Pnp transistor with base connected to collector

AI Thread Summary
The discussion centers on the behavior of a PNP transistor with its base connected to the collector, particularly focusing on the emitter current (IE) in relation to the base (IB) and collector (IC) currents. When the voltage at the emitter (VE) is set to 0 V and the base/collector voltage (VB) decreases to -1.0 V, the simulation indicates that IE equals the sum of IB and IC. Participants clarify that while the base-collector junction is effectively shorted, internal parasitic resistances may influence the actual voltage across this junction. The consensus is that current flows from the base to the emitter and from the collector to the emitter, confirming that the internal voltage VBC is not zero due to these resistances. Practical testing methods are suggested to verify the functionality of the transistor.
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.
 
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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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