw.cap · equations.yaml entry

In addition, the current and the voltage overlap for a short time during each transition, and the diode’s stored charge must be removed when it turns off; both add loss at every switching event [Erickson & Maksimović 2020, Ch. 4].

Gate-drive power

To switch a MOSFET, the driver moves the gate charge from its supply into the gate and then back to ground, once per period. The power this takes is dissipated in the driver and the gate resistances, not in the MOSFET’s channel [Balogh 2018]:

$$P = Q_g V_\mathrm{GS} f_s$$

$P$

power

$Q_g$

total gate charge

$V_\mathrm{GS}$

gate-drive voltage

$f_s$

switching frequency

Gate-drive power✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• periodic steady state

Notes

dissipated in the driver and gate resistances, not in the power switch channel

Source

Balogh 2018

Derivation

step-by-step sympy derivation

Equation id

loss.gate · equations.yaml entry

Worked example

An example switch node, gate drive and diode:

Switch and diode loss example

Symbol	Value	How it is obtained
$C_\mathrm{node}$ switch-node capacitance	100 pF	given
$V_\mathrm{sw}$ switch-node voltage at turn-on	48 V	given
$f_s$ switching frequency	100 kHz	given
$Q_g$ total gate charge	20 nC	given
$V_\mathrm{GS}$ gate-drive voltage	10 V	given
$V_F$ diode forward voltage	500 mV	given
$I_\mathrm{avg}$ average current	1 A	given
$r_d$ diode dynamic resistance	50 mΩ	given
$I_\mathrm{rms}$ rms current	1.2 A	given
${P}_{\mathrm{sw}}$ power	11.5 mW	Capacitive turn-on loss
${P}_{\mathrm{gate}}$ power	20 mW	Gate-drive power
${P}_{\mathrm{diode}}$ power	572 mW	Diode conduction loss

At this switching frequency the capacitive turn-on loss is 11.5 mW and the gate-drive power 20 mW; both double if the switching frequency doubles. The diode’s conduction loss, 572 mW, depends on the load current instead.

Try it

▶ Open in the equation explorer: Capacitive turn-on loss
Preset from the example switch-losses

Sweep the switched voltage on its logarithmic axis: the capacitive loss grows with its square, which is why this loss matters most in converters that switch high voltages. Then open the gate-drive equation and sweep the switching frequency.

Bench exercise

On a buck evaluation board whose switching frequency can be set, measure the input and output power at a fixed load with a power analyzer or with a multimeter pair on each side, at two switching frequencies. The increase in loss divided by the increase in frequency estimates the energy lost per switching period; compare it with the capacitive-loss and gate-drive equations, using the capacitance and gate charge from the data sheets.

Gotchas

• Losses that happen once per period scale with frequency. A higher switching frequency shrinks the inductor needed for a given ripple, but the capacitive, transition and gate-drive losses rise in proportion [Erickson & Maksimović 2020, Ch. 4].
• Gate power heats the driver, not the MOSFET. Budget it in the driver’s dissipation [Balogh 2018].
• A diode drop is costly at low output voltage. The same forward voltage is a larger fraction of a low output voltage, which is why low-voltage converters often replace the diode by a transistor, a synchronous rectifier [Erickson & Maksimović 2020, Ch. 4].

Go deeper

• Book chapter Switch Realization (Ch. 4) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  How transistors and diodes realize the ideal switches, and where the switching losses come from.
• 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.
• Book chapter Steady-State Equivalent Circuit Modeling, Losses, and Efficiency (Ch. 3) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  DC transformer models and how winding and semiconductor resistances limit the conversion ratio and efficiency.

Quiz

1. At every turn-on, a transistor discharges $100\,\mathrm{pF}$ at the switch node from $48\,\mathrm{V}$, at $200\,\mathrm{kHz}$. What is the capacitive turn-on loss?

   1. $11.5\,\mathrm{mW}$
   2. $23\,\mathrm{mW}$
   3. $46\,\mathrm{mW}$
   4. $2.3\,\mathrm{mW}$
   Answer and explanation

   Correct answer: B The stored energy is half the capacitance times the square of the voltage, about $115\,\mathrm{nJ}$, lost $2 \times 10^5$ times per second.

2. A MOSFET with a total gate charge of $30\,\mathrm{nC}$ is driven with $12\,\mathrm{V}$ at $100\,\mathrm{kHz}$. What is the gate-drive power?

   1. $3.6\,\mathrm{mW}$
   2. $18\,\mathrm{mW}$
   3. $36\,\mathrm{mW}$
   4. $360\,\mathrm{mW}$
   Answer and explanation

   Correct answer: C The gate charge times the drive voltage is the energy per period, $360\,\mathrm{nJ}$, taken $10^5$ times per second.

3. The switching frequency is doubled. What happens to the capacitive turn-on loss and to the gate-drive power?

   1. Both double
   2. Both are unchanged
   3. The first doubles, the second is unchanged
   4. Both quadruple
   Answer and explanation

   Correct answer: A Each is a fixed energy lost once per period, so both are proportional to the switching frequency.

4. Where is the gate-drive power dissipated?

   1. In the MOSFET's channel
   2. In the driver and the gate resistances
   3. In the load
   4. Nowhere: the gate charge is recovered
   Answer and explanation

   Correct answer: B The gate charge flows through the driver and the gate resistances on its way in and out, and they turn its energy into heat.

5. A diode with a $0.5\,\mathrm{V}$ forward voltage and $20\,\mathrm{m}\Omega$ of slope resistance carries $2\,\mathrm{A}$ on average and $2.5\,\mathrm{A}$ rms. What is its conduction loss?

   1. $1\,\mathrm{W}$
   2. $1.125\,\mathrm{W}$
   3. $1.25\,\mathrm{W}$
   4. $0.125\,\mathrm{W}$
   Answer and explanation

   Correct answer: B The forward voltage times the average current gives $1\,\mathrm{W}$, and the slope resistance times the square of the rms current adds $0.125\,\mathrm{W}$.

6. Why do low-voltage converters often replace the output diode by a synchronous rectifier?

   1. Diodes cannot block low voltages
   2. The diode's fixed forward voltage is a large fraction of a low output voltage, so it costs much of the efficiency
   3. A synchronous rectifier needs no gate drive
   4. Diodes switch too slowly at any voltage
   Answer and explanation

   Correct answer: B The forward voltage takes a fixed voltage from the output path whatever the load; a transistor that is on drops only its on-resistance times the current.