Voltage gain of a bipolar transistor circuit

[eyeweyew]

TL;DR Summary

If base-to-collector voltage gain is negative, base-to-emitter voltage gain is positive, should emitter-to-collector voltage gain negative as a result of multiplying base-to-collector gain by the inverse of base-to-emitter gain? How come emitter-to-collector gain is positive?

For a simple bipolar transistor above, base-to-collector voltage gain ≈ -gm*Rc/(1+gm*Re) < 0, base-to-emitter voltage gain ≈ 1 > 0, should emitter-to-collector voltage gain ≈ -gm*Rc/(1+gm*Re) * 1 < 0. How come it is equal to gm*Rc > 0?

 

[eyeweyew]

I think I know why. Compute this way would only give indirect gain from emitter to collector.

 

[tech99]

Why are we multiplying the two gain figures? The output voltages are in series, so if being referenced to ground they are subtracted to obtain the signal voltage between the collector and emitter.

 

[DaveE]

Are you asking about the common base BJT amplifier?
https://www.electronics-tutorials.ws/amplifier/common-base-amplifier.html

 

[eyeweyew]

I guess a better way to rephrase my question is why base-to-collector voltage gain(common emitter) vc/vb does not equal to emitter-to-collector voltage gain(common base) vc/ve multiplied by base-to-emitter voltage gain(emitter follower) ve/vb?

I believe it is related to inspection analysis limitations for circuit with feedback.

 

[Averagesupernova]

The voltage gain of transistor circuits is determined by more than just the transistor. Resistors in the circuit have alot to do with the voltage gain.

 

[Tom.G]

I guess a better way to rephrase my question is why base-to-collector voltage gain vc/vb does not equal to emitter-to-collector voltage gain vc/ve multiplied by base-to-emitter voltage gain ve/vb?

Lets try this.

Recall that a transistor is a current amplifier, the Collector current is proportional to the base current.

Also realize that the Collector current has to flow thru the Emitter.

The result is that with similiar currents, the voltage gains at the Collector and at the Emitter are determined by the resistor values - similiar resistor values yields similiar voltage gains.

There is, however, a slight Fudge Factor.

The "Fudge Factor" needed comes when you realize that the Base current also flows thru the Emitter.

A detailed explanation can be found at:
https://www.electronics-tutorials.ws/amplifier/phase-splitter.html

(above link, and many more, found with:
https://www.google.com/search?hl=en&q=phase+splitter+circuit)

Cheers,
Tom

 

[LvW]

For a simple bipolar transistor above, base-to-collector voltage gain ≈ -gm*Rc/(1+gm*Re) < 0,

This gain formula for a common-emitter gain stage with negative Re-feedback is correct - however, only in case the signal source is connected DIRECTLY to the base node (via a coupling capacitor).
When there is a resistor Rb (as in your drawing) this resistor forms a voltage divider Rb-r_in and the gain will be correspondingly lower (r_in=dynamic signal input resistance at the base node).

Just for your understanding; "Negative gain" means nothing else than a phase inversion (180deg phase shift) between the signal voltage at the base node and the output voltage at the collector.

 

[DaveE]

Recall that a transistor is a current amplifier

Yes. I find these circuits much simpler to understand and solve if, at the device level, you think in terms of currents, not voltages. Of course voltages make currents and vice-versa.

Most of your questions can be clarified if you redraw your circuit using the basic T model of a BJT shown here.

Some like to move the small signal emitter resistor $r_e$ to the base, where its value becomes $(\beta + 1) r_e$. Although you'll often see $\beta r_e$, since $\beta + 1 \approx \beta$ for high gain devices.

 

[Baluncore]

To quickly identify the BJT polarity in the circuit, draw the symbolic arrow head on the emitter. Also, identify the polarity of Vs.

 

[LvW]

Most of your questions can be clarified if you redraw your circuit using the basic T model of a BJT shown here.

Some like to move the small signal emitter resistor $r_e$ to the base, where its value becomes $(\beta + 1) r_e$. Although you'll often see $\beta r_e$, since $\beta + 1 \approx \beta$ for high gain devices.

I like to mention that there are two other small-signal equivalent circuit diagrams for a bipolar transistor, which - according to my opinion - are much better and clearer than the mentioned T-model. These two models are much closer to the real transistor function (although the T-model works - if applied properly!).
Why?
* Because the T-model containes a quantity re which can be the cause for some misinterpretations and misunderstandings.
* At first, one must know that re is a dynamic quantity (not constant) which depends on the DC quiescent current .
* From the physical point of view it is not a resistance (although it has the unit V/A) and it does not belong to the emitter. Instead, it is an abbreviation (symbol) for the invers transconductance gm=1/re.
* Misinterpretations can occur when someone thinks that such a quantity (located in the emitter path) could have any influence on the DC operating point and/or could cause negative feedback.
* More than that - looking into the base node of the model (Fig. 3 in the linked doc) one could have the impression (visual inspection) that the input resistance would be identical to re (which is wrong).

 

[DaveE]

I like to mention that there are two other small-signal equivalent circuit diagrams for a bipolar transistor, which - according to my opinion - are much better and clearer than the mentioned T-model. These two models are much closer to the real transistor function (although the T-model works - if applied properly!).
Why?
* Because the T-model containes a quantity re which can be the cause for some misinterpretations and misunderstandings.
* At first, one must know that re is a dynamic quantity (not constant) which depends on the DC quiescent current .
* From the physical point of view it is not a resistance (although it has the unit V/A) and it does not belong to the emitter. Instead, it is an abbreviation (symbol) for the invers transconductance gm=1/re.
* Misinterpretations can occur when someone thinks that such a quantity (located in the emitter path) could have any influence on the DC operating point and/or could cause negative feedback.
* More than that - looking into the base node of the model (Fig. 3 in the linked doc) one could have the impression (visual inspection) that the input resistance would be identical to re (which is wrong).

Yes, that's how it is with simple models. They're wrong. That's what makes them simple.

 

[Averagesupernova]

Re' is what we called it in school. I assume this is what is referred to as re in the above posts. Normally it was set at 25 ohms for the math we did concerning the circuits we worked with. As said, it varies with emitter current. Re' is one of the things that determine input impedance at the base. Re' will also have a hand in determining signal gain especially when the emitter resistor is bypassed with a capacitor. Of course this is assuming common emitter configuration.
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The numbers don't come out perfect, but they are close

 

[LvW]

Re' is what we called it in school. I assume this is what is referred to as re in the above posts. Normally it was set at 25 ohms for the math we did concerning the circuits we worked with. As said, it varies with emitter current. Re' is one of the things that determine input impedance at the base. Re' will also have a hand in determining signal gain especially when the emitter resistor is bypassed with a capacitor. Of course this is assuming common emitter configuration.
-
The numbers don't come out perfect, but they are close

With all respect - your post demonstrates how misinterpretations (and mistakes) can occur using the T-model. In your post, the small-signal quantity 1/gm (inverse transconductance) is written with capital letters as Re´. This looks like a classical ohmic resistor.
The problem is that, in particular, beginners are using such small-signal equivalent diagrams - and a misinterpretation of this model would be fatal.
Experienced people do not need such eqivalent diagrams - they know how a transistor works and do not need any model.

 

[Averagesupernova]

My bad. It's actually r'e. So you can blame me for confusing the 'inexperienced'. So, @LvW, are you 'experienced' or 'inexperienced' in that you didn't catch my mistake with Re' vs r'e? The correct one (r'e) is right out of a text book.
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The 'experienced' folks don't need a lot of things. That doesn't change the fact that a particular property of a transistor, or anything for that matter, exists and needs to be defined. BTW, this is not the first time I've seen something besides r'e defining that particular property of a transistor.

 

[DaveE]

Experienced people do not need such eqivalent diagrams - they know how a transistor works and do not need any model.

I have had some experience in this field and, yes, I use the simplest T (or Hybrid-π) model nearly always for BJTs. I probably won't draw it that way on paper, but that is exactly what I am thinking about. Step 1 is figure out roughly what the circuit does (amplifier or oscillator, saturated or linear, high or low power, closed or open loop, etc.) More complex stuff is added later only as needed.

We all use models, real components are complicated things. I also only use about 5 digits in π.