Transistor at cutoff, conductance, or saturation?

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SUMMARY

The discussion focuses on the operational states of a bipolar junction transistor (BJT) when the base-emitter voltage (VBB) is reversed. The consensus is that reversing the VBB potentials leads the transistor into cutoff mode, where no current flows through the transistor, effectively acting as an open circuit. This conclusion is supported by the understanding that without current to the base, the transistor cannot conduct. The terminology used clarifies that the correct term for the input terminal is "base," not "gate."

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  • Knowledge of transistor states: cutoff, saturation, and active
  • Familiarity with base-emitter voltage (VBB) concepts
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Femme_physics
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Hi everyone, took a bit of a break lately...but with 6 days left to the final test of electronics, I figured I'd clear up some understandings I have

Homework Statement



http://img225.imageshack.us/img225/4474/vbbreversed.jpg

I'm asked if the VBB potentials are flipped, at what condition will the transistor be? (conductance, saturation, cutoff...)

The Attempt at a Solution



I say cutoff, because no current flows to the gate of the transistor hence no current flows in the transistor. The transistor acts like a disconnection in the circuit.
 
Last edited by a moderator:
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Good! :)

(Although you should call it the "base" instead of the "gate" for this type of transistor.)
 
Thanks ILS :)
 

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