Common Collector Circuit

[bigdogguru]

There is a big difference with the concept of FEEDBACK and NEGATIVE FEEDBACK. We are discussing NEGATIVE FEEDBACK

Perhaps, I should remind you, a closed loop gain equation of the form:

Av = G/(1+GH) ~ Gm*Re/(1+gm*Re)

Does indeed describe a negative feedback loop.

I will leave it up to you to reference the appropriate text on feedback theory.

BigDog

 

[LvW]

To understand Colin´s position (negation of negative feedback caused by Re) I strongly suspect that he still believes that the BJT is a current-controlled device (following the simple relation Ic=beta*Ib).
In this case, it is indeed not easy to understand the VOLTAGE feedback action of an emitter resistance. On the other hand, I don´t know how he is able to explain the stabilization of the dc operating point caused by Re. It would be interesting to hear about his approach to explain this effect.

 

[bigdogguru]

It would be interesting to hear about his approach to explain this effect.

Yes, I too would like to hear his explanation, as his admission of the existence of feedback, but not negative feedback is perplexing.

BigDog

 

[colin55]

Well, if you concede that feedback is involved, but still insist that there's no negative feedback, then what are you proposing? That it's positive feedback?:shock:

Well, if you concede that feedback is involved,

I have NEVER said there is FEEDBACK involved.
What I have said is the fact that there is a relationship between the resistance of the emitter resistor and the input impedance that is mainly determined by the gain of the transistor.
Feedback implies a signal taken from a section of a circuit is fed back to a previous section and this is not the case. It is just a relationship-law or a determination that involves the gain of the transistor.
And it is obviously not a NEGATIVE FEEDBACK relationship as found in the common emitter configuration nor is it a positive feedback so the term FEEDBACK should not be used.
AND IT IS NOT USED. It is only you that has brought up this absurdity.

 

[godfreyl]

...so the term FEEDBACK should not be used.
AND IT IS NOT USED. It is only you that has brought up this absurdity.

I didn't bring it up, LvW did. I just asked you to clariy what you wrote in post 19.

 

[LvW]

Feedback implies a signal taken from a section of a circuit is fed back to a previous section and this is not the case. It is just a relationship-law or a determination that involves the gain of the transistor.
And it is obviously not a NEGATIVE FEEDBACK relationship as found in the common emitter configuration nor is it a positive feedback so the term FEEDBACK should not be used.
AND IT IS NOT USED. It is only you that has brought up this absurdity.

OK - so I will explain to you this "absurdity" in detail:

* The bipolar transistor (BJT) - like the FET - is an active 4-pol with a current output and a differential voltage input (Vbe or Vgs, respectively).
The task of an emitter resistor Re (doesn´t matter if in CE or in CC configuration) is as follows: The output quantity (current) causes a voltage drop Ve across Re and, thus, partly determines the input quantity Vbe=Vb-Ve.
* This output-input retro-active effect is in full accordance with the definition of negative feedback.
* According to the feedback theory, important amplifier parameters are influenced by negative feedback (gain, bandwidth, input/output impedances, linearity, sensitivity to external influences). For ideal feedback conditions (pure voltage or pure current) these parameters are increased/decreased (depending on the kind of feedback: V-V, V-I, I-I or I-V) by the factor (1-loop gain).
* Surprisingly (?), this exactly applies to the Re influence because the gain of the feedback loop is Aloop=-gm*Re. This can be verified very easily using Black´s classical feedback model.
* Example: (a) Input resistance for Re=0: rin=rbe=h11 ; (b) for Re finite: rin=h11*(1+gm*Re)=h11+gm*h11*Re=h11+h21*Re (because gm=h21/h11). The last expression is a well known formula which can be found in all relevant textbooks. .

*How works a voltage follower (unity-gain amplifier) realized with opamps: 100% negative feedback cause the inverting input to follow the signal at the non-inverting input (with a negligible error caused by the finite opamp gain leading to a loop gain which is not infinite, as assumed for the calculations).
*What does a voltage follower realized with a BJT? Exactly the same !! (with an error larger than for opamp realizations because the loop gain is smaller).

It is really surprising that somebody who has dealt with transistors since a considerable time calls a proven fact like Re-feedback an "absurdity".

(By the way: Colin, very often, you are using the term "gain of the transistor". I like to mention that the transistor is a transconductance device which has no gain but a transconductance gm which describes the input-output relationship. I think, that is the reason you can speak only about "a relationship-law or a determination that involves the gain of the transistor" without giving a clear physical/technical verification).

- - - Updated - - -

Show me one book where it states the emitter resistor provides NEGATIVE FEEDBACK.
You are getting confused with the common emitter stage where the emitter resistor provides EMITTER DEGENERATION.

Only now I had the time and opportunity to have a look into "The art of Electronics" (Horowitz/Hill).
Here is what they say in chapt. 2-12 about the "absurdity":
"The emitter-degenerated amplifier itself uses a form of negative feedback.....So the voltage from base to emitter is the input voltage, minus a sample of the output (IE*RE). That´s negative feedback."

 

[godfreyl]

..."The art of Electronics" (Horowitz/Hill).
Here is what they say in chapt. 2-12...

That's specifically referring to emitter degeneration in a common emitter amplifier, though. I don't think Colin has any argument with that, only with applying the same logic to common collector amps.

IMHO, it's interesting to consider the question of feedback in a phase splitter circuit where outputs are taken from the emitter and the collector of the same transistor. i.e. the transistor is acting as common-emitter and emitter-follower at the same time. I don't think one could reasonably claim that one output uses feedback but the other doesn't. If one did, one would be hard pressed to explain how only the output at the collector has feedback, when it is the output voltage at the emitter which is fed back to the input.
:fight:

 

[The Electrician]

Here's what colin55 has to say about the subject on one of his own web pages. This is somewhat over half way down the page--it's the text associated with figure 57a. The web page is at:

http://www.talkingelectronics.com/p...nsistorAmplifier-P1.html#Emitter Degeneration

The text says:

"Fig 57a. In the Bridge Circuit, 4 resistors bias the transistor and Re is the EMITTER RESISTOR.
It is also a NEGATIVE FEEDBACK resistor and works like this:
When the voltage on the base rises by 10mV, the transistor turns on more and the current through the collector LOAD resistor Rc increases and the same current flows through the emitter resistor Re.
This causes a slightly higher voltage to appear across this resistor and the voltage on the emitter rises.
We have already discussed how to turn ON a transistor or turn OFF a transistor and when the voltage on the emitter increases, the transistor is turned OFF slightly. This means the 10mV rise on the base may be offset by a 2mV rise on the emitter and the transistor will not be turned on as much. This is the effect of NEGATIVE FEEDBACK."

On that same page, we find text associated with Figure 61a:

"In Fig 57a, we saw this as NEGATIVE FEEDBACK. This effect is also called EMITTER DEGENERATION as it reduces the gain of the stage. "

 

[LvW]

That's specifically referring to emitter degeneration in a common emitter amplifier, though. I don't think Colin has any argument with that, only with applying the same logic to common collector amps.

IMHO, it's interesting to consider the question of feedback in a phase splitter circuit where outputs are taken from the emitter and the collector of the same transistor. i.e. the transistor is acting as common-emitter and emitter-follower at the same time. I don't think one could reasonably claim that one output uses feedback but the other doesn't. If one did, one would be hard pressed to explain how only the output at the collector has feedback, when it is the output voltage at the emitter which is fed back to the input.
:fight:

Hi godfreyl - in response to your contribution I only can reply as follows:

Because the voltage Ve across an emitter resistance Re is identical to one of the voltages which form the differential voltage Vbe=Vb-Ve (which in turn controls Ic - that is:the cause of Ve) it does not matter at all if the whole circuit is operated in CE or CC configuration. It also does not matter if we use Ve as an output or not or together with a collector voltage.
In any case, an output quantity (current Ic) acts back to the input with a negative sign. And that´s exactly what we call "negative feedback".

Regarding your last sentence: To have negative feedback, it is not necessarily the used output quantity which acts back to the input. In CE configuration we use the collector voltage as an output, but nevertheless, we have negative feedback caused by Re.

To comment the confusion regarding the two terms "emitter degeneration" and "feedback":
FEEDBACK is a principle to be applied in some active systems - and "EMITTER DEGENERATION" is one of several methods to realize this principle. For transistor circuits there are some more methods like "collector feedback (R between C and B)" or "Two-stage feedback".

 

[godfreyl]

It also does not matter if we use Ve as an output or not or together with a collector voltage. In any case, an output quantity (current Ic) acts back to the input with a negative sign. And that´s exactly what we call "negative feedback".

Yes, agreed. That's pretty much the point I was trying to make, in a round about way, and apparently with a notable lack of success. :smile:

I just thought it may be a useful way of looking at things for those that struggle to see how an emitter follower can have feedback. Maybe I should rephrase it:

I think anyone would agree that a common emitter stage with an emitter resistor has feedback.

Now if we decide to take an output from the emitter as well as (or instead of) the collector, does the feedback suddenly disappear? Of course not - where we choose to take the output from has no effect on whether there's feedback or not.

 

[FvM]

It should be noted that a CE amplifier with emitter degeneration is implementing a different kind of feedback than the CC stage.

The former is current controlled voltage ("series-series") feedback while the latter is voltage controlled voltage ("parallel-series") feedback.

 

[LvW]

It should be noted that a CE amplifier with emitter degeneration is implementing a different kind of feedback than the CC stage.
The former is current controlled voltage ("series-series") feedback while the latter is voltage controlled voltage ("parallel-series") feedback.

Interesting remark. FvM - I know what you mean: The difference apparently comes from the different output definitions (resp. output nodes).
However, I am not quite sure if this formal definition may be applied to the present case. In both cases, the feedback signal is derived at the same node and is caused by the same quantity (voltage drop across Re caused by Ic resp. its changes). In this context one should also remember that the definition of the four feedback types is based on IDEAL conditions (ideal voltage and current sources).
In reality, we often have a mixture between two types of feedback.
Nevertheless, it´s worth to be discussed.

 

[FvM]

The original post asks about the reason for the high input impedance of a transistor amplifier in CC configuration. The answer is voltage or xx-series feedback and applies also to CE stage with emitter degeneration. In so far it's O.K. to refer to the latter in many previous posts. Generally (negative) voltage feedback increases the input impedance.

The difference, voltage versus current controlled (parallel-xx versus series-xx) feedback matters for the output impedance. Emitter degeneration increases the CE output impedance while voltage controlled feedback causes the low output impedance of the CC stage.

This general effect of the different feedback schemes can be seen also in a "real" circuit.

 

[goldsmith]

Hi and greetings to the all Gentlemen which are contributing in here !
I promised to deliver some textbooks which are dealing with the concept of negative feedback in CC amp but i see the other dudes did it hence i see , it's not required that i do it too .

This thread is very interesting because i see a lot of good explanations ! ( but i hole the original poster is not being confused ) . to answer him/her , i deliver a summarize : because of the Negative feedback , the out put impedance is reduced and the input impedance is increased ( concept of negative feedback says that , negative feedback can affect on voltage gain and current gain and input resistance and out put resistance and of course bandwidth )

Best Wishes and Best regards to all
Goldsmith

 

[LvW]

The original post asks about the reason for the high input impedance of a transistor amplifier in CC configuration. The answer is voltage or xx-series feedback and applies also to CE stage with emitter degeneration. In so far it's O.K. to refer to the latter in many previous posts. Generally (negative) voltage feedback increases the input impedance.

Yes- of course 100% agreement.

The difference, voltage versus current controlled (parallel-xx versus series-xx) feedback matters for the output impedance.

Again agreed.

Emitter degeneration increases the CE output impedance while voltage controlled feedback causes the low output impedance of the CC stage.

I would say: ...causes the lowering of the output impedance.
What is the output impedance of the CC stage without voltage controlled feedback (Re=0)? Difficult to answer, I think, because in this case, the emitter is grounded.
Hence, the only statement we can make is:
The output resistance of a CC stage is rout=(1/gm)||Re.
That means: The output resistance at the emitter node (1/gm) is further reduced due to the feedback "network" (in parallel to 1/gm) which consists of Re only.
It is obvious that the reduction factor is much smaller than the "ideal" factor (1-loop gain) because we have no ideal voltage-controlled feedback (the source of the feedback voltage is not ideal, but has an "inherent" source resistance Re). On the other hand - improving the quality of this feedback source (Re smaller), obviously further reduces the total output resistance and confirms the feedback rules.

The above considerations may show why I have qualified FvM`s reply (post#31) as "worth to be discussed".

 

[FvM]

I would say: ...causes the lowering of the output impedance.

Agreed.

I would say: ...causes the lowering of the output impedance.
What is the output impedance of the CC stage without voltage controlled feedback (Re=0)? Difficult to answer, I think, because in this case, the emitter is grounded.

I prefer a different viewpoint. The feedback is not provided by Re, it also works with a current source load. The feedback is constituted by the way how the transistor is connected to the load, which results in Vbe = Vin-Vout. In so far, a CC stage without feedback isn't possible.

In my view, the CC output impedance "without feedback" would be equal to the output impedance of a transistor in CE configuration.

 

[LvW]

Agreed.

I prefer a different viewpoint. The feedback is not provided by Re, it also works with a current source load. The feedback is constituted by the way how the transistor is connected to the load, which results in Vbe = Vin-Vout. In so far, a CC stage without feedback isn't possible.

Yes - the same applies to another unity-gain circuit: An opamp with unity gain (at low frequencies) without negative feedback is also not possible.

In my view, the CC output impedance "without feedback" would be equal to the output impedance of a transistor in CE configuration.

But at which node is this CC output impedance (without feedback) defined ?

 

[FvM]

But at which node is this CC output impedance (without feedback) defined ?

I think the question is in so far hypothectical as a CC stage without voltage feedback can't be realized. In the equivalent circuit of a CC stage, we have the BJT model comprised of rbe, gm and rce. If the input voltage is applied without feedback, directly to rbe, the output impedance is rce, or infinite if rce is ignored.

 

[LvW]

You are completely wrong in your reasoning.
I have been teaching electronics for 40 years and written 25 books with sales of over 750,000 copies. My website has reached 22,000,000 visitors.
No-one has yet said I have been wrong.

Hi Colin55, since I am interested in some of the mentioned 25 books, I started a search on Amazon´s site - but I failed.
Not a single publication from you could be found.
Please, can you tell me where I can find these books? Are they in printed form? Which publisher?

 

[colin55]

Hi Colin55, since I am interested in some of the mentioned 25 books, I started a search on Amazon´s site - but I failed.
Not a single publication from you could be found.
Please, can you tell me where I can find these books? Are they in printed form? Which publisher?

I gave you my web address but the moderator removed it because it was SPAM.

All my books have now sold out. Some are on my website as .pdfs