# Transistor Analysis with Only RsubE

#### SamR

Joined Mar 19, 2019
4,334
I'm more than a bit perplexed with this curve ball they threw at me. No RsubB or RsubC! My thought was the IsubB would be the Vbb - Vbe/RsubE which is 1.3mA. But peeking at the answer key that is what they have for IsubE! The value for IsubB in the answer key is 30uA and my model schematic in LTS using the generic NPN model gives 12uA. I'm missing something here and I don't think it is using the αsubDC, which I could use to double-check my IsubC/IsubE relationship. No model number was given for the xstr so the PDF is not available for reference. Prior problems have Assumed VsubCE(sat) is 0.2V if that is of any help? #### michael8

Joined Jan 11, 2015
296
perhaps they are using Vbb = 0.5 volts?

#### BobTPH

Joined Jun 5, 2013
5,453
Assuming a Vbe of 0.6 the emitter current is 1.4 ma. Then the collector current is 1.4 / 0.98 or 1.429 ma and the base current is 0.029 or 29 uA. Close enough for me.

• SamR

#### SamR

Joined Mar 19, 2019
4,334
Assuming a Vbe of 0.6 the emitter current is 1.4 ma
My first approach was Vbb lost 0.7V as Vbe which left 1.3V coming from Vbb across Re. But then there is also Vcc - Vce across Re. And the answer key gives Ib as 30uA. Just how do you get Ie as 1.4 or 1.3mA without knowing what Ib or Ic are? Still confused... And NONE of the section examples are without Rb and Rc and have no Re in the circuit. Am I wrong in thinking that the 2Vbb-0.7Vbe/Re gives Ib= 1.3V/Re = an Ib of 1.3mA?

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#### crutschow

Joined Mar 14, 2008
30,419
Assuming a Vbe of 0.6 the emitter current is 1.4 ma. Then the collector current is 1.4 / 0.98 or 1.429 ma
You have it reversed.
The emitter current equals the collector current plus the base current, so the collector current is 1.4mA (0.98) = 1.372ma, giving an Ib base current of 28µA.
α (alpha) of a transistor is the factor or value that an emitter current is multiplied by to give the value of the collector current.
Am I wrong in thinking that the 2Vbb-0.7Vbe/Re gives Ib= 1.3V/Re = an Ib of 1.3mA?
Yes.
See above.

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#### SamR

Joined Mar 19, 2019
4,334
Where are you getting the Ie of 1.4 mA from? I usually find the Ib=(Vbb-Vbe)/Rb and then using β * Ib = Ic. Then Ib + Ic = Ie. How did you get the Ie= 1.4mA? I understand converting α=Ic/Ie to α*Ie=Ic, I am not seeing how you found the Ie from only what was given in the schematic?

#### crutschow

Joined Mar 14, 2008
30,419
Where are you getting the Ie of 1.4 mA from?
For Vbb = 2V and an assumed Vbe of 0.6V then, assuming linear operation, the voltage across Re must be 1.4V, giving an Re (emitter) current of 1.4mA.
Using the value of alpha, from that you can calculate the collector current and thus the base current (emitter current minus collector current).

Incidentally β is approximately 1 / (1-alpha) or 50 here.

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• SamR

#### crutschow

Joined Mar 14, 2008
30,419
What about the Vce component of Ve?
There is no significant Vce component of Ve.
Why do you think there is?
A transistor acts as a current source with a high value of collector impedance so Vce has little effect on Ve.

If you look at the characteristic curves of a bipolar transistor you will see that, in the active region, the line showing collector current change with collector voltage is nearly horizontal, indicating a high impedance (below): #### SamR

Joined Mar 19, 2019
4,334
you will see that, in the active region, the line showing collector current change with collector voltage is nearly horizontal,
Yes, nearly, but there is a slight gain in current and the PDFs give a slight 0.1 - 0.2V typical value for Vce(sat) so I expected that to be in the Ve across Re. Seems I need to re-evaluate my expectations... K, so there is no Vce component in Ve. Got it now, thx!

#### crutschow

Joined Mar 14, 2008
30,419
the PDFs give a slight 0.1 - 0.2V typical value for Vce(sat)
Yes, but the transistor in this circuit is operating in the active region, not the saturation region (where it would be acting as a switch), so the saturation voltage is not in play.

To look at the less than ideal, since the active curve is not totally horizontal, there will be a slight increase in collector current with an increase in collector voltage, but the effect of that is, slightly less base current will then be required to maintain a given emitter current, so the apparent Alpha and Beta will appear some higher.
That's why Beta is always specified at a particular collector voltage.

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• SamR

#### SamR

Joined Mar 19, 2019
4,334
And this is why I'm studying xstrs. Used them as switches for many years without really understanding what they were doing. Thx for the input. Lots to digest and I'm not as quick as I used to be.

#### LvW

Joined Jun 13, 2013
1,478
And this is why I'm studying xstrs. Used them as switches for many years without really understanding what they were doing. Thx for the input. Lots to digest and I'm not as quick as I used to be.
Just an additional short hint:
If you really want to know what "they were doing" (and why), you might consider whether it is really physically possible for a very small current to control a current that is perhaps a factor of 200 larger.
The answer is no - this is not possible (even from the energy aspect).
In reality, the transistor is not a current controlled device (but in many cases, it is easier assume Ic=f(Ib))
The well-known relationship - established by W. Shockley - for the current-voltage relationship at the pn-junction always applies:
Ie=Is[exp(Vbe/Vt)-1].
And such an exponential relationship, therefore, also applies also to Ic and Ib because of Ie=Ic+Ib.
That means: All three currents are the result of Vbe and its variations.

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• SamR

#### SamR

Joined Mar 19, 2019
4,334
Yes, Ebers Moll... I have that in the to-do pile.

#### crutschow

Joined Mar 14, 2008
30,419
So even though a BJT looks like a current-controlled device, and quacks like a current-controlled device, in theory it is a voltage-controlled device.
But for some circuit calculations, such as bias and switching, it is easier to use a simple, black-box current-controlled model (which is why the data sheets give parameters for current-gain, and the characteristic curves show collector current versus base current).

#### LvW

Joined Jun 13, 2013
1,478
So even though a BJT .......quacks like a current-controlled device.......
Does it? I never have heard such a noise....where did you hear something lke that?
In contrary, there are many, many "quacks" which are a clear and unique sign for voltage control.

#### BobTPH

Joined Jun 5, 2013
5,453
You have it reversed.
The emitter current equals the collector current plus the base current, so the collector current is 1.4mA (0.98) = 1.372ma, giving an Ib base current of 28µA.
α (alpha) of a transistor is the factor or value that an emitter current is multiplied by to give the value of the collector current.
Yes.
See above.
I new that. Or I would have of had thought about it little longer.

#### crutschow

Joined Mar 14, 2008
30,419
Does it? I never have heard such a noise....where did you hear something lke that?
I have super hearing. Plus I've looked at the BJT's characteristic curves.
Nothing there about base voltage.
Notice that the characteristic curves for the voltage-controlled FET do show gate voltage.
there are many, many "quacks" which are a clear and unique sign for voltage control.
They are rather faint for a BJT. But they are there, of course, if you study the solid-state theory.

However, using a voltage-control model for BJTs is of little use for calculating transistor bias points in a linear circuit, or determining switching control parameters.
But its transconductance value can be of use for determining small-signal AC gain.

As an observation, a device with only a few kohms dynamic input impedance (at typical operating currents) is not typical of what is usually considered a voltage-controlled device, which ideally has an infinite input impedance.

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#### LvW

Joined Jun 13, 2013
1,478
However, using a voltage-control model for BJTs is of little use for calculating transistor bias points in a linear circuit, or determining switching control parameters.
But its transconductance value can be of use for determining small-signal AC gain.
Just the opposite is true.
Examples:
* The emitter resistor RE provides current-conrolled VOLTAGE feedback - VBE is reduced when IE rises.
* Most gain stages contain a base voltage divider for biasing the BJT. This divider chain is chosen so that it provides a nearly "stiff" VOLTAGE at the base. For this reason, the current through this chain is chosen to be at least 10 times the expected base curent - with the aim to be less sensitive to the huge tolerances of this base current (which is an unwanted by-product).
* What about class-AB or class-B amplifiers? Do we not select a bias voltage Vbe on the voltage control curve Ic=f(Vbe) much lower than the typical value of 0.7 volts?
* Temperature influence: All temperature-compensating measures are based on the classsical tempco of -2mV/K.
There is no corresponding figure which specifies how much the base current must be reduced per degree.
* Switching: Saturation is defined based on VOLTAGES (Vc<Vb for npn), not based on currents.
* Transconductance gm (most important parameter for voltage gain): gm=d(Ic)/d(Vbe).
That means: gm is the slope of the VOLTAGE-control characteristic curve.

Last question: How can you say that "..voltage-control model for BJTs is of little use..." ?
Would you please give me some examples which show that the current-control model is of more use (more benefit)?

As an observation, a device with only a few kohms dynamic input impedance (at typical operating currents) is not typical of what is usually considered a voltage-controlled device, which ideally has an infinite input impedance.
I think this is not a good argument.
Would you then say that "a device with only a few kohms dynamic input impedance" is a typical current-controlled device (which ideally has ZERO input impedance?). No - I dont think that such considerations are a valid method to discriminate between voltage- and current-control.

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#### crutschow

Joined Mar 14, 2008
30,419
Just the opposite is true.
Most of your examples do not use the voltage-controlled model for the BJT (its transconductance).

For example, saturation may be defined by voltages, but to insure saturation, the base current is typically made 1/10th of the maximum collector current.

The base-emitter input voltage can be viewed as part of the base-emitter diode junction, which gives the typical Vbe operating voltages, and the observed temp coefficient of that voltage.

The base-current may be a "byproduct" of the transistor operation but never-the-less it is stated in all BJT data sheets, appears in its characteristic curves, and is used in the design of BJT circuits.

I don' understand you comment about Class AB or B amplifiers.
You still use the same Vbe value when the transistor is conducting.

I see no reason to be pedantic about what model to use with BJTs (although I know several on these forums are).
In some design calculations the voltage-control model is easier to use, and in other cases the current-control (black-box) model.
No - I dont think that such considerations are a valid method to discriminate between voltage- and current-control.
Seems valid to me, but that's your opinion and you are welcome to it.

So how do you discriminate?
Is it, that as long as a current-source has a more-than-zero input impedance, it's a voltage controlled device(?).