Discussion in 'General Electronics Discussion' started by BlackMelon, Apr 17, 2016.

1. ### Ratch

1,094
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Mar 10, 2013
How do you figure that? A capacitor can be energized to a voltage. Afterwards, the voltage remains, but the current is zero.

Ratch

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,838
Jan 21, 2010
Your argument that a current can't exist in an inductor without a voltage (with some hand waving that appears to boil down to the nonexistance of perfect inductors and conductors -- even in the face of superconductivity) is simply countered by the observation that the are no perfect capacitors or insulators and thus there is always leakage current. Because the current in a capacitor never falls to zero, this is not a case where voltage exists in the absence of current.

If you think this is an absurd proposition, remember that it is simply your argument.

3. ### Ratch

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Mar 10, 2013
Capacitors have nothing to do with current in a coil. I did miswrite in post #49 the following, "There is always a voltage present while current is present." I should have written "There is always a voltage present while current is changing."

Current does indeed fall to zero in a capacitor when it is completely de-energized. A good capacitor will hold its voltage for a very long time, thereby giving credence that to the fact that voltage can exist without current. If a capacitor is not good enough, then you can consider an electrometer. https://en.wikipedia.org/wiki/Electrometer Those instruments can hold a voltage for this side of forever.

I do not know what you are saying is an absurd proposition.

Ratch

4. ### Laplace

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Apr 4, 2010
I understand the basis for your opinion, but I disagree. I have a different opinion, as stated.
The words in your description do not coalesce into any meaningful circuit model. Since the problem is obviously not on this end, you need to provide a diagram identifying all entities involved and their relationship to your description.
So how does the relation between inductor voltage and inductor current apply to each component of the composite voltage? Enquiring minds want to know.
Correct. That's what the "(no resistance)" meant in my previous post.
My statement was that when the inductor voltage is zero, the inductor current is not changing. You claim that's wrong because if the inductor current is zero, the inductor voltage (i.e., current slope) can be non-zero. How are those two statements related, if at all?
Perhaps if you diagram this as requested above it might make more sense, but I doubt it.
ibidem
So I take that to mean that the coil will not lose its superconductivity.

5. ### Ratch

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Mar 10, 2013
Current, charge and flow are well defined words. What is there about those words you do not understand? I would be at a loss to diagram their definitions.

Just like I said. An excitation voltage applied to a coil causes a current change, which in turn causes an opposing voltage from the coil, which limits the effect of the excitation voltage. That explanation can be found in any electrical book. The excitation voltage and the opposing voltage comprise the composite voltage.

At least we agree on that.

You are right, I was thinking of current when you said voltage. Sorry for that misunderstanding.

No need to do so now, is there?

I am neither a priest or a Latin scholar.

Not from any current present in it.

Ratch

6. ### Laplace

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Apr 4, 2010
Emitter degeneration is absolutely necessary to reduce the circuit's sensitivity to variations in transistor beta.
When a constant value is assumed in the design, and negative feedback handles the rest, that equates to no role.
One must be pretty desperate to have dragged in obscure non-linear circuits to the argument. As for current mirrors, they do not depend upon having an exact value of Vbe designed in, but rather upon having the identical base voltage for thermally linked matched pairs so that equal base current and identical collector current is mirrored by the matched pair.
You seem to have a definite opinion about which is the independent variable and which is the dependent variable in the Shockley diode equation. But from the perspective of the circuit designer either the base current or the base voltage can be forced to be a particular value, and the other will naturally assume its corresponding value.
So you believe that forum discussion should be limited by your misconceptions.
So you don't really understand that the purpose for creating linear circuit models is to transcend the vagaries of device physics.

7. ### LvW

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Apr 12, 2014
A newcomer would, of course, ask: So what? Why do we want to reduce the sensitivity to beta?
Is this an inportant design parametr?
Laplace - can you answer this question and explain how emitter degeneation can reduce this sensitivity (without simply repeating Ic=beta*Ib) ?

Please ecxuse me, and with all respect: I don`t expect such an answer from an experienced engineer.

We spoke about the role of the DC voltage Vbe, correct? Is it really "desperate" to remind you that this value plays a "certain" role for current mirrors and log amplifiers? Perhaps you should ask Barrie Gilbert, who has invented many attractive circuits based on the exponenetial function Ic=f(Vbe).

... from the perspective of circuit designer? Yes - if he is following a cookbook approach using formulas without knowing/understanding the physical background. And this is OK - as long as he is doing his job.
However, for discovering novel circuits he will run into problems if he does not know "which is the independent variable and which is the dependent variable in the Shockley diode equation".

No - that`s not the case. In contrary - I wait for some replies which clearly reveal my misconceptions (if any).
But in the interest of a fair discussion you should not mix two different views: Physical facts and the "perspective of the circuit designer" (as you did).
If you read my reply again you will notice that I spoke about the hybrid parameters only. And - as you know - the corresponding equivalent diagram contains a current-controlled current source. This works for cookbook designs but does NOT reflect the physical reality. You certainly know that all transistor simulation models contain - in accordance with the physics - a voltage controlled current source, don`t you?

Last edited: Apr 21, 2016
8. ### Laplace

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Apr 4, 2010
You might find the answer you seek at the following webpage:
http://www.electronics-tutorials.ws/amplifier/transistor-biasing.html
As I said before, engineers tend to make a lot of simplifications and take a lot of shortcuts.
Such name dropper you are!
No problem. I mean, how many 'Barrie Gilbert's can there be on Facebook?
So now we can add this to the list of your misconceptions.
Even non-linear differential equations can be solved with linear numerical techniques if the increment is made small enough. Would it be unusual for a computer to run through a hundred-thousand increments for a complete simulation? Try doing that with a slide-rule.

9. ### LvW

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Apr 12, 2014
Laplace - no technical arguments, but polemic answers only (and I wonder- Why?).
Therefore, I reply to one sentence only:
Laplace - in post#66 your wrote:
"Emitter degeneration is absolutely necessary to reduce the circuit's sensitivity to variations in transistor beta."

In my post#67 I have asked you to explain this statement (because I think it was wrong).

Now I have to recognize that you are not able or not willng to explain your own claims - instead you are referring to a page containing questionable "explanations". (By the way, as you probably know - it is not a problem to find pages with wrong descriptions/explanations in the internet).

So - I will explain it to you and I wonder if this will be a new information for you (to be found also in all reputable textbooks):
(1) It is the task of the emitter degeneration resistor Re to reduce the sensitivity of the required emitter current Ie to uncertainties of the base-emitter voltage Vbe=Vb-Ve (initial uncertainties and temperatur influence, temp-co: d(Vbe)/dVth=-2mV/K).
It is easy to verify this effect by looking at the required base voltage Vb=Vbe+Ve=0.7V+Ie*Re.
As you can see, the influence of the assumed Vbe value on the required DC base voltage Vb is drastically reduced if Ve=Ie*Re>>0.7V.

(2) Because the voltage Ve=Ie*Re causes a negative voltage feedback effect we require that the DC base voltage Vb is as "stiff" as possible (preferrably independent on the voltage divider load current which is the base current Ib).
For this purpose it is common practice to design the base voltage divider (for providing Vb) with relatively small resistors allowing a DC current of app. (6...12)*Ib. In this case, the load current Ib (app. 10% of the divider current only) plays a minor role only for establishing the desired "stiff" base voltage Vb.

(3) Summary: The emitter resistor Re provides negative feedback thereby reducing the effect of Vbe uncertainties and the choice of a low-resistive base circuitry reduces the influence of base current uncertainties upon the base voltage Vb. Both principles are effective only because the BJT is a voltage controlled device : Ic=f(Vbe).
Anything wrong?
_________________________
Laplace - I have one final request to you.
You wrote: "So now we can add this to the list of your misconceptions".

In order to concentrate our discussion really on the critical points of disagreement I kindly ask you to show me the list of my misconceptions (if it is not too long). Thank you.
LvW

Last edited: Apr 22, 2016