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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?

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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.

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  • \$\begingroup\$ Is collector and emitter interchanged in the link? \$\endgroup\$ – red car Mar 23 '16 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 '16 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 '16 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 '16 at 12:42
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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.

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