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

1. ### BlackMelon

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Aug 7, 2012
http://www.es.co.th/Schemetic/PDF/BC337.PDF

According to this datasheet of a BC337 NPN bipolar transistor, the Vbe is 1.2V. What I have known before is that the amount of the Vbe depends on temperature only.

However, this datasheet states that they find this Vbe at Ic = 500 mA; Vce = 1 V. In real world, if the test conditions (Ic and Vce) change, will the result (Vbe) change also?

Thank you
BlackMelon

2. ### Colin Mitchell

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Aug 31, 2014
Yes, those Philips transistors are such rubbish that the base voltage rises to 1.2v when you are trying to get 500mA through them.

Normally the base voltage is around 0.7v - 0.8v for a good quality transistor passing its near-peak current.

The base voltage is not important because you have to supply the base with a higher voltage via a safety resistor and you main concern is to deliver the necessary current.

3. ### LvW

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Apr 12, 2014
I rather think that the base-emitter voltage Vbe is of primary concern because it controls the collector current Ic. Remember Shockleys function Ic=f(Vbe/Vt).
It is a common misconception (to be found even in some textbooks) to believe that Ic would be "controlled" by Ib.
The current into the base Ib has no control function but it does exist and must be considered for designing the DC bias network.

4. ### Colin Mitchell

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Aug 31, 2014
"I rather think that the base-emitter voltage Vbe is of primary concern because it controls the collector current Ic. Remember Shockleys function Ic=f(Vbe/Vt).
It is a common misconception (to be found even in some textbooks) to believe that Ic would be "controlled" by Ib.
The current into the base Ib has no control function but it does exist and must be considered for designing the DC bias network."

This is the completely incorrect way to think about how the transistor works and the base voltage.
You have absolutely no control over the base voltage and it rises to different values according to how the transistor is fabricated.
Luckily you are not teaching electronics, although I have had similar crazy ideas from an Indian "Professor" and when I told him of his mistakes, he closed down his website.

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

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Jan 21, 2010
Stop posting for a while, I'm just getting the popcorn.

6. ### LvW

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Apr 12, 2014
How do you know? Perhaps it is shocking news for you - but I was teaching analog electronics for more than 25 years.
Colin - since a long time I know your position in this question, but it is never too late to take into consideration positions and explanations from other sources - even "crazy ideas".
("science is a culture of doubt”, R.P. Feynman)

* Is it really necessary to give you some references from leading US-universities (and other series sources) ?
* Have you ever heard about Barrie Gilbert and his explanations how a BJT works?
* Have you ever had a look into W. Shockleys patent documents?
* Have you ever tried to explain how and why a resistor RE in the emitter path of a BJT can provide negative feedback ? Can you explain this with current control?
* Have you ever tried to understand the EARLY effect
* Have you ever tried to understand what the temp-co of -2mV/K means?
* Have you ever experienced (and tried to understand) that the voltage gain of a BJT-based gain stage does not change if we replace a BJT with beta=100 with another BJT having beta=300 (same dc operating point, of course).

LvW

Last edited: Apr 17, 2016
7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
this is the most interesting, because it can be explained even assuming a constant Vbe and "current control".

But don't let me distract you, I'll go back to my popcorn.

munch munch

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8. ### LvW

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Apr 12, 2014
Good luck!

Steve, I should add the following:
As we all know - RE-feedback increases the signal input resistance at the base node. And from feedback theory it is known that such an increase is possible only if the feedback signal is a VOLTAGE.

Last edited: Apr 17, 2016
9. ### BobK

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Jan 5, 2010
I will say my piece, which everyone ignored last time, once again.

If Ib is a function of Vbe, then we can write a function for the collector current in either Vbe or Ib, no matter which one is actually controlling it at the semiconductor physics level. They are equivalent as far as user knows. It turns out that the formula for collector current as a function of Ib is simpler and easier to apply, so I choose to use that function instead of the function of Vbe.

Bob

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10. ### Ratch

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Mar 10, 2013
The BJT by itself is a transconductance amplifier. This means that the Vbe voltage controls the Ic. The nonlinearity of the curves is due to the BJT being a diffusion device. Diffusion actions produce exponential results which are nonlinear. And, the physics of a BJT and a junction diode show that those devices, by themselves, are voltage controlled devices. The attachment below shows that.

It is wrong to say otherwise. The Ib current that exists when the transistor is in its active region is waste current that is an indicator of the collector current, not the control of it. To say that Ib controls Ic is like saying that my bedside alarm clock tells the sun when to rise. Now, you can put a large resistance in the base circuit thereby driving the base as a current source. Then, you will notice that the Ic is proportional to Ib (Someone say Beta?). But, whenever you drive the base with current, you no longer have a "bare" transistor. You have instead a current amplifying circuit.

A lot of books and folks who should know better say that Ib controls Ic, but I believe I can rebut any of those assertions.

Any takers?

Ratch

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11. ### Ratch

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Mar 10, 2013
At the higher currents, the bulk resistance between the base and emitter contributes to the higher Vbe.

Ratch

12. ### Ratch

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Mar 10, 2013
Both Ic and Ib are exponentially related to Vbe. So when you divide Ic by Ib, the exponent and Vbe cancels out, and you get a simple proportion (Beta). Then you can do whatever you want to the indicator of Ic, which is Ib, and be assured of what Ic will be. That is analogous to controlling the speed of your car by reading the speedometer. The speedometer is not controlling the speed of the car, but it indicates what the car's speed is.

Ratch

13. ### LvW

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Apr 12, 2014
1.) Of course, you can write a function between two or more parameters in several different forms. But this writing has no meaning at all while discussing physical properties.
However, in the following I don`t want to stress again the physical working principle of the BJT. To me, this is clear and does not need further to be discussed. But I want to describe a surprising contradiction which sometimes can be observed while discussing the method for designing a simple BJT-based amplifier stage: Some people believe to apply current-control, but they don`t.

2.) Bob - with all respect, do you know what you are doing? It is a surprising (funny) situation that all "defenders" of current-control believe they would use this method (because it would be "simpler and easier to apply") but this is not the case! In fact - during design of an amplifier stage they perform exactly the same steps as the "voltage-control party". That is no surprise, because there is - in principle - one way only:
* Selection of Vcc, Ic, Rc and Re
* Calculation of the DC bias circuitry taking into account (of course!) the base current Ib=Ic/beta and a base-emitter DC voltage of app Vbe=(0.65---0.7)V.
Which steps indicate current-control? Do you even think that we are forgetting the existence of a finite base current?

3.) Bob - please, excuse my introductory provoking words (do you know what you are doing) but I kindly ask you to answer the following question:
* How do you choose the resistors for the resistive voltage divider at the base?
* Don`t you follow the rule "as low as possible with respect to other constraints (input impedance, power consumption)" ?
* Why "as low as possible" ? I suppose, because you are trying to realize a DC voltage source at the base node as good as possible. As you probably know, only in this case Re-feedback works as desired! (Re produces a feedback VOLTAGE).
* This is an example for my opinion that current-control defenders do not realize that they always follow the rules of voltage control - to me, a situation I never will understand.
* The same applies to other examples: Current control "believer" use, of course, the transconductance gm=d(Ic)/d(Vbe)=Ic/26mV for calculating the voltage gain of a stage. Don`t they know what they are doing?
Isn`t it a contradiction to believe on current-control and to use the transconductance at the same time?

4.) Summary: Some people (and even some textbooks) claim that the BJT would be physically current-controlled or - at least - that the assumption of current-control would make the design of BJT circuits simpler and easier. But - up to now - nobody was able to proove this assertion with one simple example.
I really have problems to understand this phenomenon.
Is it really possible to imagine that three moving charged carriers from the base could be able to release 1000 charged carriers from the emitter (assuming a beta value of 333) ?

LvW

Last edited: Apr 18, 2016
14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
So you're saying that some part of the Vbe is not controlling Ic?

15. ### LvW

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Apr 12, 2014
OK - I think, that is one of two possible interpretations (assuming that such thing like a current source does exist)..
Another explanation/interpretation is as follows:
"Feeding a current into the base" means nothing else than to calculate a resistor RB which is part of a non-linear voltage divider. Then the value of RB is calculated to fulfill the following condition: RB=(Vcc-Vbe)/Ib .
In words: We need a resistor RB which allows a DC voltage of app. Vbe=0.7V which is derived from a DC source Vcc when the current through this two-element chain is Ib=Ic/B.
This interpretation does not need such an "artificial" product like a current source (which, in fact, always is a voltage source with a large source resistance).

16. ### duke37

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Jan 9, 2011
I agree with Bob that it does not matter what the mechanism is working inside the transistor.
Perhaps those advocating a voltage controlled model can explain the simplest bias circuit possible of a grounded emitter, a collector resistor and a resistor between collector and base. I just use beta.

This argument has been visited in the past without any conclusion. If you do get some agreement, please decide whether the chicken or the egg came first.

17. ### LvW

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Apr 12, 2014
Everybody uses beta. And - everybody uses Vbe=0.65...0.7 V for calculating this resitor RB.
So - what does this proove? Nothing. (Read my post#15).

.
This is, certainly, not a chicken-egg question.
The voltage is always first; no current without driving voltage.

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

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Jan 21, 2010
I agree entirely. I = V / R

Any other arrangement of ohms law is invalid because no current flows without a driving voltage.

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19. ### Ratch

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Mar 10, 2013
Correct, only the voltage across the E-B junction is effective in controlling Ic.

Ratch

20. ### Ratch

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Mar 10, 2013
I think we have two subjects being discussed.

1) Whether a BJT by itself is a current amplifier or a transconductance amplifier.

2) What are some methods used for circuit design using BJTs.

Describing how a circuit is designed using a particular topology does not prove or disprove subject (1) above. Neither does quoting a model. Models describe what happens to output as a result of input, but models do not show how a device works. What does prove how a device works is the physics of operation. Thus far, no one has talked about the physics of a junction diode or BJT in this thread.

With regard to (2) above, any resistors or components attached to the collector, base, or emitter turns the BJT from a single device into a functional circuit. We can talk all day about methods used to design circuits, but that does not highlight the role of Vbe, which I believe is the subject of this thread.

Ratch