Emitter Follower Help: Understand Physics & Art of Electronics

[perplexabot]

Hi all. So I have been trying to understand how a emitter follower (common collector) works. But I can't seem to grasp the physics behind it. I know it is used as a buffer, it has high input impedance and low output impedance. I don't know what properties cause the output voltage to follow the input voltage. Also if you have "The Art of Electronics," on page 67 where it says "Important points about followers," I can't understand why clipping occurs for the signal. For some reason this whole thing is kind of vague to me. Please help. Thank you.
http://upload.wikimedia.org/wikipedia/commons/b/b8/NPN_emitter_follower.svg

 

[sophiecentaur]

Hi
Without any Maths, you can say that current will pour into the base and, hence, through the transistor (current amplification factor of at least several tens) from C to E, until Vbe is around 0.7V. When that happens, Vout is 0.7V below Vin. If you vary Re, the current through the transistor will change to re-establish the 0.7V Vbe. There is so-called Voltage Feedback because any small change in Vout will cause a change in Ib and hence a large change in Ie - and this will maintain Vout to be what it was. This gives an effectively low output impedance - as the Volts are independent of the load resistance. If Vin is raised, the current that will be drawn through the base will not change appreciably because Ve 'chases Vb', remaining 0.7V below. So the current into the base of the Emitter follower will not change as Vin increases - which is what you would get with a very high resistance - hence the high input resistance.
Why does it 'limit'? If you try to take Vb too high then the transistor will saturate and no more current can pass through Re, so the volts cannot increase. The feedback mechanism runs out. Vbe need only increase by a miniscule amount beyond this and the forward biased be diode will pass loads of current. The input impedance will drop as Vb increases until it approaches that of a forward biased diode in series with Re.
There you are - no Maths and plenty of arm waving. Does it help?

 

[Ratch]

perplexabot,

But I can't seem to grasp the physics behind it.

The physics are the same as any BJT. Do you mean the topology of the common collector?

I know it is used as a buffer, it has high input impedance and low output impedance.

Its use won't help you understand how it works.

I don't know what properties cause the output voltage to follow the input voltage.

The property of any transistor in the active region. The emitter terminal voltage is around 0.7 volts different than the base terminal. So the emitter follows the base or vice versa voltage wise.

Also if you have "The Art of Electronics,

I don't, but clipping occurs whenever the input is overdriven or the operating point is chosen badly.

Ratch

 

[sophiecentaur]

It isn't clear at what level he's asking the question. I think the confusing phrase is "the physics behind it". Is it a real Physics question or a simple application question? I interpreted it as the latter. Very little 'Physics' is needed beyond V=IR, in a first stab at transistors. (Plus remembering the rules of how they behave).

 

[Jony130]

Emitter follower is a very simply circuit. The output voltage is always 0.6V lower the the input voltage.
See some examples

Also notice that the base current is (β+1) smaller then emitter current (load current).
So our base current source B1 see our load ( Re resistor) not as 100Ω resistor. But B1 see (β+1)*Re load.
Here you have anther example.
Spouse we have a 1K resistor voltage divider supply from 10V battery.
Without any load connect to the output terminal of our voltage divider the output voltage is equal 5V. Now we connect a 100Ω load resistor across the output terminal.
And now our voltage divider output voltage drops from 5V to 0.83V. So we ruin our circuit.
To fix this issue we add a buffer (emitter follower) see the diagram

Now I hope that you see why we say that emitter follower has a high input impedance and low output impedance. It's all thanks to BJT and his current gain.

 

[perplexabot]

Thank you all for your great replies.

There is so-called Voltage Feedback because any small change in Vout will cause a change in Ib and hence a large change in Ie - and this will maintain Vout to be what it was. This gives an effectively low output impedance - as the Volts are independent of the load resistance.

Why will a change in Vout cause a change in Ib? Why would Vout change in the first place? Isn't Ib the one that is being "controlled" and so is subject to change?

I don't, but clipping occurs whenever the input is overdriven or the operating point is chosen badly.

May you please clarify this phrase? I don't understand what you mean by "overdriven or the operating point is chosen badly." Do you mean if Ib is low enough to cause Vbe to be lower than .7V then clipping shall occur?

See some examples

For the first image, second figure, the current through Re, should it be 54mA?

 

[perplexabot]

I still don't understand page 67 of "The Art of Electronics" under "Important points about followers," number 1. If anyone can help that would be great.

 

[Jony130]

For the first image, second figure, the current through Re, should it be 54mA?

Yes, my mistake.

As for the clipping. The clipping can occur when transistor change his state eg. conduction to cut-off. See the example form The Art of Electronics
I slightly change the diagram

And ask yourself one question. What voltage you need to provide to open BJT in this circuit. Also look at figure 2.9 in Art of Electronics.

Why will a change in Vout cause a change in Ib? Why would Vout change in the first place?

Vout can change for example if you change load resistance.

 

[perplexabot]

Thank you for your help.

Yes, my mistake.

As for the clipping. The clipping can occur when transistor change his state eg. conduction to cut-off.

Cut-off happens at voltages less than .7v?

And ask yourself one question. What voltage you need to provide to open BJT in this circuit.

Is the answer .7V ?

Vout can change for example if you change load resistance.

That was what I was thinking, but I wasn't sure. Thanks.

On page 67 of "The Art of Electronics" it says:
The output can swing to within a transistor saturation voltage drop of Vcc, but it cannot go more negative than -5 volts. That is because on the extreme negative swing, the transistor can do no more than turn off, which it does at -4.4 volts input (-5V output).

I don't understand why clipping occurs at -5V (or -4.4V input) when cut-off happens at .7V. In other words shouldn't clipping occur at .7V?

 

[Jony130]

Thank you for your help. Cut-off happens at voltages less than .7v? Is the answer .7V ?That was what I was thinking, but I wasn't sure. Thanks.

On page 67 of "The Art of Electronics" it says:
The output can swing to within a transistor saturation voltage drop of Vcc, but it cannot go more negative than -5 volts. That is because on the extreme negative swing, the transistor can do no more than turn off, which it does at -4.4 volts input (-5V output).

I don't understand why clipping occurs at -5V (or -4.4V input) when cut-off happens at .7V. In other words shouldn't clipping occur at .7V?

No 0.7V Is a Vbe voltage not the base voltage or load voltage.

In this imagine without BJT the Ve voltage is equal 5V so to open the BJT Vb voltage must be higher then 5.6V. So If input voltage is greater than 5.6V the BJT is active mode and the voltage across the load resistor is Vout = Vin - Vbe. But if Vb is smaller than the 5.6V the BJT is in cut-off stage. And the Ve voltage is constant 5V. Input voltage don't affect the VE voltage any more because BJT is cut-off.

 

[perplexabot]

No 0.7V Is a Vbe voltage not the base voltage or load voltage.

In this imagine without BJT the Ve voltage is equal 5V so to open the BJT Vb voltage must be higher then 5.6V. So If input voltage is greater then 5.6V the BJT is active mode and the voltage across the load resistor is Vout = Vin - Vbe. But if Vb is smaller than the 5.6V the BJT is in cut-off stage. And the Ve voltage is constant 5V. Input voltage don't affect the VE voltage any more because BJT is cut-off.

Thank you so much that makes so much sense. Continuing on your example (assuming Ve is 5v when BJT is off), if Vin is 7v then Ve will be 6.3V, is that correct? Also does it saturate at 20V?

 

[Jony130]

Thank you so much that makes so much sense. Continuing on your example, if Vin is 7v then Ve will be 6.3V, is that correct? Also does it saturate at 20V?

Yes for Vb= 7V The output voltage Ve ≈ 6.3V
As for the saturation for Vb = 20V bjt is almost enter the saturation region.
To enter saturation Vb must be higher the Vcc.

Ok So now if we change the GND position as you have in Art of Electronics

I hope that now you see that without BJT we have VE = -5V. And now if you want to open the BJT the Vb voltage must we greater (lees negative) than -4.4V. If Vb is lower (more negative) than -4.4V the BJT is in cut-off mode.

 

[perplexabot]

Yes for Vb= 7V The output voltage Ve ≈ 6.3V
As for the saturation for Vb = 20V bjt is almost enter the saturation region.
To enter saturation Vb must be higher the Vcc.

Ok So now if we change the GND position as you have in Art of Electronics

I hope that now you see that without BJT we have VE = -5V. And now if you want to open the BJT the Vb voltage must we greater (lees negative) then -4.4V. If Vb is lower (more negative) the -4.4V the BJT is in cut-off mode.

Truly amazing explanation. I finally understand. Thank you so much.
Sorry for my many questions. With the BJT in cut-off mode, can it be treated as an open circuit (or infinite resistance)?

 

[Jony130]

With the BJT in cut-off mode, can it be treated as an open circuit (or infinite resistance)?

You can even delete BJT from the diagram.

 

[perplexabot]

You can even delete BJT from the diagram.

That simplifies a lot.

Once again thank you so much, you have been a great deal of help.

 

[perplexabot]

How did you solve for Vb or Ib in the second circuit?

 

[Jony130]

I use a circuit theory and solve the circuit.

From the II Kirchhoff's law we can write

Vcc = I1*R1 + I2*R2 (1)

I1 = Ib + I2 (2)

I2*R2 = Vbe + Ie*Re (3)

And Ib = Ie/(β+1) (4)

And we can solve this for Ib.

\LARGE {I_b = \frac{R2Vcc - Vbe(R1+R2)}{(\beta +1)Re(R1+R2) + (R1R2)}}= 419.047619\mu A I assume β = 99 and Vbe = 0.6V

Knowing Ib we can easy solve for Ie. Ve and Vb

But there is also a simpler way to solve this circuit by using thevenin's theorem.
We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit


Vth = Vcc * R2/(R1+R2)

Rth = R1||R2 = (R1*R2)/(R1+R2)

And now we can solve for Ib

Vth - Ib*Rth - Vbe - Ie*Re = 0

And we also know that

Ie = Ib*(β +1)

so we end up with


Vth - Ib*Rth - Vbe - Ib*(β +1)*Re = 0

\LARGE Ib = \frac{Vth-Vbe}{Rth+(\beta +1)*Re}

And this is the end.

 

[perplexabot]

Most of what you wrote made sense but I do have 3 questions.

From the II Kirchhoff's law we can write

Vcc = I1*R1 + I2*R2 (1)

I1 = Ib + I2 (2)

I2*R2 = Vbe + Ie*Re (3)

And Ib = Ie/(β+1) (4)

How did you get equation (1)? Is Vcc = 10V, if Yes, then I think I get it.

But there is also a simpler way to solve this circuit by using thevenin's theorem.
We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit


Vth = Vcc * R2/(R1+R2)

Rth = R1||R2 = (R1*R2)/(R1+R2)

I am familiar with thevenin's theorem, but I don't know how you got Vth = Vcc * R2/(R1+R2)? Once again if Vcc = 10V, then I think I understand.

And now we can solve for Ib

Vth - Ib*Rth + Vbe + Ie*Re = 0

Isn't this supposed to be: Vth - Ib*Rth - Vbe - Ie*Re = 0 ?
Or are you assuming the emitter current going from ground to transistor?

I am sorry if I am asking for too much. You are helping me a lot.

 

[Jony130]

Vcc is a supply voltage Vcc = 10V in our example.

Isn't this supposed to be: Vth - Ib*Rth - Vbe - Ie*Re = 0 ?
Or are you assuming the emitter current going from ground to transistor?

No I made a typing error. Point for you.
I edit my post and correct the error.

 

[perplexabot]

Thank you.

I have one last question for you (I hope). For the image above (top figure), did you just "copy/paste" Vin from collector to ground?

 

[Jony130]

I don't understand you question.
In our case Vin = Vcc = 10V. But sometimes Vin has a different value than supply voltage.

 

[perplexabot]

In this figure you have only one battery

In the top figure you have two batteries. How did you get the second one? Or are these two figures NOT equivelent?

 

[Jony130]

This two circuit are equivalent only when Vin = Vcc.
And yes I just just "copy/paste" Vin from collector to ground.
But also kept in mind that this second circuit with two batters is more universal for education purpose. Because Vin need not be equal to Vcc (supply voltage).

 

[perplexabot]

This two circuit are equivalent only when Vin = Vcc.
And yes I just just "copy/paste" Vin from collector to ground.
But also kept in mind that this second circuit with two batters is more universal for education purpose. Because Vin need not be equal to Vcc (supply voltage).

Thank you very much for your help.

 

[sophiecentaur]

Thank you all for your great replies.
Why will a change in Vout cause a change in Ib? Why would Vout change in the first place? Isn't Ib the one that is being "controlled" and so is subject to change?

This is basic approach to feedback theory.
An emitter follower is a circuit with Feedback. That accounts for the way that the output impedance is a lot lower than the value of the emitter load Re. If you want to analyse how feedback works (this is a general principle) you introduce an error signal and see the effect of the circuit on correcting this error. A reason that Vout could change could be that Re changes (or even that the current gain changes a bit). The feedback mechanism (involving the very high current gain of the transistor) corrects this by adjusting the Emitter current and bringing it back to (very near) what it was. When an output voltage is controlled to be more or less independent of the load, this is known as a voltage source or low impedance source. This basic approach may be unfamiliar to you but it is the way that we can see how circuits can be linearised, their frequency response flattened and their input and output impedances controlled.

You can use the methods described elsewhere in this thread to analyse what the circuit will do and they will give answers but you would need to do the sums, piecewise, for a range of values of Re, say. Approaching it as a feedback problem, you get the effective output impedance (etc.) irrespective of the value of Re. The same approach is used, of course, in amplifiers using Op Amps and simplifies many amplifier circuit problems significantly by omitting many of the less relevant circuit values.

 

[perplexabot]

This is basic approach to feedback theory.
An emitter follower is a circuit with Feedback. That accounts for the way that the output impedance is a lot lower than the value of the emitter load Re. If you want to analyse how feedback works (this is a general principle) you introduce an error signal and see the effect of the circuit on correcting this error. A reason that Vout could change could be that Re changes (or even that the current gain changes a bit). The feedback mechanism (involving the very high current gain of the transistor) corrects this by adjusting the Emitter current and bringing it back to (very near) what it was. When an output voltage is controlled to be more or less independent of the load, this is known as a voltage source or low impedance source. This basic approach may be unfamiliar to you but it is the way that we can see how circuits can be linearised, their frequency response flattened and their input and output impedances controlled.

You can use the methods described elsewhere in this thread to analyse what the circuit will do and they will give answers but you would need to do the sums, piecewise, for a range of values of Re, say. Approaching it as a feedback problem, you get the effective output impedance (etc.) irrespective of the value of Re. The same approach is used, of course, in amplifiers using Op Amps and simplifies many amplifier circuit problems significantly by omitting many of the less relevant circuit values.

Hello again. I have dealt with feedback before in op-amps, however the idea was hard for me to think about. I can understand the fact that feed back "corrects" changes of Vout so that it goes back to what it was but I don't know how this happens. That is why I am happy that you brought this subject up. When I dealt with op-amps, I treated them as black boxes and didn't think about how this can happen. Here is my reasoning of why feedback occurs in a transistor, please tell me if I am right/wrong or am missing something:

Vout changes due to noise, or a change in Re or some other reason and so Ie changes and since Ib = Ie + Ic, Ib will change as well, the change in Ib will change Vbe which in turn increases/decreases the Base-Emitter depletion region width causing more/less carries to pass through which changes Vout back to what it was.

What do you think?

 

[Jony130]

Vout changes due to noise, or a change in Re or some other reason and so Ie changes and since Ib = Ie + Ic, Ib will change as well, the change in Ib will change Vbe which in turn increases/decreases the Base-Emitter depletion region width causing more/less carries to pass through which changes Vout back to what it was.

Your description is quite correct except Ib equation.
In BJT Ie = Ib + Ic and Ic = Ib*β
So we have
Ie = Ib + Ib*β = Ib *(β+1)
And we solve for Ib
Ib = Ie/(β+1)

Also you can look at this circuit that Vbe is a cause of a Ib changes.
If Vout changes and Vin remains unchanged the Vbe also change because
Vbe = Vin - Vout. And change in Vbe Ib also change.

And in deep op amp analysis you can find here
https://www.physicsforums.com/showpost.php?p=3866238&postcount=3
I recommend you to read this

 

[perplexabot]

Thank you for the reply and the informative link.

In the link you say:

Now suppose the output voltage tried to drift in a positive direction. Can you see that this positive change would be felt through R2 and would cause the inverting pin (-) of the op amp to become slightly positive Since essentially no current flows in or out of the op amp input and the (+) input of the op amp is at ground potential. This causes VD to be greater than 0 with the (-) terminal being the most positive.

Shouldn't VD be LESS than 0? Because VD = V"+" - V"-" where V"+" = 0 and V"-" is positive so VD = 0 - (POSITIVE #) = a negative VD < 0

 

[Jony130]

Shouldn't VD be LESS than 0? Because VD = V"+" - V"-" where V"+" = 0 and V"-" is positive so VD = 0 - (POSITIVE #) = a negative VD < 0

Well you right VD is negative, but this description corresponds to this drawing
https://www.physicsforums.com/attachment.php?attachmentid=46245&stc=1&d=1334484384
But as you can see both of these methods gives the correct (the same) answer.
Negative VD forces the output voltage to decrease its value.

 

[perplexabot]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

 

[Jony130]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

I don't know, you must ask some physics. I'm not interested in this subject. And it is irrelevant from the application point of view.

 

[perplexabot]

Thank you for all your help.

Anyone?

 

[perplexabot]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

A change in Ib causes a change of carriers in the Base region, this causes an attraction/repulsion of the carriers in the Emmiter region, which causes the Base-Emitter depletion region to change width, this width correlates to a change in Vbe, but why?

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

I still don't know the answer to this?

 

[Ratch]

perplexabot,

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

A BJT is a voltage controlled current source. Whatever voltage you put on the base-emitter will determine the current of the collector. The voltage input/current output is exponential, so designers use a lot of resistance in the base to swamp out the exponential variation. That is why they use a current source (high resistance) in the base to drive the transistor. A lot of folks think that a BJT is controlled by the base current, but it is not. It is controlled by Vbe in an exponential way according to Schockley's diode equation. The base current is also controlled by Vbe in a exponential manner. So the base current is related to the collector current is a somewhat linear manner through Vbe, but it is Vbe that is controlling both. So the base current is an indicator of the collector current, but does not control it. The reason for the exponential relationship is that a BJT operates by diffusion, whereas a FET and a vacuum tube operates by electrostatic means.

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

The emitter-base junction is just like a diode in that if you put too much voltage across it, it will burn out. That will happen before you can close the depletion region.

Ratch

 

[perplexabot]

perplexabot,

A BJT is a voltage controlled current source. Whatever voltage you put on the base-emitter will determine the current of the collector.

I think I am lost, doesn't the voltage drop across base-emitter have to be .7? Just like a diode? So if you apply a voltage of 3V, .7V will drop across the Base-Emitter, but happens to the other 2.3V?

The emitter-base junction is just like a diode in that if you put too much voltage across it, it will burn out. That will happen before you can close the depletion region.
Ratch

By "burn out," does it dissipate heat until it literally burns itself out?

 

[Ratch]

perplexabot,

Originally Posted by perplexabot
Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

A change in Ib causes a change of carriers in the Base region, this causes an attraction/repulsion of the carriers in the Emmiter region, which causes the Base-Emitter depletion region to change width, this width correlates to a change in Vbe, but why?

Are you answering your own questions now? The Ib does not cause the depletion region to change, the Vbe does. Ib change is just a consequence of the Vbe change.

Originally Posted by perplexabot
Also as you increase the forward biased voltage across the Base-Emitter depletion region

the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

I still don't know the answer to this?

I believe that question was answered.

I think I am lost, doesn't the voltage drop across base-emitter have to be .7? Just like a diode? So if you apply a voltage of 3V, .7V will drop across the Base-Emitter, but happens to the other 2.3V?

You will burn out the diode if you apply 2.3 volts to it. If you apply 2.3 volts to a diode in series with a resistor, and do not exceed the current rating of the diode, then the voltage across the diode will be around 0.7 volts for a silicon diode, and 0.4 for a germanium diode. You should study Schlockey's ideal diode equaton.

By "burn out," does it dissipate heat until it literally burns itself out?

Correct.

Ratch

 

[perplexabot]

Thank you for your help.

perplexabot,
Are you answering your own questions now?

No need for ridicule.

Since Ib doesn't control Vbe then I am assuming my previous post is wrong:

Here is my reasoning of why feedback occurs in a transistor, please tell me if I am right/wrong or am missing something:

Vout changes due to noise, or a change in Re or some other reason and so Ie changes and since Ib = Ie + Ic, Ib will change as well, the change in Ib will change Vbe which in turn increases/decreases the Base-Emitter depletion region width causing more/less carries to pass through which changes Vout back to what it was.

Please help me fix my reasoning!

 

[perplexabot]

I just edited my above post, I think i should have made a new post. Sorry for the pointless post, can anyone help?

 

[sophiecentaur]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

Also as you increase the forward biased voltage across the Base-Emitter depletion region the depletion width decreases, right? Well, can you increases the voltage so much that the depletion width is effectively gone (is this even a valid question?) ?

Your original question about the emitter follower is covering three areas. It covers the (Physics of the) internal operation of the transistor, it covers the use of the current amplification in a circuit (circuit theory) and it also covers Feedback. All three of these are pretty vast subjects and best discussed separately. You may get on better if you can break your thinking up into black boxes, which is what Engineers have to do all the time. It is lucky that electronics is very suitable for this approach but also it can be a problem when people know nothing of one particular area.

 

[perplexabot]

Your original question about the emitter follower is covering three areas. It covers the (Physics of the) internal operation of the transistor, it covers the use of the current amplification in a circuit (circuit theory) and it also covers Feedback. All three of these are pretty vast subjects and best discussed separately. You may get on better if you can break your thinking up into black boxes, which is what Engineers have to do all the time. It is lucky that electronics is very suitable for this approach but also it can be a problem when people know nothing of one particular area.

Hi, thanks for the reply. I am assuming you are suggesting that I treat the transistor as a black box and not think of the internal operation due to my lack of knowledge of the physics behind it. Now I am not disagreeing with you, I do not have the knowledge a physicist has, but I would definitely like to have an idea of how Feedback works, specifically for a transistor, since it is the building block of almost everything else. If the discussion of this topic is too vast, please refer me to a link.

Thanks again.

 

[dlgoff]

Another question about a BJT, how does a change in Ib produce a change in Vbe (on a microscopic level)?

Here's a link that I think you'll find interesting.

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

A base emitter voltage of about 0.6 v will "turn on" the base-emitter diode and that voltage changes very little, < +/- 0.1v throughout the active range of the transistor...

 

[perplexabot]

Here's a link that I think you'll find interesting.

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

Thank you, that is something that will keep me busy for a while.

 

[sophiecentaur]

Hi, thanks for the reply. I am assuming you are suggesting that I treat the transistor as a black box and not think of the internal operation due to my lack of knowledge of the physics behind it. Now I am not disagreeing with you, I do not have the knowledge a physicist has, but I would definitely like to have an idea of how Feedback works, specifically for a transistor, since it is the building block of almost everything else. If the discussion of this topic is too vast, please refer me to a link.

Thanks again.

No, not at all because I don't know how much you know. What I am suggesting is that you can choose not to bother with the Physics when designing a circuit (except for very clever stuff that is pushing the envelope). You can use a whole hierarchy of approximations, starting with a very few parameters to describe how the transistor actually behaves in a circuit. When you are concerned with detailed frequency response and linearity, you may find it worth while but that is way down the line.
Also, where feedback is concerned, the basic feedback formulae often consider very high gain devices (Op Amps are often assumed to have infinite gain) and are very simple. But even a single, modern BJT exhibits high enough gain to treat it as 'infinite' when making a first stab at circuit design. Basically, in many circuits with feedback, the performance is 'defined' by (some of) the surrounding Resistors and Capacitors rather than the performance of the active devices.
Just splitting the problem up into three, will make the whole thing much easier to grasp. The whole of Engineering is based on 'just enough' or 'appropriate' accuracy. (Same thing applies to cost, too.)
Life's hard enough!

 

[Ratch]

perplexabot,

Since Ib doesn't control Vbe then I am assuming my previous post is wrong:

Correct, Ib doesn't control Vbe, but it tracks Vbe in a one to one exponential relationship according to Schockley's diode equaton. If you drive a diode with a current, the voltage across it will automatically set itself to the correct value to conform to Schockley's equation.

Here is my reasoning of why feedback occurs in a transistor, please tell me if I am right/wrong or am missing something:

Vout changes due to noise, or a change in Re or some other reason and so Ie changes and since Ib = Ie + Ic, Ib will change as well, the change in Ib will change Vbe which in turn increases/decreases the Base-Emitter depletion region width causing more/less carries to pass through which changes Vout back to what it was.

Please help me fix my reasoning!

Feedback occurs because the output (emitter) has some voltage and current in common with the input (base). If you want to get involved further, you should start a new thread called "feedback" or something like that. Ie = Ic + Ib, not the way you have it written.

Ratch

 

[perplexabot]

Thank you all for your wonderful help. I will need to read up on transistor feedback or maybe start a new thread in the near future as Ratch suggested.

 

[sophiecentaur]

Here is my reasoning of why feedback occurs in a transistor

You are right to say that feedback occurs in a transistor (the Capacity between Collector and Base, for instance, will limit the high frequency gain, for instance). BUT the feedback that is at work in a basic emitter follower is not "in the transistor". The feedback is there because of the presence of the emitter resistor. It would be there for a totally ideal BJT, with a 'step function' for the junction characteristic and infinite current gain. The feedback is due to the position of the transistor in the circuit and is due to the volts that develop across the emitter load. This distinction is very relevant and I'm not sure you are taking it on board (please correct me if I'm wrong). Think of the transistor as a not-very-good version of an ideal device. The way the circuit behaves is affected very little by the details of the transistor's performance. Whenever possible, circuit design is based on the values of the passive components and assumes that they will compensate for the details of the inadequacies of the active components. Just look at a dozen designs of (linear) transistor amplifier design. You will see that there is nearly always a load resistor (or current source) in the emitter circuit. This is to introduce feedback and to remove the effect of the non-linearity in the base current. I strongly urge you to revisit Op Amp circuits with feedback and to figure out how the various forms of feedback work. That all segues nicely into the way discrete 'real' transistors are operated and the way feedback is applied in non-ideal devices.

There is a parallel with the way we tend to treat passive devices like resistors and capacitors. Every resistor you pull out of the drawer is, in fact, a complicated combination of resistance, inductance and capacitance (revealed in its very high frequency performance) but we ignore that in nearly every case and blithely rely on V=IR to tell us how it will work. We don't get down to the smartarse level until circumstances drive us to it - or when the circuit starts playing us up!

 

[perplexabot]

The feedback is due to the position of the transistor in the circuit and is due to the volts that develop across the emitter load.

I need to give that more thought.

Thank you for your correction and the great info. I will be reading about BJTs, OP-AMPs and feedback for the next couple of weeks. Hopefully by then I will be in a better state than I am now.

 

[sophiecentaur]

I need to give that more thought.

This borders on Philosophy - so beware of brain ache!

 

[perplexabot]

This borders on Philosophy - so beware of brain ache!

Let the brain ache begin.

 

[perplexabot]

Hello again. I'm not sure If I should post my Emitter Follower question here or make a new thread. I will try posting it here first. So in "The Art of Electronics" it says the gain for an emitter follower is Gv = RL/(RL + re), where RL is the resistance of the load and re is the impedance looking back into the emitter. Isn't the gain Vout/Vin where Vout is the voltage across the Load and Vin is the voltage applied at the base? It seems as though that is not what is going on here. Please tell me how that equation they got that equation.

My approach was (which is obviously wrong) Vo/Vi = [Rs/(hfe + 1) + re] / [RL(hfe + 1) + re]
Where Rs is the Impedance of the source.

Thank you for your help.