# Solved Problems on Amplifiers with Negative Feedback

### Solution :  ### Solution :  ### Solution :  ### Solution :  ### Solution :   ### Solution :  ### Solution : Note that without negative feedback, the change in gain is 20%. However, when negative feedback is applied, the change in gain (4.7%) is much less. This shows that negative feedback provides voltage gain stability.

### Solution : ### Solution : Fig. 1

Fig. 1 shows the equivalent circuit of an amplifier along with the feedback circuit.

Voltage gain of the amplifier without feedback is:  ### Solution : ### Q11. Fig. 2 shows the negative voltage feedback amplifier. If the gain of the amplifier without feedback is 10,000, find : (i) feedback fraction (ii) overall voltage gain (iii) output voltage if input voltage is 1 mV. Fig. 2

### Solution :  ### Q12. Fig. 3 shows the circuit of a negative voltage feedback amplifier. If without feedback, Av = 10,000, Zin = 10 kΩ, Zout = 100 Ω, find :(i) feedback fraction (ii) gain with feedback (iii) input impedance with feedback (iv) output impedance with  feedback. Fig. 3

### Solution : ### Solution :

Gain without feedback, Aν = 150
Distortion without feedback, D = 5% = 0.05
Feedback fraction, mv = 10% = 0.1
If Dvf  is the distortion with negative feedback, then, It may be seen that by the application of negative voltage feedback, the amplifier distortion is reduced from 5% to 0.313%.

### Solution : The new lower cut-off frequency with feedback is The new upper cut-off frequency with feedback is Note the effect of negative voltage feedback on the bandwidth of the amplifier. The lower cut-off frequency is decreased by a factor (1 + mν Aν) while upper cut-off frequency is increased by a factor (1 + mν Aν). In other words, the bandwidth of the amplifier is increased approximately by a factor (1 + mν Aν). where BW = Bandwidth of the amplifier without feedback
BW(f) = Bandwidth of the amplifier with negative feedback

### Solution :  ### Solution :  ### Solution :  ### Solution :  ### Q19. For the emitter follower circuit shown in Fig. 4, find VE and IE. Also draw the dc load line for this circuit. Fig. 4

### Solution :  By joining points A and B, d.c. load line AB is constructed [See Fig. 5]. Fig. 5

### Q20. Determine the voltage gain of the emitter follower circuit shown in Fig. 6. Fig. 6

### Solution :  ### Solution :

When a load of 5 kΩ is added to the emitter follower, the circuit becomes as shown in Fig. 7. Fig. 7

The coupling capacitor acts as a short for a.c. signal so that RE and Rare in parallel.
Therefore, the external emitter resistance RE changes to R’E where ### Q22. For the emitter follower circuit shown in Fig. 8, find the input impedance. Fig. 8

### Solution :  ### Q23. Determine the output impedance of the emitter follower shown in Fig. 9. Given that r’e = 20 Ω. Fig.9

### Solution :  Note that output impedance of the emitter follower is very low. On the other hand, it has high input impedance. This property makes the emitter follower a perfect circuit for connecting a low impedance load to a high-impedance source.

### Q24. Determine (i) d.c. value of current in RE (ii) input impedance of the Darlington amplifier shown in Fig. 10. Fig. 10

### Solution : ### Q25. For the Darlington amplifier in Fig. 11, find (i) the d.c. levels of both the transistors and (ii) a.c. emitter resistances of both transistors. Fig. 11

### Solution : 