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MULTISTAGEANDFEEDBACKAMPLIFIERS

Learning Objectives

General Amplifier Coupling RC-coupled Two-stage

Amplifier Impedance-coupled Two-

stage Amplifier Advantages of Impedance

Coupling Transformer-coupled Two-

stage Amplifier Advantages of Transformer

Coupling Frequency Response Direct-coupled Two-stage

Amplifier Using SimilarTransistors

Direct-coupled AmplifierUsing ComplementarySymmetry of Two Transistors

Darlington Pair Advantages of Darlington

Pair Comparison Between

Darlington Pair and EmitterFollower

Special Features of aDifferential Amplifier

Common Code Input Differential Amplifier

2316 Electrical Technology

61.1. General

Often, the voltage ampli-fication or power gain or fre-quency response obtained witha single stage of amplificationis insufficient to meet the re-quirements of either a compos-ite electronic circuit or a loaddevice. Hence, two or moresingle stages of amplificationare frequently used to achievegreater voltage or current am-plification or both. In suchcases, the output of one stage serves as input of the next stage as shown in Fig. 61.1. Such amplifiersmay be divided into following two categories :

(a) Cascaded AmplifiersIn these amplifiers, each stage as well as the type of interstage coupling used are identical.

(b) Compound AmplifiersIn these amplifiers, each stage may be different from the other (one may be CE and the other

may be CC stage) and also different types of interstage couplings may be employed.

As stated above, in cascaded amplifiers, the output ac voltage of the first stage becomes theinput voltage of the second stage and the ac output of the second stage becomes the input of the thirdstage and so on. The overall voltage gain of the cascaded amplifier is equal to the product (not thesum) of the gain of the individual stages.

∴ A v= Av1 × A v2 × A v3 × . . . . . . . . . . .

However, when the voltage gain is expressed in decibels (dB), then the overall decibel gain ofthe multistage amplifier is equal to the sum of the dB gains of the individual stage i.e.

G = G1 + G2 + G3 + . . . . . . . . . .

Similarly, the overall current amplification is given by

A i = A i1 × A i2 × A i3 × . . . . . . . . . .

The overall power gain is given by

A p = Av . Ai and Gp = 10 log10 A P dB

Suppose in a two-stage cascaded amplifier, first stage has a voltage amplification of 2000 (dBgain of the 20 log10 2000 = 66 dB) and second stage has corresponding values of 1000 (60 dB). Ifthe ac output of first stage is fed into the second stage, the overall amplification would theoreticallybecome = 1000 × 2000 = 2 × 106 which corresponds to a dB gain of (60 + 66) = 126 i.e. 20 log102 × 106 = 20 × 6.3 = 126. The above result would be true only when we neglect the loading effectof first stage by the second stage. It would be approximately true so long as the impedance lookinginto the input of second stage is much greater than the output impedance of the first stage. Other-wise, the overall gain would be much less.

61.2. Amplifier Coupling

All amplifiers need some coupling network. Even a single-stage amplifier has to be coupled tothe input and output devices. In the case of multistage systems, there is interstage coupling. Thetype of coupling used determines the characteristics of the cascaded amplifier. In fact, amplifiers areclassified according to the coupling network used.

Fig. 61.1

Multistage and Feedback Amplifiers 2317

Fig. 61.2

The four basic methods of coupling are :1. Resistance-Capacitance (RC) Coupling

It is also known as capacitive coupling and is shownin Fig. 61.2 (a). Amplifiers using this coupling are knownas RC-coupled amplifiers. Here, RC coupling networkconsists of two resistors RC1 and RC2 and one capacitorC. The connecting link between the two stages is C. Thefunction of the RC-coupling network is two-fold :

(a) to pass ac signal from one stage to the next,(b) to block the passage of dc voltages from one

stage to the next.

2. Impedance Coupling or Inductive Coupling

It is also known as choke-capacitance coupling and is shown in Fig. 61.2 (b). Amplifiers usingthis coupling are known as impedance-coupled amplifiers. Here, the coupling network consists ofL1, C and RB. The impedance of the coupling coil depends on (i) its inductance and (ii) signalfrequency.

3. Transformer Coupling

It is shown in Fig. 61.2 (c). Since sec-ondary of the coupling transformer conveysthe ac component of the signal directly to thebase of the second stage, there is no need fora coupling capacitor. Moreover, the second-ary winding also provides a base return path,hence there is no need for a base resistance.Amplifiers using this coupling are called trans-former-coupled amplifiers.4. Direct Coupling

It is shown in Fig. 61.2 (d). This coupling is used where it is desirable to connect the loaddirectly in series with the output terminal of the active circuit element. The examples of such load

Modern coupling transformer

R.C. Coupled two-stage amplifier


devices are (i) headphones (ii) loud-speakers (iii) dc meters (iv) relays and (v) input circuit of atransistor etc. Of course, direct coupling is permissible only when

(i) dc component of the output does not disturb the normal operation of the load device,(ii) device resistance is so low that it does not appreciably reduce the voltage at the electrodes.

61.3. RC-coupled Two-stage AmplifierFig. 61.3 shows a two-stage RC-coupled amplifier which consists of two single-stage transistor

amplifiers using the CE configuration. The resistors R2 and R3 and capacitor C2 form the couplingnetwork. R2 is collector load of Q1 and R4 is that of Q2. Capacitor C1 couples the input signalwhereas C3 couples out the output signal. R1 and R3 provide dc base bias.

Fig. 61.3

(i) Circuit OperationThe brief circuit operation is as under :

1. the input signal ni isamplified by Q1. It isphase reversed (usualwith CE connection);

2. the amplified output ofQ1 appears across R2;

3. the output of the firststage across R 2 iscoupled to the input atR3 by coupling capaci-tor C2. This capacitor isalso sometimes referredto as blocking capaci-tor because it blocks the passage of dc voltages and currents;

4. the signal at the base of Q2 is further amplified and its phase is again reversed;5. the ac output of Q2 appears across R4;

6. the output across R4 is coupled by C3 to load resistor R5;

7. the output signal v0 is the twice-amplified replica of the input signal vi. It is in phase withvi because it has been reversed twice.

(ii) AC Equivalent CircuitThe ac equivalent circuits for the two stages have been shown separately in Fig. 61.4 and Fig.

61.5 respectively.

If Fig. 61.4, ri.1= R1 || β1.re.1

Fig. 61.4

FirstStage

CouplingElement

R1

R2 R3R4

VCE

R5

C1

C2 C3

vi

v0

Q1 Q2

SecondStage


It is the input impedance ofthe first stage and not rin(base).

The output impedance ofthe first stage is

r0.1= R2 || ri.2It is so because the input

of the second stage forms a partof the output of the first stage.As seen from Fig. 61.5.

ri.2 = R3 || β2. re2 ≅ β2 re.2,where re.1 and re.2 are ac junction resistances of the two transistors and are given by

.1 .2.1 .1 .2 .2

25mV 50mV 25mV 50mVor and ore e

E E E E

r rI I I I

= =

The output impedance of Q2 is r0.2 = R4 || R5

(iii)Voltage Gain0.2 0.2

1.2 2 .2 2 22 1.2 .2

Now.e v

e

r rr r A

r rβ β

β= = =

Also, 0.2 0.2

.2 2 .2 2 22 .2 .2.i e v

e e

r rr r A

r rβ β

β= ∴ = =

The voltage gain of the first stage is also given by a similar equation

0.1

.1vi

e

rA

r=

Example 61.1. For the two-stage RC-coupled low-level audio amplifier shown in Fig. 61.6,compute the following :

(i) ri (ii) Av1 (iii) A v2 and (iv) Av in dB. Neglect VBE and take re = 25 mV/IE.(Electronic Circuits, Mysore Univ.)

Solution. The input impedance of the cascaded amplifier is(i) ri = R1 || βl.re.1

For finding re.1, we need IE.1which approximately equals IC.1Now, IC.1 = β. IB.1Also, IB1 = 12/R1= 12/0.6 M

= 20 µA ∴ IC1 = 100 × 20

= 2000 µA = 2 mA ∴ IE.1= 2 mA

∴ re.1 = 25/2 = 12.5 Ω ; β1 re.1= 100 × 12.5 = 1250

∴ ri = R1 || β1.re.1 = 0.6 M || 1250 Ω

≅ 1250 Ω

(ii) 0.1

1 0.1 2 .2.1

Now, ||v ie

rA r R r

r= =

ri.2 = R3 || β2 re.2 = 0.6 M || 1250 Ω ≅ 1250 Ωr0.1 = 5 K || 1250 Ω = 1000 Ω ; re.1 = 12.5 Ω ∴ A v.1 = 1000/12.5 = 80

Fig. 61.6

Fig. 61.5


(iii) 0.2

2.2

ve

rA

r=

Now, r0.2 = R4 || R5 = 5 K || 20 K = 4 K, IE.2 = 2 mA – same as IE.1

re.2 = 25/2 = 12.5 Ω ∴ A v2 = 4000/12.5 = 320(iv) Av = A v.1 × A v.2 = 80 × 320 = 25,600(v) Gv = 20 log10 A v dB = 20 log1025,600 = 88 dB.

Example 61.2. For the two-stage RC-coupled amplifier shown in Fig. 61.7 compute thefollowing :

(i) ri , (ii) Av.1 , (iii) Av.2 , (iv) Av in decibels.Take β1 = β2=100. Neglect VBE and use re = 25 mV/IE.

(Applied Electronics-I, Punjab Univ. 1991)Solution. (i) The input impedance of the stage is ri = R1 || β1.re.1It should be noted that R6 does not come into the picture because it has been ac grounded by C4.

However, it would affect the dc emitter current.

.1 66 1 1

251mA

/ 10,000 1.5 10 /100CC

E

VI

R R= = =

+ β + ×re.1 = 25/1= 25 Ω; β1.re.1 = 100 × 25 = 2500 Ω ri = 1.5 M || 2500 Ω ≅ 2500 ΩΩΩΩΩ

(ii) 0.1

.1.1

v

e

rA

r=

Now, r0.1 = R2 || ri.2 and ri.2 = R3 || β2 re.2

Now, .2 6

251mA

10,000 1.5 10 /100EI = =+ ×

∴ re.2 = 25/1 = 25 Ω ; β2 .re.2 = 2500 ΩΩΩΩΩri.2 = 1.5 M || 2500 Ω ≅ 2500 Ωr0.1=5 K || 2.5 K = 1,667 Ω

∴ A v.1 = 1667/25 = 66.7

(iii) 0.2

.2.2

,ve

rA

r=

Now, r0.2 = R4 || R5 = 5 K || 20 K = 4 K

∴ A v.2 = 4000/25 = 160(iv) A v = A v.1 × A v.2

= 66.7 × 160 = 10,672 ; Gv = 20 log10 10,672 = 80.3 dB

Example 61.3. Compute the overall voltage amplification for the two-stage RC-coupled am-plifier shown in Fig. 61.8. Express the answer in decibels. Neglect VBE and use re = 50 mV/IE. Takeβ1 = β2 = 100. (Electronics-I, Allahabad Univ. 1990)

Solution. For finding the overall gain, we will have to find each stage gain.

(i)0.1

1.1

,v

e

rA

r= Now, r0.1= R3 || ri.2 and ri.2 = R5 || R6 || β2.re.2

Now, re.2 = 50/IE.2

Fig. 61.7


Drop across

R6 = 5

20 2V5 45

× =+

It also represents the approxi-mate drop across R8.

∴ IE2 = 2/1 K = 2 mA∴ re.2 = 50/2=25 Ω.β2.re.2 = 100 × 25 = 2.5 K

ri.2 = 45 K || 5 K|| 2.5 K=1.6 K; ∴ r0.1 = 5 K || 1.6 K =1.2 K

∴ re.1 = 50/IE.1= 50/2=25 Ω∴ A v.1 = 1200/25=48

.2.2 0.2 7 9

.2

; || 5K || 20K 4Kov

c

rA r R R

r= = = =

∴ A v.2 = 4000/25 = 160 ∴ Av = 48 × 160 = 7,680 ; Gv = 20 log10 7,680 = 77.6 dB

61.4. Advantages of RC Coupling

1. It requires no expensive or bulkycomponents and no adjustments.Hence, it is small, light and inex-pensive.

2. Its overall amplification is higherthan that of the other couplings.

3. It has minimum possible nonlin-ear distortion because it does notuse any coils or transformerswhich might pick up undesirablesignals. Hence, there are no mag-netic fields to interfere with thesignal.

4. As shown in Fig. 61.9, it has a veryflat frequency versus gain curvei.e. it gives uniform voltage amplification over a wide range from a few hertz to a few mega-hertz because resistor values are independent of frequency changes.As seen from Fig. 61.9, amplifier gain falls off at very low as well as very high frequencies. At

low frequencies, the fall in gain (called roll-off) is due to capacitive reactance of the couplingcapacitor between the two stages. The high-frequency roll-off is due to output capacitance of thefirst stage, input capacitance of the second stage and the stray capacitance.

The only drawback of this coupling is that due to large drop across collector load resistors, thecollectors work at relatively small voltages unless higher supply voltage is used to overcome thislarge drop.

61.5. Impedance-coupled Two Stage AmplifierThe circuit is shown in Fig. 61.10. The coupling network consists of L, C2 and R3. The only

basic difference between this circuit and the one shown in Fig. 61.3 is that inductor L has replacedthe resistor R2.

Fig. 61.9

f1 f2f

Gai

n

FlatResponse

HighFrequencyRoll off

LowFrequencyRoll off

Fig. 61.8


(i) AC Equivalent CircuitThe ac equivalent circuit (at mid-

frequency) of the cascaded amplifierhas been shown in Fig. 61.11. Becauseof mid-frequency range, effects of allcapacitances have been ignored.

(ii) Circuit OperationThe operation of this circuit is the

same as that of the RC-coupled circuitdescribed earlier.

(iii) Voltage GainIt is given by the product of two stage gains A v = A v1 × A v2

Now,0.1

.1 0.1 .2 .2 3 2 .2.1

where || and || .v L i i ee

ZA Z X r r R r

rβ= = =

In case, X L » ri.2, then, Z0.1 ≅ ri.2

Fig. 61.11

∴.2 0.2

.1 .2.1 .2

andiv v

e e

r rA A

r r= =

∴ Av =Av.1 × Av.2

Example 61.4. For the impedance-coupled two-stage amplifier shown in Fig. 61.12, computethe values of

(i) Av.2 , (ii) Av.1 at 4 kHz and (iii) Av in dB.Neglect VBE and use re = 25 mV/IE. Take β1 = β2 = 100.

(Industrial Electronics, Calcutta Univ. 1991)

Solution. (i) 0.2

.2.2

ve

rA

r=

Now, r0.2 = R4 || R5 = 8 K || 24 K = 6 KIB2 = 12/1.2M = 10 µA, IC.2 = β2 . IB.2 = 100 × 10 =1000 µA = 1 mA∴ IE2 ≅ 1 mA∴ re.2 = 25/1=25 Ω

.2

6,000

25VA = =2 240

(ii) 0.1 .2.

.1.1 .1

iv

e e

Z rA

r r= ≅

Now, re.1 = 25 Ω – equal to re.2 XL = 2π f L

= 2π × 4 × 103 × 1= 25, 130 Ω ri.2 = R3

Fig. 61.12

Fig. 61.10


|| β2.re.2 =1.2M || 2500 Ω ≅ 2500 ΩObviously, X L » ri.2 thus justifying the above approximation.

∴ A v.1= 2500/25 = 100

(iii) A v = 100 × 240=24,000, Gv = 20 log10 24,000 = 87.6 dB

61.6. Advantages of Impedance CouplingThe biggest advantage of this coupling is that there is

hardly any dc drop across L so that low collector supply volt-ages can be used.

However, it has many disadvantages :1. It is larger, heavier and costlier than RC coupling.2. In order to prevent the magnetic field of the coupling

inductor from affecting the signal, the inductor turnsare wound on a closed core and are also shielded.

3. Since inductor impedance depends on frequency, thefrequency characteristics of this coupling are not asgood as those of BC coupling. The flat part of the frequency versus gain curve is small (Fig.61.13).

At low frequencies, the gain is low due to large capacitance offered by the coupling capacitorjust as in RC coupled amplifiers. The gain increases with frequency till it levels off at the middlefrequencies of the audio range.

At relatively high frequencies, gain drops of again because of the increased reactance.Hence, impedance coupling is rarely used beyond audio range.

61.7. Transformer-coupled Two Stage AmplifierThe circuit for such a cascaded

amplifier is shown in Fig. 61.14. T1 isthe coupling transformer whereas T2 isthe output transformer. C1 is the inputcoupling capacitor whereas C2, C3 andC4 are the bypass capacitors. ResistorsR1 and R2 as well as R4 and R5 form volt-age divider circuits whereas R3 and R6are the emitter-stabilizing resistors.

(i) Circuit Operation

When input signal is coupledthrough C1 to the base of Q1, it appears in an amplified form in the primary of T1. From there, it ispassed on to the secondary by magnetic induction. Moreover, T1 provides dc isolation between theinput and output circuits. The secondary of T1 applies the signal to the base of Q2 which appears inan amplified form in the primary of T2.

From there, it is passed on to the secondary by magnetic induction and finally appears acrossthe matched load R7.(ii) Voltage Gain

20.1.1 0.1 .2 1 2 1

.1

Now — where / forv ie

rA r a r a N N T

r= = =

ri.2 = R4 || R5 || β2.re.2

Fig. 61.14

Fig. 61.13


Similarly,0.2

.2.2

ee

rA

r=

where re.2 = a2R7

Example 61.5. For the trans-former-coupled two-stage amplifiershown in Fig. 61.15, calculate (i) Av.1(ii) Av.2 and (iii) Av in dB.

Neglect VBE and use re = 50 mV/IE. Take β1 = β2 = 50 and treat the trans-formers as ideal ones.

(Electronics, Indore Univ.)

Solution. Here, for each trans-former a = 5

Drop across R2 = 9 × 4/24 = 1.5 VDrop across R3 ≅ 1.5 V and IE.1= 1.5/1= 1.5 mA

re.1 = 50/1.5 = 33.3 Ωand re.2 = 33.3 Ω — same as re.1 β2. re.2 = 50 × 33.3 = 1665 Ω ri.2 = R4 || R5 || β2.re.2

= 20 K || 4 K || 1665 Ω =1110 Ω r0.1 = a2.ri.2 = 52 × 1110 = 27,750 Ω

(i) 0.1.1

.1

27,750

33.3ve

rA

r= = =8 830

(ii) 0.2.2

.2

25 100

33.3ve

rA

r

×= = = 750

(iii) Av = 830 × 750 = 622,500 ; Gv = 20 log10 622,500 = 116 dB

61.8. Advantages of Transformer Coupling

1. The operation of a transformer-coupled system is basically more efficient because of low dcresistance of the primary connected in the collector circuit,

2. It provides a higher voltage gain,3. It provides impedance matching between stages which is desirable for maximum power

transfer. Typically, the input impedance of a transistor stage is less than its output imped-ance. Hence, secondary impedance of the interstage (or coupling) transformer is typicallylower than the primary impedance.

This coupling is effective when the final amplifier output is fed to a low-impedance load. Forexample, the impedance of a typical loud-speaker varies from 4 Ω to 16 Ω whereas output imped-ance of a transistor stage is several hundred ohms. Use of an output audio transformer can avoid thebad effects of such a mismatch.

Disadvantages1. The coupling transformer is costly and bulky particularly when operated at audio frequen-

cies because of its heavy iron core,2. At radio frequencies, the inductance and winding capacitance present lot of problems,3. It has poor frequency response because the transformer is frequency sensitive. Hence, the

frequency range of the transformer-coupled amplifiers is limited.4. It tends to introduce ‘hum’ in the output,

Q1 Q2Vi

R1

R6R5

R4 R7

T2T1

VCC

R2 R3

20 K

4 K 1 K 4 K 1 K

20 K

5:1

1 K

9 V

5:1

Fig. 61.15


61.9. Frequency Response

The characteristics of a coupling transformer arethat (i) it introduces inductances in both the input andoutput circuits (ii) leakage inductance exists between theprimary and secondary windings and (iii) both windingsintroduce shunting (distributed) capacitance especiallyat high frequencies.

A typical gain versus frequency curve for a trans-former-coupled amplifier is shown in Fig. 61.16.

It is seen that1. There is decrease in gain at low frequencies and2. Also there is decrease in gain at high frequencies except for the resonant rise in gain at

resonant frequency of the tuned circuit formed by inductance and winding capacitance in thecircuit.

The output voltage is equal to ac collector current multiplied by the primary reactance of thecoupling transformer. Since at low frequencies, primary reactance is small, the gain is less.

At high frequencies, the distributed capacitance existing between different turns of the wind-ing acts as a bypass capacitor and so reduces the output voltage and hence the gain. The peak orexaggerated gain occurs due to resonance or tuning effect of inductance and distributed capacitancewhich form a tuned circuit.

Moreover, there is frequency distortion i.e. all frequencies are not amplified equally. In fact,the flat response part of the curve is small as compared to RC coupling. However, transformer-coupled amplifiers can be designed to have a flat frequency response curve and excellent fidelityover the entire audio frequency range.

61.10. ApplicationsTransformer coupling is often employed in the last stage of a multistage amplifier where con-

certed effort is made to maximise power transfer by perfect impedance matching.

Example 61.6. In a multistage transformer-coupled amplifier, the output impedance of thefirst stage is 5 K and the input impedance of the second stage is 1 K. Determine the primary andsecondary inductances of the tranformer for perfect impedance matching at f = 2000 Hz. If oneturn gives an inductance of 10 µH, find the number of primary and secondary turns.

(Electronic Engg.-I, Osmania Univ. 1991)

Solution. It should be clearly understood that primary has to match with the output impedanceof the first stage and secondary with the input of the second stage.

∴ X Lp = output of 1st stage or 2πf Lp = 5000∴ Lp = 5000/2π × 2000 = 0.4 HAlso, X Ls = input of 2nd stage∴ 2πfLs = 1000 ; Ls = 1000/2π × 2000 = 0.08 H

Now, inductance of a coil varies as the square of its turns.∴ L ∝ N2= kN2

When, N = 1, L = 10 µH ∴ 10 × 10–6 = k × l22 or k = 10–5

For primary winding 0.4 = 10–5 Np2 × 5 or Np = 632For secondary winding, Ls = kNs

2 or 0.08 = 10–5 Ns2 or Ns = 89As seen, it is a nearly 7 : 1 step-down transformer.

Fig. 61.16

ResonantRise

Vo

ltag

eG

ain

0 fof


61.11.61.11.61.11.61.11.61.11. Dir Dir Dir Dir Direct-coupled ect-coupled ect-coupled ect-coupled ect-coupled TTTTTwwwwwo-stage o-stage o-stage o-stage o-stage AmplifAmplifAmplifAmplifAmplifier Using Similar ier Using Similar ier Using Similar ier Using Similar ier Using Similar TTTTTransistorransistorransistorransistorransistorsssss

These amplifiers operate without the use of frequency-sensitive components like capacitors,inductors and transformers etc. They are especially suited for amplifying

(a) ac signals with frequencies as low as a fraction of a hertz,(b) change in dc voltages.Fig. 61.17 shows the circuit of such

an amplifier which uses two similar tran-sistors each connected in the CE mode.Both stages employ direct coupling (i) col-lector of Q1 is connected directly to thebase of Q4 and (ii) load resistor R2 is con-nected to the collector of Q2. The resistorR1 establishes the forward bias of Q1 andalso indirectly that of Q2.

Any signal current at the base of Q1is amplified β1 times and appears at the collector of Q1 and becomes base signal for Q2. Hence, it isfurther amplified β2 times. Obviously, signal current gain of the amplifier is

Ai = β1 × β2 = β2 — if transistors are identical(i) AC Equivalent Circuit

The ac equivalent circuit isshown in Fig. 61.18(ii) Voltage Gain

The first stage i.e. Q1 doesnot produce any voltage gain i.e.Av.1 = 1 as proved below :

0.1 2 .2.1

.1 .1

. ev

e e

r rA

r rβ

= = ... (i)

As seen from Fig. 61.17, IE.2 ≅ β2 . IB.2

Also IB.2 = IC.1 = IE.1 ∴ IE.2 = β2.IE.1 or .22

.1

E

E

II

β =

Also .1.1

25e

Er

I= and .2

.2

25e

Er

I=

Substituting all these values in Eq. (i) above, we get

.2 .1.1

.1 .2

25 125

E Ev

E E

I IA

I I= × × =

It proves the statement made above.

Now,0.2

.2.2

ve

rA

r= and Av = Av.1 × Ae.2 = Av.2

(iii) Advantages1. The circuit arrangement is very simple since it uses minimum number of components.2. It is quite inexpensive.3. It has the outstanding ability to amplify direct current (i.e. as dc amplifier) and low-fre-

quency signals.

Fig. 61.17

Fig. 61.18


4. It has no coupling or by-pass capacitors to causea drop in gain at low frequencies. As seen from Fig. 61.19,the frequency-response curve is flat upto upper cut-offfrequency determined by stray wiring capacitance and in-ternal transistor capacitances.

(iv) Disadvantages1. It cannot amplify high-frequency signals.2. It has poor temperature stability.It is due to the fact that any variation in base current

(due to temperature changes) in one stage is amplified inthe following stage (or stages) thereby shifting the Q-point.However, stability can be improved by using emitter-sta-bility resistor (Fig. 61.21).

(v) ApplicationsSome of the applications of direct-coupled amplifiers are in

1. regulator circuits of electronic power supplies,2. pulse amplifiers 3. differential amplifiers,4. computer circuitry, 5. electronic instruments.Example 61.7. For the direct-coupled amplifier of Fig. 61.20, calculate(a) current gain , (b) voltage gain of first stage,(c) voltage gain of second stage, (d) overall voltage gain in dB,(e) overall power gain in dB, (f) input resistance.Neglect VBE and use re = 50 mV/IE.Solution. (a) A i = β1β2=100 × 50 = 5000 (b) A v.1 = 1 — Art 61.11

(c) 0.2

.2 .2.2 .2

50Now,v e

e E

rA r

r I= =

Let us find the value of IE.2 starting from the value of IB.1. As seen from Fig. 61.20, IB.1 = 12/1.2M = 10 µA IC.1 = β1. 1β1 = 100 × 10

=1000 µA IE.1 = IC.1 = 1000 µA

= 1 mA IB.2 = IC.1 = 1 mA IC.2 = β2. IC1 = 50 × 1

= 50 mA∴ IE.2 = 50 mAand re.2 = 50/50 = 1 ΩAlso, re.2 = R2 = 200 Ω

∴ .2200

1vA = = 200

(d) A n =1 × 200 = 200 ; Gv = 20 log 10200 = 46 dB(e) A p =Av . Ai = 200 × 5,000 = 106 ; Gp = 20 log10 106 = 60 dB(f) ri =R1 || β1.re.1 — Fig. 61.18

.1.1

50 5050

1eE

rI

= = = Ω ∴ ri = 1.2 M || (50 × 100) ≅ 5 K

Fig. 61.19

f2

Gai

n

f

Fig. 61.20


Example 61.8. For the emitter-stabilized direct-coupled amplifier of Fig. 61.21, find(i) Av.1, (ii) Av.2, (iii) Av and (iv) ri.Neglect VBE and use re = 50 mV/IE.

(Electronics-I, Mysore Univ. 1992)Solution. As seen, emitter resis-

tors R3 and R5 have been used to im-prove temperature stability. The volt-age divider R1 – R2 together with R3 de-termines the emitter current of Q1. Theresistor R4 has dual function (i) it actsas load resistor for Q1 and (ii) it estab-lishes base bias of Q2.

(i) Since unbypassed resistor R3is present

0.1.1

.1 3v

e

rA

r R=

+As seen from the ac equivalent diagram of Fig. 61.22

Fig. 61.22

r0.1 = R4 || β2. (re.2 + R5)

Now .2.2

50e

E

rI

=

For finding IE.2, let us first find IE.1

Drop across 220

20 2 V180 20

R = × =+

Same is the drop across R3 since V BE has been neglected.∴ IE.1 = 2V /1 K = 2 mA

Now,.1 4

.25

20 2 82mA

2CC E

EV I R

IR

− − ×= = =

∴ re.2 = 50/2 = 25 Ω and β2 (re.2 + R5) ≅ 200 K

0.1 .150

8 ||200 7.7 252i er K K K r= = = = Ω

∴ .17,700

7.5(25 1000)vA = =

+

(ii)0.2

.2 0.2 6 .2.2 5

6,000; 6000 3

( ) (25 2000)v Ve

rA r R A

r R= = = Ω∴ = ≅

+ +(iii) A v =7.5 × 3 = 22.5(iv) As seen from Fig. 61.22

ri = R1 || R2 || β1 (re.1 + R3) = 180 K || 20 K || 100 K —neglecting re.1 = 15.25 K

Fig. 61.21


Example. 61.9. In the circuit of Fig. 61.21, find dc voltage at points marked 1, 2, 3 and 4.Neglect VBE.

If a 1-V dc signal at input 1 changes by 0.01 V, what would be the voltage variation at theoutput?

Solution. V 1 = drop across R2 = 2 VV 2

≅ V 1= 2 V; V 3 = V CC – IC.1 R3= 20 – 2 × 8 = 4 VV4

≅ V3 = 4 V; V 5 = V CC – IC.2 R6 = 20 – 2 × 6 = 8 V∆ v0 = A v . ∆ vi = 22.5 × 0.01= 0.225 V = 225 mV.

61.12. Direct-coupled Amplifier Using Complementary Symmetry of TwoTransistors

In this case, an NPN transistor is directly-coupled to its complementary i.e. a PNP transistor.Fig. 61.23 shows a two-stage cascadedamplifier using two complementary tran-sistors connected in CE configuration.

The circuit differs from that shownin Fig. 61.17 in the following three ways :

1. It uses complementary transis-tors rather than similar ones,

2. Instead of V CC, V EE power batteryhas been used,

3. Output is taken directly from ter-minal of load resistor R2.

More Practical CircuitA more practical circuit of the above type is shown in Fig. 61.24. Here, bias current of Q1 is

determined by voltage divider R1 – R2 and R3. As before, R4 performs two functions :1. It acts as load for Q1 and 2. Establishes bias voltage for Q2

The emitter resistors R3 and R6, as usual, meant to improve amplifier stability.

Fig. 61.24

(i) Circuit OperationWhen a positive-going signal is applied to the base of Q1, then1. its base current increases,

Fig. 61.23


2. hence, its collector current increases (IC = βIB),voltage drop across R4 increases,

4. consequently, voltage at the collector of Q1 and the base of Q2 becomes less positive or inother words, more negative,

5. hence, a negative-going signal is applied to the base of Q2.The negative-going signal applied to the base of Q2 causes,1. an increase in its forward bias (remember, it is a PNP transistor),2. an increase in collector current 3. an increase in the voltage developed across R6,4. an amplified positive going output signal at R6.Hence, it is seen that a signal applied to the input of a two-stage complementary amplifier

appears at the output in an amplified form and of the same polarity.

Voltage GainVoltage GainVoltage GainVoltage GainVoltage Gain

It is the same as for the circuit of Fig. 61.17.Example 61.10. For the complementary symmetry circuit of Fig. 61.25, find (a) Av (b) V1, V2,

V3, V4 and V5. Neglect VBE and assume R3, R5 » re.

Fig. 61.25

Solution. (a) The overall voltage gain is

Av = Av.1 × Av.2 ≅ 64

3 5

6 62 8

RRR R

× = ×

(b) 2

11 2

2EER

V V VR R

= =+ ∴ V2 ≅ V1 = 2V

∴ IE.1 = 3

2 2 1mA2

V VR K

= = , IC1 ≅ IE.1=1mA

V3 = VEE – IC.1 R3 = 20 – 1 × 6 = 14 V, V4 ≅ V3 = 14 V ;

.220 –14

1mA6KCI =

20 14

1mA6K

−= =

∴ .2 .2 1mAE CI I≅ =


5 .2 6EV I R∴ = × = 1 × 8 = 8 V

61.13. Darlington PairIt is the name given to a

pair of similar transistors soconnected that emitter of oneis directly joined to the baseof the other as shown in Fig.61.26 (a). Obviously, theemitter current of Q1 becomesthe base current of Q2.

Darlington pairs arecommercially mounted in asingle package that has onlythree leads : base, collector and emitter as shown in Fig. 61.26 (b). It often forms a double CC stagein multistage amplifiers. It is so because a Darlington connection can be considered equivalent totwo cascaded emitter followers.

Main Characteristics(i) Current GainIt can be proved that current gain of a Darlington pair is (1 + β1) (1 + β2) = (1 + β)2 ≅ β2

if thetransistors are identical (i.e. β1 = β2).

ProofIB.2 = IE.1=(1 + β1) IB.1 ≅ β1 IB.1

IE.2 ≅ β2 . IB.2= β1 β2 IB.1

2.21 2

.1

Ei

B

IA

I∴ = =β β =β

It means that a Darlington pair behaves like a single transistor having a beta of β2.

(ii) Input ImpedanceIn Fig. 61.26 (a), the input impedance seen from the base of Q2 is

ri.2 = β2 (re.2 + RE) ≅ β2.RE

Input impedance as seen from the base of Q1 is

ri.1 = β1(re.1 + ri.2) = β1 re.1 + β1 ri.2 = β1 re.1 + β1 β2 RE ≅ β1β2 RE

or rin(base) of Q1 = β2 RE

Note. If there is a load resistance RL coupled to the emitter of Q2, then

ri.1 – β2(RE || RL) = β2.rE

As seen, load impedance RE has been transformed into β2RE. Obviously, a Darlington pair iscapable of high input impedance. In fact, whenever a load causes a severe loss in voltage gain(loading effect), it is usual to step up load impedance via a FET stage, a single CC stage or Darlingtonpair when much greater impedance transformation is required.

(iii) Voltage GainAssuming re.1 = re.2 = re we have

11

1

Ev

ee E

E

RA

rr RR

≅ = ≅+ + —as in an emitter follower

VCC

Q1 Q2

RE

(a) (b)

B

C

E

Fig. 61.26


61.14. Advantages of Darlington Pair

1. It can be readly formed from two adjacent transistors in an IC.2. It has enormous impedance transformation capability i.e. it can transform a low-imped-

ance load into a high impedance load. Hence, it is used in a high-gain operational amplifierwhich depends on very high input impedance for its operation as an integrator or summingamplifier in analogue applications.

3. It uses very few components.4. It provides very high β-value.

61.15. Comparison Between Darlington Pair and Emitter Follower

We will refer to Fig. 9.28 and Fig. 61.26.

1. Input impedance of Darlington pair is β2RE whereas that of emitter follower is βRE (Art.9.8).

2. Current gain of Darlington pair is β2 whereas that of emitter follower is β.

3. However, voltage gains of the two are identical.

Example. 61.11. For the Darlington pair shown in Fig. 61.27, calculate the value of 1. β,2.input impedance,3. voltage amplification.

Assume β1 = β2 = 100 and RL » (re.1 + re.2 ).

Solution. The approximate values areas under :

1. β of Darlington pair

= β1 × β2

=100 × 100 = 10,000

2. rE = RE || RL

=10 K || 500 Ω = 475 Ωrin(base) of Q1 ≅ β2rE

= 10,000 × 475=4.75 M

ri = input impedance of the pair

= RB || rin(base) of Q1

= 1 M || 4.75 M = 0.826 M = 826 K

3. Av ≅ 1

Example 61.12. A CE amplifier stageshown in Fig.61.28 is to drive a 100 Ω load to10 Vp–p level. An input signal of 1 Vp–p is avail-able. Find out if the stage is overloaded or not.Also, find out if this overloading has been avoidedby using a Darlington pair as a buffer betweenQ1 and the load as shwon in Fig. 61.29. Takeβ1= β2 = 50.

Solution. The approximate voltage gain ofQ1 is given by

Fig. 61.27

Fig. 61.28


Fig. 61.29

|| 5K||100 1

500 5C LL

vE E

R RrA

r R

Ω= = = ≅Ω

01

1 0.2 mV5

v∴ = × =

Fig. 61.30

Obviously, due to overloading of the stage,severe attenuation has occurred, Hence, the ampli-fier stage cannot, by itself, drive the heavy load.

By using a Darlington pair, a 100 Ω load hasbeen transformed to

β2RL = 2500 × 100 = 250 K

Now, rL= RC || β2RL = 5 K || 250 K

5K5 10

500vK A≅ ∴ = =Ω

This also represents total voltage gain becausevoltage gain of Darlington pair is one.

∴ v0 =A v × vi = 10 × 1= 10 Vp–pAs seen, now the stage would be able to drive the load since the design goal of 10 V P–P has been

met.

Fig. 61.31


61.16. Special Features of a Differential AmplifierIt consists of two basic CE amplifiers having their emitters directly-coupled to each other.

Fig. 61.30 shows both the block diagram and the circuit diagram of such an amplifier.As seen, it has two separate input termi-

nals 1 and 2 and two separate output terminals3 and 4. Voltages may be applied to either orboth input terminals and output may be takenfrom either or both output terminals.

There are certain specific phase relation-ships between both input and both output ter-minals as discussed below.

(a) Single-ended OperationIn Fig. 61.31, input signal vi.1 is applied

to terminal 1 with terminal 2 grounded.It is seen that an amplified and inverted

output signal is obtained at terminal 3 (phaseinversion of a CE amplifier) but an equally-am-plified and in-phase signal appears across out-put terminal 4. The differential output voltage has the polarity shown in the figure.

As shown in Fig. 61.32, when in-put signal v1–2 is applied to input termi-nal 2, an amplified and inverted signalappears at output terminal 4 whereasequally-amplified but in-phase signalappears at terminal 3.

In summary, we can say that inputat any of the two terminals causes out-puts at both terminals 3 and 4. The twooutput are opposite in phase but of equalamplitude.

(b) Double-ended OperationFig 61.33 illustrates the double ended mode of operation when two input signals of opposite

phase are applied to the two input terminals.Input signal at each input terminal causes signals to appear at both output terminals. The re-

sultant output signals have a peak value of 2 V – twice the value for single-ended operation.However, if two in-phase and equal signals were applied at the two input terminals, the result-

ant output signal at each output terminal would be zero as shown in Fig. 61.34. It means that outputbetween the collectors would be zero.

Fig. 61.32

Fig. 61.33 Fig. 61.34

Differential amplifier


If vi.1 and vi.2 change by exactly the same amount, even then output voltage between terminals4 and 3 remains zero because of symmetry. Only when vi.1 and vi.2 differ from each other, we get anoutput voltage. When vi.1 is more positive than vi.2, the output terminal 4 is more positive thanterminal 3.

(c) Inverting and Non-inverting InputsWhen positive vi.1 acts alone, it produces a differential output voltage with terminal 4 positive

with respect to terminal 3 as shown in Fig. 61.35. That is why the input terminal 1 is called non-inverting input terminal. How ever, when positive vi.2 acts alone, the output voltage is inverted i.e.terminal 3 becomes positive with respect to terminal 4. That is why input terminal 2 is called invert-ing terminal.

61.17. Common Mode Input

Fig. 61.36 illustrates the common-mode input of a differential amplifier i.e. when similar orsame input signal is applied to both inputs. If the two halves of the diff-amp are identical, the acoutput voltage will be zero. The diff-amp is then said to the perfectly balanced.

Fig. 61.35 Fig. 61.36

The common-mode rejection ratio (CMMR) is defined as.

.

i cm

o cm

A vCMMR

v

×=

Suppose, vi.cm = 1 V. In a perfectly-balanced and symmetrical diff-amp, the output should bezero. But, in practice, there is always a small output signal because of non-symmetry. Suppose,A = 100 and vno.em = 0.01V. Then

10100 1V

10,000 20 log 10,0000.01V

CMMR×= = = = 80 dB

Larger the value of CMMR (i.e. smaller the value of vo.cm), better the diff-amp. Ideally, acommon-mode input should produce zero output voltage. Hence, an ideal diff-amp has a CMMR ofinfinity.

Example 61.13. Calculate the approximate output voltage for the diff-amp of Fig. 61.37 whichuses only a single-ended non-inverting input of n-1 = 1 mV. Take re = 25 mV/IE and neglect VBE.

Solution. IE ≅ V EE/RE = 12/6 = 2 mA


The current is divided equally between the two transistors so that IE.1 = IE.2 = 1 mA.Voltage amplification of each half is approximately given by A = RC /re

Now, re = 25/1= 25 Ω

∴12,000

48025

A= =

∴ v0 = A × v i.1= 480 × 1

= 480 mV

The ac output signal is in phase with the input.

Example 61.14. If a differential input signal of 1 mV is applied to the diff-amp shown in Fig.61.38, calculate the output voltage. Neglect VBE and take re = 25 mV/IE.

Solution.

Here, vi.1 – vi.2 = 1 mV

As seen from Ex. 61.13,

A = 480

∴ vo = A (vi.1 – vi.2) = 480 × 1 = 480 mV

Fig. 61.37

Fig. 61.38


Example 61.15. A differential inputsignal of 1 mV is applied to the diff-amp ofFig. 61.39 when used in single-ended out-put mode. Calculate the approximate valueof output voltage. Neglect VBE and take re =25 mV/IE.

Solution. Here, again 1 mV is the dif-ferential input signal i.e. vi.1 – vi.2 = 1 mV

Since single-ended mode is being used,A = 480/2 = 240

∴ v0 = A(vi.1 – vi.2)

= 240 × 1= 240

61.18. Differential Amplifier

Fig. 61.40 shows the circuit of a differential or difference amplifier. As seen

1. it contains two CE amplifiers,

2. it uses only resistors and transistors3. it is directly-coupled (emitter-to-emitter) amplifier,4. it can accept two inputs by means of T1 and ground

and also T2 and ground,5. it can provide two separate outputs by means of

T3 and ground and T4 and ground,6. it can provide a single output between T3 and T4

i.e. differential output.

Circuit Operation

We will consider a balanced differential amplifierin which Q1 and Q2 are identical and their associated com-ponents are matched. In that case, each amplifier stageproduces same voltage gain. A = r0/re

In Fig. 61.40. ro = R2 or R3

The output voltage between terminals T3 and T4 is

v0 (T3 – T4) = A(vi.1 – vi.2)

where, A = voltage gain of each stage

Advantages1. It uses no frequency-dependent coupling or bypassing capacitors. All that it requires is

resistors and transistors both of which can be easily integrated on a chip. Hence, it is exten-sively used in linear IC5.

2. It can compare any two signals and detect any difference. Thus, if two signals are fed intoits inputs, identical in every respect except that one signal has been slightly distorted, thenonly the difference between the two signals i.e. distortion will be amplified.

3. It gives higher gain than two cascaded stages of ordinary direct coupling.

4. It provides very uniform amplification of signal from dc upto very high frequencies.

Fig. 61.40

Fig. 61.39

1

RC1

12 V

RC2

3 4

10 K 10 K

2

0.5mA 0.5mA

1mA


5. It provides isolation between input and output circuits.

6. It is almost a universal choice for amplifying dc.

7. It finds a wide variety of applications such as amplification, mixing, signal generation,amplitude modulation, frequency multiplication and temperature compensation etc.

Example 61.16. Calculate the single-ended anddifferential gain of the diff-amp shown in Fig. 61.41.Use re = 25 mV/IE.

Solution. The current of 1 mA from a constant-current source divides into two equal parts.

IE.1 = IE.2 = 1/2 = 0.5 mA

re.1 = re.2

2550

0.5= = Ω

Hence, single-ended voltage gain is

0 10

50C

e e

r R KA

r r= = = =

Ω 200

Hence, each stage has a voltage gain of 200.

If we consider differential (double-ended) gain, its value is twice i.e. 2 × 200 = 400.Example 61.17. For the diff-amp shown in Fig. 61.41, voltage gain of each stage is 200,

vi.1 = 30 mV and vi.2 = 20 mV. Find the voltages between

(i) T3 and ground, (ii) T4 and ground,(iii) T3 and T4 .

What are the polarities of output terminals T3 and T4 ?

(Basic Electronics, Bombay Univ.)

Solution. As stated earlier in Art. 61.18.

(i) v0 (T3) =A1 × vi.1 = 200 × 30 mV = 6 V

(ii) v0(T4) =A2 × vi.2 = 200 × 20 mV = 4 V

(iii) v0 (T3 – T4) =A (vi.1 – vi.2) = 200 (30 × 20) mV = 2 V

Since vi.1 > vi.2 ; ic.1 R2 > i.2 R3, hence T4 will be positive with respect to T3.

Example 61.18. Calculate the over-all voltage gain of the two-stage RCcoupled amplifier shown in Fig. 61.42.Neglect VBE and take β1 = β2 = 100.

Solution. We will first find gain ofQ2 and then multiply it with that of Q1 tofind the overall gain.

DC voltage from base to ground forQ2 = drop across 40 K

= 30 × 40/(40 + 80) = 10 V

Hence, IE.2 ≅ 10 V/10 K = 1 mA

Assuming silicon transistorre.2 = 50/1 mA = 50 Ω

Fig. 61.41

Q1

T1

R1R4

T2

Q2

R2

T3 T4

+VCC

R3

Fig. 61.42


Fig. 61.43

22

.2

10K||10K100

50L

ve

rA

r∴ = = =

Fig. 61.44

For finding A v1, we must first calculate the ac load resistance rL as seen by Q1. It equals theparallel combination of 10 K, 80 K, 40 K and βre = 100 × 50 = 5 K (because it forms part of the loadon Q1)*.

Fig. 61.45

∴ rL1 =10 K || 80 K || 40 K || 5 K ≅ 3 K

∴ A v1 = rL1/Re1=3 K/50 Ω = 60

∴ A = A v1 × A v2 = 100 × 60 = 6000

* (Emitter resistance does not come into the picture because it has been ac grounded by the capacitor.


Tutorial Problems No. 61.1

1. For the two-stage RC-coupled amplifier shown in Fig. 61.43 calculate the approximate values of(a) voltage gain for the first stage, (b) voltage gain for the second stage,(c) voltage gain for the amplifier, (d) ac input resistance of the amplifier.Neglect V BE and use re = 25 mV/IE [(a) 76.2 (b) 256 (c) 19,500 or 85.8 dB (d) 1250 ΩΩΩΩΩ]

Fig. 61.462. For the two-stage RC-coupled amplifier using base and emitter bias and shown in Fig. 61.44, calcu-

late approximate values of(i) A v.1(ii) A v.2 (iii) Av (iv) riNeglect V BE and use re = 25 mV/IE [(i) 61.6 (ii) 128 (iii) 7,900 or 78 dB (iv) 2.6 K]

3. Fig. 61.45 shows the circuit of a two-stage impedance-coupled amplifier. Compute the approximatevalues of(i) voltage gain of 1st stage,(ii) voltage gain of the 2nd stage ; both at f = 2 kHz,(iii) voltage gain of the amplifier.Neglect V BE and take re = 25 mV/IE. [(i) 80 (ii) 420 (iii) 33,600 or 90.5 dB]

4. For the direct-coupled amplifier shown in Fig. 61.46, calculate the approximate value of(i) A i, (ii) A v.1, (iii) A v.2, (iv) A v, (v) A p and (vi) r4Neglect V BE and assume re = 25 mV/IE. Also, β1 = 50 and β2 = 25.

[(i) 1250 (ii) 1 (iii) 240 (iv) 240 (v) 300,000 or 55 dB (vi) 1250 ΩΩΩΩΩ]

1. The decibel gain of a cascaded amplifierequals the(a) product of individual gains(b) sum of individual gains(c) ratio of stage gains(d) product of voltage and current gains.

2. If two stages of a cascaded amplifier havedecibel gains of 60 and 30, then overall gainis ..................... dB.(a) 90 (b) 1800(c) 2 (d) 0.5

3. Cascading two amplifiers will result in(a) reduction in overall gain and increase

in overall bandwidth.(b) reduction in overall gain and reduction

in overall bandwidth(c) increase in overall gain increase in over-

all bandwidth

(d) increase in overall gain and reductionin overall bandwith

4. The overall bandwidth of two identical volt-age amplifiers connected in cascade will(a) remain the same as that of a single stage(b) be worse than that of a single stage(c) be better than that of a single stage(d) be better if stage gain is low and worse

if stage gain is high.5. RC coupling is popular in low-level audio

amplifiers because it(a) has better low frequency response(b) is inexpensive and needs no adjust-

ments(c) provides an output signal in phase with

the input signal(d) needs low voltage battery for collector

supply.

OBJECTIVE TESTS – 61


6. Frequency response characteristic at a single-stage RC coupled amplifier is shown in Fig.61.47. The fall in gain at both ends of the char-acteristic is due to

(a) transistor shunt capacitances

(b) bypass and coupling capacitances of thecircuit

(c) transistor shunt capacitances of the lowerend and bypass and coupling capaci-tances at the higher end

Fig. 61.47

(d) transistor shunt capacitances at thehigher end and bypass and coupling ca-pacitances at the lower end.

7. The most desirable feature of transformercoupling is its(a) higher voltage gain(b) wide frequency range(c) ability to provide impedance matching

between stages(d) ability to eliminate hum from the out-

put.8. A transformer coupled amplifier would give

(a) maximum voltage gain(b) impedance matching(c) maximum current gain(d) larger bandwidth

9. In multistage amplifiers, direct coupling isespecially suited for amplifying(a) high frequency ac signals(b) changes in dc voltages(c) high-level voltages(d) sinusoidal signals.

10. The outstanding characteristic of a direct-coupled amplifier is its(a) utmost economy(b) temperature stability

(c) avoidance of frequency-sensitive co-mponents

(d) ability to amplify direct current and low-frequency signals.

11. A signal may have frequency componentswhich lie in the range of 0.001 Hz to 10 Hz.Which one of the following types of couplingsshould be chosen in a multistage amplifierdesigned to amplify this signal ?(a) RC coupling(b) transformer coupling(c) direct coupling(d) double-tuned transformer

12. Darlington pairs are frequently used in lin-ear ICs because they(a) do not require any capacitors or in-

ductors(b) have enormous impedance transfor-

mation capability(c) can be readily formed from two adja-

cent transistors(d) resemble emitter followers.

13. When same input signal is applied to boththe inputs of an ideal diff-amp, the output(a) is zero(b) depends on its CMMR(c) depends on its voltage gain(d) is determined by its symmetry.

14. The common-mode rejection ratio of an idealdiff-amp is(a) zero(b) infinity(c) less than unity(d) greater than unity.

15. One of the advantages of a Darlington pair isthat it has enormous .............. transformationcapacity.(a) voltage(b) current(c) impedance(d) power

16. A Darlington pair and an emitter followerhave the same(a) input impedance(b) current gain(c) voltage gain(d) power gain.

17. Which of the following mode of operation ispossible with a differential amplifier ?


ANSWERS

1. (b) 2. (a) 3. (c) 4. (c) 5. (b) 6. (d) 7. (c) 8. (b) 9. (b) 10. (d) 11. (c)12. (c) 13. (a) 14. (b) 15. (c) 16. (c) 17. (d) 18. (d) 19. (b)

(a) single-ended input(b) differential input(c) common-mode input(d) all of the above.

18. The gain of a Darlington amplifier pair is de-termined by ........................... the beta val-ues.(a) subtracting(b) adding(c) dividing

(d) multiplying19. The amplifier in which the emitter and col-

lector leads of one transistor are connected tothe base and collector leads of a second tran-sistor is called an ................... amplifier.

(a) push-pull(b) Darlington(c) differential(d) complementary

feedback amplifiers - wordpress.com and feedback amplifiers learning objectives general amplifier...

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