Determine Instant Time on IGBT
- Thread starter Laz31
- Start date
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- Tags
- igbt power electronics
These are the same formulae that apply to any switching device, not just IGBTs, that go through from a high resistance to a low resistance state, or vice versa. In the course of a switching cycle the total dissipation/energy is the sum of:
\( E_{total} = E_{on} + E_{on-static} + E_{off} \)
where
\( E_{on-static} = i_c × v_{CEsat} × (t3 - t2) \)
the losses due to the on-resistance of the device.
As to the timing, this should explain it...
For a first approximation the energy in the switch on or switch off phase is:
\( E_{on} = 0.5 t_{rise} × i_{on} × v_{CE-off} \)
and
\( E_{off} = 0.5 t_{fall} × i_{on} × v_{CE-off} \)
\( E_{total} = E_{on} + E_{on-static} + E_{off} \)
where
\( E_{on-static} = i_c × v_{CEsat} × (t3 - t2) \)
the losses due to the on-resistance of the device.
As to the timing, this should explain it...
For a first approximation the energy in the switch on or switch off phase is:
\( E_{on} = 0.5 t_{rise} × i_{on} × v_{CE-off} \)
and
\( E_{off} = 0.5 t_{fall} × i_{on} × v_{CE-off} \)
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