KCC's quizzes AQQ235 about a bipolar common emitter amplifier - a kind proposal from our colleague Martin Walker

1.  Is there enough information?  I assume that we have the values of the resistors, within some range like Rb1 = 460k and Rb2 = 100k, and something reasonable for Vs, like between 1 and 2 volt for example. In fact, with these values, I found that we can get the beta value of the transistor by reading Vout:  beta = 1200 - 200 Vout.  (Not bad, I was impressed of the linearity from beta between 200 and 700 ! ) So, I would answer yes to this question, given "reasonable" assumptions. As counter-example, we can easily enter in deep saturation of the transistor too with Rb2 >= Rb1, that is why I assumed among other things that we are able to get a reading of 3 volt for Vout, which implies that the NPN is not blocking, not in deep saturation, but in active mode). And clearly, beta cannot become negative, so the linearity is definitively limited to some "range".
2. Within what I consider reasonable variations, the formula for beta that I gave up here is independent of Vs ( at least, for values of Vs between 1 and 2 volt). I don't think that we can speak of (constant) gain in this case, if the gain is computed as Vout /  Vs.
3. I won't describe this circuit as an amplifier, but as a beta-reader (given the linearity).
4. It is surely very dependent of temperature since it seems controlled by the Vbe diode behavior. I am not sure on what direction to go in. Do we wish to keep the linearity of the beta value versus Vout, do we wish an amplifier, or something else.

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Additional: You can experiment with the CircuitLab simulation of the circuit (we obtain a lightly different equation, with a less ideal model for the NPN): /https://www.circuitlab.com/circuit/sd2453n4byu6/betareader/

First let's assume by "gain" you mean small signal voltage gain. In that case for the common emitter configuration without an emitter degeneration resistor the gain Av is gm * Rc (a dimensionless ratio) . For a BJT device we know that gm = Ic/Vt (where Vt is the thermal voltage which is about 26 mV at room temperature). We are also given that half of Vcc or 3 V is dropped across Rc so Ic = 3/Rc. Using these equations and solving for Av we get 3V/26mV or about 115. It turns out that the absolute value of Rc drops out in the calculation and all we need to know is that Rb1 and Rb2 are such that the voltage at the collector is as stated in the problem statement.

-DM

1. Is there enough information to work out the gain of this circuit?

Yes, there is enough information to work out the gain of this circuit.

2. If so, what is the voltage gain of the circuit?

The gain of  this circuit can be calculated by RC/Re, where Re is the  internal resistance of the transistor at BE junction.

3. Is it a good amplifier?

It can be used as a very small signal high gain amplifier but it is not stable.  No, it is not  good as amplifier.

4.  If not, what would you do to mitigate for its limitations?

This amplifier configuration is highly  sensitive to temperature variations as Re is temperature dependent.

The gain will also be very high due to low Re and hence would be unstable.

To get over these limitations , a resistor is RE is added from emitter to ground of the transistor.

This Resistor will add negative feedback at emitter and stabilise the transistor but  reduce the gain.

The Gain now would be RC / (RE + Re).

To make the gain independent of Re, RE is made bigger than Re to make its effect negligible , hence getting over temperature variations.

If a higher gain is still required , for AC signals, a capacitor CE can be added across RE, this will bring back the gain to RC/Re.

1. Given the conditions, the transistor should be in its linear region. Assuming small signal voltage gain is what we mean by gain "Av= vout / vs", then we need the Rc value to evaluate Av= beta * Rc / hie (hie is the input impedance in the BJT hybrid parameters model)

2. Av= beta * Rc / hie ( Rs considered negligeable)

3. Very sensitive to temperature variations

4. You can add a resistor between emitter and ground (that will introduce a negative feedback for stabilization) and maybe a capacitor in parallel. 

1. No, the amplification depends on the current gain of the Transistor an on its Base-Emitter diode characteristics. 

2. -

3. No, its amplification is very dependent on the actual transistor, and its temperature. The operating point will not be stable and thus the available voltage range at the output is not guaranteed 

4. The amplifier should be enhanced by an Emitter resistor to stabilize operating point of the amplifier. If a high gain is desired, an additional R-C in parallel to this emitter resistor will enhance gain. 

1. No, the amplification depends on the current gain of the Transistor an on its Base-Emitter diode characteristics. 

2. -

3. No, its amplification is very dependent on the actual transistor, and its temperature. The operating point will not be stable and thus the available voltage range at the output is not guaranteed 

4. The amplifier should be enhanced by an Emitter resistor to stabilize operating point of the amplifier. If a high gain is desired, an additional R-C in parallel to this emitter resistor will enhance gain. 

Thanks Michel for your extensive  explanation on this circuit! Answer to question 1 is correct! A lot can be said when you know the Vout is set at the middle of VCC.

Concerning questions 2, 3 and 4, some different interpretations / appreciations can be made; we can wait for other feedback to compare their arguments. But your remark related to the temperature effect is true.

Thanks Doug! Yes, the "gain" mentioned here is of course related to the small signal gain! Your reasoning conducting to the gain estimation is brillant! We will see later how other will describe this circuit! May be some comment from the original author...

Thanks Rajesh for your prompt reply! Correct to say temperature variation is the weak point of this circuit, which can still be seen as an amplifier since it is capable to magnify a few tens of mV (at Vs) to several volts at Vout with very standard BJTs such as BC107, BC108 or BC109. We will hear from others to conclude further...

Merci Gaetano! Answers 1, 3 and 4 are fully correct! On question 2, may be if you elaborate further you can come to a gain expression that gives Ac independent of beta (if beta is large, above 80; which is the case with standard BJTs like BC107, BC108 or BC109 having beta of several hundreds. But let's see our coming other feedback!