Pnp transistor with base connected to collector

  • Thread starter DS2016
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  • #1
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Hello all,
I have a problem to understand a pnp transistor with base (B) connected to collector (C).
When I set the VE (voltage of emitter) at 0 V and the VB, which is the voltage of base and collector nodes, since they are tied together), decrease from 0 V to -1.0 V,the simulation result shows me that the emitter current (IE) is equal to | IB + IC |.
I can understand the result regarding the equivalent circuit of pnp, but i can not understand the physics behind this, because VBC=0 V. Does operate the diode of BE junction in forward-biased? what is about the diode of BC junction, when VBC=0 V?

I would be thankful for any reply
 

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  • #2
cnh1995
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I'm no expert but from your wording, this is what I think:
since they are tied together),
Means B-C junction is shorted.
what is about the diode of BC junction, when VBC=0 V?
Since BC diode is shorted, no current would flow through the BC junction.
 
  • #3
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Thanks for your answer.
The simulation results show me that a current (IC), which is a little lower than IB (defined for base current ), flows through the collector node and the sum of both currents | IB+IC| is equal to the emitter current IE.
In my case, the external nodes of base and collector are shorted, but each of this layer has internal parasitic resistance. I think the internal VBC is not 0 V.
 
  • #4
cnh1995
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which is a little lower than IB (defined for base current ), flows through the collector node and the sum of both currents | IB+IC| is equal to the emitter current IE.
Yes. But there is no current "through" the BC junction. Current flows from B to E and C to E such that Ib+Ic=Ie..
. I think the internal VBC is not 0 V.
It should be 0 since B and C are externally connected by a conductor.
 
Last edited:
  • #5
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Many thanks for the information. It took me time to understand that, but now it's clear for me.
Thanks again.
 
  • #6
Tom.G
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Try a real-world experiment. This is a "Quick-and-Dirty" test to check if a bipolar junction transistor (BJT) is functional.

1) Connect an Ohmmeter to forward bias the B-E junction with the Collector open. You should get a fairly low resistance reading.
2) Now connect the Collector to the Base. The Ohmmeter reading will decrease by a small amount.
3) If you then disconnect the Base, you should get a very high reading between Collector and Emitter.
Edit: 4) Connect E-C and Ohmmeter to reverse bias the Base junctions. Should be a very high reading.
 
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