Understanding Transistors: Amplifying AC Signals

[Bassalisk]

I am trying to understand transistors. Here is what I got.

Assume we have a forward biased transistor(emitter-base direct, collector-base inverse).
The point of transistor, in this mode, is to amplify a signal from base. That signal will vary the current in collector.

So if i have a AC source sending sinusoid signals to the base, assuming that sinusoid doesn't have very big amplitude, I will have as an output, an amplified sinusoid.

Is this case that amplifying of current in base to get very big currents in collector?

 

[dlgoff]

Is this case that amplifying of current in base to get very big currents in collector?

Yes.

where β can be called the current gain. The value of β is not highly dependable since it depends on I_c, V_ce, and the temperature.

For more reading on this, check out this fine explanation:

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html"

 

[Bassalisk]

Yes.



For more reading on this, check out this fine explanation:

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html"

Took me 2 WHOLE months to understand what the hell was getting amplified. Nobody in my material said that the POINT of transistor is to amplify small signals that come through base and in the end come out of collector.

Thank you very much.

 

[dlgoff]

You're very welcome.

 

[cabraham]

The best way to understand "reansistor action" is with the equation
Ic = alpha*Ie.
When the b-e jcn is forward biased, the emitter emits electrons (npn type of bjt) towards the base, & the base emits holes towards the emitter. The holes from the base recombine with electrons in the emitter.

But the electrons from the emitter incur only a very small number of recombinations in the base. Nearly all keep moving into the collector region. The b-c jcn is reverse biased, & its electric field attracts electrons from the b-e regions. Thus when forward biased, emitter electrons transit through the base & into the collector, becoming collector current.

But to emit electrons from the emitter, the b-e jcn must be forward biased. This requires base current Ib, & b-e voltage, Vbe. The amount of Ib needed for a given Ic is determined by the device current gain beta, i.e. Ic = beta*Ib.

The amount of Vbe needed for a given Ic is given by Ebers-Molls per
Ic = alpha*Ies*(exp((Vbe/Vt)-1), or subsequently Vbe = Vt*ln((Ic/Ies)+1).

In a nutshell, Ie, the emitter current, consists of emitted electrons. Nearly all of them transit through & beyond the base region, & enter the collector, becoming collector current. The number of collected electrons is determined by the number of emitted electrons.

In an amplifier, the signal source being amplified outputs a varying current & voltage. This results in a varying number of emitter electrons, & varying number of collected electrons. The bjt is not literally amplifying Ib, Vbe, or Ie. Rather these input variables determine the charges put into transit from the emitter which ultimately get collected. The transistor outputs a current in its collector, which is controlled by the emitter current, which is controlled by some signal source such as a transducer (microphone, photodetector), or generator (d/a converter, phono cartridge, Hall effect device).

Did I help?

Claude

 

[Bassalisk]

The best way to understand "reansistor action" is with the equation
Ic = alpha*Ie.
When the b-e jcn is forward biased, the emitter emits electrons (npn type of bjt) towards the base, & the base emits holes towards the emitter. The holes from the base recombine with electrons in the emitter.

But the electrons from the emitter incur only as very small number of recombinaztions in the base. Nearly all keep moving into the collector region. The b-c jcn is reverse biased, & its electric field attracts electrons from the b-e regions. Thus when forward biased, emitter electrons transit through the base & into the collector, becoming collector current.

But to emit electrons from the emitter, the b-e jcn must be forward biased. This requires base current Ib, & b-e voltage, Vbe. The amount of Ib needed for a given Ic is determined by the device current gain beta, i.e. Ic = beta*Ib.

The amount of Vbe needed for a given Ic is given by Ebers-Molls per
Ic = alpha*Ies*(exp((Vbe/Vt)-1), or subsequently Vbe = Vt*ln((Ic/Ies)+1).

In a nutshell, Ie, the emitter current, consists of emitted electrons. Nearly all of them transit through & beyond the base region, & enter the collector, becoming collector current. The number of collected electrons is determined by the number of emitted electrons.

In an amplifier, the signal source being amplified outputs a varying current & voltage. This results in a varying number of emitter electrons, & varying number of collected electrons. The bjt is not literally amplifying Ib, Vbe, or Ie. Rather these input variables determine the charges put into transit from the emitter which ultimately get collected. The transistor outputs a current in its collector, which is controlled by the emitter current, which is controlled by some signal source such as a transducer (microphone, photodetector), or generator (d/a converter, phono cartridge, Hall effect device).

Did I help?

Claude

Yes! So basically, with this current that is going in base, you control how much recombinations take place? e.g. make the current in base bigger, ergo more recombinations happen

But this lead me to a another question, how does saturation region takes it part in transistor? I vaguely understand what it is, I know that both E and C are forward biased.

 

[Studiot]

See also this thread

https://www.physicsforums.com/showthread.php?t=401308&highlight=transistor

 

[dlgoff]

But this lead me to a another question, how does saturation region takes it part in transistor? I vaguely understand what it is, I know that both E and C are forward biased.

A NPN transistor in saturation will still have a collector voltage a few tenths of a volt above emitter voltage and will provide a large current useful for "switch on" applications like this:

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/transwitch.html#c2"

 

[Bassalisk]

I will check those links out. Thanks, will come back with hopefully better understandings.

 

[Bassalisk]

From a circuit point of view saturation is rather like a water tap (faucet). There comes a point where you cannot turn in on any further.

Does this mean you push the current through base so much, that the resistor on the collector wire picks up so much current that it drops the voltage so the collector becomes forward biased?

 

[dlgoff]

Notice that the equation for determining the collector current from the base current in post 2 above using β is dependent on the base to emitter voltage, the collector current, and the temperature for any particular NPN transistor. The actual equation for determining the collector current is

.

So yes, there is a limiting collector current which is equal to the saturation current. Of course one could increase the potentials so as to "blow" the junctions; hence no collector current.

 

[cabraham]

Notice that the equation for determining the collector current from the base current in post 2 above using β is dependent on the base to emitter voltage, the collector current, and the temperature for any particular NPN transistor. The actual equation for determining the collector current is

.

So yes, there is a limiting collector current which is equal to the saturation current. Of course one could increase the potentials so as to "blow" the junctions; hence no collector current.

Well. the equation you quote, Ebers-Moll, is also very temp dependent. With a temp span of -55 to +125 C, beta will vary by a factor of 2 or more. With collector current, we can factor in another factor of 2 or more, & a factor of 3 or more w/ speciman. Hence beta can vary from 50 to 500 typical, a factor of 10.

In the E-M equation, "Ies" can vary by well over a million with temp alone. Ies also varies w/ speciman. So the Ebers-Moll equation for determining Ic in relation to Vbe is more unpredictable than the beta equation.

If you doubt me, then I suggest setting up a simple experiment on the bench. Place a bjt in a 0 deg C chamber. Then plot Ic vs. Vbe. Repeat at a temp of 25 C, then 75 C. Over this limited range, for a given Vbe, Ic will vary over a factor of hundreds to thousands.

This is why we avoid relying on Vbe or upon Ib to control Ic. But if we establish a specific Ie value, then Ic = alpha*Ie. The alpha factor is about 0.98 give or take 0.01 over the full temp range, full Ic range, & considering device variation. We control Ic w/ Ie because alpha is very predictable & stable.

Also you mentioned that the "actual equation for determining Ic" is Ebers-Moll. If a constant current source is connected to the base side of the base-emitter junction, then Ic = beta*Ib. In this configuration, Ic is determined solely by Ib. But we avoid this configuration due to beta dependency. But the same current source connected on the emitter side results in Ic = alpha*Ie. The result is very stable over full temp range w/ any device.

If we connected a voltage source across the b-e terminals, the Ic is determined by alpha*Ies*exp((Vbe/Vt)-1). But Ies is too unstable & the resulting Ic is less predictable than in the case w/ Ib & beta.

Also, driving Vbe w/ a voltage source is thermally unstable. The bjt will likely be incinerated. Thus the equation which is used to describe bjt action is Ic = alpha*Ie. We design using this relation. I've been an EE for 33 years & this is gospel.

Also, the E-M eqn you stated, does not hold for saturation condition. You gave the short 1-term eqn. Saturation requires the full 2-term eqn as follows:

Ic = alpha*Ies*exp((Vbe/Vt)-1) - Ics*exp((Vbc/Vt)-1).

Ies is the b-e jcn sat current, whereas Ics is that for the b-c jcn. Also the terms subtract. As forward bias on the b-e jcn increases, the b-c jcn voltage increases from negative to positive. Then the 2nd term in the equation subtracts from the 1st term. Both terms are increasing but they nearly cancel due to sign difference. Hence once the bjt enters saturation, increasing Ib/Vbe results in very little increase in Ic.

Make sense? BR.

Claude

 

[Bassalisk]

I think i got, give me a few more hours to draw you a picture of what I understand.

 

[Studiot]

Since I know you are an electrical/electronic (which?) engineering student this reply is for several of your current threads.

Since your stated aim is to get ahead in basic electronic stuff, including practical work I suggest you look at the following older book.

http://www.abebooks.co.uk/servlet/B...&bt.y=0&sts=t&tn=electronic+circuits+handbook

Be aware that there have been several more modern versions. All the more modern ones will lead you to do a lot of small microprocessor programming rather than the basics of electronics.

The above version is very cheap and all the components are basic and easy and cheap to obtain or substitute.
If you follow the projects section you will build (and understand) some useful test equipment - power supply - function generator - voltmeter etc.

OK back to the transistor thread.

Another fact you will learn is that (electronic) circuit analysis is a separate subject from physics. Obviously EEs have set things up to make analysis and calculation easy, rather than using physical characteristics such as carrier density and so on. In general they generate mathematical models instead.

Here is the simplest for transistors.

For the electronics I observe that the following are basic maths and physics, not models.

A transistor is a three terminal device. See attached sketch.
There are three possible terminal currents.
label these Ib, Ie, Ic
There are three possible interterminal voltages.
label these
Vbe, Vcb, Vce.

From basic physical laws

Vbe + Vcb + Vce = 0

Ib + Ie + Ic = 0


From basic maths we have six variables and two equations.
Therefore we have four independent variables and two dependent ones that can be determined from a knowledge of the others plus the two equations.


If we introduce models of the transistor action, such as Ebbers-Moll or others we can deduce further relationships between the variables to assist calculation.

The above is true for all models, but once we introduce a model, subsequent theory is only true for the model validity conditions.

In particular, Ebbers-Moll models the amplifying part of the action, as does the part below.

We can also define three ratios with these currents, however only two are independent, we can always get the third by cross cancelling.

Originally the two that were chosen were called alpha and beta, after the first two letters of the Greek alphabet. So

alpha = Ic/Ie

Beta = Ic/Ib

For a simple model of the non linear (switching) look here.

http://forum.allaboutcircuits.com/showthread.php?t=9146

 

[dlgoff]

Also, the E-M eqn you stated, does not hold for saturation condition. You gave the short 1-term eqn. Saturation requires the full 2-term eqn as follows:

Ic = alpha*Ies*exp((Vbe/Vt)-1) - Ics*exp((Vbc/Vt)-1).

Ies is the b-e jcn sat current, whereas Ics is that for the b-c jcn. Also the terms subtract. As forward bias on the b-e jcn increases, the b-c jcn voltage increases from negative to positive. Then the 2nd term in the equation subtracts from the 1st term. Both terms are increasing but they nearly cancel due to sign difference. Hence once the bjt enters saturation, increasing Ib/Vbe results in very little increase in Ic.

Make sense?

Makes sense to me. I only hope it makes sense to the OP at his present stage of learning.

 

[cabraham]

Another fact you will learn is that (electronic) circuit analysis is a separate subject from physics. Obviously EEs have set things up to make analysis and calculation easy, rather than using physical characteristics such as carrier density and so on. In general they generate mathematical models instead.
----------------------
If we introduce models of the transistor action, such as Ebbers-Moll or others we can deduce further relationships between the variables to assist calculation.

The above is true for all models, but once we introduce a model, subsequent theory is only true for the model validity conditions.

In particular, Ebbers-Moll models the amplifying part of the action, as does the part below.----------------------
Originally the two that were chosen were called alpha and beta, after the first two letters of the Greek alphabet. So

alpha = Ic/Ie

Beta = Ic/Ib

For a simple model of the non linear (switching) look here.

http://forum.allaboutcircuits.com/showthread.php?t=9146

Carrier density etc. is not necessary when operating the device in the active region at frequencies well below "ft", the transition frequency. The OP is in the learning stages of electronics & getting into carrier density, doping profiles, geometry, excess stored carriers, propagation time, diffusion capacitance, etc. is way too much.

What I gave was a most basic explanation of transistor action using charge carriers & fields present in the semicon material. This is the best approach at the early stages. Later on, when more advanced detail is needed, one can examine the charge control model, & then eventually, get into the realm of QM (quantun mechanics).

For the beginner, that is too deep. What I gave him, as well as others, is a good start. For a bjt in the saturation region, the best way to understand what is happening is the charge control model. But I don't know if the OP is adanced enough to explore that.

One thing I wanted to emphasize is that the E-M eqn (Ebers-Moll) has 2 terms, not just one. In the cutoff & active regions, the 1st term suffices. But in saturation, the b-c jcn is fwd-biased, so that the 2nd term is large enough & cannot be ignored. When discussing the bjt in the saturated state, the full 2-term expression for E-M must be used.

As far as E-M modeling "the amplifying part of transistor action", please keep in mind that a bjt (or a FET as well) provides BOTH current & voltage amplification. The beta equation describes the current gain per Ic = beta*Ib, i.e. the needed input base current swing for a given collector current swing.

The E-M eqn can be differentiated for small signal ac variation to yield ic = gm*vbe. Here gm = Ic/Vt. This relation describes the transconductance gain of the bjt, i.e. the amount of input voltage swing needed for a given collector current swing.

The current gain of the amp stage cannot exceed beta, the current gain of the raw device. Likewise the stage transconductance cannot exceed gm. Make sense?

Claude

 

[Studiot]

Claude, my comments were not meant as a criticism of your excellent presentation of the physical side of this.

But remember, all this models the transistor as a three terminal device.

Real world electronic circuit theory (needs to) use two port, four terminal networks. Bassalisk is at least a year from meeting these officially.

For circuit conceptual purposes I like to model a transistor as a water tap which is a device to control the flow of water by a mechanism ( the handle).

A transistor is a device to control the flow of electricity by an (electrical) mechanism.

This may be current or voltage in which case E-M may apply or it may be illumination/irradiation in which case E-M does not apply.

go well

 

[Bassalisk]

I understand it all well now. All makes sense. Thank you all.

 

[Bassalisk]

Carrier density etc. is not necessary when operating the device in the active region at frequencies well below "ft", the transition frequency. The OP is in the learning stages of electronics & getting into carrier density, doping profiles, geometry, excess stored carriers, propagation time, diffusion capacitance, etc. is way too much.

What I gave was a most basic explanation of transistor action using charge carriers & fields present in the semicon material. This is the best approach at the early stages. Later on, when more advanced detail is needed, one can examine the charge control model, & then eventually, get into the realm of QM (quantun mechanics).

For the beginner, that is too deep. What I gave him, as well as others, is a good start. For a bjt in the saturation region, the best way to understand what is happening is the charge control model. But I don't know if the OP is adanced enough to explore that.

One thing I wanted to emphasize is that the E-M eqn (Ebers-Moll) has 2 terms, not just one. In the cutoff & active regions, the 1st term suffices. But in saturation, the b-c jcn is fwd-biased, so that the 2nd term is large enough & cannot be ignored. When discussing the bjt in the saturated state, the full 2-term expression for E-M must be used.

As far as E-M modeling "the amplifying part of transistor action", please keep in mind that a bjt (or a FET as well) provides BOTH current & voltage amplification. The beta equation describes the current gain per Ic = beta*Ib, i.e. the needed input base current swing for a given collector current swing.

The E-M eqn can be differentiated for small signal ac variation to yield ic = gm*vbe. Here gm = Ic/Vt. This relation describes the transconductance gain of the bjt, i.e. the amount of input voltage swing needed for a given collector current swing.

The current gain of the amp stage cannot exceed beta, the current gain of the raw device. Likewise the stage transconductance cannot exceed gm. Make sense?

Claude

I think Claude should be promoted to a senior member. His knowledge and way of explaining has been most valuable to me, and I am sure to others.

As well as Studiot.

Thank you very much both.