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

current shunt feedback amplifier circuit diagram , voltage shunt feedback amplifier theory
Read this article before it –
https://sarthakata.com/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_II_I ………………………(25)
With I_S = 0,        I_I = – I_F = – BI_O = BI
Hence,                I = V/Z_O – BA_II_I
Or                    I (1 + BA_I) = V/Z₀ ……………………(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_OFZ_L /Z_OF + Z_L = Z_O (1+BA_I) Z_L /Z_O (1+BA_I) + Z_L
= Z_OZ_L /Z_O + Z_L [1 + BA_I /  1 + BA_IZ_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 > Z₀.
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 20⁰ 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 log₁₀  |A_F/A| = 20 log₁₀ |1/6| = – 15.56 dB
(b) AB = 100 x 0.05 = 5
(c) V_O = A_FV_I = 100 x 0.05 = 5 V
(d) V_F = BV_O 0.05 x 5 = 0.25 V
Intro Exercise – 2

1. 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

2. 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.

3. 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

4. The small signal voltage gain A_V = V₀/V_S is

(a) – 4.38
(b) 4.38
(c) – 1.88
(d) 1.88

5. 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

6. 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

7. 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

8. 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

9. 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

10. In a CB amplifier, the maximum efficiency could be

(a) 25%
(b) 85%
(c) 99%
(d) 50%

11. consider an amplifier circuit shown in figure below. if the transistors Q₁ and Q₂ has parameters g_m1‘r₁ and g_m2,r₂ respctively then voltage gain |A_v| is

(a) g_m1r₂ / 1+g_m2r₂
(b) g_m2r₂ / 1+g_m2r₂
(c) g_m1r₁ / g_m2r₂
(d) g_m1r₂ / 1+g_m2r₂

12. The transistor parameters of Q₁ and Q₂ of circuit shown in figure are (g_m1 , r₁) and (g_m1r₂), respectively. the voltage gain |A_v| of the following circuit is

(a) g_m1r₁
(b) g_m2r₁
(c) g_m1r₂
(d) g_m2r₂

13. 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^-3V, 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

14. The value of B is

(a) 103.4
(b) 135.5
(c) 134.5
(d) 102.5

15. The value of V_CE is

(a) 6.4 V
(b) 4.7 V
(c) 1.3 V
(d) 4.2 V
 

16. 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

17. 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

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

(a) B
(b) V_BE
(C) I_CO
(d) All of these

19. 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

20. 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

21. 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

1. (a) I_C = (75/75+1) I_E = 75/76 (1) = 0.987 mA

R_C 5.2/0.987 = 3.04 K

2. (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_BR_C
= 5 – 760 x4.62 x 10^-3 = 1.49 v

3. (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, r₀ = ₀₀

4. (a) A_V = – g_mR_C (r/r + R_B) = BR_C / r + R_B

= – (24) (4)  (5/5+250) = – 1.88

5. (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

6. (d) Z_IF = Z_I (1 + AB)

Z_OF = Z_O / 1 + AB

7. (C)
8. (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.

9. (a) s = I_C/I_CO 51

which means I_C increases 51 times as fast as I_CO.

10. (c)
11. (a) By small signal equivalent circuit analysis,

In the circuit,
g_m1 V₁ + V_O /r₂ = g_m2V₂
V₁ = V_M,V_O = – V₂
So,   g_inV_M + V_O / R₂ = – g_m2V_O
g_m1 V_in = – (g_m2 + 1/r₂) V_O
Voltage gain
|A_V| = V_O/V_IN = g_m1r₂ / 1 + g_m2r₂

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

V_O = V₂‘g_m1V₁ = V₂/r₂
V₁ = V_in
so,     g_m1 V_in = V₂/r₂ = V₂ = (g_m1r₂) V_in
V_O = (g_m1r₂)V_in
|A_V|= |V_O/V_in| = g_m1r₂

13. (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 (10³) = 0
1.5 – 26 x 10^-3 In( I_C/I_S) – I_C (10³) = 0
As I_E = I_C’
so,   i.5 = 26×10^-3 In (I_C/I_S) + I_C x10³
I_C = 775 x 10^-6 A
So, voltage
V = I_E x10³ = I_C x 10³
= 775 x 10^-6 x 10³ = 775 x 10^-3
= 775 mV

14. (c) V_B = – I_BR_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

15. (b) V_CE 3-V_E = 3-(-1.7) = 4.7 V
16. (c) I_E = 5.4 /2 = 0.5 mA

4 = 0.7 + I_BR_B + I_CR_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

17. (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.
18. (d) In BJT as temperature increase B increases, V_BE increases by 2.5 mV/⁰c and I_CO doubles for every 10⁰c increase in temperature.
19. 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
10⁶ (200 + 1)
= 201 MHz

20. I_E = V_EE + V_BE / R_E

= 4 – 0.7 / 1.2 = 2.75 mA
V_CB = V_CC – I_CR_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

21. (C) Using formula for voltage divider bias,

R_Th = R₁ ||R₂ = (39) (3.9) / 39 + 3.9 = 3.55
V_CC = 22 V
E_TH = R₂V_CC / R₁ + R₂
= 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