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

1. ### Ratch

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Mar 10, 2013
Vbe controls Ic according to the physics of the BJT. I can elucidate that in more detail if you so desire. Circuit analysis is an approximation of what is happening. As you can see below, this transistor's Ic increases almost one amp for a for a increase of only 150 millivolts. The Ic really starts to take off at 0.7 volts, and Vbe doesn't increase much afterwards even for larger currents. Notice the mislabel of the horizontal axis. In the middle of its active region, the Vbe is approximately 0.7 volts.

Both the large DC bias and the small signal voltage are applied to the B-E same terminals. The response of the signal voltage vbe is determined from the slope (also known as transconductance) of the DC bias curve above. This slope is considered to be linear at voltages > 0.7 volts.

As I explained in a previous posting, the question of what controls Ic was raised. Many folks think Ib controls Ic, but I aver that Vbe is the real control of Ic. The physics of the transistor prove that Vbe controls Ic. This does not mean that design techniques should be based on trying to discern the small changes of Vbe. Likewise no one should say that Ic is determined by Ib. Design methods exist which do not depend of the minute changes in Vbe, while acknowledging that Vbe controls Ic.

Ratch

2. ### Laplace

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Apr 4, 2010
But your post itself explains why Vbe is irrelevant - it is only the derivative of Vbe that gives useful information for small-signal analysis. Also, in reality the transconductance value is obtained by other means than differentiating the Ic vs Vbe curve. Where Vbe is sometimes useful is in large-signal analysis where Vbe is treated as a constant value, for instance, when doing biasing calculations or when driving the transistor into saturation (as was the case for the original post in this thread with Ic=500mA.) The bottom line, again, is that Vbe is not very interesting nor very useful for electronic circuit design except where it is assumed to be a constant value that can vary over time and temperature. Trying to read anything more into the value of Vbe is inappropriate.

3. ### BlackMelon

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Aug 7, 2012
After reading your comments and some external information, I think I have got what I want to know why the datasheet states a test condition for Vbe. This might resolve your conflicts.

According to Ratch's latest graph, it is the Vbe that controls Ic, and that's quite a scientific view. However, in engineering view, even the Ic rises up high, the Vbe varies very little from 600mV to 850mV. The variation is so small compared to Vcc of a transistor (Says like Vcc=12, 24, 48V). Therefore, some textbooks say it is always constant (0.7V) and use the base current instead to predict the Ic (it might be easier to observe).

Agree or disagree? Just give a reply

BlackMelon

4. ### Laplace

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Apr 4, 2010
Engineers tend to make a lot of simplifications, and take a lot of shortcuts.

5. ### Ratch

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Mar 10, 2013
As far as I can see, other than being helpful in knowing the physics of the BJT, Vbe will only affect the bias point and the transconductance in designing circuits. For the higher Ic currents in a BJT, Vbe is going to be around 0.7 volts and won't change much. For very small currents, the Vbe can change a lot as shown in the curve above, and can change the bias point quite a lot if the bias point is set to a low value. You are right in that small signal transconductance (gm) is not found by differentiation, but by Ic/Vt, where Vt is the thermal voltage of 25.2 millivolts at room temperature.

6. ### Ratch

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Mar 10, 2013
You are going to run into trouble if you use the base current to predict the collector current. That is because there is a large beta variation within the same transistor designation, let alone those of different designations. If you want consistent Ic, apply a constant voltage across Re like I suggested to BobK in post #37 above. Then Ic will be the same no matter what the temperature or transistor of the same designation you use.

Ratch

7. ### LvW

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Apr 12, 2014
Disagree!
For designing circuits we do not "predict" Ic because we want to realize a circuit with some specific properties (for example: Voltage gain). And these properties depend on Ic and NOT on Ib.
Consequently, we SELECT resp. REQUIRE a certain value for Ic and use the beta value in order to know whow much base current Ib=Ic/beta we must take into account for calculating the base bias circuitry. And this Ic value is determined by Vbe only.
However, because we do not know the corresponding exact value for Vbe we make use of negative feedback, which makes the circuit less sensitive to the momentary Vbe value (as shown in my former post#40) - and we select a suitable (but fixed) value - for calculation purposes only.

Hence - first Ic=f(Vbe) and then Ib.

Last edited: Apr 19, 2016
8. ### LvW

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Apr 12, 2014
Irrelevant? Is it really necessary to explain the role of Vbe again and again?
Do you know that the DC value of Vbe solely determines the quiescent current Ic which - in turn - determines the point where the derivative is taken?

May I ask you by which means? The result of the process of differenciating is gm=Ic/Vth.

9. ### Ratch

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Mar 10, 2013
Sorry I did not catch this sooner. You can indeed have a voltage without a current. A capacitor, for instance, can be energized up to its working voltage, and no charge will flow after it reaches its whatever voltage. Therefore, voltage is present, but no current.

Ratch

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

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Jan 21, 2010
so then an inductor will allow a current without a voltage?

11. ### Ratch

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Mar 10, 2013
No, when a current is present in an inductor without an external voltage being applied, the voltage to drive the current comes from the inductor itself. Specifically, the energy released from the collapsing magnetic field.

Ratch

12. ### Laplace

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Apr 4, 2010
Yes, if it is a pure inductor (no resistance). Short out a super-conducting coil that has current already flowing and the current will keep circulating forever (with no voltage across the coil).

13. ### Ratch

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Mar 10, 2013
Current does not flow, remember? It is either present or it exists. Charge can flow, however.

Current will exist forever in any conductor, not just coils, if superconductivity is present. There is nothing to take away the energy of the charge carriers. It is analogous to moving through an empty region of interstellar space. But causing a current in a superconductor is an empty feat, because no energy can be removed from the circuit without it losing its superconductivity.

Ratch

14. ### Laplace

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Apr 4, 2010
Of course not. That would be a fool's errand simply because Vbe has no role in transistor circuit analysis & design other than as being one constant diode voltage drop. The exact value assumed by Vbe is a concept from device physics that has no role in discrete transistor circuit analysis. Simple amplifier circuits are designed using linear circuit models of the transistor instead, the first such linear model usually learned is the transistor hybrid model using the four h parameters for a two-port network. As stated in INTEGRATED ELECTRONICS by Millman & Halkias, "The basic assumption in arriving at a transistor linear model or equivalent circuit is that the variations about the quiescent point are assumed small, so that the transistor parameters can be consisered constant over the signal excursion." Note that the current generator at the collector of the hybrid model has a magnitude of hfe∙ib.

There are more sophisticated small-signal linear circuit models of the BJT intended to give better results at higher frequencies, including transconductance models, but there is no specific incorporation of Vbe in any of them.
Do you know that the base current which is forced to flow in the transistor from the bias network determines the DC value of Vbe?

15. ### Laplace

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Apr 4, 2010
Current flows, charge moves.
So is the correct answer to @(*steve*) 's question about coils YES or NO?
What if only half the energy stored in the magnetic field is bled off and the coil shorted again? What happens to the remaining energy? Does a lesser amount of current then keep flowing in the coil?

16. ### Ratch

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Mar 10, 2013
Charge movement is current. Current flow means charge flow flow, which is redundant and ridiculous.

No, of course. The current that exists when the external voltage is zero comes from the voltage produced by the magnetic field collapse. There is always a voltage present while current is present.

The same thing that happens to more energy happens to less energy. Moving charge carriers cause a magnetic field to form and store energy. If the energy is less, the current will be less also.

Ratch

17. ### Laplace

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Apr 4, 2010
It is what it is. Current flows, charge moves.
That statement is nonsense. How many voltages are there across the terminals of a coil?
A superconducting coil is nothing more than an ideal inductor. The inductor voltage is L∙di/dt. If the inductor voltage is zero, then the current flowing in the inductor is not changing, nor is the magnetic field changing. If the magnetic field is collapsing then the external voltage across the coil is proportional to the rate at which the field collapses. There is always a voltage present when the current is changing. If the inductor voltage is forced to be zero (by shorting the terminals) then it is impossible for the current to change and the magnetic field will be static.

18. ### LvW

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Apr 12, 2014
No role?
Did you ever try to analyze (or simulate) a gain stage without emitter degeneration?
Again: Vbe plays the major role and determines the main properties of the circuit.
And exactly THIS is the reason we are trying to reduce the circuit´s sensitivity to Vbe uncertainties using the concept of negative feedback.
It is a severe misconception to believe that the assumption of a constant Vbe value (during the design of a gain stage) would mean that Vbe - as you wrote - would play "no role".

And what about current mirrors, Gilbert cells, log-amplifiers,..? It is the DC value of Vbe which defines the function of these circuits!

Sorry, but this is simply false. You again are mixing up cause and effect. Of course - as in each conductive body (linear or non-linear) - it is the voltage across this part that allows a current. Physically spoken - a current can never "produce" a voltage. It is only the electrical field strength (caused by the applied voltage) within the conductive material that enables the charges to move.
I am aware that for calculating purposes we assume that a current (for example in a simple voltage divider) produces a voltage drop across each resistor. And that`s OK.
But - as far as I understand - in this thread we are discussing physical properties of an active device and not the various tools and practical methods for analyzing circuits.
This applies, for example, also to the equivalent small-signal diagram based on hybrid parameters. It is an equivalent diagram that can be used for calculating purposes but it does not reflect the physical working principles of the BJT.

19. ### Ratch

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Mar 10, 2013
Current is already charge flow. It does not flow again. "Current flow" is syntactically incorrect.

How so?

One composite voltage, usually composed of a excitation voltage and a reactive coil voltage.

Which means the coil has no resistance.

That's wrong. A point on the current curve of a coil can be at zero, but the slope at that point can something other than zero.

That is true for the voltage produced by the coil. The composite voltage across the terminals will depend of the excitation voltage plus the voltage produced by the coil.

There is always a voltage produced by the coil when the current is changing. The voltage at the terminals might be modified by the excitation voltage.

The current can be changed by any number of ways, but as long as it is constant, so will the magnetic field be constant.

Ratch

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

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Jan 21, 2010
And of course one can never have a capacitor in which no current is (redacted).

so its not an example of voltage without current.