# Another transistor discussion

Discussion in 'General Electronics Discussion' started by dorke, Oct 20, 2015.

1. ### dorke

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Jun 20, 2015
A beginner is thinking like this : "same current flowing here" :
My words are chosen to explicitly explain the composition of currents.

You can contradict me any time you want.
and,
thanks for letting me know the meaning of Ib,
what would I have done without you Kid...

Last edited by a moderator: Oct 20, 2015
2. ### BobK

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Jan 5, 2010
I did catch it, but you has posted the correction before I got a chance to, damn it.

Not trying to argue here, but why do consider "current from base to the emitter" to be an inaccurate description of Ib, at least for an NPN transistor using conventional current? Surely all the current flowing into the base is flowing out of the emitter, as demonstrated in your Ie = Ib + Ic equation. So, why is the quoted phrase wrong?

Bob

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3. ### BobK

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Jan 5, 2010
I agree that he was confused about the relationship of base current and collector current.

But I still disagree that "what goes into the base comes out the emitter" is false. Just because there is also current coming from the collector out through the emitter does not make that statement false.

Bob

4. ### dorke

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Jun 20, 2015
Well,
The best I can say for it is being "partly true"...
The only fully true statement is the current composition ie=ib+ic.
Putting it in words : The emitter current is the sum of the base and collector currents.

In the extreme case of large hfe,
like in a darlington transistor(hfe=1000 or more) we can totally ignore ib
and say ie=ic.

5. ### BobK

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Jan 5, 2010
Actually, even that is not true.

Try connecting the base to 5V through a 1K resistor, the collector to ground and the emitter to +5V. Now the equation is:

Ic = Ib + Ie

Bob

6. ### dorke

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Jun 20, 2015
That is true,Tr in reverse active mode.
But :
1.It is kind of a "trick": reversing the rolls of collector and emitter.
2.It isn't the playground of beginners.
3. I have never used it,have you?

7. ### Ratch

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Mar 10, 2013
To summarize what happens in a BJT biased to be in the active region.

1) The base current Ib is leakage or waste current that does not contribute to the amplification function of the BJT. It is current from the base-emitter diode that does not get sent across the base collector diode. Careful design of a BJT can greatly reduce this leakage current, but it cannot be completely eliminated.

2) The collector current Ic is proportional to the base current Ib within a moderate range. This proportion is called BETA.

3) Although the base current is proportional to Ic, it does not control Ic. Vbe controls Ic.

4) A BJT is a transconductance amplifier (voltage controls current) when operated in the active region.

5) The base current forms part of the emitter current.

Ratch

Last edited: Oct 21, 2015
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8. ### dorke

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Jun 20, 2015
Ratch,
Agree with you on all.
but number 4)A BJT is a transconductance amplifier-
that is true only in C.E ,but not in C.C or C.B.

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May 12, 2015

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The transistor doesn't know how it's connected

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11. ### dorke

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Jun 20, 2015
Agree it doesn't,
it only feels the pain...

12. ### AnalogKid

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Jun 10, 2015
I completely and totally agree. And, it always surprises me how difficult this is to teach (or hammer into, depending...) young'ns.

ak

13. ### AnalogKid

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Jun 10, 2015

For common emitter circuits, maybe, but not for emitter followers. In the output stage of a power supply or audio power amp, base current flows directly into the load, and can account for 5% to 10% of the load current in some applications. Semantically, that might not be covered by the phrase "contribute to the amplification function", but I think that characterizing base current as wasted is a misleading generalization.

ak

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14. ### BobK

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Jan 5, 2010
Only by accident

(Gee, this transistor does not seem to have enough gain )

Bob

15. ### Ratch

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Mar 10, 2013
I stand by what I said. No matter what configuration you set up, you are adding external components to the transistor. This means that you are no longer working with just a device. You are instead working with a transistor circuit. I aver that the transistor within the circuit will still act like a transconductance amplifier, whereby the voltage across its base-emitter will affect the collector current. This is analogous to a op-amp which can be made to do a multitude of electrical functions by adding external components while the op-amp itself still is functioning as a dual high gain voltage amplifier.

That's right. It still functions the same way, but its input and output is modified by the external components..
For a CC circuit, base current does not contribute to the current amplification. That in itself is a "waste". If we cut out the collector, then all the base current would drive the load and we would not have any current amplification. Furthermore, it would load down the base network whose purpose is to provide the desired voltage. So the fact that the base current still goes through the load on a CC configuration does not mean that the circuit is operating optimally. In a perfect world, no base current would exist and the current would depend entirely on the base voltage across the load.

Ratch

Last edited: Oct 21, 2015
16. ### Martaine2005

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May 12, 2015
Hi guys to all involved in the transistor wars...
Can I ask a really stupid question? No, stupider than my usual questions..!!
When talking about transistors, the physics plays its part on how it works.
But when people talk about a circuit that uses a transistor, the usual response is "but that's in a circuit, so outside components will affect it.". That is the point I don't understand.
I may be not thinking this through properly, so forgive me. But what can a transistor do without outside circuit components?
Feel free to call me all the names under the sun..And above.

Martin

17. ### BobK

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Jan 5, 2010
Well, since there are no external components, it would just sit there looking pretty. You have to connect the terminals to some voltages (often through resistors) before it does anything.

The external components do not change the physics of the transistor, they just set up the conditions under which it is working.

Bob

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18. ### Ratch

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Mar 10, 2013
Hi.
I hope we can answer it.
Definitely
You know how to connect up an op-amp to be an amplifier. Why are you not asking how two external resistors are setting the gain of the op-amp?
You are forgiven and forgotten.
Why are you not asking what can an op-amp do without outside components?
Very well, DS (Discerning Scholar).

19. ### Martaine2005

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May 12, 2015
Hi Ratch,
The reason I didn't mention op-amps is because we are talking about transistors.
As already laid bare, the physics of the active region and leakage from base to emitter etc.
Totally above my head...As Bob said "sitting pretty" all doled up and nowhere to go!.
So my real point is, if one knew all the physics and characteristics of said transistor, what does that have to do with a circuit that needs a transistor?
I 'honestly' cannot work out what a lone transistor is good for..

Martin

20. ### Ratch

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Mar 10, 2013
Makes no difference as far as their integration into other circuits. Just like a capacitor, resistor, or inductor. By themselves, they are just components. But combined properly into a circuit, they are part of a useful development.
Knowing the physics of a BJT tells you why the Vbe is related to the Ic exponentially (physics of diffusion). What determines thermal runaway? What determines the Early effect? What determines a lot of other things? Those characteristics are not explained from empirical data gleaned by measurements or parameter charts.
.
Neither can you work out what use is a lone resistor by itself. It is a building block, not a complete entity by itself.

Ratch

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