# What exactly is meant by the current gain of a transistor?

Discussion in 'General Electronics Chat' started by midnightblack, Mar 17, 2012.

1. ### midnightblack Thread Starter Member

Feb 29, 2012
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If I understand this correctly, a base emitter current of a certain level turns on a transistor.

The part I don't understand is the current gain part. In my understanding, when a transistor turns "on", it means that it allows conduction between the collector and emitter, so it would not matter if the current going between the emitter and collector is 50mA or 500mA. Right? So if Current gain hfe=Ic/Ib, how is constant and computable?

There are holes in my understanding which I'm looking to fill.

What is the use of a darlington pair? If a current in the base of the first transistor switches "on" the transistor, then it would simply conduct fully--> right?

Or is it that the base current is only high enough to get the transistor TR1 to the active mode? --> So we get the base current of the transistor TR2 high enough to set it into saturation mode by connecting it to the emitter of TR1?

2. ### #12 Expert

Nov 30, 2010
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Your idea of "on" is faulty. Speaking in terms of a common emitter circuit, If we decide that we have some transistors with a current gain of 50, then a milliamp through the base-emitter junction will enable 50 milliamps through the collector-emitter circuit if there is enough voltage available at the collector.

Transistors are not very reliable to have exactly one amount of current gain. That is why it is specified as a range of possibilities. That range is also limited by the C-E saturation voltage of any transistor.

When you make a Darlington circuit, the current gain is 50 squared...if there is enough voltage available to keep the circuit out of the C-E saturation region.

When you want the transistor saturated to achieve minimum voltage from Collector to emitter, you try to give the B-E junction a tenth of the current through the Collector.

That's a start. Ask again if you have a specific.

Sep 7, 2009
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4. ### hobbyist AAC Fanatic!

Aug 10, 2008
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If I'm remembering this correctly, from my course material on the subject of transistor physiscs,

the bjt is made up of 3 doped stuctured regions, the base is made very thin with respect to the emitter and collector, so when a transistor is assembled, the majority charge carriers in both the emitter and collector, are attracted over into the base region to recombine with the minority charge carriers, this recombination, causes a net charge effect very close to the base of zero potential, this is called I think the depletion region.

This is where no more recombination can take place due to the base and emitter and collector having a potential barrier, that repels rather than attracts carriers from crossing over.

This potential barrier is the inherent Vbe that must be applied to break down this barrier.

So when a Vbe is applied to the base emitter junction, the depletion region, (barrier potential is overcome as charge carriers are withdrawn from the base, when this happens the barrier potential is smaller so recombination can take place again, however with a stronger voltage applied at the collector and emitter, a small amount of the majority carriers cross over into the base and recombine, while a majority (beta value) of the carriers cross thru the base into the collector region and flow out into the circuit.

So a small amount of base current has got to flow out the base terminal, in order for the barrier potential to deplete to the point where a (beta value) more current can flow from the emitter thru the base into the collector region for a greater gain of current output.

Depending on the make up of the transistor, as current flow out of the base is increased the greater the barrier potential depletes allowing more current to flow from emitter to collector.

The transistor is acting as a variable resistance between emitter and collector, due to the amount of charge carriers are withdrawn from the base, (base current) so a small base current will result in a smaller depletion of the barrier potential, so a smaller (beta value) of collector current can flow from emitter to collector, as a larger base current is drawn out, then the barrier potential has a larger depletion, resulting in greater amount of (beta value) collector current to flow from emitter to collector.

When the base current is saturated, the barrier potential is depleted completely so maximum curent regardless of (beta) will flow from emitter to collector, because the potential barrier resistance is near zero. That is now in saturation.

When the transistor is turned off, then the barrier potential builds back up again, and you now have the depletion region at maximum width, and greateat amount of resistance between emitter and collector, known as cutoff.

This is written from memory, so someone can correct me where needed in the different ternms I used in the explanation.

5. ### midnightblack Thread Starter Member

Feb 29, 2012
31
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Thanks for your explanation. I want to fully understand how the transistor works and I realised just now that actually I have a problem with the fact that its described in terms of "current gain". This is very confusing to me, when I see it in terms of current gain, I think of the transistor as a sort of device which is like a variable current source.

In my understanding, I think of the transistor as an attenuator with a gain varying between 0 and 1 rather than an amplifier with gain >1. If this is wrong, can you please explain why it is wrong?

If this is correct, I still don't fully understand why darlington pair is needed. Is it that the Vbe of TR1 is so small that the output (i.e. emitter voltage in relation to ground) of TR1 is too small to be useful, but since its output is dependent on its Vbe, if we connect the output to base of TR2, we are ultimately making the Vce of TR2 dependent on the Vbe of TR1 which finally makes the total output big enough to be useful?

EDIT: I think I am understanding it better now, current gain hfe only applies during active mode. So at which points of Vbe are cutoff, saturation and active mode?

Last edited: Mar 18, 2012
6. ### midnightblack Thread Starter Member

Feb 29, 2012
31
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Thanks for your detailed answer, and for your time. I think I understand it more now, though I am still slightly shady over the whole beta/hfe/current gain of an amplifier. :/

7. ### studiot AAC Fanatic!

Nov 9, 2007
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A simple way to understand it is:

A transistor forces the collector current to be equal to the gain times the base current, regardless of the collector load impedance.

So if the gain is 100 and the base current is 1 milliamp the collector current is 100 milliamps.

Driving 100 milliamps through a collector load resistor of say 100 ohms means that there is 10 volts across that resistor.
The transistor adjusts its collector voltage to cause this to happen.

A darlington can be considered as a super gain transistor so that we can get edit: - increased gain allowing the same collector current from say 1 microamp base curent instead of 1 milliamp.

A transistor is only 'on or off' if when used in switching mode, in which case the gain is low and irrelevant.

In amplifying or 'active mode' it can be turned more an more 'on' - like a water tap in the bath.

Last edited: Mar 18, 2012
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8. ### #12 Expert

Nov 30, 2010
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The way the circuit is configured around the transistor determines if it has a gain of less than one. To achieve a gain of less than one you would use an emitter follower circuit and the gain of less than one is a Voltage gain of less than one. The current gain (of a lot more than one) can still be very useful in an emitter follower circuit.

A transistor circuit can be configured as a very predictable variable current source.

Your question about the Darlington pair does not make sense to me. Volts base to emitter too small to be useful? I have never thought in those terms.

I get the feeling that you are trying to get one set of answers to fit all circuits that a transistor can be used in. The circuit is more important than the transistor! For any particular question you must choose a configuration to get reasonable answers.

Meanwhile, I have a very confusing fact to share. I graphed Collector Current (from 1 nanoamp to 1 milliamp) vs Volts base to emitter from .24V to .60V on log-log graph paper. The results were a straight line such that for every 60 mv increase in Vbe, the collector current increased by a factor of 10. This indicates to me that a transistor might have zero collector current for zero volts base to emitter, but there is no "on" point. The instant Vbe becomes non-zero, collector current is flowing and it increases continuously. It is not a step function!

Last edited: Mar 18, 2012
9. ### studiot AAC Fanatic!

Nov 9, 2007
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Note this refers to voltage gain which is a property of the circuit as a whole, not current gain.

Current gain is primarily a property of the transistor itself.
However since this varies from transistor to transistor the circuit design should not rely on this parameter.

That is one good reason why we normally work input and output in voltage terms rather than current terms.

10. ### midnightblack Thread Starter Member

Feb 29, 2012
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By gain of <1, I meant the gain of the transistor itself not of the whole circuit to which it is attached. Is this not correct?

I am unsure if I'm right, but I thought this was why people use a darlington pair. If the Vbe is too small, the transistor output is not big enough so we connect it to another transistor's base and drive the output using the whole thing. E.g. drive a motor using darlington pair instead of a single transistor.

Something like that; I am trying to understand the transistor first before I go into circuits using them. The transistor's behaviour is constant regardless of what circuit it is in isn't it? Its only the Vbe, Vce etc etc that changes each time which changes other values. No?

I am not sure what data you used to plot this? Did you use an equation or experimental data? How can the collector current increase per Vbe stay constant? Doesn't it depend on whether it is in the saturation region, cutoff or active region?

EDIT: I just realised you said the data was between 0.24V and 0.60V, so this is the active region?

Last edited: Mar 19, 2012
11. ### midnightblack Thread Starter Member

Feb 29, 2012
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I would like to know, exactly which variables and what values of them dictate which mode it is in?

I would assume it is dependent on Vbe and Vce, but I am unsure what values of them make the transistor act in cutoff, saturation and active mode.

12. ### #12 Expert

Nov 30, 2010
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1) Transistors exist because they can control a greater current through their collector-emitter circuit than that current which is applied to their base-emitter circuit. That is the definition of a current gain of more than one, and transistors have it.

2) If the APPLIED Volts base to emitter is too small, then insufficient current gain will be the result, and creating a Darlington arrangement will fix this problem. For a moment I thought you meant, "If the inherent base to emmitter voltage of the transistor is too small". You can see why I was confused.

3) A transistor is a three legged animal such that a small current through the base-emitter circuit will enable a larger current to flow through the collector-emitter circuit, as long as sufficient voltage is available for the transistor to operate. There are circuits which intentionally make use of the limitations imposed by the collapse of this relationship by starvation of voltage. The operation of a transistor is NOT constant, regardless of the circuit in which it is employed.

4) I used more than one voltage source and more than one meter to measure a transistor's behavior while giving it more than enough collector voltage to stay out of its saturation region. The results do not say that the collector current increases while the Vbe stays constant.

13. ### crutschow Expert

Mar 14, 2008
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A darlington pair is used to increase current gain (reduce the base current for a desired collector current). The base-emitter voltage of a darlington when on is actually the sum of the two base-emitter voltages so it is always larger than a single transistor, not less.

A transistor is not a variable resistor, it acts as a variable current regulator with gain from base to collector of greater then one, and with a very high collector impedance.

A transistor has three main operating regions: cutoff region (no base or collector current), active region (collector current proportional to base current), and saturation region (Vce is less than Vbe, transistor is fully on, acting like a switch). Vbe is 0V in the cutoff region and around 0.6-0.7V in the active and saturated regions. Vbe has a logarithmic relationship between voltage and base current.

The data sheet beta value only applies to the active region. If you connect a transistor directly to the supply voltage with no collector resistor then it will be either in cutoff, with no base or collector current, or will be in the active region where the collector current equals the base current times the current gain (beta or Hfe).

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
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BJT is OFF (cut-off) when Vbe is lees then 0.5V.

And the boundary between active region and saturation is determined by supply voltage VCC and Rc resistor value.
In saturation increase the base current causes no further change in collector current

Check this imagines

And kept in mind that BJT is a active device thanks to its property, the ability of a amplifying the power.
Amplifier is a device that allow as control the flow of "high power" by helps of a low power. To the Amplifier effect occurred, two things are necessary: source of energy and a device for controlling the flow of this energy - > the amplifier.

15. ### studiot AAC Fanatic!

Nov 9, 2007
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I would be happy to discuss my earlier post when I see that you have read and understood it.

However your response to various posts here suggests you still do not have a proper working model of a transistor.

Do you understand what I said about a tap (american = faucet) ?