Module 4 Unit 4 Feedback in Amplifiers

Transcription

1 Module 4 Unit 4 Feedback in mplifiers eview Questions:. What are the drawbacks in a electronic circuit not using proper feedback? 2. What is positive feedback? Positive feedback is avoided in amplifier circuits. Discuss. 3. Explain negative feedback. In what ways does it modify amplifier performance? 4. What are the different types of negative feedbacks? Draw appropriate circuits. 5. Write the basic expression for voltage gain of an amplifier with feedback and explain the terms used in the expression. 6. What type of feedback will you use in an amplifier to (i) Increase input impedance (ii) Decrease input impedance 7. Draw the gain-frequency response of an C coupled amplifier. Discuss fall in gain at very low and at very high frequencies. 8. Two identical amplifier stages having gains of 50 each and phase shift between input and output signals of 80 o each are cascaded. How much is the over all gain and phase shift of 2-stage amplifier? 9. What are the different methods of coupling signals used in amplifiers? Discuss their advantages and disadvantages. 0. Two transistors having current gains of 00 and 80 respectively form Darlington pair. How much is current gain of the Darlington pair?

2 Problems:- 4. In the series shunt feedback amplifier shown in fig. calculate the voltage gains without feedback,, and with feedback F. + 5μv + v o (=v) - - ~ 99k v i k 2

3 Solution:- The open loop gain, is defined as the ratio of output voltage, v o. to the error voltage, v e, which is input to the basic amplifier. Therefore, v V o 5 = ve 5 0 V Voltage gain of feedback amplifier, F, is expressed as, F Where is gain of the feedback network. Since >> F nd is, 2 k k 99k Therefore, F F

4 4.2 Common emitter (CE) amplifier shown in fig. has voltage gain of 200 when E = 0. Stability is brought through negative feedback by adding resistor E. Calculate the value of resistor E using feedback concepts so that final voltage gain (= F ) is equal to V 50k 6k v o v i 50k k E

5 Solution:- The value of gain of feedback network, can be obtained using the basic feedback relation, F Since F = 00 and = 200, we have or 00 Further, the value of is given by the ratio, E C or, E C E 60

6 4.3 The feedback amplifier shown in fig. makes use of an op amp with internal gain (open loop gain) = 0 5. How much is the output voltage for input signal v s = 2mV in the circuit shown. +2V v s ~ k k V 49k k 2 v o L =0k

7 Solution:- The gain of feedback amplifier, F, is F ecause >> nd, gain of feedback network,, is 2 k k 49k 50 Therefore, F 50 nd the output voltage, v 0, is v o = F X v s = 50 X 2 mv or v 0 = 00 mv

8 4.4 n amplifier exhibits distortion in the form of voltage fluctuations of ~ 0%. These fluctuations are to be restricted to ~ % by incorporating negative feedback in the amplifier circuit. If ultimate gain is desired to be 20, what should be the open loop gain of the amplifier? Solution:- The distribution, D F in an amplifier with feedback and distortion, D without feedback in the amplifier are related as, D F D ( ) Where is open loop gain of amplifier and is the gain of feedback network. Now, D and D F 0% % ( given) ( given) Therefore, using above equation, 0 or, ( or, ( Now, or, ) ) F ( F 0 0 ) 0 20 ( F 20)

9 4.5 Figure shows an op amp circuit with voltage series feedback through resistors and 2. The internal gain of op amp is 5 X 0 4 and input impedance is 00kΩ. Find out the gain and input impedance of the amplifier with feedback. +2V + v o - v s ~ -2V 2 98k 2k k

10 Solution:- The gain of amplifier with feedback, F, is F Where is open loop gain of amplifier and is gain of feedback network. Since >> F Now, 2 2k 2k 98k 00 Therefore, F F The input impedance with and without feedback in a voltage-series circuit are related as, Z i(f) = ( + ) Z i Z i Or Z i(f) = 5 X 0 4 X 0-2 X 00k Or Z i(f) = 50 MΩ

11 4.6 n op amp has open loop band width of 5Hz at gain 2 X 0 5. Feedback is incorporated and resistors and 2 form the feedback network. Find the bandwidth of the amplifier shown in fig. k k +9V 00k 2 v i V v o

12 Solution:- The gain of feedback network,, in the circuit of fig. is 2 k k 00k The bandwidth with feedback, f (F) and without feedback (Open loop bandwidth), f 2, are related as f (W) = ( + ) f 2 Where is open loop gain of op amp or f (W) f 2 or f (W) = 2 X 0 5 X 0-2 X 5 H Z or f (W) = 0 kh Z

13 4.7 Two C coupled amplifiers are connected to form a 2-stage amplier. If the lower and upper cutoff frequencies of each individual amplifier respectively are 00 H Z and 20 kh Z, What these frequencies are for the 2-stage amplifier? Solution:- For multi-stage amplifier consisting of n identical amplifying stages, the lower cut off f (multi) is expressed as, f ( multi) f / n 2 () Where f is lower cut-off of individual stages. nd the upper cutoff f 2(multi) is expressed as, f f / n 2( multi ) 2 2 (2) Where f 2 is upper cut-off of individual stages. Therefore, the lower cut-off of 2-stages amplifier (Eq()) is f (2 stage) or f (2 stage) 2 00 / H Z Similarly, the upper cutoff of the 2-stage amplifier (Eq 2), is f 2(2 stage) or f 2(2 stage) f / kH 2.8 kh Z Z Thus the band width gets reduced in a multi-stage amplifier.

14 4.8 The two-stage amplifier shown in fig uses transistors Q and Q 2, both having current gain β of 80 and dynamic emitter resistance, r e, of 25 Ω each. Find out the over all voltage gain of the amplifier. +2V C 0k 200Ω Q 2 Q v o v i 3k

15 Solution:- In a multi-stage amplifier, it is convenient to analyse the last stage (output stage) first. In the present case, the last stage (Q 2 stage) is emitter follower which has voltage gain of unity. Then the over all gain of two-stage amplifier(product of gains of individual stages) is same as gain of first stage which is common emitter amplifier in fig. The gain of CE amplifier is V V r E r c r' e Since E = 0 (for Q transistor), r E =0 Then, V r r' c e Where r c is the effective impedence seen by the collector of transistor Q. That is, r c = C Z i(base)2 The impedance, Z i(base)2 at the base of transistor Q 2 is, Z i(base)2 = β. E2 = 80 X 3k or Z i(base)2 = 240kΩ Therefore, c = C Z i(base)2 = 0k 240k or r c = 9.6 kω Therefore, over all gain of amplifier of fig is

16 V or rc r' e V 3 384