0
\$\begingroup\$

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?

\$\endgroup\$

2 Answers 2

Reset to default
0
\$\begingroup\$

Consider it with a load.

schematic

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.

\$\endgroup\$
4
  • \$\begingroup\$ Is collector and emitter interchanged in the link? \$\endgroup\$
    – red car
    Mar 23, 2016 at 3:56
  • \$\begingroup\$ 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. \$\endgroup\$
    – Spehro Pefhany
    Mar 23, 2016 at 4:12
  • \$\begingroup\$ What happens to the voltages at the base and collectr of Q1 when M1 is off? Can they be any voltage? \$\endgroup\$
    – red car
    Mar 23, 2016 at 5:01
  • \$\begingroup\$ 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). \$\endgroup\$
    – Spehro Pefhany
    Mar 23, 2016 at 12:42
1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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