Transistors / hfe / Active region conflicts

[Laplace]

It remains your secret...

It's probably just too simple to understand. Ib controls Vbe, Vbe controls Ic. We design bias networks to maintain constant Ic by having Ic control Ib. Vbe is assumed to be a constant diode forward voltage drop (~0.6V), therefore its particular value is not relevant. However, when we use a transconductance model for signal analysis, then the Vbe used is the signal Vbe (i.e. delta-Vbe) and not the actual (bias) Vbe.

 

[LvW]

Every thing was answered already:

https://www.electronicspoint.com/th...e-region-conflicts.275168/page-3#post-1661658
https://www.electronicspoint.com/th...e-region-conflicts.275168/page-3#post-1661733
https://www.electronicspoint.com/th...e-region-conflicts.275168/page-3#post-1661809

That`s why I have stated already that we turn around in circles.

 

[Ratch]

It's probably just too simple to understand. Ib controls Vbe,

No, the physics of the BJT cannot explain how base charge flow controls Vbe. A base current source contains the voltage that will control the BJT. If you set the base current, then Vbe will conform to that current. The base current is set by external means, whereas the Vbe is applied to the depletion region within the device itself. The internal control of the BJT does not change just because you apply a voltage (by an indirect current method) to the base, as opposed to a direct voltage application. It is the voltage applied to the BJT by whatever method, that controls Ic, despite biasing methods that indirectly set that voltage. Therefore, Vbe controls both Ic and Ib. If you say that Ib controls Ic through Vbe, then you are talking about a current amplifying circuit, not a BJT by itself.

Ratch

 

[Laplace]

If you say that Ib controls Ic through Vbe, then you are talking about a current amplifying circuit, not a BJT by itself.

Yes, I was talking about biasing a transistor, so that would be in a circuit. This is, after all, an Electronics Forum where the usual process is to put the transistor into a circuit.

Nor does is really matter whether Vbe is applied directly or set indirectly by other means, in a transistor circuit Vbe is assumed to be one forward diode drop generated by Ib and the particular value of Vbe is not relevant.

 

[LvW]

Nor does is really matter whether Vbe is applied directly or set indirectly by other means, in a transistor circuit Vbe is assumed to be one forward diode drop generated by Ib and the particular value of Vbe is not relevant.

This statement was commented already in post#45:
https://www.electronicspoint.com/th...e-region-conflicts.275168/page-3#post-1661658

 

[Ratch]

Yes, I was talking about biasing a transistor, so that would be in a circuit. This is, after all, an Electronics Forum where the usual process is to put the transistor into a circuit.

Nor does is really matter whether Vbe is applied directly or set indirectly by other means, in a transistor circuit Vbe is assumed to be one forward diode drop generated by Ib and the particular value of Vbe is not relevant.

Most of the current present in the emitter-base junction (Ie) passes on thru the base collector junction and into the collector circuit. Only a small fraction of this current passes out of the base terminal, and is considered waste current. Are you saying that this waste current is determining Vbe? Or is this waste current determined by Vbe? And is this waste current not an indicator of Ic?

Ratch

 

[Laplace]

This statement was commented already in post#45:

If Vbe were of fundamental importance, then circuit design procedures would not assume that it is a standard value equal to one diode drop. But in fact they do exactly that, and use negative feedback so the collector current can modify (control) the base current in order to stabilize the collector current. All the while considering Vbe to be a constant value. Because the value of Vbe is not relevant. I will concede that Vbe can become very important when a poor design has been done, because the object of a good design is to make the value of Vbe not relevant.

 

[Ratch]

If Vbe were of fundamental importance, then circuit design procedures would not assume that it is a standard value equal to one diode drop. But in fact they do exactly that, and use negative feedback so the collector current can modify (control) the base current in order to stabilize the collector current. All the while considering Vbe to be a constant value. Because the value of Vbe is not relevant. I will concede that Vbe can become very important when a poor design has been done, because the object of a good design is to make the value of Vbe not relevant.

Yes, feedback and other design techniques are used in design to mitigate the exponential response of Ic to Vbe. For that reason, it could be said in a contrarian manner that Vbe is considered of primary importance. Otherwise, so much design effort would not be lavished on it. However, that is not the point. The point is that Vbe controls both Ic and Ib, and Ib does not control Vbe. I am looking at the physics aspect of this statement, and am not considering design methods that do not apply to this point.

Ratch

 

[Laplace]

The point is that Vbe controls both Ic and Ib

Yes, Vbe determines Ib if one drives the base-emitter junction with a voltage source, since that makes Vbe the independent variable and Ib dependent. But a circuit designer does not drive the base-emitter junction with a voltage source. So what importance does the original statement have for a circuit designer using discrete components? It should be a warning to not allow Vbe to control Ib & Ic, rather design the circuit so that Ib controls Ic and use negative feedback for stability. Then Vbe is not relevant.

 

[Ratch]

Yes, Vbe determines Ib if one drives the base-emitter junction with a voltage source, since that makes Vbe the independent variable and Ib dependent. But a circuit designer does not drive the base-emitter junction with a voltage source. So what importance does the original statement have for a circuit designer using discrete components? It should be a warning to not allow Vbe to control Ib & Ic, rather design the circuit so that Ib controls Ic and use negative feedback for stability. Then Vbe is not relevant.

You can never control Vbe with Ib. You can only fool yourself into thinking you do. Connect a voltage source across a one ohm resistor. Set the voltage to one volt. You should be assured that the voltage source is controlling the current present in the resistor. Now insert an ammeter in series with the resistor and voltage source. Crank up the voltage until the ammeter reads one amp. Although you set the current by a ammeter, it is still the voltage that controls the current and not the ammeter. The ammeter is an indicator of the current, not its control. No matter to what value you adjust the ammeter current, it is the voltage source that controls it. Same with the BJT, only it is a variable resistor whose value depends on the current. That still makes no difference with what is controlling the current, specifically it is the voltage Vbe.

Ratch

 

[Laplace]

That still makes no difference with what is controlling the current, specifically it is the voltage Vbe.

OK, so now take a similar setup with a current source in series with a resistor of some random value, an ammeter, and a diode. Crank up the current source until the ammeter reads one ampere. Does one know the voltage being supplied by the source? No. Does one know the voltage drop across the resistor? No. Does one know the voltage drop across the diode? No (but it will be close to 0.7V). Do we know the current through the diode? Yes. Exactly one ampere because the current source is forcing one ampere to flow. And the voltage is not relevant.

 

[Ratch]

OK, so now take a similar setup with a current source in series with a resistor of some random value, an ammeter, and a diode. Crank up the current source until the ammeter reads one ampere. Does one know the voltage being supplied by the source? No. Does one know the voltage drop across the resistor? No. Does one know the voltage drop across the diode? No (but it will be close to 0.7V). Do we know the current through the diode? Yes. Exactly one ampere because the current source is forcing one ampere to flow. And the voltage is not relevant.

I anticipated you would present an argument like that. The answer is that the current source is supplying the voltage which lowers the back-voltage of the depletion region and controls the current that is caused by the diffusion process. Therefore, it is a voltage that controls the current no matter where the voltage comes from. It is not necessary to know all the values at every point in the circuit to acknowledge that voltage is controlling the current. If voltage is controlling the current, then it is, in fact, relevant no matter whether we know the voltage value or not.

Ratch

 

[Laplace]

Therefore, it is a voltage that controls the current no matter where the voltage comes from.

There is certainly a correspondence between the diode voltage and diode current. But in this case the current is the independent variable. The independent variable controls the dependent variable.

 

[LvW]

Yes, Vbe determines Ib if one drives the base-emitter junction with a voltage source, since that makes Vbe the independent variable and Ib dependent. But a circuit designer does not drive the base-emitter junction with a voltage source.

I cannot resist to reply to this (false) statement again. Didn`t you read my earlier posts - disproving this claim?
Did you ever hear about the commonly used method to use a voltage divider for biasing the BJT with a VOLTAGE?
And this applies to all three basic configurations - common emitter (CE) as well as CC and CB.
For CB and CC this is the ONLY method that is used.
More than that - with the aim to make the produced voltage as "stiff" as possible the divider is made as low-resistive as possible.
Didn`t you know that?
Here, "as possible" means "as allowed" with respect to other constraints - in particular with respect to the amplifiers input resistance, which shouldn`t be too low.
Of course, for calculating both resistors, the current IB has to be taken into account - simply because it exists (principle of loaded voltage divider).
However, IB has absolutely no controlling properties. If the base current IB would be only 1% of the current through the voltage divider (or smaller) you even could forget the base current at all for during calculating the corresponding two resistors.

 

[Ratch]

There is certainly a correspondence between the diode voltage and diode current. But in this case the current is the independent variable. The independent variable controls the dependent variable.

Before you can make that claim, you have to explain how a current can influence the depletion region of the diode junction. The depletion region current is influenced primarily by the diffusion process and the voltage applied across it. You can calculate what the voltage would be if such and so current existed, but that is not the same as claiming that current controls the diode voltage.

Edit: Sorry for the misstatement. The last word should have been voltage.

Ratch

 

[Laplace]

More than that - with the aim to make the produced voltage as "stiff" as possible the divider is made as low-resistive as possible.
Didn`t you know that?

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.

 

[Laplace]

Before you can make that claim, you have to explain how a current can influence the depletion region of the diode junction.

Not really. I just have to look up the I-V characteristic curve for the diode.

 

[Ratch]

Not really. I just have to look up the I-V characteristic curve for the diode.

Those curves were made by empirical methods. Therefore, they do not show cause and effect. Any attempt to use those curves to show the physical relationship of current and voltage is bogus. Only a physics explanation can determine cause and effect.

Ratch

 

[LvW]

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.

`...emitter feedback for Ib? It is slowly becoming funny.
As you do not want to "repeat that explanation again" I kindly ask you to tell me where you have already explained such a surprising BJT property.

By the way: Have you ever heard that RE provides VOLTAGE feedback which works only in case the base is biased with a voltage (as "stiff" as possible) ?
The current Ib plays no role at all in this feedback scheme. Do you need prooves?

 

[Laplace]

Any attempt to use those curves to show the physical relationship of current and voltage is bogus.

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.