current shunt feedback amplifier circuit diagram , voltage shunt feedback amplifier theory

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feedback amplifier is also called as | feedback amplifier solved problems pdf

**Voltage-shunt feedback**

Proceding in the similar manner, we heve

Z_{OF} = Z_{O} / 1 + BZ_{M} ………………………..(23)

and Z’_{F} = Z’_{O} / 1 + BZ_{M} ……………………………(24)

For voltage sampling, it is clear that Z_{OF} < Z_{O}

**Current-shunt feedback**

In fig. 47, replacing V_{O} by V_{,} we have

I = V/Z_{O} – A_{I}I_{I} ………………………(25)

With I_{S} = 0, I_{I} = – I_{F} = – BI_{O} = BI

Hence, I = V/Z_{O} – BA_{I}I_{I}

Or I (1 + BA_{I}) = V/Z_{0} ……………………(26)

Z_{OF} = V/I = Z_{O} (1 + BA_{I}) ……………………(27)

E.q. (27) is the expression for the output impedance with feedback and without load resistance R_{L}. To find R’_{OF}

Z’_{OF} = Z_{OF}Z_{L} /Z_{OF} + Z_{L} = Z_{O} (1+BA_{I}) Z_{L} /Z_{O} (1+BA_{I}) + Z_{L}

= Z_{O}Z_{L} /Z_{O} + Z_{L} [1 + BA_{I} / 1 + BA_{I}Z_{O} / ( Z_{O} + Z_{L}) ……………………………….(28)

Using eq. (27) and with Z’_{O} = Z_{O}||Z_{L}, We have

Z’_{OF} = Z’_{O} [1 + BA_{I} / 1 + BA_{I}] …………………….(29)

**Current-series feedback**

Proceeding in the similar manner, we have

Z_{OF} = Z’_{O} (1 + BY_{M}) …………………….(30a)

and Z’_{OF} = Z’_{O} (1 + BY_{M} / 1 + BY_{M}) …………………..(30b)

From eqs. (30a) and (30b), we see that for current samppling Z_{OF} > Z_{0}.

**Effect of positive Feedback **

If|1 + BA|> 1, it is considered to be negative feedback and If|1+ BA|< 1. it is considered to be positive feedback. In second case, the resultant transfer gain A_{F} will be greater than A, the nominal gain without feedback, since

|A_{F}| = |A| / |1 + BA| > |A|

positive feedback increases the amplification but at the cost of reduced stability.

**Example 11. ** An amplifier has an open-loop gain of 500 and a feedback of 1. if open-loop gain changes by 20^{0} c due to the temperature, fiud the percentage change in the closrd-loop gain.

**Sol. ** Given, A = 500, B = 0.1, dA / A = 20

Change in closed-loop gain

dA_{F} / A_{F} = dA / A x 1 / 1 +BA

= 20 x 1 / 1 + 500 x 0.1

= 0.3921 = 39.21%

**Example 12. **An amplifier has a voltage gain of 100, the feedback ratio is 0.05. find

(a) the voltage gain with feedback in dB.

(b) feedback factor.

(c) the output voltae, if input voltage is 1 V.

(d) the feedback voltage.

**Sol. ** (a) voltage gain

A_{F }= A /1 + BA = 100/1+100×0.05

= 100 / 6 = 16.67dB

F = 20 log_{10 }|A_{F}/A| = 20 log_{10} |1/6| = – 15.56 dB

(b) AB = 100 x 0.05 = 5

(c) V_{O} = A_{F}V_{I} = 100 x 0.05 = 5 V

(d) V_{F} = BV_{O} 0.05 x 5 = 0.25 V

**Intro Exercise – 2**

- The common-emitter current gain of the transistor is B = 75. The voltage V
_{BE}in on state is 0.7 V. the values of I_{C}and R_{C}are

(a) 0.987 mA, 3.04 K

(b) 1.013 mA, 2.96 K

(c) 0.946 mA, 4.18 K

(d) 1.057 mA, 3.96 K

- The common-emitter current gain of the transistor is B=75, the voltageV
_{BE}in on state is 0.7 V. the value of V_{C}is

(a) 1.49 V

(b) 2.9 V

(c) 1.78 V

(d) 2.3 V

**statements for linked answer questions 3 and 4**

Consider the circuit shown below. the transistor parameters are B=120 and V_{N} =_{00.}

- The hybrid parameter vaules of g
_{in}, r and r_{o}are

(a) 24 mA/V, _{00}, 5 k

(b) 24 mA / V, 5 K, _{00}

(c) 48 mA / V, 10 K, 18.4 K

(d) 48 mA / V, 18.4 K 10 K

- The small signal voltage gain A
_{V}= V_{0}/V_{S}is

(a) – 4.38

(b) 4.38

(c) – 1.88

(d) 1.88

- The nominal quiescent collector current of a transistor is 1.2 mA if the renge of B for this transistor is 80<B<120 and if the quiescent collector current changes by = 10% the range in value for r is

(a) 1.73 k < r < 2.59 k

(b) 1.93 k < r < 2. 59 k

(c) 1.73 k < r < 2.59 k

(d) 1.56 k < r < 2.88 k

- In the circuit shown in figure, the feedback causes

(a) an increase in the input impedance but a decrease in the output impedance

(b) an increase in both the input and output impedances

(c) a decrease in both the input and output impedances

(d) a decrease in the input impedance but an increase in the output impedance

- In the circuit shown in the given figure, R
_{F}provides

(a) current-series feedback

(b) current-shunt feedback

(c) voltage-shunt feedback

(d) voltage-series feedback

- while choosing operating point Q which that following factors of BJT are considered?

(a) power supply

(b)AC and DC load

(c) maximum transistor ratings

(d) all of the above

- when we say, s =51 which means that

(a) I_{C} increases 51 times as fast at I_{CO}

(b) I_{C} is 51 times greater than I_{CO}

(c) I_{C} is 51 times greater than I_{B}

(d) none of the above

- In a CB amplifier, the maximum efficiency could be

(a) 25%

(b) 85%

(c) 99%

(d) 50%

- consider an amplifier circuit shown in figure below. if the transistors Q
_{1}and Q_{2}has parameters g_{m1}‘r_{1}and g_{m2},r_{2}respctively then voltage gain |A_{v}| is

(a) g_{m1}r_{2} / 1+g_{m2}r_{2}

(b) g_{m2}r_{2} / 1+g_{m2}r_{2}

(c) g_{m1}r_{1} / g_{m2}r_{2}

(d) g_{m1}r_{2} / 1+g_{m2}r_{2}

- The transistor parameters of Q
_{1}and Q_{2}of circuit shown in figure are (g_{m1}, r_{1}) and (g_{m1}r_{2}), respectively. the voltage gain |A_{v}| of the following circuit is

(a) g_{m1}r_{1}

(b) g_{m2}r_{1}

(c) g_{m1}r_{2}

(d) g_{m2}r_{2}

- consider the circuit shown in figure below. what is the value of voltage V? Assume, I
_{S}= 6×10^{-16 }A, V_{R}= 26×10^{-3}V, B>>1

(a) 775 mV

(b) 800 mV

(c) 695 mV

(d) 215 mV

**Statement for linked answer question 14 and 15 **

In the circuit shown below V_{B} = – 1V

- The value of B is

(a) 103.4

(b) 135.5

(c) 134.5

(d) 102.5

- The value of V
_{CE}is

(a) 6.4 V

(b) 4.7 V

(c) 1.3 V

(d) 4.2 V

- In the circuit shown below, voltage V
_{E}= 4 V. The values of a and B are respectively

(a) 0.943, 17.54

(b) 0.914, 17.54

(c) 0.914, 11.63

(d) 0.914, 10.63

- For the output characteristics of BJT the transistor will enter into cut-off, if

(a) I_{C} reaches I_{C max}

(b) I_{B} is increased by 40 uA

(c) I_{B} is decreased by 40 uA

(d) V_{CE} is reduced below V_{CE,Q}

- In BJT, which of the following factors changes with temperature?

(a) B

(b) V_{BE}

(C) I_{CO}

(d) All of these

- An n-p-n transistor has a beta cut-off frequency f
_{B}of 1 MHZ and emitter short circuit low frequency current gain B_{O}of 200. the unity gain frequency f_{r}and the alpha cut-off frequency f_{ar}respectively are

(a) 200 MHz, 201 MHz

(b) 201 MHz, 200 MHz

(c) 200 MHz, 199 MHz

(d) 199 MHz, 200 MHz

- The voltage V
_{CB}and the current I_{B}for the CB configuration of the figure shown below will be

(a) 3.4 v and 45.8 uA

(b) 34 v and 4.5 mA

(c) 6 vand 2.75 mA

(d) none of these

- For the voltage divider bias circuit the DC bias voltage and current values are

(a) V_{CE} = 15.05 V, I_{C} = 6.05 mA

(b) V_{CE} = 12.22 V, I_{C} = 0.85 mA

(c) V_{CE} = 12.22 V, I_{C} = 0.85 mA

(d) V_{CE} = 15. 05 V, I_{C} = 6.05mA

**Answers with Solutions**

- (a) I
_{C}= (75/75+1) I_{E}= 75/76 (1) = 0.987 mA

R_{C} 5.2/0.987 = 3.04 K

- (a) 5 = (1+B) 10 I
_{B}+ 20 I_{B}+ 0.7 +B2 I_{B}

5 = (760 + 20 +150) I_{B} + 0.7

I_{B} = 4.62 uA

I_{C} BI_{B} = O.347 mA

V_{C} = 5 – (B + 1) I_{B}R_{C}

= 5 – 760 x4.62 x 10^{-3} = 1.49 v

- (b) I
_{BQ}= 2 – 0.7 = 5.2 uA

I_{CQ } = BI_{B} = (120) (5.2) =0.642 mA

g_{m} I_{CQ} / V_{I} = 0.624 / 0.0259 = 24 mA/V

r = BV_{I}/I_{CQ} = B/g_{m} = 120/24 = 5 k, r_{0} = _{00}

- (a) A
_{V}= – g_{m}R_{C}(r/r + R_{B}) = BR_{C}/ r + R_{B}

= – (24) (4) (5/5+250) = – 1.88

- (d) r = BV
_{R}/I_{CQ}

r_{max }= (120) (0.0259) / 1.08 = 2.88 k

r_{max} = 180|10.0259|/1.32 = 1.56 k

- (d) Z
_{IF}= Z_{I}(1 + AB)

Z_{OF} = Z_{O} / 1 + AB

- (C)
- (d) selection of Q point depends upon following factors

(a) AC and DC loads on the stage

(b) the availble power supply

(c) the maximun transistor ratings

(d) the peak signal excursions to be handled by the stage

(e) tolerable distortion.

- (a) s = I
_{C}/I_{CO}51

which means I_{C} increases 51 times as fast as I_{CO}.

- (c)
- (a) By small signal equivalent circuit analysis,

In the circuit,

g_{m1 }V_{1} + V_{O} /r_{2} = g_{m2}V_{2}

V_{1} = V_{M,}V_{O} = – V_{2}

So, g_{in}V_{M} + V_{O} / R_{2} = – g_{m2}V_{O}

g_{m1} V_{in} = – (g_{m2} + 1/r_{2}) V_{O}

Voltage gain

|A_{V}| = V_{O}/V_{IN} = g_{m1}r_{2} / 1 + g_{m2}r_{2}

- (c) By small-signal AC analysis of the circuit,

V_{O} = V_{2}‘g_{m1}V_{1} = V_{2}/r_{2}

V_{1} = V_{in}

so, g_{m1 }V_{in} = V_{2}/r_{2} = V_{2} = (g_{m1}r_{2}) V_{in}

V_{O} = (g_{m1}r_{2})V_{in}

|A_{V}|= |V_{O}/V_{in}| = g_{m1}r_{2}

- (a) I
_{E}= I_{C}+I_{B}, I_{C}= BI_{B}

As, B >> 1, I_{E} ~I_{C}, I_{C} = I_{S} e^{VBE/VR}

So, V_{BE} = V_{T} in (I_{C}/I_{S})

= 26 x 10^{-3} In (I_{C}/I_{S})

Also by applying KVL in base-emitter loop, we get

1.5 -V_{BE} – I_{E} (10^{3}) = 0

1.5 – 26 x 10^{-3} In( I_{C}/I_{S}) – I_{C} (10^{3}) = 0

As I_{E} = I_{C’}

so, i.5 = 26×10^{-3} In (I_{C}/I_{S}) + I_{C} x10^{3}

I_{C} = 775 x 10^{-6} A

So, voltage

V = I_{E} x10^{3} = I_{C} x 10^{3}

= 775 x 10^{-6} x 10^{3} = 775 x 10^{-3}

= 775 mV

- (c) V
_{B}= – I_{B}R_{B}

I_{B} = -V_{B} / R_{B} = 1 / 500 = 2.0 uA

V_{E} = – 1 – 0.7 = – 1.7 V

I_{E} = V_{E} – (-3) / 4.8

= – 1.7 + 3 / 4.8 = 0.271 mA

I_{E} / I_{B} = (B + 1) = 0.271 /2

B = 134.5

- (b) V
_{CE}3-V_{E}= 3-(-1.7) = 4.7 V - (c) I
_{E}= 5.4 /2 = 0.5 mA

4 = 0.7 + I_{B}R_{B} + I_{C}R_{C }– 5 (I_{C}~I_{E})

8.3 = 100I_{B} + 0.5 x 8

I_{B} = 43 uA

I_{E}/I_{B} = 1 + B = 0.5/43 = 11.63

B = 10.63, a = B/1+B a = 0.914

- (c) From the give output characteristics, it is clear that input signal can swing 40 uA about Q, because if the base current decreases by more than 40 uA, the transistor is driven off.
- (d) In BJT as temperature increase B increases, V
_{BE}increases by 2.5 mV/^{0}c and I_{CO}doubles for every 10^{0}c increase in temperature. - f
_{r}= B_{O}f_{b}= (200) x 1

= 200 MHz

and f_{a} = f_{b} / 1-a = f_{b –}= f_{a}(B + 1)

1 – B / 1 + B

10^{6} (200 + 1)

= 201 MHz

- I
_{E}= V_{EE}+ V_{BE}/ R_{E}

= 4 – 0.7 / 1.2 = 2.75 mA

V_{CB} = V_{CC} – I_{C}R_{C} (assuming I_{C~}I_{E})

= 10 – (2.75)x2.4 = 3.4 V

Then, I_{B} = I_{C}/B = 2.75/60 = 45.8 uA

- (C) Using formula for voltage divider bias,

R_{Th} = R_{1 }||R_{2} = (39) (3.9) / 39 + 3.9 = 3.55

V_{CC} = 22 V

E_{TH} = R_{2}V_{CC} / R_{1} + R_{2}

= 3.9 x 22 / 39 x3.9 = 2 v

V_{BE} = 0.7

I_{B} = E_{TH} – V_{BE} / R_{TH} + (1 + B) R_{E} = 6.05 uA

I_{C} = BI_{B}

I_{C} = 0.85 mA

R_{C} = 10 K R_{E} = 1.5 K

V_{CE} = V_{CC} – I_{C} (R_{C} + R_{E}) V_{CE} = 12.22 V