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Voltage Drop across NPN BJT C&E?

Ratch

Mar 10, 2013
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You are correct, of course, but can you change voltage without changing current? I draw a specific current through the base to get a certain voltage.

No, that base current is waste leakage current that comes from the charge carriers that do not make it from the emitter to collector, and are instead lost to the base due to the voltage on the base. It is a phenomenon that can be minimized, but not eliminated. Ib is exponentially proportional to Vbe, as is Ic also exponentially proportional to Vbe. Therefore, Ic and Ib are linearly proportional to each other within a range of currents. This proportion is called beta. That makes Ib a indicator of Ic, but it is still Vbe that controls both Ic and Ib. The Ib is a nuisance current that has to be taken into account.

With the same base voltage and current can I have different collector conditions?

When you say base voltage, you do not mean Vbe, do you? What I am saying is that if you set a current on a BJT with no Rb or Re and note the Vbe, the Vbe will be the same if you add Rb and Re and adjust the base voltage to show the same Ic.

I cannot calculate collector voltage knowing base voltage, or can I? I can take collector current, divide it by gain and get base current. Can I calculate base voltage know collector voltage?

Since a BJT is a reasonable current generator when operated in the active region, the collector voltage will not influence the Ic very much. That can be seen from the Ic vs Vc curves. Since betas vary considerable between BJTs unless they are selected, it is not a good idea to rely on beta for design. Anything can be calculated if the parameters are measured and known, but that is not always practical.

Ratch
 

LvW

Apr 12, 2014
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With the same base voltage and current can I have different collector conditions? I cannot calculate collector voltage knowing base voltage, or can I? I can take collector current, divide it by gain and get base current. Can I calculate base voltage know collector voltage?

Yes - the underlined sentence is correct, of course, because you have performed the right sequence:
It is the classical procedure - starting with Ic and calculate the resulting current Ib.
However, this does not imply any controlling function. In contrary!.

Furthermore, may I direct your attention to the attachement to my post#48?
I think, the shown drawings - together with my comments under 3) of pos#48 - can answer your questions and clarify your doubts.
EDIT: For your convenients, I have enclosed an updated pdf-version of the diagram.
 

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Herschel Peeler

Feb 21, 2016
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Yes - the underlined sentence is correct, of course, because you have performed the right sequence:
It is the classical procedure - starting with Ic and calculate the resulting current Ib.
However, this does not imply any controlling function. In contrary!.

Furthermore, may I direct your attention to the attachement to my post#48?
I think, the shown drawings - together with my comments under 3) of pos#48 - can answer your questions and clarify your doubts.
EDIT: For your convenients, I have enclosed an updated pdf-version of the diagram.

What you say makes sense and I can see it working in a current mirror of a differential amplifier, but I must be missing something. I I hold Vbe and Ib constant and change Ic Vce changes (and beta). So how can we say Vbe controls Vce? What is controlling what? Ice controls Vce and beta. Vbe seems irrelevant. Yes, if we change Vbe Ice and Vce change, but is it Ice that determines beta?
 

LvW

Apr 12, 2014
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What you say makes sense and I can see it working in a current mirror of a differential amplifier, but I must be missing something. I I hold Vbe and Ib constant and change Ic Vce changes (and beta). So how can we say Vbe controls Vce? What is controlling what? Ice controls Vce and beta. Vbe seems irrelevant. Yes, if we change Vbe Ice and Vce change, but is it Ice that determines beta?

I must admit that I have problems to understand everything in your response.
* For VBE and IB constant I can change Ic (a little bit) by changing VCE only. This is due to the well-known Early effect.
However, this effect does not touch at all the question if VBE or IB are (mainly) controlling Ic.
* More than that, you are asking "how can we say Vbe controls Vce"?
Who has claimed that? Vbe only controls Ic. The voltage Vce is determined by the collector resistor RC and the current IC (Ohms law).
* Here is a short summary:
**VBE determines/controls IE - and IE is split into IE=IB+IC. Hence, VBE controls also IC (and, of course, IB).
** VCE is determined by IC (Ohms law).
** Secondary effect: IC slightly increases with VCE (Early effect) - but this is no controlling function (not caused by the input). It is an additional effect only.
** The split ratio between IB and IC is not absolutely constant - it somewhat depends on the value of IE. This results in a small change of the value B=IC/IB. However, this is also a secondary effect and can be neglected in most cases if compared with huge tolerances connected with the absolute B value.
 

Ratch

Mar 10, 2013
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What you say makes sense and I can see it working in a current mirror of a differential amplifier, but I must be missing something. I I hold Vbe and Ib constant and change Ic Vce changes (and beta). So how can we say Vbe controls Vce? What is controlling what? Ice controls Vce and beta. Vbe seems irrelevant. Yes, if we change Vbe Ice and Vce change, but is it Ice that determines beta?
Unless you factor in secondary effects, you will not be able to change Ic while holding Vbe constant, while the BJT is operating in its active region. Look at the BJT data sheets and see how independent Ic is compared to the collector voltage. The BJT is a transconductance device, meaning voltage controls current. To make it a voltage or current amplifier, you need to add external components. This makes the total a current/voltage amplifier a circuit, not a single active component.

Ratch
 

Herschel Peeler

Feb 21, 2016
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Design 899 Vbe Vce Ie Ic beta test.PNG
I must admit that I have problems to understand everything in your response.
* For VBE and IB constant I can change Ic (a little bit) by changing VCE only. This is due to the well-known Early effect.
However, this effect does not touch at all the question if VBE or IB are (mainly) controlling Ic.
* More than that, you are asking "how can we say Vbe controls Vce"?
Who has claimed that? Vbe only controls Ic. The voltage Vce is determined by the collector resistor RC and the current IC (Ohms law).
* Here is a short summary:
**VBE determines/controls IE - and IE is split into IE=IB+IC. Hence, VBE controls also IC (and, of course, IB).
** VCE is determined by IC (Ohms law).
** Secondary effect: IC slightly increases with VCE (Early effect) - but this is no controlling function (not caused by the input). It is an additional effect only.
** The split ratio between IB and IC is not absolutely constant - it somewhat depends on the value of IE. This results in a small change of the value B=IC/IB. However, this is also a secondary effect and can be neglected in most cases if compared with huge tolerances connected with the absolute B value.

Yes, changing Vbe does change Ic ,,, which changes Vce.

Is Vbe an indicator of Ic? In the example attached I can change Ic which changes Vce, but Vbe does not change.
We cannot derive Ic or Vce from Vbe. Certainly Ib or Vb changes Ic and Vce, but I don't see how we can calculate Ie or Vce by knowing Vbe,
Ib times gain does not give us Ic. Yes, Ic divided by gain always gives us Ib, but Ic or Vce are not predictable from Ib or Vbe.
Ib times gain gives us a maximum Ic, but Ic is determined by the collector load. Ic can change with no change in Vbe or Ib.
There is certainly something here I am missing.
.
 

Ratch

Mar 10, 2013
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View attachment 29685


Yes, changing Vbe does change Ic ,,, which changes Vce.
As said before, Vbe control both Ic and Ib.

.
Is Vbe an indicator of Ic?.
Certainly, Vbe controls IC.

.
In the example attached I can change Ic which changes Vce, but Vbe does not change.
Ic changed only 0.07 ma when Vce changed 2.03 volts. Using the definition of resistance we get 2.03/0.07E-3 = 29k ohms. That indicates the BJT is a pretty good current generator. As you can see from the characteristic curve enclosed by attachment, the Ic vs Vce curve is quite horizontal in the region you are operating the BJT. That indicates a high resistance.

.
We cannot derive Ic or Vce from Vbe. Certainly Ib or Vb changes Ic and Vce, but I don't see how we can calculate Ie or Vce by knowing Vbe,.
You can't very easily compute Ic that way. Unless you know some of the internal parameters like the bulk resistances of the base and emitter slabs. But, who cares what the Vbe is? The idea is to swamp out the variations of Vbe by putting lots of resistance in the emitter and base circuits so that Re, Rb, and Vb set the quiescent Ic point. Vbe will take care of itself and conform to whatever current you desire. Even though Vbe controls the transistor itself, the external components control the circuit which includes the BJT.

.
Ib times gain does not give us Ic. Yes, Ic divided by gain always gives us Ib, but Ic or Vce are not predictable from Ib or Vbe..
If you are referring to a current amplifier circuit, which consists of a BJT with external components, then yes, Ib times the circuit gain will equal the Ic. Otherwise, it makes no sense because a BJT by itself is not a current amplifier. It is a transconductance amplifier.

.
Ib times gain gives us a maximum Ic, but Ic is determined by the collector load. Ic can change with no change in Vbe or Ib..
If the BJT operating within its active region, then Ic changes little because it is an output from a current generator.

.
There is certainly something here I am missing.
.
Don't be confused between the "naked" BJT and the transistor circuit wherein the BJT resides. The BJT circuit has different characteristics than the BJT alone by itself.

Ratch

Peeler.JPG
 

Herschel Peeler

Feb 21, 2016
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As said before, Vbe control both Ic and Ib.

.
Certainly, Vbe controls IC.

.Ic changed only 0.07 ma when Vce changed 2.03 volts. Using the definition of resistance we get 2.03/0.07E-3 = 29k ohms. That indicates the BJT is a pretty good current generator. As you can see from the characteristic curve enclosed by attachment, the Ic vs Vce curve is quite horizontal in the region you are operating the BJT. That indicates a high resistance.

.You can't very easily compute Ic that way. Unless you know some of the internal parameters like the bulk resistances of the base and emitter slabs. But, who cares what the Vbe is? The idea is to swamp out the variations of Vbe by putting lots of resistance in the emitter and base circuits so that Re, Rb, and Vb set the quiescent Ic point. Vbe will take care of itself and conform to whatever current you desire. Even though Vbe controls the transistor itself, the external components control the circuit which includes the BJT.

.If you are referring to a current amplifier circuit, which consists of a BJT with external components, then yes, Ib times the circuit gain will equal the Ic. Otherwise, it makes no sense because a BJT by itself is not a current amplifier. It is a transconductance amplifier.

.If the BJT operating within its active region, then Ic changes little because it is an output from a current generator.

.Don't be confused between the "naked" BJT and the transistor circuit wherein the BJT resides. The BJT circuit has different characteristics than the BJT alone by itself.

Ratch

View attachment 29687

Ah, yes. At as low a base current as I was working Vbe changes so little I couldn't measure it.
 

LvW

Apr 12, 2014
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View attachment 29685
There is certainly something here I am missing.
.

Yes.
In your example, you have altered the current gain as well as the collector resistor - and you are surprised that Ic and Vce have changed in spite of constant Vbe and Ib ? There was no change of the input condidtion at all - and you think this "test" could be a counterexample against Vbe-control?
Please, realize that there is a direct exponential relationship between Vbe and Ie resp. Ic. This is the well-known Shockley equation, which is the basis for the Gummel-Poon transistor model incorporated into all SPICE-based circuit simulators.
So - how can you say "We cannot derive Ic or Vce from Vbe" ???

It is another question if we make use of this equation during hand-made calculations. This is a problem because of a strong temperature dependence (and large tolerances) of this relationship. That is the reason we use voltage feedback (Re) which allows us to use a rough estimate for Vbe (0.65...0.7V). See the stabilization line in my pdf attachement (former post).
 
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Herschel Peeler

Feb 21, 2016
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Yes.
In your example, you have altered the current gain as well as the collector resistor - and you are surprised that Ic and Vce have changed in spite of constant Vbe and Ib ? There was no change of the input condidtion at all - and you think this "test" could be a counterexample against Vbe-control?
Please, realize that there is a direct exponential relationship between Vbe and Ie resp. Ic. This is the well-known Shockley equation, which is the basis for the Gummel-Poon transistor model incorporated into all SPICE-based circuit simulators.
So - how can you say "We cannot derive Ic or Vce from Vbe" ???

It is another question if we make use of this equation during hand-made calculations. This is a problem because of a strong temperature dependence of this relationship. That is the reason we use voltage feedback (Re) which allows us to use a rough estimate for Vbe (0.65...0.7V). See the stabilization line in my pdf attachement (former post).

Got it. Thank you for your words on this and patience.
 

LvW

Apr 12, 2014
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You are welcome. Answering such questions (and clarifying contradictory "explanations") is the main purpose of this forum.
 
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