Calculator
MOSFET power loss calculator
A buck converter's high- and low-side MOSFET losses from datasheet values, with each junction temperature solved against its own loss, thermal runaway, and derating.
How it works
Conduction loss
Each switch conducts for its share of the period, the high side for the duty cycle and the low side for the rest. Its conduction loss is its RMS current squared times its on-resistance. The on-resistance rises with junction temperature, so it is read at the temperature the switch actually reaches.
Switching loss
The high side turns on against the full input voltage and off against the full load current. For a few nanoseconds it carries both at once. The calculator times those edges through the gate loop. The driver's resistance and the MOSFET's internal gate resistance set the gate current, and the Miller plateau sets how long the drain voltage takes to move. With Qgs, Qgd and a threshold voltage it can place that plateau. Without them it falls back to the rise and fall times, then to total gate charge alone, and says so.
Output capacitance and recovery
When the high side turns on, it charges the low side's output capacitance from the input rail through its own channel, and that loss lands in the high side. It also sweeps out the low side's body-diode recovery charge. That energy is shared between the two switches, about half each unless the datasheet gives the recovery current or time. So a low-side part with a large Qoss or Qrr heats the high side too.
Dead time
Between one switch turning off and the other turning on, the low side's body diode carries the load current. That costs the diode's forward voltage times the current for the dead time, twice per period. The low side itself switches at near zero volts, so it has no switching overlap loss.
Where gate charge goes
Charging and discharging the gates takes Qg × Vg × f_sw from the gate supply. Most of that is dissipated in the driver and the gate resistors, not in the MOSFET. The calculator reports it on its own line and keeps it out of each switch's total.
Junction temperature and runaway
Loss raises the junction temperature, and a hotter junction has a higher on-resistance and so more loss. The calculator solves the two together, T_j = T_a + P(T_j) × Rθ, for each switch. If the loss rises faster with temperature than the thermal path can carry it away, no stable temperature exists and the result is thermal runaway. The derating figures are the load current and the ambient temperature at which the first switch reaches its own Tj(max), or runs away.
What it does not model
- Ringing and the package's own inductance.
- Two switches sharing one heatsink. Each switch has its own thermal resistance here.
- Several MOSFETs in parallel.
- Converters other than a synchronous buck.
The design tool models shared heatsinks, paralleled devices and other converters.
Reading the numbers off a datasheet
- Rds(on) and Qg come from the static and dynamic tables. Use the rows at the gate drive you will run.
- Qgs and Qgd are in the gate charge table, usually at a stated drain voltage and current.
- The Rds(on) ratio comes from the graph of on-resistance against junction temperature. Divide the value at 125 °C by the value at 25 °C.
- Qoss and Eoss come from the dynamic table or the Eoss graph. The output-capacitance loss needs Eoss. Enter the drain voltage they were measured at.
- Rθ junction to ambient is in the thermal table. It depends on the board, so use a figure for a layout like yours.
FAQ
How do you calculate MOSFET switching loss?
Switching loss is half the input voltage times the load current times the time the switch spends in transition, per edge, times the switching frequency. The transition time comes from the gate loop. It is the charge the gate must move through the Miller plateau, divided by the current the driver can push through the gate resistance.
Why is the low-side MOSFET's switching loss zero in a synchronous buck?
The low side turns on and off while its body diode is already carrying the current, so its drain voltage is only a diode drop. It still has conduction, dead-time and gate-drive losses.
Why does my high-side MOSFET run hotter than its conduction loss suggests?
It charges the low side's output capacitance at every turn-on, and that loss is all its own. It also takes about half of the low side's body-diode recovery energy. At high input voltage and frequency those terms can outweigh conduction.
What is thermal runaway in a MOSFET?
On-resistance rises with temperature. If the extra loss from a one degree rise heats the junction by a degree or more through its thermal path, the temperature cannot settle. The calculator reports that instead of a temperature.
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