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

By   July 3, 2020

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

ZOF = ZO / 1 + BZM ………………………..(23)

and                     Z’F = Z’O / 1 + BZM ……………………………(24)

For voltage sampling, it is clear that ZOF < ZO

Current-shunt feedback

In fig. 47, replacing VO by V, we have

I = V/ZO – AIII ………………………(25)

With IS = 0,        II = – IF = – BIO = BI

Hence,                I = V/ZO – BAIII

Or                    I (1 + BAI) = V/Z0 ……………………(26)

ZOF = V/I = ZO (1 + BAI) ……………………(27)

E.q. (27) is the expression for the output impedance with feedback and without load resistance RL. To find R’OF

Z’OF = ZOFZL /ZOF + ZL = ZO (1+BAI) ZL /ZO (1+BAI) + ZL

= ZOZL /ZO + ZL [1 + BAI /  1 + BAIZO / ( ZO + ZL)  ……………………………….(28)

Using eq. (27) and with Z’O = ZO||ZL, We have

Z’OF = Z’O [1 + BAI / 1 + BAI] …………………….(29)

Current-series feedback

Proceeding in the similar manner, we have

ZOF = Z’O (1 + BYM) …………………….(30a)

and            Z’OF = Z’O (1 + BYM / 1 + BYM) …………………..(30b)

From eqs. (30a) and (30b), we see that for current samppling ZOF > 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 AF will be greater than A, the nominal gain without feedback, since

|AF| =  |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

dAF / AF = 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

AF = A /1 + BA = 100/1+100×0.05

= 100 / 6 = 16.67dB

F = 20 log10  |AF/A| = 20 log10 |1/6| = – 15.56 dB

(b) AB = 100 x 0.05 = 5

(c) VO = AFVI = 100 x 0.05 = 5 V

(d) VF = BVO 0.05 x 5 = 0.25 V

Intro Exercise – 2

  1. The common-emitter current gain of the transistor is B = 75. The voltage VBE in on state is 0.7 V. the values of IC and RC 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

  1. The common-emitter current gain of the transistor is B=75, the voltageVBE in on state is 0.7 V. the value of VC 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 VN =00.

  1. The hybrid parameter vaules of gin, r and ro 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

  1. The small signal voltage gain AV = V0/VS is

(a) – 4.38

(b) 4.38

(c) – 1.88

(d) 1.88

  1. 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

  1. 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

  1. In the circuit shown in the given figure, RF provides

(a) current-series feedback

(b) current-shunt feedback

(c) voltage-shunt feedback

(d) voltage-series feedback

  1. 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

  1. when we say, s =51 which means that

(a) IC increases 51 times as fast at ICO

(b) IC is 51 times greater than ICO

(c) IC is 51 times greater than IB

(d) none of the above

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

(a) 25%

(b) 85%

(c) 99%

(d) 50%

  1. consider an amplifier circuit shown in figure below. if the transistors Q1 and Q2 has parameters gm1‘r1 and gm2,r2 respctively then voltage gain |Av| is

(a) gm1r2 / 1+gm2r2

(b) gm2r2 / 1+gm2r2

(c) gm1r1 / gm2r2

(d) gm1r2 / 1+gm2r2

  1. The transistor parameters of Q1 and Q2 of circuit shown in figure are (gm1 , r1) and (gm1r2), respectively. the voltage gain |Av| of the following circuit is

(a) gm1r1

(b) gm2r1

(c) gm1r2

(d) gm2r2

  1. consider the circuit shown in figure below. what is the value of voltage V? Assume, IS = 6×10-16 A, VR = 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 VB = – 1V

  1. The value of B is

(a) 103.4

(b) 135.5

(c) 134.5

(d) 102.5

  1. The value of VCE is

(a) 6.4 V

(b) 4.7 V

(c) 1.3 V

(d) 4.2 V

 

  1. In the circuit shown below, voltage VE = 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

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

(a) IC reaches IC max

(b) IB is increased by 40 uA

(c) IB is decreased by 40 uA

(d) VCE is reduced below VCE,Q

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

(a) B

(b) VBE

(C) ICO

(d) All of these

  1. An n-p-n transistor has a beta cut-off frequency fB of 1 MHZ and emitter short circuit low frequency current gain BO of 200. the unity gain frequency fr and the alpha cut-off frequency far respectively are

(a) 200 MHz, 201 MHz

(b) 201 MHz, 200 MHz

(c) 200 MHz, 199 MHz

(d) 199 MHz, 200 MHz

  1. The voltage VCB and the current IB 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

  1. For the voltage divider bias circuit the DC bias voltage and current values are

(a) VCE = 15.05 V, IC = 6.05 mA

(b) VCE = 12.22 V, IC = 0.85 mA

(c) VCE = 12.22 V, IC = 0.85 mA

(d) VCE = 15. 05 V, IC = 6.05mA

Answers with Solutions

  1. (a) IC = (75/75+1) IE = 75/76 (1) = 0.987 mA

RC 5.2/0.987 = 3.04 K

  1. (a) 5 = (1+B) 10 IB + 20 IB + 0.7 +B2 IB

5 = (760 + 20 +150) IB + 0.7

IB = 4.62 uA

IC BIB = O.347 mA

VC = 5 – (B + 1) IBRC

= 5 – 760 x4.62 x 10-3 = 1.49 v

  1. (b) IBQ = 2 – 0.7 = 5.2 uA

ICQ  = BIB = (120) (5.2) =0.642 mA

gm  ICQ / VI = 0.624 / 0.0259 = 24 mA/V

r = BVI/ICQ = B/gm = 120/24 = 5 k, r0 = 00

  1. (a) AV = – gmRC (r/r + RB) = BRC / r + RB

= – (24) (4)  (5/5+250) = – 1.88

  1. (d) r = BVR/ICQ

rmax  = (120) (0.0259) / 1.08  = 2.88 k

rmax = 180|10.0259|/1.32  = 1.56 k

  1. (d) ZIF = ZI (1 + AB)

ZOF = ZO / 1 + AB

  1. (C)
  2. (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.

  1. (a) s = IC/ICO 51

which means IC increases 51 times as fast as ICO.

  1. (c)
  2. (a) By small signal equivalent circuit analysis,

In the circuit,

gm1 V1 + VO /r2 = gm2V2

V1 = VM,VO = – V2

So,   ginVM + VO / R2 = – gm2VO

gm1 Vin = – (gm2 + 1/r2) VO

Voltage gain

|AV| = VO/VIN = gm1r2 / 1 + gm2r2

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

VO = V2‘gm1V1 = V2/r2

V1 = Vin

so,     gm1 Vin = V2/r2 = V2 = (gm1r2) Vin

VO = (gm1r2)Vin

|AV|= |VO/Vin| = gm1r2

  1. (a) IE = IC+IB, IC = BIB

As, B >> 1, IE ~IC, IC = IS eVBE/VR

So,          VBE = VT in (IC/IS)

= 26 x 10-3 In (IC/IS)

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

1.5 -VBE – IE (103) = 0

1.5 – 26 x 10-3 In( IC/IS) – IC (103) = 0

As IE = IC’

so,   i.5 = 26×10-3 In (IC/IS) + IC x103

IC = 775 x 10-6 A

So, voltage

V = IE x103 = IC x 103

= 775 x 10-6 x 103 = 775 x 10-3

= 775 mV

  1. (c) VB = – IBRB

IB = -VB / RB = 1 / 500 = 2.0 uA

VE = – 1 – 0.7 = – 1.7 V

IE = VE – (-3) / 4.8

= – 1.7 + 3 / 4.8 = 0.271 mA

IE / IB = (B + 1) = 0.271 /2

B = 134.5

  1. (b) VCE 3-VE = 3-(-1.7) = 4.7 V
  2. (c) IE = 5.4 /2 = 0.5 mA

4 = 0.7 + IBRB + ICRC – 5                 (IC~IE)

8.3 = 100IB + 0.5 x 8

IB = 43 uA

IE/IB = 1 + B = 0.5/43 = 11.63

B = 10.63, a = B/1+B  a = 0.914

  1. (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.
  2. (d) In BJT as temperature increase B increases, VBE increases by 2.5 mV/0c and ICO doubles for every 100c increase in temperature.
  3. fr = BO fb = (200) x 1

= 200 MHz

and          fa = fb / 1-a = fb –= fa(B + 1)

1 – B / 1 + B

106 (200 + 1)

= 201 MHz

  1. IE = VEE + VBE / RE

= 4 – 0.7 / 1.2 = 2.75 mA

VCB = VCC – ICRC                        (assuming IC~IE)

= 10 – (2.75)x2.4 = 3.4 V

Then,   IB = IC/B  = 2.75/60 = 45.8 uA

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

RTh = R1 ||R2 = (39) (3.9) / 39 + 3.9 = 3.55

VCC = 22 V

ETH = R2VCC / R1 + R2

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

VBE = 0.7

IB = ETH – VBE / RTH + (1 + B) RE = 6.05 uA

IC = BIB

IC = 0.85 mA

RC = 10 K RE = 1.5 K

VCE = VCC – IC (RC + RE)   VCE = 12.22 V

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