IGBT图解

le flows through MOS channel while Ih flows across PNP transistor Ih= a/(1-a) le, IE-le+lh=1/(1-a)' le Since IGBT has a long base PNP, a is mainly determined by ar si0 2

ar= 1/cosh(1/La), La: ambipolar diff length a-0.5 (typical value)

p MOSFET channel current (saturation), le=U"Cox"W(2"Lch)"(Vc-Vth)

le Thus, saturated collector current Ic, sat=1/(1-a)"le=-1/(1-a)"UCox"W/(2Lch)"(Vo-Vth)2

Also, transconductance gm, gm= 1/(1-a)"u' Cox W/Lch*(Vo-Vth)

Turn-On

1. Inversion layer is formed when Vge>Vth

2. Apply positive collector bias, +Vce

3. Electrons flow from N+ emitter to N-drift layer providing the base current for the PNP transistor

4. Since J1 is forward blased, hole carriers are injected from the collector (acts as an emitter).

5. Injected hole carriers exceed the doping level of N-drift region (conductivity modulation)

. Turn-Off

1. Remove gate bias (discharge gate)

2. Cut off electron current (base current, le, of pnp transistor)

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