# IGBT naming question

I was looking at this site http://www.electronics-tutorials.ws/power/insulated-gate-bipolar-transistor.html

Here an IGBT is an NMOS with a BJT> But the BJT looks to be a pnp device. When the NMOS turns on, Vbe =0 , how will the pnp conduct? Making the gate negative, mean no channel exists in the NMOS, and the NMOS is off. The 2 ends could actually be at any potential. How does this circuit actually work?

Consider it with a load.

simulate this circuit – Schematic created using CircuitLab

The base current of Q1 is determined by the hFE of Q1. $V_{BE}$ is about -0.75V, not 0V.

• Is collector and emitter interchanged in the link? Mar 23, 2016 at 3:56
• I drew it the same as your link- the collector (C) of the approximate hybrid device is the emitter of Q1. This is only an approximation that illustrates the operation. Mar 23, 2016 at 4:12
• What happens to the voltages at the base and collectr of Q1 when M1 is off? Can they be any voltage? Mar 23, 2016 at 5:01
• The collector is grounded, so it's going to be at 0V by definition. The base will be Vbe of Q1 with leakage current through M1, so a couple hundred mV below the emitter of Q1 (and the latter will be at about +200V in the circuit shown). Mar 23, 2016 at 12:42

Consider the following two states:

• When $V_{GE}$ is less than the threshold voltage of the MOSFET, $I_B = 0$, and therefore $I_C = 0$. The IGBT is off.

• When $V_{GE}$ is greater than the threshold voltage of the MOSFET, $I_B$ is set by the $h_{FE}$ of the BJT, and flows through the MOSFET. Therefore $I_C = h_{FE} I_B$; the IGBT is on.

Also, keep in mind that it is a simplified equivalent circuit. An actual IGBT is a PNPN (four-layer) semiconductor device. It's not just a BJT and MOSFET glued together in a box.