PNP BJT....Collector and emitter currents very different

F

Flyback

Well-Known Member

• Apr 23, 2018

• #1

Hello,

In the attached common base schematic, the collector current should equal the emitter current. However, collector current is very low at 800uA.
In this configuration, and with the FMMT560Q PNP transistor, is there any situation where , say, high leakage current would mean that collector and emitter currents were not almost equal to each other?

FMMT560Q datasheet
https://www.diodes.com/assets/Datasheets/FMMT560Q.pdf

 

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• PNP.jpg

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audioguru

Well-Known Member

Most Helpful Member

• Apr 23, 2018

• #2

The collector current is almost the same as the emitter current which is also very low.

 

Attachments

• common base transisor.png

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F

Flyback

Well-Known Member

• Apr 23, 2018

• #3

Thanks, though we are worried that when the PNP saturates, the collector current will be much less than the emitter current.

 

alec_t

Well-Known Member

Most Helpful Member

• Apr 24, 2018

• #4

If you run a simulation with the two voltage sources changing there is very little difference between the collector and emitter currents. Try adding some leakage in the sim and see if the difference becomes greater than whatever you consider acceptable.

 

F

Flyback

Well-Known Member

• Apr 24, 2018

• #5

Thanks again, and I mean, the PNP in the schematic of the top post cannot saturate because its Base-Collector junction cannot get forward biased in that schematic and setup.

 

audioguru

Well-Known Member

Most Helpful Member

• Apr 24, 2018

• #6

Flyback said:

Thanks, though we are worried that when the PNP saturates, the collector current will be much less than the emitter current.

Click to expand...

How can the PNP saturate? You show NO input signal. Is V1 the input signal? What is its voltage range?
An NPN works exactly the same as a PNP in that circuit except the polarities are reversed.

 

MikeMl

Well-Known Member

Most Helpful Member

• Apr 24, 2018

• #7

With the resistors you chose, V1 has to be >58V for Q1 to saturate.



Notice the difference between I(R1) and I(R2) = -Ib(Q1). Note the increase of V(c)

 

MikeMl

Well-Known Member

Most Helpful Member

• Apr 24, 2018

• #8

Flyback

Conversely, if V1=5V, then as R1 is reduced below 421Ω, Q1 saturates: