S.Y. Diploma : Sem. III [EJ/EN/ET/EX/EV/ED/IU/DE/IS/IC/IE...

54

S.Y. Diploma : Sem. III [EJ/EN/ET/EX/EV/ED/IU/DE/IS/IC/IE/MU] Electronics Devices and Circuits

Time: 3 Hrs.] Prelim Question Paper [Marks : 100

Q.1(a) (i) A circuit used for establishing Quiscent operating point (Q) in the centre of active region to avoid distortion is defined as transistor biasing circuit. For this purpose, normally circuit uses one external d.c. supply, few resistors and may be 1 capacitor. By choosing their values properly, B E junction is forward biased and B C junction is reverse biased.

Some of the transistor biasing circuits used are 1) Fixed Biased circuit 2) Base biased with collector feedback 3) Base biased with emitter feedback 4) Voltage (or potential) divider 5) Emitter biased

Q.1(a) (ii)

The regions of operation are shown in characteristics drawn for VGS = OV.

1) Ohmic region OA : ID v/s VDS and obeys ohm’s law. 2) Pinch off or saturation region BC : ID remain constant at maximum value

even though VDS is increased. Current ID in this region is given by schockly’s equation

2

GSD DSS

P

VI I 1

V

3) Breakdown region CD : ID starts increasing very rapidly due to breakdown of gate to source junction due to avalanche effect. This region of operation should be avoided to reduce damage to JFET.

Q.1(a) (iii) Multi-Stage Amplifier

A circuit in which number of single stage amplifiers are connected in cascade (in series) such that output of the previous amplifier is connected to the input of the next amplifier is defined as multi-stage amplifier.

VDS

ID (mA)

IDSS

A

BC1

D

VP O

VGS = 0V

VDS 5 10 15 208 250

ID (mA)

1

2

3

4

5 VGS = 0V

1V

2V

3V

4V

Vidy

alank

ar

the centre otre circuit. For this For

resistors and may nd ma E junction is forward tion is forward

The regions of operation are sho regions of operation 1) Ohmic region OAOhmic regio : I : IDD v/s V

2) PinchP off or saturation ff or saturationeven though Veven though DS is isschockly’s equatioschockly’s e

D DSSI ID IDSS

3) Breakdobreakdreg

) (iii) MuA

ID (mA) mA)

anaya

lIDSS

A

B

dyaVya5 10 15 208 250 10 15 yyayaa

V

2V

3V3V

4Vyalalalllyayaalalalaa

Prelim Question Paper Solution

55

A single stage voltage amplifier is not capable of giving too large voltage gain. Sometimes input signals are very weak, in such cases they are required to be amplified more. For this purpose multi-stage amplifiers are used.

Q.1(a) (iv) Intrinsic stand-off ratio ( ) is one of the important characteristics of UJT and given by

= 1

E

B

BB I 0

r

r

= 1

1 2 E

B

B B I 0

r

r r

It is defined as the ratio of internal resistance between B1 and E to the total internal resistance between 2 base terminals provided emitter current IE = 0. The value of is given by 0.5 < < 0.8 and typically = 0.65 or o.7.

Q.1(a) (v) UJT i) From the construction it is clear that the

device has one (uni) P N junction. ii) Like transistor it is 3 terminal device.

The terminals are known as Base 1, Base 2, Emitter.

Due to the above 2 reasons it to called uni-junction transistor.

Q.1(a) (vi) Barkhausen Criterion

For an oscillator the input voltage Vs is absent i.e. Vs = 0 and the feedback signal Vf is supposed to maintain the oscillations. Therefore substitute Vs = 0 into Equation (5) to get,

Vi(1 A ) = 0 or A = 1

This condition must be satisfied in order to obtain sustained oscilations. With an inverting amplifier introducing a 180 phase shift between Vi and

Vo, the feedback network must introduce another 180 phase shift to ensure that Vi and Vf are in phase.

These two conditions which are required to be satisfied to operate the circuit as an oscillator are called as the "Barkhausen criterion" for sustained oscillations.

B2 (Base 2)

B1(Base 1)

E (Emitter)

Vidy

alank

ar

s of

etween Between B11 and E to the total d E to the t provided emitter current Irovided emitter current IEE = 0.

ypically = 0.65 or o.7. = 0.65 or o.7.

r that the he nction.

minal device. device. wn as Base 1, Base 1

sons it to called uni-sons it to called uni-

iterion n

Vidy

Vid For an osci

Vidsig

Vid

into E

Vi V VVV

knkknn B2 (B

E (Emitter)(Em

Vidyalankar : S.Y. Diploma EDC

56

Q.1(a) (vii) Depending upon the type of 3 semiconductor regions used, transistor is classified into following 2 types. Their symbols are also given.

Q.1(b) (i) Comparison between ve feedback and +ve feedback +ve feedback ve feedback 1) VIN and Vf are in

phase hence effective input voltage increases

VIN and Vf are 180 out of phase hence effective input voltage decreases.

2) Voltage gain increases

Afb = v

v

A1 A

Voltage gain decreases

Afb = v

v

A1 A

3) Distortion increases. Distortion decreases. 4) Noise in the output signal

increases. Noise in the output signal decreases.

5) Circuit becomes unstable, starts oscillating and produces new output signal of different frequency.

Stability of amplifier improves since Afb remains constant because it becomes independent of transistor parameter.

6) It is used in all oscillators. It is used in amplifiers.

Q.1(b) (ii) Voltage at the gate terminal VG = IGRG but IG = 0 RIN of JFET very high. VG = 0 Now Vs = VG VGS = VGS

By Ohm’s law ID = S GS

S s

V VR R

VGS = IDRs Thus voltage drop across Rs reverse biases gate to source P N junction.

GS

DQS

VI

R … (i)

Veff

Vf

VIN Veff. VIN

Vf

ID RD

+VDD

IG = 0

RG

VG VD

VS

VDS

VGS

ID RS

Transistor

2) P N P1) N P N

(a) Structure

N P N

B

E CP N P

B

E C

(a) Structure

(b) SymbolCollector Emitter

Base Base

(b) Symbol

Collector Emitter

Vidy

alank

ar

and +ve feedbackve feedbacknve feedback ve feednkVIN and V and f are 180 180out of phase hence out of phaseffective input effective inpvoltage decreasesvoltage dna increases ses

Afb = alvv

v

AA1 vAA

Voltage gaVolt

alaaortion increases. ncreases. D

yaoise in the output signal output signal increases. increase yad

5) Circuit becomes unstable, star Circuit becomes unoscillating and produces newoscillating and produoutput signal of different foutput signal of differedy

aidid6) It is used in all oscilla It is used in all oscillaidyVid

Vid

Vid

VidQ.1(b) (ii) (b) Voltage at the gate teVoltage at the

R RININ of JFET ve o VVGG = 0 =

Now V s = VG

By Ohm’ B

T

lallallaaaVeff

VVf

VVININ

llaarararrraaaB B

C

Base Base

bol

kakakakakakkkaakk

Collector ector Emitter


57

A graph can be plotted (straight line) between ID and VGS. On the same graph, transfer characteristics is plotted and as shown Q point can be selected in the centre by selecting proper value for Rs. Similarly applying K.V.L. to the output we get All voltages 0

+ VDD IDRD VDS IDRs = 0 VDD IDQ (RD + RS) = VDSQ … (ii)

Equations (i) & (ii) establish Q point.

This is one of the best biasing network for JFET and is used because of the following reasons. (1) Uses only one d.c. supply + VDD (2) Requires few component (3) Q point is stable because of following two reasons.

(i) Q point independent of FET parameter. (ii) Due to negative feedback introduced by voltage drop IDRS. Assume

that due to temperature variation, ID increases. This increase voltage drop IDRs. Now VGS = IDRs & hence reverse bias on gate source junction increases. This reduces ID and is brought back to original value. Exactly reverse action takes place if ID decreases. Hence due to this negative feedback ID and Q point remains stable.

Q.1(b) (iii) Input characteristics :

CBE EE V constantI v/s V

1) Transistor is working in active region hence its input P N (B E) junction is forward biased. Hence the input characteristics is exactly identical to forward characteristics of P N junction.

2) By convention, VEB and IE are negative but the curve is plotted in 1st quadrant.

3) characteristics are plotted by keeping output voltage VCB constant. Hence 2 curves are plotted by keeping VCB = 0 & then VCB = 2V as constant.

4) As VCB increases in positive direction then the curve shifts slightly towards left since less input voltage is now required for sending same input current.

IDSS

VGS (OFF)

VGS

ID

Q

2

VCB = +2 V

VCB = 0V

VEB (Volts)0.1 0.3 0.5 0.7 0.9

4

6

8

10

IE (mA)

Scale :X axis : 1 cm = 0.1 V Y axis : 1 cm = 2 mA

Vidy

alank

arnd is used because of the d because of the

two reasons. ns. rameter. er.

troduced by voltage drop Ied by voltage drop DRS. Aariation, I IDD increases. This increas increases. This i

IDRs & hence reverse bias on gence reverse biahis reduces Iduces D and is brought ba and is

erse action takes place if Ition takes place if ID decrea eedback I eedback IDD and Q point remains st and Q point re

s :

alaCBCB

E EEE V constantV constantCBCBI v/s VVE E

1) Transistor is wo1) Transistois forward biais foforward charw

2) By convquadr

3) cha3H

4

VGS (OFF) GS (OFF)

arararrrar

Qddyddd

alal

22

VVCBCB = +2 V = +2 V

dydyyy0.1 0.3 00

44

66

8


58

Q.2(a) Voltage Amplifier AV = voltage gain of the amplifier RIN = input resistance of the amplifier RO = output resistance of the amplifier VS = A.C. signal from function (or signal) generator VO > VIN without distortion

Wave form :

Requirements of good voltage amplifier : i) Voltage gain (AV ) : It should be as high as possible or It must be sufficient

which depends upon the application. ii) Input resistance (RIN ) : It is measured in ohms. Ideally it must be infinite,

practically it must be as high as possible. This avoids the loading (decreasing) of input signal.

iii) Output resistance (RO ) : It is measured in ohms. Ideally it must be zero, practically it must be as low as possible. This avoids the loading (decreasing) of the voltage signal present at the output terminals.

iv) Frequency response (or Band width) : It is measured in Hz or KHz or MHz. Ideally it must be infinite, practically it must be sufficient for the required application. Basically Band width represents the range (group) of frequencies which are amplified properly by the amplifier.

v) Distortion : It must be as low as possible. If the shape of the amplified output voltage V0 is different from the shape of input voltage then we say distortion is present. This must be avoided in any amplifier.

vi) Stability : This must be good so that Q point remains stable in the centre of active region under D.C. conditions.

Q.2(b) Comparison between 3 configurations of transistor

Parameter Common Base Common Emitter Common collector 1) Input

impedance Lowest 25 Medium 1k Highest = D.C RE

2) Output Impedance

Highest 1M Medium 50k to 100k

Lowest = RE/ D.C.

VS

RIN

i/pVoltage Amplifier

AV

VIN

RS

+

+VCC

VO

RO

o/p RL

VIN

t

VO

t

Vidy

alank

ar VO > VINN without distortion without distortion

d voltage amplifier : amplifier : ) : It should be as high as possi should be as high as

s upon the application. the application. ance (RINN ) : It is measured in o) : It is measure

y it must be as high as posmust be as high aasing) of input signal. nput signal.

put resistance put resistance (R(RO ) : It is measu : It is ractically it must be as low as possctically it must be as

of the voltage signal present at th voltage signal prev) Frequency response equency response (or Band (or B

Ideally it must be infinite, peally it must be infiniteapplication. Basically Bplication. Basically Bfrequencies which are afrequencies whi

v) Distortion Distortion : It must be: It mvoltage Voltage 0 is differis present. This esen

vi) vi) Stability : Thiyactive regioa

2(b) ComparisPaVVV

) In

nkntt

laalantt

lanan


59

3) Current gain D.C 1 lowest 100 < D.C. < 500 high

( D.C. + 1) Highest

4) Voltage gain High Highest Less than 1 lowest 5) Power gain Moderate Highest Medium 6) Leakage

current ICBO = ICO lowest 5 A for Ge, 1 A for Si

ICEO = D.C.ICBO High 500 A for Ge, 20 A for Si

ICEO High 500 A for Ge, 20 A for Si

7) Phase shift 0 (In phase) 180 (out of phase)

0 (In phase)

8) Cut-off frequency

High Lower than CB Depends upon load

9) Thermal stability

High Low High

10) Applications For high frequency For audio frequency For impedance matching as a buffer.

Q.2(c) Controlled Series Voltage Regulator

Important blocks and their functions are : i) Reference voltage : Properly reverse biased zener diode is used as

constant D.C. reference voltage.

ii) Sampeling network : Two resistors are connected as potential divider across the output terminals for sampeling the output voltage.

iii) Comparator : Op amp or transistor compares the actual sampled output voltage with constant reference voltage. The difference between them known as error voltage is present at its output.

iv) Error Amplifier : It amplifies the error voltage given to it. This is now used as control voltage to adjust the voltage drop across control element. Normally in most of the circuit same transistor or same op amp is used as both comparator and error amplifier.

v) Control element : A transistor in active region is working as emitter follower. This transistor is working as a variable resistance. The voltage drop VS across it is automatically adjusted by control signal till error voltage becomes zero. Due to this V0 is kept constant under all conditions.

Error Amplifier

Comparator

Reference voltage

Control Element

Sampeling Network

VIN

(D.C.) unregulated

VS+

+

V0 (Regulated) RL

Vidy

alank

ar

500A for Si i r (In phase) ) rrDepends upon load ends upon load ark

r

High High

kako frequency For impedance equency For impedance matching as a matching abuffer. buffer. kanknknk

nt blocks and their functions are : nt blocks and their functionReference voltage :erence voltage : Properly r constant D.C. reference voltage. tant D.C. reference v

ii) Sampeling network :ampeling network : Tw Tacross the output terminaross the output termina

iii) Comparator : Comparator : Op avoltage with consoltage known as error vwn a

iv)iv) Error Amplifcontrol vocmost of and e

v) C

Amplifier plifier laaComparator parato

alaaReference voltage rence voltage

yal

yayalanaaananlalaSam

alaaann


60

Q.2(d) There are 4 types of ve feedback amplifier : 1) Voltage series 2) Voltage shunt 3) Current series 4) Current shunt

1) Voltage series ve feedback amplifiers

2) Voltage shunt

3) Current Series

4) Current Shunt

Amplifier Av

Feedback network

+

+ VIN

+ Vid V0

Vf V0

+

RL +Vf

+

(a) RIN (b) R0

Fig. (a)

Fig. (b)

Amplifier Av

Feedback network

+

+ VIN

+Vf V0

Vf

+

RL

+

(a) RIN (b) R0

V0

Fig. (c)

AmplifierAv

Feedback network

+

+ VIN

Vf

v0

Vf

+

RL

+

(a) RIN (b) R0

+ Vid

Fig. (d)

AmplifierAv

FeedbackNetwork

+

+ VIN

+Vf V0

Vf

+

RL

+

(a) RIN (b) R0 Vi

dya

4) Current Shunt Current Shun

Vidy

alank

arries

aa

+

kkakk0

+

(a) R RININ

(b) R0 (b) R0

Fig. (b)Fig. (

ananalaalaFeedback network an

kan

kkk+

alanannk

ank

aa+

R

+ laanknknn

yayaAyyyadydydydyaVVff yy++yaya

++

yayViVViViViViViV

+ VIN


61

Note : (i) An ammeter is connected in series while voltmeter is connected in parallel. (ii) Hence at the output if it is parallel connection then it is voltage feedback and

if it is series connection then it is current feedback. (iii) Similarly at the input if it is series connection then it is series feedback and if

it is parallel connection then it is shunt feedback. (iv) If 2 resistors are connected in series then RS increases. (v) If 2 resistors are connected in parallel then RP decreases.

Q.2(e) (a) UJT relaxation oscillator Circuit diagram

(b) Waveforms

1BV positive pulses are used for triggering S.C.R.

Working : At the start (t = 0) when d.c. supply of VBB volts is given to the circuit, there is no charge on capacitor and hence VC = VE = 0. This is anode of internal P – N junction which is at 0 volts. Due to internal potential divider of UJT, the cathode voltage VK of P-N junction is given by Vk = VBB where = Intrinsic stand-off ration = 0.7. Since VA < Vk, internal P-N junction is reverse biased and hence UJT remains off.

B

AVBB + 0.7= Vp

Vv

0 t

t

t

1BV

2BV

+VBB

Tr Ts

Ts > > TrVE = VC

+

C

R1 = 47

R2 = 150

B1

B2

+VBB

R

VE = VC

1BV

2BV

UJT = 2N246 or 2N2647

ack

11BBV positive puV p11BB

Working there isP –

yayayal

yal

yal

yalalallank

yy B

AA 0.7= Vp

VVvv

0 0

11BBVVBB

2BVB

+VB

aVE = VCC an

knknnaalalaaalalalalalalaRR11 = 47 = 4

500

B11BVB

2BVB

UJT = 2N246 or 2N2647 = 2N246 or 2N264


62

OFF UJT acts like open switch and allows capacitor C to charge towards +VBB through variable resistor R. VC = VE now starts increasing exponentially and when VE = ( VBB + 0.7) = VP then internal P-N junction is forward biased and starts conducting. UJT is now switched on and the charged capacitor starts discharging through ON UJT and resistor R1. VC = VE starts decreasing exponentially and when VE = Vv valley voltage, UJT goes automatically off.

OFF UJT once again acts like open switch and allows capacitor to charge. The entire waveform (O-A-B) respects itself thus producing continuous periodic sweep voltage oscillations.

In the above circuit charging time constant = R C While discharging time constant = R1 C Since R > > R1, we get Ts > > Tr which is the condition required for sweep voltage.

Q.2(f) D.C. amplifier Frequency response To improve the low frequency response, all the capacitors in the circuit are removed. Hence output of 1st stage (i.e., collector of TR1) is directly connected to the base of TR2. No coupling network is used and hence it is known as direct coupled amplifier.

This is the only amplifier capable of amplifying very low f A.C. signal as well as D.C. signal of zero frequency and hence it is called A.C. amplifier. Though it is mainly used for amplifying D.C. signal of zero frequency, it is also used for amplifying A.C. signal upto fH which is sufficiently high compared to audio frequency of 20 kHz.

Q.3(a) Comparison between voltage amplifier and power amplifier Voltage Amplifier Power Amplifier

1) Amplifies voltage & voltage gain.

Av = O

IN

VV

Amplifies power & power gain.

0p

IN

PA

P

2) Small signal amplifier. VIN in mV. Large signal amplifier. VIN in volts. 3) Class A amplifier. Class B or Class C amplifier. 4) Distortion low. Distortion high. 5) Physical size of transistor used is

small and has plastic package. Physical size of transistor used is large and has metal package.

6) D.C. > 100 20 D.C 50 7) Collector current 1 to 5 mA. IC > 100 mA

VIN

R1

R2 RE

RC

+VCC

RE

RC'

Q2

VOQ1

f

VO

(or AV)

fH

0.707 Vmax

Vmax

Bandwidth

Vidy

alank

ar

ecrey off.

o charge. The ge. Ttinuous periodic period

quired for sweep voltage. for sweep voltage.

cy response response

w frequency response, all the cauency response output of 1 of 1sts stage (i.e., collector o stage (i.e., co

2. No coupling network is used coupling network plifier.

the only amplifier capable of amp the only amplifier capable o signal of zero frequency and hengnal of zero frequen

ainly used for amplifying D.C. sigsed for amplifying amplifying A.C. signal upto ying A.C. signal upto fHf wfrequency of 20 kHz. frequency of 20 kHz.

Q.3(a) Comparison between vol3(a) Comparison betweVoltage AV

Vid

ViVid1) Amplifies voltage1) Amplifie

Av = O

IN

VVO

VIVid

ViVid

2) Small si2) SVVV3) ClassVVV4) DisVVV

5)

ank

lanaankVO

or AV)

an0.707 Vmax

Vmax

anBandwidthBa


63

8) A.C. output power is low in mW. PAC > 1 watt 9) Output impedance is high 10 k to

12kOutput impedance low 200

10) Normally R C coupled. Normally transformer coupled or direct coupled.

11) Heat sinks not required. Heat sinks must be used with power transistor.

12) Impedance matching poor. Excellent impedance matching. 13) Transistor used BC 147, 148, 547,

548. 549. Power transistor used SL100 SK100, AC187 AC188, 2N3055.

14) Transistors have thin base since it handles low current.

It has thick base to handle large current.

Q.3(b) Thermal run-away

i) D.C. CE CP V I ii) Temp. PD.C.

iii) ICBO Temp. iv) ICEO 100 ICBO

v) IC = IB + ICEO

When D.C. voltage is applied to transistor, collector current IC starts flowing. This produces voltage drop VCE on the transistor.

Due to this D.C. power is dissipated in Transistor and its temperature increases. This increases ICBO and since ICEO = 100 ICBO and IC= D.C. IB + ICEO, both ICEO and IC increase. This again increases PD.C. and temperature of transistor. This is a closed cycle due to which temperature of transistor continuously goes on increasing. “The damaging of a transistor due to continuous rise in its temperature is defined as the thermal run away”. It can be avoided by (i) Selecting proper transistor biasing circuit. (ii) By using transistor biasing stabilization circuits.

IC

VCE PDC = VCE Ic

Temp.

IC

ICEO

ICBO

PDC = VCE IC

Thermal run away

Vidy

alank

ar

with power wer r matching. g rrsed SL100100 SK100, K100, 2N3055. 55. arrase to handle large andle large arka

r ii) Temp. ii) Temp. PD.C.

iv) I iv) ICEOCEO 100 IEO

. voltage is applied to transistor, co s applied to transs voltage drop Vs voltage drop VCEC on the transiston the tr

Due to this D.C. power ise to this D.C. pThis increases IThis increa CBO anIICC increase. This ag increase. closed cycle dueclosincreasing. incre

“The damaas the thIt can

yyyyyyyady

IC

VVCE C

lanknaalaala

nnnknkICBO

ermal run run away away


64

Q.3(c) Multi-stage amplifier

Let 1 2 nV V VA , A , ........ ,A be the voltage gains of each amplifier connected in

cascade. A.C. input voltage VS to be amplified is given to the input terminals of 1st amplifier. The output voltage VO is taken across RL connected to the output terminals of nth amplifier. By definition of voltage gain, we have

AV = output voltageinput voltage

Hence, 1 2 3 n 1 n

32 4 n n 1V V V V V

1 2 3 n 1 n

VV V V VA ; A ; A ; ...... ; A ; A

V V V V V and over-all

voltage gain OV

S

VA

V. Multiplying individual voltage gains of each amplifier, we get

1 2 3 n 1 nV V V V VA A A ..... A A = 32 4 n 1 n n 1

1 2 3 n 1 n

VV V V V V.....

V V V V V

= n 1

1

VV

But Vn + 1 = VO and V1 = VS

= O

S

VV

= AV

Hence AV = 1 2 3 nV V V VA A A ..... A .

Hence overall voltage gain AV of multi-stage amplifier is obtained by multiplying the voltage gains of individual amplifiers connected in cascade.

Q.3(d) (i) Voltage regulation or Load regulation Let VNL = output D.C. voltage at no load (open circuit) VFL = output D.C. voltage when full rated load current is flowing

Then % of V.R. = % of L.R. = N.L. F.L.

F.L.

V V100

V

It is defined as the ratio of change in D.C. output voltage when load current changes from 0 to full rated load, to the D.C. output voltage at full load.

Ideally its value should be zero and practically it must be as low as possible. (Note : Since the change in V0 is produced due to change in load current, it is known as voltage regulation or load regulation).

(ii) Line (source) regulation Let VHL = Load (output) D.C. voltage with high line (A.C.) voltage

VLL = Load (output) D.C. voltage with low line voltage VN = Normal D.C. output voltage.

V1

+

VS

1st

Amplifier

1VA

V2

2nd Amplifier

2VA

V3 Vn

nth Amplifier

nVA

Vn + 1 = V0 RL

Vidy

alank

armplifier connected in r connected in

to the input terminals of nput terminals of RL connected to the output nected to the output

in, we have in, we havekn 1nV

n 1 nn 1

V VV Vn nnnnnV 11

nn

V VV Vn 1 n1n 1 n1n 1 nn 1

n 11 V; A ; A; AnnV and o

ndividual voltage gains of each ampal voltage gains of ea

1 nV1 nA = = la

2 4 n433

1 2 32 3

VVV V VV V332 4 n 1433VV3333 ......V V VV V V1 2 32 32

1133 4 n4433 ...... 1

= =

alan 11

11

Vn

VV1

But V But Vn + 1 =

= = aOO

S

VVO

VVS

= AV

1 2 3 n3V V V VV V1 2 3 n3A A A ..... AA AV V VVA AAAV VVV .

overall voltage gain A overall voltage gain AV of multi-sta of moltage gains of individual amplifiersage gains of individu

(i) Voltage regulltage regulation or Load ration or Lo Let V Let VNL = output D.C. voltagutput D.C. volt

V VFLF = output D.C. voltput D.C. volt

Then % of V.R. = % Then % of V.R

It is defined as t It is definchanges from

Ideally its v Id((Note : is kno

(ii) L

rraarrrrr


65

Then, % of S.R. = H.L. N.L.

N

V V100

V

It is defined as the change in output D.C. voltage to the change in A.C. input voltage and is expressed in mV or percentage of output voltage.

Ideally it should be zero and practically as low as possible. (iii) Ripple Rejection

It denotes the regulators ability to reject unwanted ripple voltage. It is normally expressed in dB.

Ripple rejection in dB = ripple(output)10

ripple(input)

V20log

V

It is defined as 20 times the common logarithm of voltage ratio obtained by dividing A.C. ripple voltage at the output of regulator to the A.C. ripple voltage at the input of regulator. Ideally it should be zero and practically as small as possible.

(iv) Output impedance (Z0)

Z0 = 0

L

VI

It is defined as the ratio of incremental change in output D.C. voltage V0, to the incremental change in output load current IL. It is measured in ohms. Z0 should be as small as possible. Since when IL flows through it a voltage drop IL Z0 is produced on it. The final D.C. output voltage reduces by this amount which should be ideally zero (i.e., Z0 = 0)

Z0 can be reduced by 1) Using transistor as emitter follower. 2) Using ve feedback in the circuit

Q.3(e) Sweep speed = cdvdt

i.e. it is defined as the rate of change of sweep voltage

w.r.t. time. For sweep voltage to be linear then sweep speed must be constant. For this constant current charging of capacitor is used

where vc = 1

i dtC

= I

dtC

vc = ItC

cdvdt

= IC

= constant

There are 3 types of errors which can be present in a sweep voltage. 1) Slope or sweep speed error (es) 2) Displacement error (ed) 3) Transmission error (et)

1) The slope or sweep speed error (es)

es = Difference in slope at beginning and end of sweep

Initial value of slope

Vidy

alank

ar

le voltage. It is . It

m of voltage ratio obtained by m of voltage ratio obtained of regulator to the A.C. ripple regulator to the A.C. r

as small as possible. all as possible

f incremental change in output D.C incremental change in oue in output load current in output load current IIL. It is

ll as possible. Since when I possible. Since wh L flowuced on it. The final D.C. output The final D.C.

hould be ideally zero (i.e., Z be ideally zero (i.e., Z0 = 0) uced by 1) Using transistoy 1) Using t

2) Using 2) Usin ve fe

peed = ycdvdt

i.e. it is defined as i.e. it is defin

time. For sweep voltage to be lime. For sweep voltaor this constant current charging of constant current cha

where vwhere c = = dy1i dt

CC = = yI

v vcc = = dItIC

Vi

ccdvdvdtdt

= I

There are 3 typeTher1) Slope or s1) S2) Displac) 3) Tran

)


66

For ideal linear sweep, the slope remains same because it is a straight line and hence es = 0 for ideal sweep.

2) The displacement error (ed)

vs = Instantaneous value of actual sweep voltage

sv = Instantaneous value of linear sweep voltage

Vs = Maximum value of actual sweep voltage

Then ed = s s max

s

(v v )V

. Thus it is defined as the ratio of maximum

difference between the actual sweep voltage and the linear sweep voltage to the maximum value of actual sweep voltage.

3) The transmission error (et)

VS = Actual sweep voltage

sV = Linear sweep voltage

Normally R C coupling network is used at the output due to which transmission error is introduced.

Transmission error et = s s

s

V V

V

It is defined as the ratio of difference between uncompensated output and compensated output to the uncompensated output.

Q.3(f)

(a) By potential divider formula

Vf = 10

1 2

RV

R R = 0

2V

2 8

Vf = 0.2 V0 …(1)

= 0.2 2 = 0.4 volts [ V0 = 2V] Vf = 0.4 volts

Voltage

vs

TS t

s sV V

sv

VS

TS t

SV

Without RC network

Vf

+

V0

8k

2k R

Vin

+

Vf 2K

8mV A 2V RL = 1K

8K

ned as the ratio of maximum ed as the ratio of maxim

tage and the linear sweep voltage e linear sweep voltage.

ge

g network is network is due to which to which

s introduced. ed.

error e et t = =als ss

ss

V VV Vs ss

VVss

ned as the ratio of difference bet the ratio of differennsated output to the uncompensatutput to the uncom

(a) By poten(a) B

kakaa

TTSS

t

sv

anananaVVS

SSVVSS

anWithout

idyidydya

dyayydyydydydydddydy

Vid

VidVin

+ + d

8mV Ayydy


67

(b) From equation (1) Vf = 0.2 V0 But Vf = V0 Hence feedback factor = 0.2

(c) A = 0

id

VV

where V0 = 2 103 mV and Vid = 8 mV

Open loop gain, A =32 10

8 = 250 = A

(d) Afb = A

1 A =

2501 250 0.2

= 25051

= 4.902

Q.4(a) Class B push-pull amplifier

Circuit Diagram

Working : The two NPN power transistors are connected in push-pull, i.e., when one transistor is ON the other is OFF and vice-versa. Emitters of both transistors are connected to ground while base of both transistors is connected to ground through centre tap secondary of driver transformer. A.C. input voltage VIN to be amplified is given to the primary of driver transformer. Its centre-tapped secondary is used for producing two equal and opposite signals V1 and V2 which are given to the base of both transistors. Similarly, collectors of both transsitor are connected to the primary of output transformer. A +VCC D.C. supply is given to the centre top of primary. Speaker of 8 is connected to the secondary of output transformer. The turns ratio N1 : N2 is adjusted for proper impedance matching.

2

1L L L

2

NR R [R 8 ]

N

When A.C. input signal VIN is not given then both Q1 and Q2 are in cut-off because base and emitter of transistors is at O D.C. volts. As shown above V1 is in phase with and V2 is output of phase with A.C. input signal VIN. For the first +ve half cycle of VIN, the ve half cycle of V2 drives Q2 more into cut-off while +ve half cycle of V1 drives Q1 into conduction when V1 becomes + 0.7 volts. Similarly for the next half cycle Q2 conducts and Q1 is driven into cut-off.

+

VIN

V1 =

V2 =

C.T.

+

+Q1

Q2

N1

N1

N2

Speaker 8

output C.T.+

VCC

ICQ2Ci

ICQ1Ci Output transformer

Vidy

alank

ar

e two NPN power transistors are c NPN power transistoor is ON the other is OFF and vice the other is OFF

ected to ground while base of bound while bash centre tap secondary of driver t centre tap second

plified is given to the primary o is given to the econdary is used for producing twoary is used for produc

are given to the base of both traven to the base of both are connected to the primary oare connected to the primary to the centre top of primaryo the centre top of primaryoutput transformer. The tput transformermatching. matching.

2

LR RL1NN1NN

R1NN11

2NN

When A.C. Whebase andwith an

V

laV2 =

alllan

alank

alaalaalalalaalalanknnnnknknkkk

QQ2 2

NN1

N1

N2 C.T.Caaaaananananan

++VCCClalaaICQ

2Ciinn

ICQCQ

Output transformer Output tran

nknnannnnnnnknnkannknknk


68

Q1 thus produces 1st half cycle while Q2 produces 2nd half cycle. These are transferred to speaker by output transformer and hence both the half cycles are produced at the output, thus reducing distortion.

Q.4(b) Comparison between BJT and JFET

Parameter BJT FET 1) Control Element Current controlled device

input current IB controls output current IC

Voltage controlled device. Input voltage VGS controls output current ID.

2) Device type Bipolar : Current flow due to both majority and minority carriers

Unipolar : Current flow due to majority carriers only

3) Input junction B E junction is forward biased

G S junction is reverse biased.

4) Input resistance Very low compared to JFET of the order of few K .

Very high of the order of several M .

5) Thermal stability Less hence thermal run-away possible.

More hence thermal run away not present

6) Thermal Noise More noisy Less noisy 7) Gain Bandwidth

product High Low

8) Switching speeds High Low 9) Cut off frequency Low High 10) Size Bigger than JFET less

suitable for I.C. fabrication Smaller and hence more suitable for I.C. fabrication.

11) Type and symbols

(a) (b)

(a) (b)

12) Application Low frequency amplifiers and oscillators

(1) High f application (2) Impedance matching to

avoid loading effect

Q.4(c) CROSS OVER DISTORTION For class B operation base and emitter of power transistors are kept at 0 volt D.C. Hence when A.C. input signal are not given both the transistors are in cut off region.

D

S

G N Channel

G D

N

P Channel

C

E

BN P N

C

E

BP N P

Vidy

al

) Type and symbols e and symbols

alank

ar

T ntrolled device. devi

tage VGS controls ontro current Int IDD. . rrpolar : Current flow Current flow

ue to majority carriers ajority carriers only ark

r

G S junction is reverse junction is reversbiased. biased. kak to JFET

w K . Very high of the order of Very high of the oseveral Mseveral M . kankermal run-n-

le. More hence thermaMore hencerunun away not preaway nknky Less noisy s nonnLow L

anngh gh Low anLow High laBigger than JFET less than JFET less suitable for I.C. fabrication uitable for I.C. fabric

Slaa(a) (a

(b) (b)

al

ViVidy

a

12) Application 12) Applica

Vid

ViVVid

VQ.4(c) CROSSROVOFor clasD.C. Hff r

C

BN N P P N N

P P


69

A silicon transistor requires + 0.7 V (for NPN transistor) and 0.7 V (PNP transistor) w.r.t. emitter to conduct. Due to this only when a.c. input signal reaches 0.7 V, then only the two transistors start conducting. As shown above, when VIN 0.7V, then two transistors do not conduct resulting in the distortion of output signal. This distortion occurs whenever A.C. signal is crossing from +ve to ve or ve to +ve cycle and hence it is known as cross over distortion.

This can be avoided by using class AB power amplifier. Q point is selected slightly in active region due to which transistor conducts for slightly more than half cycle thus avoiding cross over distortion.

Q.4(d) Since < 1. Normally = 0.7 R = 10 k = 10 103 and C = 0.1 10 6 F

By formula T = e1

RClog1

Now 1

1 =

11 0.7

= 1

0.3 = 3.33

Hence elog (3.33) = 1.203

T = 610 0.1 10 1.203 = 31.203 10 sec

f = 1T

= 3

1

1.203 10 =

3101.202

= 10001.202

f = 831.95 Hz

Q.4(e) For common emitter configuration The exact relation between output current IC and input current IB is given by IC = DC IB + ICEO …(i)

Similarly for common base configuration The exact relation between output current IC

and input current IE is given by IC = D.C. IE + ICO [ICO or ICBO]

Now put IE = IC + IB IC = D.C. (IC + IB) + ICO

= D.C. C D.C. B COI I I

VIN

VO

t

cross over distortion

+0.7 V

0.7 Vt

IC

IB

C

E

B

C E

B

IC IEVidy

alank

aransistor) and ansistor) and 0.7 V (PNP 0.7 V (PNs only when a.c. input signal only when a.c. input si

art conducting. As shown above, nducting. As shown aot conduct resulting in the distortio resulting in the

never A.C. signal is crossing from A.C. signal is cr s known as crossn as cross over distortion. over dis

ss AB power amplifier. Q point iower amplifier. Qhich transistor conducts for slightansistor cond

ver distortion. tortion.

0.7 7 and C = 0.1 d C = 0.1 10 10 66 F

= eeRCloglog1

1111

y

1 = =

ya

111 0.71 0.7

=

ya

110.30.

= 3

eg (3.33) = 1.203 = 1.2e

T = T = 10 0.1 10 0 1 100.1 100 1 10

f = f y11T = =y1. f = 831.95 H f = 831.95

Q.4(e) 4(e) For common emitter confir common emitteThe exact relation betwThe exact rand input current Iand input cur B i I C = DC IB + I

Similarly for coSimilaThe exact rThe

and in IC

distortion on


70

Fig. 1 : RC phase shift oscillator using transistor

C D.C. C D.C. B COI I I I

D.C. C D.C. B CO(1 )I I I

COC B

II I

1 1 …(ii)

Companies equations (i) & (ii) we get

= 1

and CEO1

I1

ICO

Now by formula = 1

CEO1

I1

1

ICO = CO CO1

I 1 I1

1

But normally 100 < < 500. Also neglecting 1 we can write + 1 100 ICEO = 100 ICO

Hence reverse leakage current in common emitter is 100 times greater than the reverse leakage current in common base configuration.

Q.4(f) RC Phase Shift Oscillator using Transistor

A typical RC phase shift oscillator using transistor as an active device is shown in figure 1.

The Circuit consists of a single stage amplifier in C.E. configuration and the RC phase shifting network.

The resistors R1, R2 and RE are connected for transistor biasing CE is the emitter bypass capacitor.

Operation : As shown in figure 1 the output

Vo of the single stage CE amplifier has been connected as an input to the RC phase shifting network.

The output of the phase shifting network is connected at the input of the amplifier.

As the amplifier is C.E. type, it introduces a phase shift of 180 between its input and output. The phase shifting network will introduce an additional 180 phase shift to make the phase shift around the loop equal to zero.

The phase shift around the loop will be precisely equal to 0 only at one frequency "f" which is the frequency of operation. If the gain of the amplifier

Vidy

alank

ar we can write rite + 1 + 1 100 100

mmon emitter is 100 times greate emitter is 100 timemmon base configuration. ase configuration

g Transistorsistor

anor using transistor

lanla single stage amplifier in

alaaland RE re connected

alacitor.

yaure 1 the output

dyaof the single stage CE

dyaplifier has been con

dy

nput to the RC

dyfting netwo

idy

e output of the phase s

Vidytwork is connected

Vidyinput of the amp

Vid As the amplifi

Vidtroduc

Vieen VViVpVV


71

and feedback factor are adjusted properly to have a loop gain |A | 1 the sustained sinusoidal oscillations will be obtained at the oscillator output.

Note that the RC feedback network of figure 1 is slightly different from the one we have discussed in section 6. The resistance R3 of Figure 1 is selected in such a way that,

R3 + Zi = R ... (1) where, Zi = Input impedance of the CE amplifier. It is given by, Zi = hie || (R1 || R2)

But as R1 and R2 are large enough. We can neglect them. Hence, Zi hie

Substituting this value into Equation (1) we get, R3 + hie = R or R3 = R hie ... (2)

And if we do not neglect the resistors R1 and R2 then the value of R3 is given by, R3 = R [R1 || R2 || hie] ... (3)

Q.5(a) Av = open loop voltage gain; Afb = Closed loop voltage gain;

= feedback factor then by formula Afb = v

v

A1 A

Now 1 i.e. normally is always less than 1 and its maximum value can be only one. But the open loop voltage gain Av is very high and hence Av >> 1. Neglecting 1 in the denominator we get

Afb = v

v

AA

=1

Thus closed loop gain Afb is equal to the reciprocal of feedback factor . In ve feedback amplifiers only resistors are used whose values are fixed once

they are selected. = Ratio of these resistors 1

1 f

RR R

which is constant.

Thus Afb remains constant and is independent of the parameter of the transistor ( a.c or hfe). Because of this reason, stability of voltage amplifier improves when ve feedback is used.

Q.5(b) Comparison between series and parallel resonant circuit

Parameter Series Resonant Circuit Parallel Resonant Circuit 1) Circuit

i.e. R, L and C are in series.

i.e. coil and C are in parallel.

2) f0 f0 =

1

2 LC f0 =

2

2

1 1 R2 LC 4L

1

2 LC

where R 0 and can be neglected.

R L C R L

C Vidy

alank

arrrarar

arkar

kaka

... (2)

ka hen the valu is givenka ... (3)

Afb = an

v

v

A11 vA

ways less than 1 and its maximu ess than 1 and its p voltage gain A voltage gain Avv is very high and is very h

inator we get tor we get

=

a1

gain Afbb is equal to the reciprocal is equal to the recck amplifiers only resistors are uslifiers only resistors

elected. = Ratio of these resisto = Ratio of thes

Afbb remains constant and is indep remains constant c or hfe). Because of this reason, ). Because of this

ve feedback is used. dback is us

b) Comparison between series ) Comparison between series

Parameter Series Parameter

Vidy

ViViddy1) Circuit Circuit

Vid

ViVidd

2) fVVV


72

Parameter Series Resonant Circuit Parallel Resonant Circuit 3) Phase angle = 0 i.e. ‘I’ and

Vin are in phase at f0. Phase angle = 0 at f0.

4) p.f. p.f. = cos = 1 p.f. = cos = 1 5) Reactance XL = XC at f0

X = reactance BL = BC at f0 B = Susceptance

6) Zeff Zeff = R and is minimum Zeff = L

RC and is maximum

7) I0 max. at f0 and I0 = INV

R I0 = INV

L / RC = INRCV

L

and is minimum 8) Q0

Q0 = 0

0

L 1 1 LR RC R C

and is known as voltage magnification factor

Q0 = 0

0

L 1 1 LR RC R C

and is known as current magnification factor.

9) At f0, VL = VC = Q0 VIN i.e. i.e. VL and VC are more than VIN.

At f0, IC = Q0 IT i.e. IC is more than IT.

10) Resonance Curve

11)

B.W. = 0

0

fQ

B.W. = 0

0

fQ

Q.5(c) Circuit diagram :

21

L L2

NR R

N where LR is the impedance transferred to primary side.

V0

ff1 f0 f2

Vmax

0.7 Vmax B.W.

I

f f0

+

R2

R1

RE

VIN

2V p to p

N1 N2

+VCC Speaker

RL = 8

Output transformer

CE

CIN Vidy

alank

ar

s maximum um rarC = aININRCVVII

LL minimum um arar0 = = kkaaaka0

00

L 1 1 L1R RC R CR RC R C00

and is known as current known as currenmagnification factor. magnification fakan

a i.e. i.e.

than VINN..At fAt f0, I ICC = Q = Q0 I ITT i.e. I i.e C is mothan Ith T. nknn

yan

B.W. = W. ya00

0

f0Q

yadydydya

rcuit diagram :agram

aalalaalaff1 ff00 ff2allB.W.Ban

I


73

Working : The resistors R1 and R2 are voltage divider biasing network which establishes Q point in the centre of a active region. RE stabilizes the Q point. By-pass capacitor CE by-passes A.C. output signal from the emitter. This removes ve feedback which improves the gain of the amplifier. Input capacitor CIN blocks D.C. voltage and allows A.C. input voltage VIN to pass to the base of transistor. The collector resistor RC is replaced by the primary winding of output transformer. Its resistance is very small which reduces power losses in it thus improving power efficiency. The maximum improves from 25% (for RC) to 50% (for transformer coupled). The transformer gives good D.C. isolation as well as impedance matching. Speaker impedance of 8 (RL) on secondary side is transferred to LR on primary

side, where2

1L L

2

NR R

N. Hence by using step-down (N2 < N1) transformers,

impedance of speaker is matched to the output impedance of amplifier. Due to this maximum power is transferred to the loud-speaker.

Q.5(d) Advantages : (i) Bandwidth of the amplifier increases (B.W.)new = (1 + Av )(B.W.)old (ii) (Amplitude, frequency, phase harmonic) distortion of the amplifier reduces.

D0 (new) = old

v

D(1 A )

(iii) Noise signal in the output reduces.

0(old)0(new)

v

NN

1 A

(iv) Stability of the amplifier improves due to which voltage gain becomes independent of transistor parameters and hence it remains almost constant.

(v) (a) RIN for voltage amplifier should be high. (b) R0 of voltage amplifier should be low.

By using voltage series ve feedback amplifier, RIN increases and R0 decreases.

(new)INR = (1 + Av ) RIN(old)

RO(new) = 0(old)

v

R

1 A

Disadvantages :

The only disadvantage is that voltage gain of the amplifier decreases

Afb = v

v

A1 A

But if more voltage gain is required then multistage amplifiers with ve feedback can be used.

Vidy

alank

ar

e ba

tput transformer. sform it thus improving roving

(for RCC) to 50% (for ) to 50% (for

ell as impedance matching. ell as impedance matchin is transferred to s transferre LLR on primary on prim

g step-down (Ndown (N22 < N < N1) transforme) tran

e output impedance of amplifier. ut impedance of am the loud-speaker. -spea

increases (B.W.) increases (B.W.)newnew = (1 + A = (1 + v )(B phase harmonic) distortion of the ahase harmonic) distortion

alold

v1 A )vAv

the output reduces. utput reduces. a0(old)w)

vv

N

1 vvAAity of the amplifier improves du amplifier impro

ependent of transistor parameters ependent of transistor paraa) R RININ for voltage amplifier should for voltage amp

(b) R 00 of voltage amplifier shoul of voltage amplifie

By using voltage series By using voltage serie ve feee fe

(new)new)INR = (1 + AR = (1 + A(new)new)IN v ) R RI

R RO(new) O(new) = d0(old)RR

1 A

Disadvantages Disa

The only disa The o

A


74

Q.5(e)

Applying K.V.L. at the input of the amplifier We get All voltages = 0

+VIN vid vf = 0 VIN Vf = vid.

By definition of feedback factor f

0

VV

hence f 0V v . Putting this value in

above equation we get VIN v0 = vid …(1)

By definition of open loop voltage gain AV = 0

id

vv

v0 = AV vid from equation (1) v0 = AV (vIN v0) = AV vIN AV v0

V0 + AV v0 = Av vIN V0 (1+ Av ) = Av vIN

0

IN

vV

= V

V

A1 A

But by definition 0fb

IN

vA

V closed loop voltage gain.

Hence Afb = v

v

A1 A

In the above expression denominator > 1. Hence Afb < Av i.e. voltage gain of the amplifier reduces when ve feedback is used.

Q.5(f) I.C. 723 is basically a series voltage regulator. Its important blocks and their brief description is given below : i) Voltage Reference Amplifier : A temperature compensated 6.2 V zener is

biased with a constant current source. An op-amp A1 is used as a buffer amplifier which provides reference output voltage of 7.15 V which is capable of supplying current upto 15 mA.

ii) Comparator and error amplifier : Op-amp A2 is working both as comparator and error amplifier. The reference output (or fraction of it) is given to the non-inverting input terminal. Similarly the output voltage V0 (or

Voltage Amp.Av

Feedback network

+

+ VIN

+ Vid V0

Vf V0

+

RL +Vf

Vidy

alank

ar

ence f 0f 0v .v .00VVff Putting this va Putting

…(

age gain Aage gain AV = an00

id

vvvv

vid from equation (1) from equa AV (vIN v0) )

= AV v vININ A AVV v0

v0 = Avv v vININ Av ) = A v v vININ

y0

IN

vVI

= = yaV

V

A11 VA

by definition efinition

dy

00fbfb

IN

vvAA

VIfbfbAA closed lo

Hence AHence Afbfb = =yv

v

A1 vvAA

In the above expression den the above expresHence Ace Afbfb < A < Av i.e. volta iused. u

Q.5(f) Q.5(f) I.C. 723 is basicaI.C. description is gdescri) Voltage i) V

biasedams


75

fraction of it) is given to the inverting input terminal of op-amp A2. These two voltages are compared with each other and the difference between them known as error voltage is produced. This error voltage is amplified by op-amp A2 and is available at its output as control voltage.

iii) Output Stage : It consists of series pass control transistor Q2. The control

voltage at the output of op-amp. A2 is given to the base of Q2. This control voltage adjusts the voltage drop VCE of Q2 till error voltage becomes zero and V0 remains constant.

iv) Over-Current protection circuit : Transistor Q1 connected at the output works as current protection circuit. An external resistor RCL connected between its base (current limit) and emitter (current sense) fixes the maximum value of the load current. Under short circuit condition also the load current does not exceed this value.

v) How to vary the output voltage : A potentiometer R2 is varied to change the output voltage to the required value. This R2 is connected in potential divider circuit across output terminals or it is connected to reference output terminal.

Q.6(a) Zener Diode in Parallel to RL

Circuit Diagram :

Equations : (i) V0 = VZ hence if VZ is constant then V0 is constant. (ii) By K.C.L. at A , IT = IZ + IL (iii) By K.V.L., All voltages = 0 IN T S 0V I R V 0

+

A2

Q1

Q2

VZ

V0

VC

C.S. (Current sense)

C.L. (Current

limit)

Frequency compensation

A1

IConstant current source

Temperature Compensated zener

A1 : voltage reference amplifier

Vref.

Inverting input

Non-inverting input

A2 : Comparator and Error Amplifier

Booster terminal

Current protection circuit

Output Stage

V+

V

RS

ITVZ = V0 RL

IL

IZ

A

VIN

Vidy

alank

arf series pass control transistor Qs pass control t

-amp. A A22 is given to the base of is given to the bge drop Vge drop VCECE of Q Q22 till error voltage till error

ection circuit :ircuit Transistor Q Transisto 1 ct protection circuit. An externaection circuit. An e

base (current limit) and emitter current limit) andalue of the load current. Under the load current.

nt does not exceed this value. s not exceed this vao vary the output voltage :he output voltag A po

output voltage to the required vale to the requirvider circuit across output terminader circuit across o

erminal. nal.

Zener Diode in Parallel to R Zener Diode in Parallel to RL L Circuit Diagram : Circuit Diagram :

Equations qu (i) V0 =

(ii) Byi)

naraarararraraaaaaarrararaakakkakakakakkkakakka

Q1

2

VVZ

VV0

VC

C.S. C.S. (Current (Current sense)sens

C.L. C.L. (Current ent

limit)lim

parator and nd or Amplifier er nk

a

Boterminn

Current protection Current protectiocircuit circu

O

dyydddddyyRRSS

VVIN


76

V0 = VZ = VIN IT RS and ITRS = VIN VZ

RS = IN Z

T

V VI

= IN Z

Z L

V VI I

(iv) Thus by selecting the value of RS properly the condition (knee) (max)Z Z ZI I I is

satisfied for all conditions due to which VZ = V0 remains constant. Working : VIN varying IL constant : It is assumed that RS is so selected that V0 = VZ = constant.

A B 1. Assume VIN increases. Assume VIN decreases. 2. This increases VIN V0 This decreases VIN V0 3. IT increases and IT = IZ + IL. But IL is

constant since V0 is constant IT decreases and IT = IZ + IL. But IL is constant since V0 is constant.

4. IZ increases but IZ < max.ZI hence V0 is

constant

IZ decreases but IZ > (knee)ZI hence V0

is constant. Conditions necessary for this circuit to work properly are

(i) VIN > VZ, so that zener diode is properly reverse biased. (ii) Value of RS is so selected that under all conditions;

(knee) (max.)Z Z ZI I I is

satisfied. Hence zener diode operates in proper breakdown condition and VZ = V0 remains constant.

Q.6(b) 2 stage transformer coupled amplifier

TR1 : Input transformer TR2 : Driver transformer TR3 : Output transformer

+VCC

+

RE' CE'

VO

VS

TR1

R1

R2

RE CE

R1'

R2'

RL

TR3

TR2

A.C. ground

C1Vidy

alank

ar

(mZ

ted that V0 = V VZZ = =

B arrrN decreases. eases.

kacreases cre VVININ V0 kareases and Iases and ITT = IZ + I ILL. But I. Bu L is nstant since Vnce V00 is constant. is constantkaZ decreases but Ieases but IZ Z >

(knee)ZI henceZ

is constant.nstant. knkt to work properly are perly a is properly reverse biased. perly reverse bias

ed that under all conditions; under all condition Z(IZ

r diode operates in proper break diode operates in propestant.

er coupled amplifier pled amplifier

TRTR1 : Input TR2 : DrivTR3 : O

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Vid

Vid

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Vid

Vid

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R1

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77

Frequency response

A transformer is used for coupling the A.C. signal from previous stage to next stage. The main advantages of transformer are : (i) very good D.C. isolation, (ii) very good impedance matching, (iii) resistance of primary is less hence D.C. losses are less which increases efficiency.

In the above circuit, if

1 2V VA andA are the voltage gains of individual stage then

overall voltage gain 1 2V V VA A A . The functions of each component are given

below. (i) R1 and R2 are voltage divider biasing network which establishes Q point in

the centre of active region. (ii) RE stabilizes Q point. (iii) CE by-passes A.C. signal at the emitter. This removes ve feedback and

improves voltage gain. (iv) Transformer : (a) Gives D.C. isolation, (b) Provides very good impedance

matching by selecting number of turns of primary and secondary. (v) C1 : This capacitor provides A.C. ground due to which maximum A.C. voltage

is applied to the 2nd stage.

As shown in frequency response, this amplifier is capable of amplifying one particular resonant frequency f0. Hence they are used as tuned voltage amplifiers by connecting capacitors across primary and secondary winding. These are used for amplifying one particular intermediate frequency in communication circuits like Radio, TV, etc.

Q.6(c) The gain of an amplifier is denoted by "A" and A = output signalinput signal

.

(i) Voltage gain OV

IN

VA

V : It is defined as the ratio of output voltage VO to the

input voltage VIN. It is a pure number and has no unit of measurement AV > 1 since VO > VIN.

(ii) Current gain OI

IN

IA

I : It is defined as the ratio of output current IO to the

input current IIN. It is a pure number and has no unit of measurement AI > 1 since IO > IIN.

f f0

V0

(or AV) Low frequency

Flat regio

Resonant size

High frequency

Vidy

alank

aral from previous stage to next from previous stage to :

pedance matching, (iii) resistance atching, (iii) resi which increases efficiency. ncreases efficie

e the voltage gains of individual stoltage gains of indiv

2. The functions of each componee functions of each c

divider biasing network which esta ivider biasing network whigion. n

t. C. signal at the emitter. This renal at the emitter. T

ge gain. n. : (a) Gives D.C. isolation, (b) Pives D.C. isolation

by selecting number of turns of priecting number of turns capacitor provides A.C. ground dtor provides A.C.

plied to the 2 2ndnd p p p stage. stage.

hown in frequency response, thiswn in frequency reticular resonant frequency f resonant frequen 0.

amplifiers by connecting capacitoers by connecting caThese are used for amplifyiThese are used for amplicommunication circuits like Racommunication circuits like Ra

Q.6(c) c) The gain of an amplifier The gain of an

(i) Voltage gain (i)

input voinsince

(ii) C


78

(iii) Power gain OP

in

PA

P : It is defined as the ratio of output power PO to the

input power PIN. It is a pure number and it has no unit of measurement. Relation between AV, AI and AP Multiplying the values of AV and AI, we get

AV AI = O O

IN IN

V IV I

= O O

IN IN

V IV I

= O

IN

PP

= AP

Since VO IO = output power PO and VIN IIN = input power Thus AP = AV AI

Q.6(d) (a) Circuit diagram : (i) Capacitively coupled (ii) Inductively coupled VIN = Modulated high frequency (radio frequency) carrier signal (b) Frequency Response : Av = voltage gain (i) f0 = Resonant frequency = fIN

= carrier frequency (ii) B.W. = f2 f1

(iii) B.W. = 0

0

fQ

(iv) Av = AC L

i

rR

(v) rL = effZ = L

RC (max.)

Working : Resistors R1 and R2 are voltage divider biasing network which establishes ‘Q’ point in the centre of active region. RE stabilizes the ‘Q’ point. CE by passes A.C. signal at the emitter to the ground. This removes ve feedback due to which Av increases. Cin and Cc block d.c. voltage and allows A.C. voltage to pass.

+

R1

R2

V0 CC

VIN

+ VCC

CIN

RL

C L

+

R1

R2

V0

VIN

+ VCC

CIN

CP

R C

LP LS RL

AV = Voltage gain

ff1 f0 f2

AV(max.)

0.7 AV(max.)

B.W.Vidy

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aructively coupled ely coupled

Modulated high frequency ated high frequency (radio frequency) carrier signaio frequency) car

equency Response : quency Response Av = voltage gain = voltage gain

WorkinResi

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nka

+ VCC

CIN nknknknnnknkkkkkknkkCP kkkkkkkkkk

dydydyAA VV

= Voltage gain= Voltage gain

A V( )V( 0.7 A0.7 A

V(max )V(max )


79

In the above circuit Rc is replaced by single tuned (parallel resonant) circuit. Either L or C or both are variable. Their values are so adjusted (tuned) that resonant

frequency 01

f2 LC

of the tuned circuit becomes exactly equal to incoming

signal frequency. At f0, Zeff of parallel circuit becomes maximum and is given by

Zeff = L

RCwhere R = resistance of inductor. As shown above Av Zeff and hence

voltage gain becomes maximum at f0 as shown in the frequency response.

B.W. = 0

0

fQ

and hence as Q0 increases, frequency response becomes more

narrower and bandwidth decreases. Hence circuit becomes more selective.

Thus the circuit is capable of amplifying one particular high frequency (f0) or a very narrow band of high frequencies (f2 f1). Since only 1 tuned circuit is used. It is known as single tuned voltage amplifier. The output voltage can be either capacitively coupled or inductively coupled.

Q.6(e) Advantages : (i) Maximum power, = 78.4% which is higher than Class A amplifier.

(ii) Even harmonic components are eliminated.

(iii) More A.C. power output per transistor than Class A.

(iv) D.C. current of both transistor flow in opposite direction in the primary winding of output transformer. Hence magnetic fluxes produced by them cancel each other due to which core of transformer does not saturate.

(v) It is possible to eliminate both the transformers which reduces the cost and improves the frequency response.

Disadvantages : (i) Two power transistors are required when increases the cost.

(ii) The two transistors must be a matched pair, i.e., their parameters must be identical to avoid distortion. This also increases the cost of power transistor.

(iii) Driver stage required for producing two equal and opposite A.C. input signals.

(iv) Distortion is more than Class A amplifier.

(v) Because of Class B operation, cross-over distortion present.

(vi) Transformers are bulky and occupy more space.

(vii) Cost increases because of 2 transformers.

Note : Since two transformers can be eliminated and A.C. power output is more, Class B push-pull power amplifier are used maximum.

Q.6(f) Comparison of class A, B and C power amplifier Parameter Class A Class B Class C 1) Diagram of A.C.

load line. Refer fig. 1(a). Refer fig. 1(b). Refer fig. 1(c).

2) Q-point Centre of active region

On the border of cut-off and active region

In the cut-off region.

3) A.C. collector current, iC

Flows for complete cycle (0 to 360 )

Flows for half cycle (0 to 180 )

Flows for less than half cycle.

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s given

Zeff and hence henf

quency response. ponse

ponse becomes more becomes more

omes more selective. more selectiv

ticular high frequency (fcular high frequency (f00) or a ) Since only 1 tuned circuit is used. only 1 tuned circuit is

r. The output voltage can be eithput voltage can

hich is higher than Class A amplifier than Class A a are eliminated.minated.

er transistor than Class A.stor than Class A.ransistor flow in opposite directansistor flow in opposit

nsformer. Hence magnetic fluxermer. Hence magnet ue to which core of transformer doch core of transfo

eliminate both the transformers wate both the transform requency response. ncy respons

es : ower transistors are required when sistors are require

two transistors must be a matche two transistors must be a entical to avoid distortion. This alsntical to avoid distor

Driver stage required for produr stage required signals.nals.

(iv) Distortion is more than Classstortion is more than Cla(v) Because of Class B opera(v) Because of Class B opera(vi) Transformers are bulky vi) Transformers ar(vii) Cost increases becau) Cost increasesNote :N Since two trans SinceClass B push-pull poClass B push

Q.6(f) Comparison oQ.6(f) ComParamVVV1) DiagloaVVV

2)


80

4) D.C. collector current, IC

Flows even if A.C. input signal is not given.

Does not flow if A.C.input signal absent.

Does not flow if A.C.input signal absent.

5) Maximum power,

25% for RC 50% for TX (Lowest)

78.4% (Medium)

> 80% (Highest)

6) Distortion Lowest Medium Highest 7) Output power to

load. Lowest High Highest

8) Cross-over distortion

Not present Present Not present

9) Hum (Noise signal)

Absent Low High

Vidy

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st rhest rrNot present present arkr

High High

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S.Y. Diploma : Sem. III [EJ/EN/ET/EX/EV/ED/IU/DE/IS/IC/IE...

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