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Laplace

Apr 4, 2010
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The current Ib plays no role at all in this feedback scheme.
The base current Ib is controlled by the collector current in this feedback scheme. The collector current flows through the emitter resistor to develop the emitter voltage. The emitter voltage plus one diode drop (Vbe) subtracted from the Thevenin equivalent bias voltage, divided by the Thevenin equivalent bias resistor gives the base current. So, Yes - I need to see your explanation.
 

LvW

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The base current Ib is controlled by the collector current in this feedback scheme.

The base current Ib is "controlled by the collector current" ?
In post#61 you wrote exactly the opposite: It's probably just too simple to understand. Ib controls Vbe, Vbe controls Ic.
Are you able to realize the contradiction?
This prooves that you have a certain lack of understanding for the phenomenon called "control".

Here is the answer to your question:
RE provides voltage feedback and the loop gain can be calculated as LG=-gmRE. That`s all. I suppose you know that the transconductance gm is independend on Ib.
This prooves that the base current, of course, has no influence on the feedback action.

By the way: You forgot to tell us where you have already explained how Ib influences feedback (quote: Biasing with a stiff base voltage is done with emitter feedback for Ib which makes the value of junction voltage not relevant. I'm not inclined to repeat that explanation again.)
 
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Ratch

Mar 10, 2013
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Don't be so quick to dismiss empirical methods. If the theoretical prediction did not agree with the empirical result, then the physicists would need to develop a better theory. The empirical result is definitive.

Empirical methods are great for determining what a device does or using a device. But, it does not disclose how a device works. You need to understand the underlying physics to do that. That is why just looking at a Ib vs Ic graph does not tell you what is in control (neither is controlling the other). Only by studying the physics of the device can you determine that Vbe controls both simultaneously.

Ratch
 
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Laplace

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The base current Ib is "controlled by the collector current" ? In post#61 you wrote exactly the opposite: It's probably just too simple to understand. Ib controls Vbe, Vbe controls Ic. Are you able to realize the contradiction?
Where is the contradiction? One statement described operation of a transistor in isolation. The other statement addressed the operation of a feedback bias network for a transistor.
This prooves that you have a certain lack of understanding for the phenomenon called "control".
Nah, it just proves that taking things out of context comes too easily for you.
RE provides voltage feedback and the loop gain can be calculated as LG=-gmRE. That`s all. I suppose you know that the transconductance gm is independend on Ib.
Recall that in Post#74 you brought up the subject of biasing a BJT with a voltage: "Did you ever hear about the commonly used method to use a voltage divider for biasing the BJT with a VOLTAGE? ... with the aim to make the produced voltage as 'stiff' as possible." And I stated that was only ever done with emitter feedback for Ib because emitter degeneration allows Ic to control Ib in a negative feedback loop. And I have provided the large-signal analysis to illuminate the emitter feedback for Ib. Now you present a small-signal analysis result which is out of context and inapplicable for explaining the feedback biasing scheme.
This prooves that the base current, of course, has no influence on the feedback action.
Nah, it just proves that making arguments which are out of context comes too easily for you.
By the way: You forgot to tell us where you have already explained how Ib influences feedback.
I only ever alluded to emitter feedback for Ib as the biasing scheme using emitter degeneration for Ic to control Ib. You made up the rest of it yourself, so you will have to explain it to yourself. I don't care to see your explanation.
 

Laplace

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Only by studying the physics of the device can you determine that Vbe controls both simultaneously.
So here you are at the beginning, again. Since you have some conceptual difficulty accepting that a current source controls the sourced current, there is no point to this discussion.
 

Ratch

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So here you are at the beginning, again. Since you have some conceptual difficulty accepting that a current source controls the sourced current, there is no point to this discussion.

A current source is not part of the device. It is a external to the device, and its addition makes a current amplifying circuit. This was pointed out to you earlier. You cannot put a device into a circuit and say the device has the same characteristics as the circuit to which it is embedded. I am fully aware that some devices like a magnetic amplifier are current amplifier devices, but not BJTs, FETs or vacuum tubes by themselves. You still have not explained how the physics of the BJT allow it to be a current amplifier. Before you make any other statement about how a BJT works, you should at least give a physics explanation of your pronouncement.

Ratch
 
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LvW

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And I stated that was only ever done with emitter feedback for Ib because emitter degeneration allows Ic to control Ib in a negative feedback loop. And I have provided the large-signal analysis to illuminate the emitter feedback for Ib. Now you present a small-signal analysis result which is out of context and inapplicable for explaining the feedback biasing scheme.
...emitter feedback for Ib...
Sorry - it`s pure nonsense.
Don`t you have any access to a textbook which explains to you how "emitter degeneration" works?

In short
: Ic increase (due to temperature or tolerance effects) increases the dc voltage Ve across Re.
That means: For a fixed voltage at the base node Vb the difference Vb-Ve=Vbe reduces correspondingly.
(By the way: That is the reason this feedback scheme needs a dc voltage Vb as stiff as possible).
And because Vbe determines Ic there will be a corresponding Ic reduction (nearly) back to the former value. It`s that easy!
That is the only feedback effect in this circuit and there is only one feedback loop (may be you are not able to identify it).
And the loop gain (do you know the meaning of "loop gain"?) is LG=-gmRe.
If you think that this feedback loop with the loop gain is "out of context" and "inapplicable explaining the feedback scheme",
it only prooves my assumption that you have a severe lack of understanding of the feedback principle.
That is OK, of course. But in such a case, you should be able to recognize your own deficiencies and accept explanations from other people (and not describe it as "inapplicable").


Quote: "Where is the contradiction? One statement described operation of a transistor in isolation. The other statement addressed the operation of a feedback bias network for a transistor."
So you really think that the transistor will behave differently - in isolation or within a circuit?
You do not see the contradiction? ("base current Ib is controlled by the collector current" versus "Ib controls Vbe, Vbe controls Ic").

Laplace - don`t you think it`s time to stop? Don`t you recognize that - more and more - you are going to lose your way within the labyrinth of (false) arguments?
 
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Laplace

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A current source is not part of the device. It is a external to the device, and its addition makes a current amplifying circuit. You cannot put a device into a circuit and say the device has the same characteristics as the circuit to which it is embedded.
One must surely realize that adding a current source does not make much of a circuit given the infinite source resistance providing isolation. However, a voltage source has zero source resistance - there is no isolation at all. A voltage source is not part of the device. It is external to the device, and its addition makes ...what? And who says the device has the same characteristics as the circuit to which it is embedded? My claim is that good transistor circuit design makes the value of Vbe not relevant.
Post#26 I aver that at the same temperature, the base current, Vbe, and collector current of a BJT will be the same in a circuit as it would be if the transistor were isolated from the circuit and run only on voltage sources.
It is also true that at the same temperature, the base current, Vbe, and collector current of a BJT will be the same as it would be if the transistor were isolated from the circuit and run only on current sources providing the same current as provided by the voltage sources. How could it be any different? The relationship between base current and collector current will still be the same. The current being forced to flow in the junction generates the same voltage as before, and that cannot be explained as a voltage divider since voltage division in general does not work when a resistance is infinite. Also, the most recognizable and useful (since it provides the basis for constructing the load line) is the static characteristic where the base is driven by a stepped current source that clearly shows the relationship between between base current and collector current. You might say this really only shows the operation of the measurement circuit; however, it may also reflect the manufacturers' desire that the device will actually be used in a real circuit.
You still have not explained how the physics of the BJT allow it to be a current amplifier. Before you make any other statement about how a BJT works, you should at least give a physics explanation of your pronouncement.
The terminal characteristics of a device are sufficient information to make the device useful in circuit design. For large-signal analysis the only useful model is the transistor as a current amplifier. If as you say above, the addition of a current source to control the base current makes a current amplifying circuit, then that is fine because large signal circuits drive the base with a current anyway. For small-signal high-frequency analysis the most useful model is the transistor as a transconductance amplifier because the math just works out to be simpler that way.
2015-09-05.png
 

LvW

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For large-signal analysis the only useful model is the transistor as a current amplifier.
View attachment 21793
Simply false. Do you know how the output of a large-signal push-pull stage looks like (class-B operation)?
It is a clean copy of the Ic=f(Vbe) characteristics - including the "dead zone" (cross-over distortions).

More than that - the characteristic curves in Fig. 1 show a mathematical relation between Vce, Ic and Ib. That`s all. How can you derive from such a set of curves that the BJT is used as a current amplifier? Do you know the meaning and the purpose (and the limitations) of curves to be found in data sheets?
 
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Ratch

Mar 10, 2013
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One must surely realize that adding a current source does not make much of a circuit given the infinite source resistance providing isolation. However, a voltage source has zero source resistance - there is no isolation at all. A voltage source is not part of the device. It is external to the device, and its addition makes ...what? And who says the device has the same characteristics as the circuit to which it is embedded? My claim is that good transistor circuit design makes the value of Vbe not relevant.

It is also true that at the same temperature, the base current, Vbe, and collector current of a BJT will be the same as it would be if the transistor were isolated from the circuit and run only on current sources providing the same current as provided by the voltage sources. How could it be any different? The relationship between base current and collector current will still be the same. The current being forced to flow in the junction generates the same voltage as before, and that cannot be explained as a voltage divider since voltage division in general does not work when a resistance is infinite. Also, the most recognizable and useful (since it provides the basis for constructing the load line) is the static characteristic where the base is driven by a stepped current source that clearly shows the relationship between between base current and collector current. You might say this really only shows the operation of the measurement circuit; however, it may also reflect the manufacturers' desire that the device will actually be used in a real circuit.

The terminal characteristics of a device are sufficient information to make the device useful in circuit design. For large-signal analysis the only useful model is the transistor as a current amplifier. If as you say above, the addition of a current source to control the base current makes a current amplifying circuit, then that is fine because large signal circuits drive the base with a current anyway. For small-signal high-frequency analysis the most useful model is the transistor as a transconductance amplifier because the math just works out to be simpler that way.
View attachment 21793

Still no physics explanation from you on how a current can influence the depletion region and thereby the charge flow across the PN junction. Isolation of the BJT input is not a factor. You are still hung up on design techniques and empirical reasoning to explain how a transistor works. In other words, you cannot separate what a BJT does from how it does it. Only the physics of the BJT can explain how it works internally.

Ratch
 

LvW

Apr 12, 2014
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Perhaps you should consult a tutorial on class-B amplifiers where the output is a direct function of the base current.
http://www.electronics-tutorials.ws/amplifier/amp_6.html
View attachment 21796

Laplace - do you uncritical believe everything you find in the net?
I can tell you that the drawing you have posted is completely nonsense (by the way: the whole site "electronic-tutorials" contains many errors).
1.) Do you recognize that the "input signal" is across the Vce axis? And the output signal has Ib as a parameter.
2.) Another "surprising " fact: The corresponding circuit shows transformer input and output coupling - and the figure shows a DC load line. Didn`t you "stumble" over this interesting property of a transformer?
3.) In the text they correctly speak about an input signal which must be "greater than the base-emitter voltage" !
So - what is your intention? What did you want to show us? Did YOU understand this funny picture?
 
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Laplace

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Still no physics explanation from you on how a current can influence the depletion region and thereby the charge flow ...Only the physics of the BJT can explain how it works internally.
Why do you frequent an electronics forum and expect explanation of internal device physics? The terminal characteristics of devices are what make circuits work. Perhaps if one were to hang out in a physics forum instead, then topics of internal junction physics would be more relevant. And as soon as I finish building the micro-lithography, wafer etching, and diffusion oven facility in my basement, I'll be a frequent visitor at the physics forum too. See you there!
 

Ratch

Mar 10, 2013
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Why do you frequent an electronics forum and expect explanation of internal device physics?

Because internal device physics correctly describes how a device really works.
The terminal characteristics of devices are what make circuits work.

The terminal characteristics are determined by the internal device physics and the external components.
Perhaps if one were to hang out in a physics forum instead, then topics of internal junction physics would be more relevant.

Internal device physics are relevant here, too.
And as soon as I finish building the micro-lithography, wafer etching, and diffusion oven facility in my basement, I'll be a frequent visitor at the physics forum too. See you there!

Perhaps.
Ratch
 

LvW

Apr 12, 2014
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I think it is quite obvious that you have no idea what a load line is or how it works.
Yes - most probably, you are right.
Therefore: I would be very happy if you could teach me how a transformer-coupled amplifier can work along a DC load line. I am curious for your anwer.
Do you still rely on the shown graph?
Surprising situation.
 
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