Reasons for power losses: https://www.dynexsemi.com/assets/downloads/DNX_AN6156.pdf

http://web.eecs.utk.edu/~dcostine/ECE481/fall2013/lectures/L7_slides.pdf

Calculatio of power losses from Datasheet: https://www.element14.com/community/docs/DOC-20553/l/igbt-power-losses-calculation-using-the-data-sheet-parameters

You can also find the losses, calculated from the producer, directly reported in the Datashhet. For example a Infineon IGBT modules datasheet reports:

Turn-on energy loss per pulse

IC = 300 A, VCE = 600 V, LS = 30 nH

VGE = ±15 V, di/dt = 6000 A/μs (Tvj = 150°C)

RGon = 1,8

Tvj = 25°C 16.5 mJ

Tvj = 125°C 25.0 mJ

Tvj = 150°C 30.0 mJ

Turn-off energy loss per pulse

IC = 300 A, VCE = 600 V, LS = 30 nH

VGE = ±15 V, di/dt = 4000 A/μs (Tvj = 150°C)

RGoff = 1,8

Tvj = 25°C 23.5 mJ

Tvj = 125°C 35.0 mJ

Tvj = 150°C 39.0 mJ

As you are aware an IGBT is a semiconductor device which is capable of controlling the current flow through the device via control of the electric field within the device. Hence why it is a voltage controlled device.

In effect any wires / conductors connected to and from the terminals of the device serve as waveguides for the propagating em fields and within which the electrical energy sourced from the supply is transmitted via the device to the load.

In the off state the device has a high impedance and behaves much like a capacitor / insulator / dielectric whereby energy is stored in the electric field of the polarised dielectric material. As such this prevents the em fields from propagating through the device.

In the on state the device impedance is lower and behaves more like a conductor / resistor allowing the em fields to propagate through the device like any conductor and in doing so the propagating magnetic field penetrate the device material resulting in heat losses like any normal conductor.

What in effect the IGBT is doing is controlling the storage of energy in and between the em fields within the device. In the off state most if not all energy is stored in the Electric Field and in the on state most if not all energy is stored in the Magnetic Field. In many ways it is doing what a transformer does namely exchanging energy between the propagating electric and magnetic fields and in doing so is able to make it appear like a voltage controlled "switch".

The switching losses are due to the transient exchange and transfer of energy between the propagating em fields where the em field penetration goes from high to low during switching from an insulator to a conductor.

Such energy is dissipated as heat within the dielectric material as it becomes unpolarised and polarised and similarly heat is dissipated due to penetration of device materials by the magnetic field. These losses are due to the device time constant required to store and remove em energy like any RC or RL circuit.

What I have noted above is not normally explained but rather reference is made to the switching characteristics of the device as others have noted as it is more easily observable and understandable and uses the circuitry approach as you correctly note.

The reasons for switching losses is the energy consumption by the device in the on state, in the off state and when it is converted from the on to the off sate and vice versa. So, the power dissipated in the swtich ca be expressed by:

P= Pon + Poff+ Pdyn, with Pdyn is the dynamic losses

Pon = Ion Von ton/T, with Ion the on current, Von the on voltage, and ton is the on current and T is the cycle time.

Poff= Voff Ioff tof/T, same for the off time toff,

The dynamic power = Integral v i dt over the transition on and transition off period. If the transitions are linear then Pdyn = Voff Ion (tron + troff)/4 T,

tron and troff are the transition times from off to on state and vice verse.

Best wishes

Converter losses are divided into two physical factors

conduction loss and

switching loss

Conduction loss is the voltage drop in the circuit when the device is conducting

Switching loss is loss that follows from the fact that the converter sees full voltage and current when it switches on or off.

Note: despite the conduction and switching losses, converters usually have 90% - 95% efficiency. And there are various techniques for reducing the losses mentioned above.

Yes you can calculate the losses using info from datasheets. But be careful when using the datasheet (depending on the application) as precision design may warant that you test the switching devices to obtain real-world data.

So, you can calculate the switching loss and this will depend on the load. Is it inductive or resistive or capacitive load? That means you need parameters like

Vcc, Tri, Tfi, Trv, Tfv and T, where Tri=current rise time, Tfi=current fall time, Trv=voltage rise time, and Tfv=voltage fall time, and T=period

use this formula

Psw=Vcc x I (Tri + Tfv +Trv + Tfi)/2T (in this case, an inductive resistive load is assumed). Where Psw=switching power loss.

Thank you all.

As Simone said, we can attribute the switching losses of a switch like MOSFET to its body capacitor. However, this can only justify the losses during the turn-on process. Because, in turn-on process the switch is shorted and the stored energy in the drain-source capacitor dissipates in the switch resistor.

But, what happens during turn-off transition interval. what is the source of energy which dissipates?

By design the gate voltage is designed to change or influence the charge distribution / density and hence electric field strength / distribution within the doped layers of the IGBT materials (see link below)

https://www.semanticscholar.org/paper/Simulation-of-the-IGBT-switching-off-with-the-model-Schumann-Bohmer/a08f041c24c0198c95be120c97856ca06529703f/figure/0

By design these are meant to block in the off state with no gate voltage applied. When a gate voltage is applied as you noted it dynamically alters the internal charge distribution / density and electric field strength / intensities within the various doped layers as the link above demonstrates.

At turn off and without an applied gate voltage the charge densities and electric field intensities must return to what they were in the off state.

Thus at turn on the applied gate voltage causes the charge distribution, /density and electric field strength / intensity to change or become distorted and in doing so stores energy within the device to do so.

Thus when the gate voltage is removed this stored energy is returned (akin to a spring) to allow the charge distribution / density and electric field intensity to revert to their off state.

It is during these dynamic transitions that internal losses occur due to energy being dissipated in the device and gate drive resistors which are around 50% of the stored energy as per conventional RC circuits.

Also see

https://www.semanticscholar.org/paper/Modern-high-voltage-IGBTs-and-their-turn-off-Eckel-Bakran/01579f4fa4b147cb0123461cfab82e3a08b64665

Here is a useful reference

http://scitechconnect.elsevier.com/wp-content/uploads/2016/02/IGBT-structural-design-chapter-3.pdf