How do i calculate the switching power loss in sinusoidal pulse width modulation SPWM in a H bridge inverter (4 IGBTs), and precisely is it linearly proportional to the frequency switching or to the number of switching actions ?


switghing_loss_wattage = energy_per_transition x number_of_transitions_per_second

If you are doing PWM using an H-bridge... During each switch, current flows in two diodes during the dead time, two transistos are turning on, two are turning off, and the capacitance of all 4 of them is either charged or discharged. Therefore...

energy_per_transition = 2 x (E_dead + E_cond_on + E_cond_off) + 4 x (E_cgs + E_cgd + E_cds)

Capacitive losses are typically more significant at higher voltages or higher frequencies and can sometimes be ignored.

When using an IGBT the datasheet will often quote an E_on or E_off value that lumps together the various parts of the turn on/turn off losses. The quoted values are typically on valid at specific voltages or currents.

The approximate energy lost in each transistor during each switch comes from the following sources...

1) Each time a you turn on a high-side transistor in an H-bridge you must ensure that the low-side transistor is turned off first and vice versa because if they were both on at once then power and ground would be shorted through the transistors and destroy them. Therefore a small amount of dead-time is usually added between turning off a transistor and turning on the other member of the pair. IGBTs typically have turn off times in the hundreds of ns to microseconds range. During the dead time if driving an inductive load the load current remains about constant and must have a path. The path will either be through the body diode of the transistor or through a separately added Schottky diode. In either case the energy dissipated during the dead time will be

E_dead = diode_voltage x load_current x dead_time.

2) When a transistor in an H-bridge is off it has the entire supply voltage across it but no current, so no power loss. When the transistor is on it has about 0V ocross it and all the load current, but very little power loss. When transistioning from off to on the transistor current remains about constant when driving an inductive load but the voltage across the transsitor moves in an approximately linear fashion between the full supply voltage and 0V. It will take some finite time for the transistor to make the transistion. The turn on conduction losses are approximately

E_cond_on = 1/2 x supply_voltage x load_current x turn_on_time.

Transitioning from on to off will take some finite time. The turn off conduction losses are approximately

E_cond_off = 1/2 x supply_voltage x load_current x turn_off_time.

3) There is losses due to the gate source capacitance charging.

E_cgs = 1/2 x gate_source_capacitance x gate_source_voltage^2.

4) There is losses due to the gate source capacitance charging.

E_cgd = 1/2 x gate_drain_capacitance x gate_drain_voltage^2.

5) There is losses due to the drain source capacitance charging.

E_cds = 1/2 x drain_source_capacitance x drain_source_voltage^2.

In an H-bridge each transistor turns on once and turns off once per switching cycle, and there are 4 transistors.

energy_per_transition = 2 x (E_dead + E_cond_on + E_cond_off) + 4 x (E_cgs + E_cgd + E_cds)

The load_current to use in the calculations should be the average load current, which for a sine-wave is 2/pi x the_peak_load_current.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.