# Transistors / hfe / Active region conflicts

Discussion in 'Electronics Homework Help' started by george2525, Aug 21, 2015.

1. ### george2525

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Jan 30, 2015
Hello,

I studied a BTEC course a couple of years ago and was taught how to determine (for a standard BJT in common emmitter config) the Hfe /Rin / Rout by using the output / input characteristic graphs for the Transistor model. This involved plotting load lines etc

for large signal analysis these relations are

Rin = deltaVBE/delta IB - this seems fine

Hfe = delta IC/ delta IB - where ''delta'' means ''change''

This last formula is now causing me problems at university. All the previous output graphs for transistors I have seen had a clear gradient in the active region (increasing left to right) so these ''deltas'' were easy. Now it seems the active region has a constant current (IC) for all base currents, with no gradient!

I am confused now

Is it the case that people like to pretend that the IC current is constant in the active region or is it the case that I may be looking at completely different output characteristics?

I do have issues with the active region - I fail do understand how increasing VCE has no effect on the current unless the effective resistance of the transistor increases in proportion to the applied voltage - I guess this would be another great question to ask -

If anyone could tell me how this IC is (apparently) constant in the active region in terms of transistor resistance it would be a massive help to me as I prefer to think of them as resistors which often works for me.

hope most of that was clear.

Thanks, G

2. ### Laplace

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Apr 4, 2010
Why does it seem the active region has a constant collector current for all base currents? What is it about the characteristic curves at university that is different from the ones used at BTEC? Please post the images.

3. ### Ratch

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Mar 10, 2013
You want to know why the collector of a BJT in the active region acts like a current generator? That is an easy question to answer. The Vbe brings the charge carriers into the base region where they are swept away into the collector circuit by the collector voltage. A small percentage of the charge carriers from the emitter go into the base circuit as waste current, but the majority of the charges form the collector current. So it is the Vbe voltage that controls the availability of the of the charge carriers that go into the collector circuit. That is why cranking up the collector voltage does not increase the collector circuit. If there are no more charges in the base for the increased collector voltage to attract, then you have a change in voltage causing no change in current. That is the definition of a current source--constant current despite the voltage change. It is all a matter of transistor physics.

Ratch

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4. ### george2525

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Jan 30, 2015
Ok heres the standard output graph that I see most often ...

Note the IC curves are all perfectly straight...

Here is an image similar to the ones I had always used - Note the clear gradients of the IC curves which therefore allow you to calculate their gradient IE Rout = delta VCE / delta IC

this calculation can only be done on the second graph. Are these graphs supposed to be the same?

Last edited: Aug 22, 2015
5. ### george2525

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Jan 30, 2015
Ok thanks. I kind of understand that concept. Could I say that as we crank up VBE, the effective resistance of the transistor decreases until it reaches the active region and after that point (eg 0.8Vbe) the effective resistance INCREASES?. I realise that a resistor is different but if we view it as a current limiter of some kind we would say that if Vcc increases yet IC stays the same it is effectively a Vcc dependant resistor in the active region. Can I get away with that ;-)

6. ### Ratch

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Mar 10, 2013
You have to understand that a BJT operating in the active region without any external base, emitter, or collector resistance is a transconductance amplifying device. That is, a change in Vbe voltage controls a change in Ic collector current. In other words, a BJT by itself is a voltage controlled current source. You can get philosophical and poetical about how a current source changes its resistance or distributes its voltage to keep its current output constant, but a current source just is what it is.

Ratch

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7. ### Laplace

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Apr 4, 2010
Of course they are the same graph, but of different components. The first graph gives the characteristic curves of an idealized transistor whereas the second graph gives the characteristic curves of a real transistor, where the current gain h.FE varies with the collector-emitter voltage and h.FE varies with the collector current. One can draw a load line on the second graph just as on the first.

Also note that while the BJT operates as a transconductance device (collector current controlled by base voltage), it is not characterized as a transconductance device. That is because the transconductance effect is highly non-linear and quite temperature sensitive thereby making Vbe a rather useless quantity to consider except as a diode forward voltage drop (~0.6V). Real circuits are designed to generate the bias point Vbe by forcing a controlled current to flow through the base-emitter junction. Perhaps this is why transistors used in circuits are thought of as being current-controlled even though a transconductance transistor model may be be used for small-signal analysis. In that case it is delta-Vbe and not Vbe that is the relevant quantity.

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8. ### Ratch

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Mar 10, 2013
Neither of the above graphs are really graphs of the device alone. Whenever you drive a transistor with a current source, you are inserting an external resistor from the current source into the base circuit. This external resistor is not part of the device. What you now have is a current amplifier or current controlled current source (CCCS) circuit, not a device. That is fine if you know what is going on, namely that a transconductance device and been converted into a current amplifier by the addition of lots of resistance to the input.

Ratch

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9. ### george2525

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Jan 30, 2015
Thanks for your time guys. That was very helpful. I have never seen the 2 graphs referred to as 'ideal' or 'actual' so that info is great to know

10. ### george2525

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Jan 30, 2015
Ok so are you saying that if Vbe could be controlled reliably and independently (without any external circuitry) then the transistor would function normally - control the value of IC?

I was under the impression the IB was necessary to provide the electrons (or receive them if you like) to the supply, and that these electrons were essential to eliminating the forward biased depletion region.

So if there was no IB, are you saying that Vbe (in imaginary land) would serve the same function?

11. ### Laplace

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Apr 4, 2010
Whenever you drive the base of a transistor with a voltage source, you are inserting an external resistor from the voltage source into the base circuit. The better the fidelity of the voltage source, the closer this external resistor (i.e., source internal resistance) approaches zero. As this external resistor approaches zero it becomes indistinguishable from a short circuit.

Whenever you drive the base of a transistor with a current source, you are inserting an external resistor from a voltage source into the base circuit. The better the fidelity of the current source, the closer this external resistor approaches infinity. As this external resistor approaches infinity it becomes indistinguishable from an open circuit. By definition an open circuit cannot be part of the transistor device. For an ideal current source, it is as though base current magically appears from nowhere.

So consider which scenario becomes a part of the transistor device, a short-circuit (voltage source) or an open-circuit (current source).

12. ### Ratch

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Mar 10, 2013
A transistor always functions "normally", unless you apply too much voltage or do not limit its current and burn it out. Even within a circuit, it functions as a transconductance amplifier, just as an op-amp functions as a dual high amplification voltage amplifier even when it is externally wired to be a current amplification circuit.

Your impression is wrong. Ib is an indicator of Ic, not a control of Ic. It is a wastage current that cannot be avoided. You need to read my explanation of what Vbe does in post #75 of this link. https://www.electronicspoint.com/th...stor-works-base-current-version.269901/page-4 , It is hard to manufacture a BJT that "leaks" only a minimal amount base current. Both the base current and collector are exponentially related to Vbe, so Ic and Ib are somewhat linearly related to each other within a range. This relationship is called BETA.

Fabrication specialists have tried to manufacture transistors with extreme low base currents (very high betas), but found it to be impractical due to low lot yields and numerous out of specification parameters that occur due to the design methods needed to keep the current out of the base circuit..

Ratch

13. ### Ratch

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Mar 10, 2013
By definition, a voltage source has no internal resistance. In practice, it can be modeled as a ideal voltage source in series with a small resistor.

Not quite so. A short circuit differs in that it does not have a voltage source in series with it. Most voltage sources have an internal impedance that is insignificant compared with the bulk resistance of the transistor. For practical purposes, the internal resistance of the voltage source can be ignored.

Not quite so. An open circuit differs in that it does not have a current source in series with it. Most current sources have an internal impedance that overwhelms the impedance of the transistor. For practical purposes, the internal resistance of the current source can be considered infinite.

Not quite so. A current source is not the same as a open circuit unless the current is turned off. Base current does not appear by magic, it appears because the current source supplies the current.

Neither one does.

Ratch

14. ### george2525

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Jan 30, 2015
''The collector voltage removes or replaces, depending the type of BJT, the charge carriers. In order to keep the charge flow going, a voltage vbe is applied to the base. This reduces the back-voltage and allows a current to exist to the collector, which is limited by a new equilibrium of the uncovered charges.''

Ratch[/QUOTE]

Ok I read that. It was Interesting thanks.

I would like to ask regarding the above - Does increasing Vbe reduce the back voltage between base and collector (and the depletion region width) because Vb>0 ?

I realise that collector / base is reverse biased so will have a wider depletion region than the base emitter junction but increasing Vb still shortens this region when compared to Vb=0V

So when you say ''reduces the back voltage'' you mean at both depletion regions?

15. ### Ratch

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Mar 10, 2013
The back voltage is caused by uncovered charges due to the diffusion process. If Vbe cancels some of this back voltage, the diffusion will increase and the depletion region will become narrower because a smaller depletion region corresponds to a lessor back voltage. I am referring to the emitter-base junction, not the base-collector junction.

Vbe does not affect the C-B depletion region. The Vcb does.

The Vbe and Vcb each have their own independent depletion regions.

Ratch

16. ### george2525

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Jan 30, 2015
''Vbe does not affect the C-B depletion region. The Vcb does.''

Ratch[/QUOTE]

yea sorry thats what I meant when I said Vb, I meant Vcb

I guess what im saying then is that as Vbe increases, Vcb decreases and therefore BOTH depletion regions become narrower. With the collector / base region becoming only slighty narrower.

17. ### Ratch

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Mar 10, 2013
Yes, you are describing base width modulation or the Early effect. It is what makes the Vc-Ic curve have a slant instead of being horizontal. You can read all about it here and elsewhere. https://en.wikipedia.org/wiki/Early_effect

Ratch

18. ### Laplace

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Apr 4, 2010
A current source has an infinite resistance; when anything is connected to a device through an infinite resistance, the device is unable to discern that anything has been connected.
And this is quite true (from an external standpoint; while from the viewpoint of the transistor seeing an infinite resistance at its base terminal, it's still magic) but the point is that the base current is controlled while Vbe is not directly controlled by an applied voltage source. Vbe is free to do what it will do.
This would be true because the base-emitter junction is a forward biased diode where the I-V relationship obeys the ideal diode equation which gives an exponential relationship between Vbe and Ib. It is not possible to develop Vbe without a corresponding base current. Even though it is forbidden to apply a voltage source directly to the base-emitter junction, doing so would create a corresponding base current; it is the base current which controls Vbe because base-emitter voltage is never directly applied from a voltage source.

So there is the full story: base current controls Vbe, and Vbe controls collector current.

19. ### Ratch

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Mar 10, 2013
An infinite resistance source does not mean that the device cannot receive current. It means than the device will receive the same set amount of current no matter if the device changes its resistance, unless it changes its resistance to compare to the current source, like turning itself off. If the device were connected to a voltage source, it would be able to control its current by changing its resistance.

No magic involved. I already explained what seeing an infinite resistance by a device means to it. Vbe can be directly controlled by applying a voltage to the base-emitter terminals. However, that is not a good idea because of the nonlinear characteristics of the transistor's transconductance.

That is true. The waste current that comes out of the base circuit is impossible to stop. Too bad it cannot be shunted to the collector where it would do some good.

You can apply a voltage directly to the base-emitter junction even though it is not a good idea due to its nonlinear voltage-current relationship.

You got that backwards. The physics of the diode shows that voltage controls the diode current, not the other way around. The voltage lowers the back-voltage caused by the depletion region and allows more current to be present. Current does not control the voltage across the depletion region. Diode current is an indicator of the applied voltage, not its control. Only voltage can lower the back-voltage of a diode allowing more charge carriers to flow. Therefore voltage controls current in a junction diode, not vice versa.

No, Vbe controls both the base current and the collector current in an exponential relationship according to the physics of the transistor. That is why the collector and base currents are related to each other by BETA.

Ratch

20. ### Laplace

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Apr 4, 2010
Exactly! No matter what the device does, nothing changes. It's as though the device were connected to nothing.
Damnable Physics!
One can do many things which are forbidden by common sense, but that leads to perdition.
When the diode is driven by a current source the current is set by the current source, and Vbe becomes whatever value it must to accommodate that current because the current source controls the base current.
The value of Vbe is totally irrelevant to the design process of setting the transistor bias current, which is why Vbe is assumed to be a standard value. If Vbe controlled the base current, it would lead to perdition.