# Solved Problems on Single Stage Transistor Amplifiers

### Q1 : Select a suitable value for the emitter bypass capacitor in Fig. 1 if the amplifier is to operate over a frequency range from 2 kHz to 10 kHz.

Fig. 1

### Solution :

An amplifier usually handles more than one frequency. Therefore, the value of C_{E} is so selected that it provides adequate bypassing for the lowest of all the frequencies. Then it will also be a good bypass ( X_{C} ∝ 1/f ) for all the higher frequencies.

Suppose the minimum frequency to be handled by C_{E} is f_{min}. Then CE is considered a good bypass if at f_{min},

### Q2 :For the transistor amplifier shown in Fig. 2, R_{1} = 10 kΩ, R_{2}= 5 kΩ, R_{C} = 1 kΩ, R_{E} = 2 kΩ and R_{L}= 1 kΩ.

(i) Draw d.c. load line (ii) Determine the operating point (iii) Draw a.c. load line.

Assume V_{BE} = 0.7 V.

Fig. 2

### Solution :

**(i) d.c. load line :**

To draw d.c. load line, we require two end points i.e., maximum V_{CE} point and maximum I_{C} point.

Maximum V_{CE} = V_{CC} = 15 V

This locates the point A (OA = 5 mA) of the d.c. load line. Fig. 3 shows the d.c. load line AB.

Fig. 3

**(ii) Operating point Q :**

The Voltage across series combination of R1 and R2 is 15 V. Applying voltage divider theorem, voltage across

R2 = 5 V.

The voltage across R_{2} (= 5 kΩ ) is 5 V i.e. V_{2} = 5 V.

∴ Operating point Q is 8.55 V, 2.15 mA. This is shown on the d.c. load line.

**(iii) a.c. load line :**

To draw a.c. load line, we require two end points viz. maximum collector-emitter voltage point and maximum collector current point when signal is applied.

This locates the point D (OD = 19.25mA) on the i_{C} axis. By joining points C and D, a.c. load line CD is constructed as shown in Fig. 4.

Fig. 4

### Q3 : In the transistor amplifier shown in Fig. 5, R_{C }= 10 kΩ, R_{L}= 30 kΩ and V_{CC}= 20V. The values R_{1} and R_{2} are such so as to fix the operating point at 10V, 1mA. Draw the d.c. and a.c. load lines. Assume R_{E }is negligible.

Fig. 5

### Solution :

**(i) d.c. load line :**

For drawing d.c. load line, two end points such as maximum V_{CE} point and maximum I_{C} point are needed.

Maximum V_{CE} = 20 V. This locates the point B (OB = 20V) of the d.c. load line on the V_{CE} axis.

This locates the point A (OA = 2 mA) on the I_{C} axis. By joining points A and B, the d.c. load line AB is constructed as shown in Fig.6.

Fig. 6

**(ii) a.c. load line :**

To draw a.c. load line, we require two end points viz maximum collector-emitter voltage point and maximum collector current point when signal is applied.

This locates the point C (OC = 2.33 mA) on the i_{C} axis. By joining points C and D, a.c. load line CD is constructed as shown in the above Fig. 6.

### Q4 : In the circuit shown in Fig. 7, find the voltage gain. Given that β = 60 and input resistance R_{in} = 1 kΩ.

Fig. 7

### Solution :

So far as voltage gain of the circuit is concerned, we need only R_{AC}, β and R_{in}

### Q5 : In the circuit shown in Fig. 8, if R_{C} = 10 kΩ, R_{L} = 10 kΩ, R_{in}= 2.5 kΩ, β =100, find the output voltage for an input voltage of 1mV r.m.s.

### Solution :

### Q6 : In a transistor amplifier, when the signal changes by 0.02V, the base current changes by 10 μA and collector current by 1mA. If collector load R_{C }= 5 kΩ and R_{L}= 10 kΩ, find: (i) current gain (ii) input impedance (iii) a.c. load (iv) voltage gain

(v) power gain.

### Solution :

**(ii) input impedance**

**(iii) a.c. load**

**(iv) voltage gain**

**(v) Power gain **

### Q7 : In Fig. 9, the transistor has β = 50. Find the output voltage if input resistance R_{in}= 0.5 kΩ.

Fig. 9

### Solution :

β = 50, R_{in }= 0.5 kΩ

### Q8 : Determine the ac emitter resistance for the transistor circuit shown in Fig. 10.

### Solution :

### Q9 : For the amplifier circuit shown in Fig. 11, find the voltage gain of the amplifier with (i) C_{E} connected in the circuit (ii) C_{E} removed from the circuit.

Fig. 11

### Solution :

**(i) With C _{E} connected :**

**(ii) Without C _{E} :**

### Q10 : For the circuit shown in Fig. 12, find (i) a.c. emitter resistance (ii) voltage gain (iii) d.c. voltage across both capacitors.

Fig. 12

### Solution :

(i) In order to find a.c. emitter resistance re′ , we shall first find D.C. emitter current I_{E }. To find I_{E}, we proceed as under:

(ii)

(iii) The d.c. voltage across input capacitor is equal to the d.c. voltage at the base of the transistor which is V_{2} = 1V. Therefore, d.c. voltage across C_{in} is 1V.

Similarly, d.c. voltage across C_{E} = d.c voltage at the emitter = V_{E} = 0.3V.

### Q11 : For the circuit shown in Fig. 13, find (i) the d.c. bias levels (ii) d.c. voltages across the capacitors (iii) a.c. emitter resistance (iv) voltage gain and (v) state of the transistor.

Fig. 13

### Solution :

**(i) D.C. bias levels :**

The d.c. bias levels mean various d.c. currents and d.c. voltages.

Therefore, all d.c. bias levels stand calculated.

**(ii) d.c. voltages across the capacitors**

**(iii) a.c. emitter resistance**

**(v) state of the transistor**

As calculated above, V_{C} = 10.4V and V_{E} = 2.3V. Since V_{C} > V_{E}, the transistor is in active state.

### Q12. An amplifier has a voltage gain of 132 and β = 200. Determine the power gain and output power of the amplifier if the input power is 60 μW.

### Solution :

### Q13. For the circuit shown in Fig. 14, determine (i) the current gain (ii) the voltage gain and (iii) the power gain. Neglect the a.c. emitter resistance for the transistor.

Fig. 14

### Solution :

In most practical circuits, the value of a.c. emitter resistance re′ for the transistor is generally quite small as compared to R_{E} and can be neglected in circuit calculations with reasonable accuracy.

(i)

(ii)

(iii)

### Q14. Determine the input impedance of the amplifier circuit shown in Fig. 15.

Fig. 15

### Solution :

### Q15. An amplifier has an open circuit voltage gain of 1000, an input resistance of 2 kΩ and an output resistance of 1Ω. Determine the input signal voltage required to produce an output signal current of 0.5A in 4Ω resistor connected across the output terminals.

### Solution :

Fig. 16 shows the equivalent circuit of the amplifier. Here A0 = 1000.

Fig. 16

### Q16. An amplifier has an open circuit voltage gain of 1000, an output resistance of 15Ω and an input resistance of 7kΩ. It is supplied from a signal source of e.m.f. 10mV and internal resistance 3kΩ. The amplifier feeds a load of 35 Ω. Determine (i) the magnitude of output voltage and (ii) power gain.

### Solution :

Fig. 17

(i)

(ii)

### Q17. An amplifier, when loaded by 2 kΩ resistor, has a voltage gain of 80 and a current gain of 120. Determine the necessary signal voltage and current to give an output voltage of 1V. What is the power gain of the amplifier ?

### Solution :

Fig. 18