# Variation of voltage gain with frequency

For a transistor amplifier, the voltage gain

(a) remains constant for all frequencies.

(b) is high at high and low frequencies and constant in the middle frequency range.

(c) is low at high and low frequencies and constant at mid frequencies.

(d) None of the above.

My attempt: probably naive, since there is more change in voltage at higher frequency it should be option (b) but this is wrong and the correct one is ____. ( I know the answer but not reason)

Any explanation at the level of high school(12th class) would be helpful.

Edit: after comments , the only amplifier circuit diagram discussed is given below

• It should be None of the above, but I'll bet the questioner is making some assumptions about what 'a transistor amplifier' is, so it won't be. – Neil_UK Mar 30 '16 at 14:57
• I found this quite helpful nptel.ac.in/courses/117107095/lecturers/lecture_25/… – JM97 Apr 1 '16 at 1:16
• Where is this schematic from? It's nonsense: the base is biased by a voltage source $V_{BB}$ that would short any input signal, i.e. $v_o$ would always be 0. – Curd Jan 26 '18 at 20:52

Depends of the schematic, for just a transistor it lowers with frequency, but in tipical amplifier it is c). At lower frequencies if coupled with capacitor gain is low because of reactance of that capacitor. At high frequencies it is lower because of transistor gain. If it is DC coupled, than have low gain only on high frequencies.

http://elearning.vtu.ac.in/P9/notes/06ES32/Unit6-MSS.pdf

EDIT: after your update, definitely it is answer C. At lower frequencies you have losses in capacitor before Rb (at frequency 0, it have infinite reactance).

You can see input section: It consist from capacitor, resistor and base-emiter diode (you can think of it as a resistor in this case). Resistance (reactance) of capacitor is $$Xc = \frac{1}{2\pi f C}$$ So resistance of capacitor changes (lowers) with frequency. At low frequencies it is very high, so input signal is divided between it, resistor and base junction and only small part of input signal is "seen" at base junction which is then amplified. Because of that you will have lower output signal at low frequencies.

At higher frequencies, input capacitor influence is negligible, so circuit amplification depends only on transistor which have lower gain at high frequencies.

In the middle reactance of the capacitor is small and also gain of transistor is high, so your gain would be high.

EDIT2: You also have capacitor on output which will also lower gain of low frequencies (similar as input capacitor because of high resistance (reactance)).

• What is the difference between just a transistor and a tipical transistor? Could you explain for reason being c the correct answer. At level of high school – JM97 Mar 30 '16 at 15:17
• I have uploaded the circuit diagram image , hope this may help – JM97 Mar 30 '16 at 15:32
• What do you mean by input voltage is divided – JM97 Mar 30 '16 at 16:25
• Divided by "resistive" divider formed by capacitor, resistor and base junction. On base junction you will get only part of voltage (Rbe / Rtotal), Rtotal = sum of resistances at given frequency. – Darko Mar 31 '16 at 8:04
• Could you please elaborate why transistor has low gain at high frequencies? – JM97 Mar 31 '16 at 16:02

The answer is highly dependent on what exactly is meant by "transistor amplifier". The question is either ambiguous, or it assumes context from something that was discussed in class that we don't know about. Transistor amplifiers can be designed for all kinds of frequency responses.

That said, all transistors stop working like transistors above some frequency. Therefore, the gain of a transistor amplifier will go down (assuming it was above 1 in the first place) at high frequencies. However, this is probably not what your teacher is looking for.

Again, this is either a bad question or it assumes context within your particular class.

• I have uploaded the circuit diagram in the question , hope this helps – JM97 Mar 30 '16 at 15:26
• @JM97: Since you already accepted a answer, there is no point going into this further. – Olin Lathrop Mar 30 '16 at 18:19

Me too in class 12. So, I would like to answer it.

At lower frequencies: As you know, the formula for Capacitive reactance since unit 4.

$$\mathrm{X_C = \frac{1}{2\pi f C}}$$ $$\mathrm{X_C \propto \frac{1}{f}}$$

As the frequency (of the input voltage signal) would be lower, the capacitor in the input circuit would offer high reactance. That is, there will be some kind of drop of voltage (AC) across the capacitor.

We know from the circuit diagram and from our knowledge of transistor amplifier,

$$\mathrm{V_{BB} + v_i(input) = I_BR_B + V_{BE} + \Delta I_B(R_B + r_i(input \ resistance))}$$

As I said, there will be a voltage drop across the input capacitor, so input voltage would decrease. As a result, the following quantities decrease:

$$\mathrm{(V_{BB}+v_i), \ I_B(Base \ current), \ I_C(Collector \ current) \ as \ I_B \propto I_C(I_C = \beta \ I_B)\ in \ the \ active\ state \ of \ transistor.}$$

Hence,

$$\mathrm{\Delta V_{CE} = V_{o}(output \ voltage) \ decreases.}$$

As for same input voltage, we are getting less output voltage, so the voltage gain decreases.

At higher frequencies: At higher frequencies, the current amplification factor of the transistor decreases(It is the nature of the transistor, it do so at higher frequencies).The performance and its capability decreases as the frequency increases. See the following figure:

As we all know, the current gain, $$\mathrm{A_v = \frac{-\beta_{ac} \ R_L}{r}; \ r = r_i + R_B; \ R_L = output\ resistance}$$

Thus, if the beta value decreases, the voltage gain decreases.

So, overall, we will get a graph like this (Voltage gain versus frequency):