Pnp transistor with base connected to collector

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Discussion Overview

The discussion focuses on the behavior of a pnp transistor with its base connected to the collector, particularly examining the implications of this configuration on current flow and junction biases. Participants explore both theoretical and practical aspects of the transistor's operation, including simulation results and real-world testing.

Discussion Character

  • Exploratory
  • Technical explanation
  • Experimental/applied

Main Points Raised

  • One participant questions the physics behind the emitter current (IE) being equal to |IB + IC| when the base (B) and collector (C) are tied together, particularly regarding the biasing of the junctions.
  • Another participant suggests that since the B-C junction is shorted, no current would flow through it, implying that the BC diode is effectively bypassed.
  • A participant notes that while the external nodes of base and collector are shorted, internal parasitic resistance may lead to a non-zero internal VBC, challenging the assumption that VBC is 0 V.
  • There is a discussion about the flow of current from B to E and C to E, with some participants agreeing that IB + IC = IE, while others emphasize that there is no current through the BC junction itself.
  • A later post suggests conducting a real-world experiment to test the functionality of the transistor, outlining steps to measure resistance and biasing conditions.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of the BC junction and the implications of the internal voltage VBC. There is no consensus on whether VBC can be considered 0 V due to the internal resistances and the nature of the connections.

Contextual Notes

Participants acknowledge the presence of internal parasitic resistances that may affect the behavior of the transistor, indicating that the discussion is limited by assumptions about ideal conditions.

DS2016
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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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I'm no expert but from your wording, this is what I think:
DS2016 said:
since they are tied together),
Means B-C junction is shorted.
DS2016 said:
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.
 
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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.
 
DS2016 said:
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..
DS2016 said:
. 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:
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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.
 
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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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