why is the BJT in cutoff region when the base is open?

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matts

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Hello Everyone,

I have only recently begun my study of the analog electronic. I just want to ask why the BJT is in its cutoff operating region when its base is open,no matter how you change the collector to emitter bias. Should some of the electrons in emitter (NPN) flow through the two junctions when Vce is large enough? Also when I simulate the circuit in Multisim, the currents through collector and emitter are not equal, why is so?


Thanks a lot in advance
 

The non equal currents are due to Multisim. Where did you think the difference in currents is going to?
When there is no injected base current, the +Ps and the -Ns at the junction balance out leaving what is called the DEPLETION ZONE (no free carriers). Any current that passes across this is due to thermal effects.
Frank
 

matts said:
Should some of the electrons in emitter (NPN) flow through the two junctions when Vce is large enough?
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Yes, indeed. There will be a current - similar to a diode that is reverse biased until breakthrough. However, I doubt if the part can be further used.
 

OK. Thanks a lot.
 

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