# Solved Problems on Amplifiers with Negative Feedback

**Q1. The voltage gain of an amplifier without feedback is 3000. Calculate the voltage ****gain of the amplifier if negative voltage feedback is introduced in the circuit. Given that feedback fraction m _{v} = 0.01.**

**Solution :**

**Q2. The overall gain of a multistage amplifier is 140. When negative voltage feedback ****is applied, the gain is reduced to 17.5. Find the fraction of the output that is fedback to the input.**

**Solution :**

**Q3. When negative voltage feedback is applied to an amplifier of gain 100, the ****overall gain falls to 50.**

**(i) Calculate the fraction of the output voltage fedback.**

**(ii) If this fraction is maintained, calculate the value of the amplifier gain required if the overall stage gain is to be 75.**

**Solution :**

**Q4. With a negative voltage feedback, an amplifier gives an output of 10 V with an ****input of 0.5 V. When feedback is removed, it requires 0.25 V input for the same output. Calculate (i) gain without feedback (ii) feedback fraction m _{v }.**

**Solution :**

**Q5. The gain of an amplifier without feedback is 50 whereas with negative voltage ****feedback, it falls to 25. If due to ageing, the amplifier gain falls to 40, find the percentage reduction in stage gain (i) without feedback and (ii) with negative feedback.**

**Solution :**

**Q6. An amplifier has a voltage amplification Av and a fraction m _{v }of its output is **

**fedback in opposition to the input. If m**

_{v }= 0.1 and Aν = 100, calculate the percentage change in the gain of the system if Aν falls 6 db due to ageing.**Solution :**

**Q7. An amplifier has a voltage gain of 500 without feedback. If a negative feedback ****is applied, the gain is reduced to 100. Calculate the fraction of the output fed back. If, due to ageing of components, the gain without feedback falls by 20%, calculate the percentage fall in gain with feedback.**

**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.

**Q8. An amplifier has an open-loop gain Av = 100,000. A negative feedback of 10 db ****is applied. Find (i) voltage gain with feedback (ii) value of feedback fraction m _{v }.**

**Solution :**

**Q9. An amplifier with an open-circuit voltage gain of 1000 has an output resistance ****of 100 Ω and feeds a resistive load of 900 Ω. Negative voltage feedback is provided by connecting a resistive voltage divider across the output and one-fiftieth of the output voltage is fedback in series with the input signal. Determine the voltage gain with negative feedback.**

**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:

**Q10. An amplifier is required with a voltage gain of 100 which does not vary by ****more than 1%. If it is to use negative feedback with a basic amplifier the voltage gain of which can vary by 20%, determine the minimum voltage gain required and the feedback factor.**

**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, Z _{in }= 10 kΩ, Z_{out }= 100 Ω, find :(i) feedback fraction (ii) gain with feedback (iii) input impedance with feedback (iv) output impedance with feedback.**

Fig. 3

**Solution :**

**Q13. The gain and distortion of an amplifier are 150 and 5% respectively without ****feedback. If the stage has 10% of its output voltage applied as negative feedback, find the distortion of the amplifier with feedback.**

**Solution :**

Gain without feedback, Aν = 150

Distortion without feedback, D = 5% = 0.05

Feedback fraction, m_{v} = 10% = 0.1

If D_{vf } 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%.

**Q14. An amplifier has a gain of 1000 without feedback and cut-off frequencies are ****f1 = 1.5 kHz and f2 = 501.5 kHz. If 1% of output voltage of the amplifier is applied as negative feedback, what are the new cut-off frequencies ?**

**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

**Q15. The current gain of an amplifier is 200 without feedback. When negative ****current feedback is applied, determine the effective current gain of the amplifier. Given that current attenuation m _{i} = 0.012.**

**Solution :**

**Q16. An amplifier has a current gain of 240 and input impedance of 15 kΩ without feedback. If negative current feedback (m _{i} = 0.015) is applied, what will be the input impedance of the amplifier ?**

**Solution :**

**Q17. An amplifier has a current gain of 200 and output impedance of 3 kΩ without ****feedback. If negative current feedback (m _{i} = 0.01) is applied; what is the output impedance of the amplifier ?**

**Solution :**

**Q18. An amplifier has a current gain of 250 and a bandwidth of 400 kHz without ****feedback. If negative current feedback (m _{i }= 0.01) is applied, what is the bandwidth of the amplifier?**

**Solution :**

**Q19. For the emitter follower circuit shown in Fig. 4, find V _{E }and I_{E}. 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 :**

**Q21. If in the above example, a load of 5 kΩ is added to the emitter follower, what ****will be the voltage gain of the circuit ?**

**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 R_{E} and R_{L }are in parallel.

Therefore, the external emitter resistance R_{E} 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 R _{E} (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 :**