MULTISTAGE AMPLIFIERS

Multistage Amplifiers

Two or more amplifiers can be connected to increase the gain of an ac signal. The overall gain can be calculated by simply multiplying each gain together.

A’v = Av1Av2Av3 ……

Introduction

Many applications cannot be handle with single-transistor amplifiers in order to meet the specification of a given amplification factor, input resistance and output resistance

As a solution – transistor amplifier circuits can be connected in series or cascaded amplifiers

This can be done either to increase the overall small-signal voltage gain or provide an overall voltage gain greater than 1 with a very low output resistance

Multistage Amplifiers

Multi-stage amplifiers are amplifier circuits cascaded to increased gain. We can express gain in decibels(dB).

Two or more amplifiers can be connected to increase the gain of an ac signal. The overall gain can be calculated by simply multiplying each gain together.

A’v = Av1Av2Av3 ……

Example

Find Vout

Av1 = 5 Av2 = 10 Av3 = 6 Avn = 8

Vin = 5 mV

6

Multistage Amplifier Cutoff Frequencies and Bandwidth When amplifiers having equal cutoff frequencies are cascaded,

the cutoff frequencies and bandwidth of the multistage circuit are found using

)(1)(2

/1

1)(1

/12)(2

BW12

12

TCTC

n

CTC

nCTC

ff

ff

ff

Introduction (cont.) Multistage amplifier configuration:

Rc1 Rc2RB

Vi Q1

Q2R2

RL

R1Vo

Vi Q1

Q2

R3

R1 R2

Vi

Vo

R1

Vi

Vo

T

Cascade /RC coupling

Cascode

Transformer coupling Darlington/Direct coupling

Q1

Q2

Q1

Q2

i) Cascade Connection-The most widely used method-Coupling a signal from one stage to the anotherstage and block dc voltage from one stage tothe another stage-The signal developed across the collectorresistor of each stage is coupled intothe base of the next stage -The overall gain = product of the individual gain

**refer page 219

i) Cascade Connection (cont.) small signal gain is: by determine the voltage gain at stages 1 & stage 2

therefore

- the gain in dB

input resistance

Output resistance - assume ,so also

Therefore

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iLCCmm

S

oV RR

RRRrRgg

V

VA

121 //// rRRRis

0SV 021 VV 0211 VgVg mm

20 CRR

21 VVV AAA

)log(20)( VdBV AA

Exercise 1:

Draw the ac equivalent circuit and calculate the voltage gain, input resistance and output resistance for the cascade BJT amplifier in above Figure. Let the parameters are:

kRRkRRkRRkRR EECC 1,2.2,7.4,15 21214231

0)(21 ,7.0,200,20 rVVVV ONBEQQCC

Solution 1 (cont.):

21

1

// RR

RB

1r

1V

2V 2r2CR1CR

11 Vgm22 Vgm43

2

// RR

RB

iV0V

1B 2B

1E

1C

2E

2CiR

0R

Ac equivalent circuit for cascade amplifier

Solution 1:

DC analysis:At Q1:

At Q2:

AC analysis:At Q1:

At Q2:

Sg

kr

m 153.0

307.1

1

1

Why the Q-point values same for bothQ1 & Q2 ?

mAI

AI

CQ

BQ

979.3

89.19

1

1

Sg

kr

m 153.0

307.1

2

2

mAI

AI

CQ

BQ

979.3

89.19

2

2

Solution 1 (cont.):

From the ac equivalent circuit:At Q1, the voltage gain is:

Where is the o/p voltage looking to the Q1 transistor

and is the i/p resistance looking into Q2 transistor

Therefore

The voltage gain at Q1 is:

2iR

10QV

)//( 2110

1 iCmQ

VQ RRgVi

VA

06.102)36.957//2.2(153.01 kAVQ

36.957307.1//7.4//15// 222 kkkrRR Bi

Solution 1 (cont.):

From the ac equivalent circuit:At Q2, the voltage gain is:

Where is the i/p voltage looking into the Q2 transistor

Therefore, the voltage gain at Q2 is:

The overall gain is then,

** The large overall gain can be produced by multistage amplifiers!!So, the main function of cascade stage is to provided the larger overall gain

2iQV

)( 22

02 Cm

iQVQ Rg

V

VA

6.336)2.2(153.02 kAVQ

353,34)6.336)(06.102(21 VQVQV AAA

Solution 1 (cont.):

From the ac equivalent circuit:

The i/p resistance is:

The o/p resistance is:

36.957

307.1//7.4//15//// 121 kkkrRRRi

kRR Co 2.22

ii) Cascode Connection-A cascode connection has one transistor on top of (in series with) another-The i/p into a C-E amp. (Q1) is, which drives a C-B amp. (Q2)-The o/p signal current of Q1 is the i/p signal of Q2

-The advantage: provide a high i/p impedance with low voltage gain to ensure the i/p Miller capacitance is at a min. with the C-B stage providing good high freq. operation

**refer page 223

ii) Cascode Connection (cont.)

From the small equivalent circuit, since the capacitors act as short circuit,by KCL equation at E2:

solving for voltage

Where

the output voltage is

or

222

211

Vg

r

VVg mm

2V

SmVgr

V 12

22 1

222 rgm

)//)(( 22 LCmo RRVgV

SLCmmo VRRr

ggV )//1 2

221

ii) Cascode Connection (cont.)

Therefore the small signal voltage gain:

From above equation shows that:

So, the cascode gain is the approximately

* The gain same as a single-stage C-E amplifier

LCmmS

V RRr

ggV

VA //

1 2

221

0

LCmV RRgA //1

111 2

2

2

22

rgm

iii) Darlington Connection-The main feature is that the composite transistor acts as a single unit with a current gain that is the product of the current gains of the individual transistors-Provide high current gain than a single BJT-The connection is made using two separate transistors having current gains of and

So, the current gain

If

The Darlington connection provides a current gain of

21 D

1 2

21

2 D

Figure 1: Darlington transistor

iii) Darlington Connection (cont.) Figure shown a Darlington configuration

refer page 222

iii) Darlington Connection (cont.) The small current gain :

Since

Therefore

Then,

The o/p current is:

The overall gain is:

** The overall small-signal current gain = the product of the individual current gains

iimm IIrgVg 11111

212 )( rIIV ii

iimm IIVgVgI )1( 12122110

211210 )1( i

i I

IA

ioi IIA /

11 rIV i

iii) Darlington Connection (cont.)

The input resistance:Known that:

So, the i/p resistance is:

The base of Q2 is connnected to the emitter of Q1, which means that the i/p resistance to Q2 is multiplied by the factor , as we saw in circuits with emitter resistor.

So, we can write: and

Therefore

The i/p resistance is then approximately

**The i/p resistance tends to be large because of the multiplication

21121 )1( rIrIVVV iii

211 )1( rrRi

1

1

CQ

T

I

Vr

2

21

CQCQ

II

212

211 rI

Vr

CQ

T

212 rRi

)1( 1