Bjt Small Signal Amplifier

Small Signal BJT Amplifi ers

3 Chapter ObjectivesX Develop h-parameter model of BJT for AC analysis.X Analyze various CE, CB and CC ampli ers

to determine input impedance, output impedance, voltage gain and current gain.

X Compare the performance of CE, CB and CC ampli ers.

X Learn method to increase input impedance such as Darlington circuit and bootstrapping technique.

3.1INTRODUCTIONThe basic construction, characteristics and DC biasing of the bipolar junction transistors were discussed in Chapter 1. The purpose of biasing is to establish a Q-point about which variations in current and voltage can occur in response to an AC input signal. The term small signal refers to the use of signals that take up a relatively small percentage of an amplifi ers operational range. The amplifi ers designed to handle these small AC signals are called small signal ampli-fi ers. The h-parameter equivalent model is commonly used in the small signal AC analysis of bipolar junction transistors. We will defi ne this model and analyze the amplifi er characteristics in terms of this model.

3.2SMALL SIGNAL BJT AMPLIFIERFigure 3.1 shows a voltage divider biased transistor amplifier. The capacitor CC1 couples the sinusoidal AC source to the base of the transistor. The capacitor CC2 couples the output to the load. These coupling capacitors block DC and thus prevents the source resistance RS and the load resistance RL from changing the DC bias voltages at the base and collector. The capacitors act as short circuits to the signal voltage. The emitter resistor RE improves the stability of the system but provides negative feedback which reduces the gain of the amplifier. This AC degeneration can be avoided by connecting the capacitor CE in parallel with resistor RE. For DC operation, capacitor CE acts as open circuit and improves the stability. For AC operation, capacitor CE acts as short circuit and bypasses the resistor RE.

M03_XXXXXXXX_XX_C03.indd 1M03_XXXXXXXX_XX_C03.indd 1 7/21/12 9:54 AM7/21/12 9:54 AM

3.2Electronic Circuits I

Hence, CE is known as the emitter bypass capacitor. The input signal voltage Vi causes the base voltage to vary sinusoidally above and below its DC bias level. The resulting variation in base current produces a larger variation in collector current. As the collector current increases, the collector voltage decreases. The collector current varies above and below its Q-point value in phase with the base current. The collector to emitter voltage varies above and below its Q-point value 180o out of phase with the base voltage, as shown in Fig. 3.2.

IC

VCC

IBQ

ICQ

VCEQ

VCE

Q

0

Fig. 3.2Small Signal Waveforms

VS

RS

RE CE

R1 RC

CC1Vo

+VCC

CC2

RLR2

Fig. 3.1Common-emi er Amplifi er


3.3h-parameter Model of BJT3.3

3.3h-PARAMETER MODEL OF BJTConsider the black box representation of the BJT as shown in Fig. 3.3.

V1

+

V2

+

I2I1

BJT

Fig. 3.3Two-port Network

The two-port network equations in terms of h-parameter are given by

V1 = h11 I1 + h12 V2I2 = h21 I1 + h22 V2

where h11, h12, h21 and h22 are h-parameters.

h-parameter(a) Case I: When output port is shorted, i.e. V2 = 0.

h11 = VI

V

1

1 02 =

It is called input impedance and is denoted by hi.

h21 = II

V

2

1 02 =

It is called forward current gain and is denoted by hf.

(b) Case II: When input port is opened, i.e. I1 = 0.

h12 = VV

I

1

2 01 =

It is called reverse voltage gain and is denoted by hr.

h22 = IV

I

2

2 01 =

It is called output admittance and is denoted by ho.

Since these four parameters represent input impedance, voltage gain, current gain and out-put admittance, these parameters are called hybrid parameters.

Let V1 = Vi, I1 = Ii, I2 = Io and V2 = Vo



Hence, these equations can be written as

Vi = hi Ii + hr VoIo = hf Ii + ho Vo

The fi rst equation represents Kirchhoffs voltage law to the input circuit which can be rep-resented as shown in Fig. 3.4.

Vi

+

Ii hi

hrVo

Fig. 3.4Input Circuit

The second equation represents Kirchhoffs current law to the output circuit which can be represented as shown in Fig. 3.5.

hoVo

+

Io

hf Ii 1

Fig. 3.5Output Circuit

Combining these two circuits, the h-parameter model is obtained, which is shown in Fig. 3.6.

hoVo

+

Io

hf Ii 1Vi

+

Ii hi

hrVo

Fig. 3.6h-parameter Model for BJT

The h-parameter model is used for representing transistor in CB, CE and CC confi gurations. The h-parameters, however, will change with each confi guration. A second subscript is added to the h-parameter notation. For the CB confi guration, the letter b is added. For CE and CC confi gurations, the letter e and c are added, respectively.



3.3.1h-parameter Model for CB ConfigurationFigure 3.7 shows the transistor in the CB confi guration.

E

B

C

B

Ie Ic

Ve

+

Vc

+

Fig. 3.7Common-base Confi guration

h-parameter equations are written as,

Ve = hib Ie + hrb VcIc = hfb Ie + hob Vc

The h-parameter model for the CB confi guration is shown in Fig. 3.8.

hob

Ic

hf bIe1

Ie hib

hrbVc

C

B

E

B

Vc

+

Ve

+

Fig. 3.8h-parameter Model for CB Confi guration

3.3.2h-parameter Model for CE ConfigurationFigure 3.9 shows transistor in the CE confi guration.

Ic

Ib

Vb

Vc

C

EE

B

+

+

Fig. 3.9Common-emi er Confi gurationh-parameter equations are written as,

Vb = hie Ib + hre VcIc = hfe Ib + hoe Vc



The h-parameter model for the CE confi guration is shown in Fig. 3.10.

hoe

Ic

hf eIb1

Ib hie

hreVc

C

E

B

E

Vc

+

Vb

+

Fig. 3.10h-parameter Model for CE Confi guration

3.3.3h-parameter Model for CC ConfigurationFigure 3.11 shows transistor in the CC confi guration.

Ie

Ib

Vb

Ve

E

C

B

C

+

+

Fig. 3.11Common-collector Confi guration

h-parameter equations are written as,

Vb = hic Ib + hrc VeIe = hfc Ib + hoc Ve

The h-parameter model for the CC confi guration is shown in Fig. 3.12.

hoc

Ie

hf cIb1

Ib hic

hrcVe

B

C

E

C

Ve

+

Vb

+

Fig. 3.12h-parameter Model for CC Confi guration



3.3.4h-parameter Conversion Generally h-parameter values for the CE confi guration are given in transistor data sheet. The h-parameter values for CB and CC confi gurations can be calculated from h-parameter values for the CE confi guration. Table 3.1 gives h-parameter conversion formulae from CE to CB and from CE to CC confi gurations.

Table 3.1h-parameter Conversion Formulae

Conversion from CE to CB Conversion from CE to CC

(i) hib = h

h1ie

fe+ (i) hic = hie

(ii) hob = h

h1oe

fe+ (ii) hfc = (1 + hfe)

(iii) h =

hh1fe

fe+ (iii) hoc = hoe

(iv) hrb = h h

h1oe ie

fe+ hre (iv) hrc = 1 hre

For CC confi guration hic = hie hfc = (1 + hfe) hfe hoc = hoe hrc = 1 hre 1

Hence, the h-parameter model of the CE confi guration is used for AC analysis of CC amplifi er.

3.3.5Advantages of h-parameter Model1. The h-parameter model consists of all linear elements. Hence, the analysis of the tran-

sistor circuit becomes easier by using Kirchhoffs voltage and current laws.2. h-parameters can easily be obtained from input and output characteristics of transistor.3. h-parameters are normally specifi ed by transistor manufacturer.4. It is useful at low frequency operations.

3.3.6Disadvantages of h-parameter Model1. It is not suitable at high frequency operations.2. h-parameter vary with device and temperature.



3.4GRAPHICAL DETERMINATION OF h-PARAMETERS The relationship between VBE and IC in terms of IB and VCE for the CE amplifi er is given by

VBE = f1 (IB, VCE)IC = f2 (IB, VCE)

The input characteristics give the relationship between the input voltage VBE and the input current IB for different values of output voltage VCE. The output characteristics give the relationship between the output voltage VCE and the output current IC for different values of input current IB. The h-parameters can be determined graphically from input and output characteristics of the amplifi er. Determination of hie: From the defi nition of hie,

hie = VIBE

B V = constantCE

= Q

1Slope of input characteristic at the -point

The parameter hie can be obtained as the change in base-emitter voltage, VBE2 VBE1 , divided by the change in base current, IB2 IB1, for a constant collector-emitter voltage VCE at the quiescent point Q. The reciprocal of the slope of input characteristic at the quiescent point Q gives the value of hie as shown in Fig. 3.13.

VCE1VCEQ

VCE2

VBE

IB

VBE

IB

0

Fig. 3.13Determination of hie

Determination of hre: From the defi nition of hre,

hre = VV

BE

CE I = constantB


3.4Graphical Determination of h-parameters 3.9

The parameter hre can be obtained as the change in base-emitter voltage, VBE2 VBE1, divided by the change in collector-emitter voltage, VCE2 VCE1, for a constant base current IB at the quiescent point Q. A horizontal line on the input characteristics of Fig. 3.14 represents a constant base current.

VCE1VCEQ

VCE2

VBE

IBQ

VBE1 VBE2

IB

0

Fig. 3.14Determination of hre

Determination of hfe: From the defi nition of hfe,

hfe = II

C

B V = constantCE

The parameter hfe can be obtained as the change in collector current, IC2 IC1, divided by the change in base current, IB2 IB1, for a constant collector-emitter voltage VCE at the quies-cent point Q. A vertical line on the output characteristics of Fig. 3.15 represents a constant collector-emitter voltage.

IC2 IB2

IB1

IBQ

IC1

0

IC

VCE VCEQ

Fig. 3.15Determination of hfe



Determination of hoe: From the defi nition of hoe,

hoe = I

VC

CE I = constantB

= Slope of output characteristic at the Q-pointThe parameter hoe can be obtained as the change in collector current, IC2 IC1 , divided by

the change in collector-emitter voltage VCE2 VCE1 for a constant base current IB at the quiescent point Q. The slope of output characteristic at the quiescent point Q gives the value of hoe as shown in Fig. 3.16.

VCE IC

IB2

IB1

IBQ

0

IC

VCE

Fig. 3.16 Determination of hoeSimilarly, common-base h-parameters can be determined graphically from input and output

characteristics of the CB confi guration as shown in Figs. 3.173.20.Determination of hib

hib = VIBE

E V = constantCB

= Q

1Slope of input characteristic at the -point

VCB2 VCBQVCB1

VBE

IE

0

IE

VBE

Fig. 3.17Determination of hib


3.4Graphical Determination of h-parameters 3.11

Determination of hrb

hrb = VV

BE

CB I = constantE

IEQ

VCB1

VBE1VBE2

VCBQVCB2

VBE

IE

0

Fig. 3.18Determination of hrb

Determination of hfb

hfb = II

C

E V = constantCB

IC1

IC

IE1

IE2

IEQ

VCBQVCB0

IC2

Fig. 3.19Determination of h



Determination of hob

hob = I

VC

CB I = constantE

= Slope of output characteristic at the Q-point

VCB IC

IC

IE1

IE2

IEQ

VCB0

Fig. 3.20Determination of hob

3.5EXACT ANALYSIS OF TRANSISTOR AMPLIFIER Figure 3.21 shows the general h-parameter equivalent circuit with load RL connected to the output.

ho RL

Io

hf Ii1

RS

VS

Ii hi

hrVo

Zi Zo

Vo

+

Vi

+

Fig. 3.21General h-parameter Equivalent Circuit

Current gain (Ai)

It is the ratio of output current Io to the input current Ii.

Ai = II

o

i


3.5Exact Analysis of Transistor Amplifi er 3.13

Applying Kirchhoffs current law to the output circuit,

Io = hf Ii + hoVoBut Vo = IoRL Io = hf Ii ho IoRL (1 + ho RL) Io = hf Ii

Io = h Ih R1f i

o L+

Ai = II

o

i

= hh R1

f

o L+

Voltage gain (Av)

It is the ratio of output voltage Vo to the input voltage Vi.

Av = VV

o

i

Applying Kirchhoffs voltage law to the input circuit,

Vi hi Ii hr Vo = 0 Vi = hi Ii + hr Vo

We know that, II

o

i

= hh R1

f

o L+

Ii = I h R

h(1 )o o L

f

+

But Io = VR

o

L

Ii = VR

h Rh

(1 )oL

o L

f

+

Hence, Vi = hV h R

h R(1 )i o o L

f L

++ hr Vo

= h h R h h h R

h Rf r L i i o L

f L

Vo



VV

o

i

= h R

h h h R h h Rf L

i i o L f r L +

Av = VV

o

i

= h R

h h h R h h Rf L

i i o L f r L

+

= h R

h h h h h R( )f L

i i o f r L

+

= h R

h h Rf L

i L+ , where h = hi ho hf hr

Input impedance (Zi)

It is the ratio of input voltage Vi to the input current Ii.

Zi = VI

i

i


Vi hi Ii hrVo = 0 Vi = hi Ii + hr Vo = hi Ii hr Io RL

= hi Ii hrh Ih R1f i

o L+

RL

= hh h R

h R1if r L

o L

+

Ii

Zi = VI

i

i

= hi h h R

h R1f r L

o L+

Output impedance (Zo)

It is the ratio of output voltage Vo to the output current Io when input voltage VS = 0.

Zo = VI

o

o V = 0s




VS hi Ii RS Ii hr Vo = 0

Ii = V h V

R hS r o

S i

+

When VS = 0,

Ii = h V

R hr o

S i+

Applying Kirchhoffs current law to the output circuit,

Io = hf Ii + ho Vo

= hf h V

R hr o

S i++ ho Vo

= hh h

R hof r

S i

+

Vo

Zo = VI

o

o

= h

h hR h

1

of r

S i

+

Voltage gain (Avs)

It is the ratio of output voltage Vo to the input voltage VS.

Avs = VV

o

S

Considering source resistance RS as shown in Fig. 3.22,RS

VS Zi Vi

Fig. 3.22Thevenin's Equivalent for the Input Circuit



VV

i

S

= Z

Z Ri

i S+

Now, Avs = VV

o

S

= VV

o

i

VV

i

S

= Av VV

i

S

= Av Z

Z Ri

i S+

Hence, voltage gain Avs is always less than Av.

Current gain (Ais)It is the ratio of output current Io to the input current IS.

Ais = II

o

S

Considering source resistance as shown in Fig. 3.23,

VS

RSZi

Ii

Fig. 3.23Input Circuit

By source transformation (Fig. 3.24),

ZiIS RS

Ii

Fig. 3.24Norton's Equivalent for the Input Circuit



II

i

s

= R

R ZS

S i+

Ais = II

L

s

= II

L

i

II

i

s

= Ai II

i

s

= Ai R

R ZS

S i+

Hence, current gain Ais is always less than Ai.

Example 3.1: Calculate the values of Ai, Av, Zi and Zo for the circuit shown in Fig. 3.25 if the CE amplifi er uses transistor with the specifi cations given in the fi gure and a resistive load of 1 k.

hoe RL

Io

hf eIi1VS

Ii hie

hreVo

Zi Zo

Vo

+

Vi

+

hfe = 220hie = 2.7 k

hre = 1.5 104

hoe = 18

Fig. 3.25Example 3.1

Solution:

RL = 1 k RS = 0

(i) Current gain Ai = hh R1

fe

oe L+

= 2201 18 10 1 106 3+

= 216.11



(ii) Voltage gain Av = h R

h h h h h R( )fe L

ie ie oe fe re L

+

= 220 1 10

2.7 10 (2.7 10 18 10 220 1.5 10 )(1 10 )

3

3 3 6 4 3

+

= 81.48The negative sign indicates a phase shift of 180 between input and output voltages.

(iii) Input impedance Zi = hie h h R

h R1fe re L

oe L+

= 2.7 103 220 1.5 10 1 101 18 10 1 10

4 3

6 3

+

= 2.66 k

(iv) Output impedance Zo = h

h hR h

1

oefe re

S ie

+

= 1

18 10 220 1.5 100 2.7 10

64

3

+

= 173.07 k

3.6EXACT ANALYSIS OF COMMON-EMITTER AMPLIFIERFigure 3.26 shows a common-emitter amplifi er.

VS

RS

RE CE

R1 RC

CC1Vo

+VCC

CC2

RLR2


To analyze the AC signal operation of an amplifi er, an AC equivalent circuit is developed. The capacitors CC1, CC2 and CE are replaced by short circuit because the capacitive reactance is approximately zero at signal frequency. The DC source is replaced by ground. The AC equivalent circuit is shown in Fig. 3.27.


3.6Exact Analysis of Common-emi er Amplifi er3.19

VS

RS

Vo

RLRCR1 R2

Fig. 3.27AC Equivalent Circuit

The BJT is replaced by the h-parameter model as shown in Fig. 3.28.

hoeRC RL

Ic

hf eIb

Io

1RS

RB

VS

Ii Ib hie

hreVo

Zi Z iZin Z o Zo

Vo

+

Vi

+

B C

E

Fig. 3.28h-parameter Equivalent Circuit

RB = R1 || R2 RL = RC || RL

Input impedance

Zi = hie h h R

h R1fe re L

oe L

+

Zi = Zi || RB Zin = Zi + RS

Output impedance

Zo = h

h hR R h

1

( || )oefe re

B S ie

+

Zo = Zo ||RC ||RL

Voltage gain

Av = h R

h h h h h R( )fe L

ie ie oe fe re L

+



The negative sign indicates a phase shift of 180 between input and output voltages.Considering voltage gain with source resistance RS,

Avs = Av Z

Z Ri

i S+

Hence, voltage gain Avs is always less than Av.

Current gain

For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.

Io = VR

o

L

Ii = VZ

i

i

Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L

Considering current gain with source resistance,

Ais = Ai R

R ZS

S i+

Hence, current gain Ais is always less than Ai.

Example 3.2: Find Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.29.

VS

1 k

1 k CE

5 k

CC1

Vo

+VCC

CC2470 k

Zin Zo

hfe = 220hie = 2.7 k

hre = 1.5 104

hoe = 18




Solution:

AC equivalent circuit (Fig. 3.30)

VS

RS

Vo

RCRB


h-parameter equivalent circuit (Fig. 3.31)

hoeRC

Io = Ic

hf eIb1

RSRB

VS

Ii Ib hie

hreVo

Zi Z iZin Z o Zo

Vo

+

Vi

+

B C

E


(i) Input impedance Zi = hie h h R

h R1fe re C

oe C+

= 2.7 103 220 1.5 10 5 10

1 18 10 5 10

4 3

6 3

+

= 2.54 k

Zi = Zi || RB = (2.54 103) || (470 103) = 2.53 k

Zin = Zi + RS = 2.53 103 + 1 103 = 3.53 k

(ii) Output impedance Zo = h

h hR R h

1

( || )oefe re

B S ie

+



= 1

18 10 220 1.5 10[(470 10 ) || (1 10 )] 2.7 10

64

3 3 3

+

= 110.12 k Zo = Zo || RC = (110.12 10

3) || (5 103) = 4.78 k

(iii) Voltage gain Av = h R

h h h h h R( )fe C

ie ie oe fe re C+

=

= 395.97

Avs = Av Z

Z Ri

i S+

= 395.97 2.53 10

2.53 10 1 10

3

3 3

+ = 283.79

(iv) Current gain Ai = Av ZR

i

C

= (395.97) 2.53 10

5 10

3

3

= 200.35

Ais = Ai R

R ZS

S i+

= 200.35 1 10

1 10 2.53 10

3

3 3

+ = 56.75

Example 3.3: Find Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.32.

VS

1 k

1 k

5 k

CE

5 k

CC1

Vo

+VCC

CC2470 k

Zin Zo

hfe = 220hie = 2.7 k

hre = 1.5 104

hoe = 18


220 5 102.7 10 (2.7 10 18 10 220 1.5 10 )(5 10 )

3

3 3 6 4 3

+



Solution:


VS

RS

Vo

RCRB

RL



hoeRC RL

Ic

hf eIb

Io

1RS

RB

VS

Ii Ib hie

hreVo

Zi Z iZin Z o Zo

Vo

+

Vi

+

B C

E


RL = RC || RL = (5 103) || (5 103) = 2.5 k


h R1fe re L

oe L

+

= 2.7 103 220 1.5 10 2.5 101 18 10 2.5 10

4 3

6 3

+

= 2.62 k Zi = Zi || RB = (2.62 10

3) || (470 103) = 2.61 k Zin = Z i + RS = 2.61 10

3 + 1 103 = 3.61 k


h hR R h

1

( || )oefe re

B S ie

+

= 1

18 10 220 1.5 10[(470 10 ) || (1 10 )] 2.7 10

64

3 3 3

+

= 110.119 k Zo = Zo || RC || RL = (110.119 10

3) || (5 103) || (5 103) = 2.44 k




h h h h h R( )fe L

ie ie oe fe re L

+

=

= 200.8

Avs = Av Z

Z Ri

i S+

= 200.8 2.61 102.61 10 1 10

3

3 3

+

= 145.17

(iv) Current gain Ai = AvZR

i

L

= (200.8) 2.61 105 10

3

3

= 104.8

Ais = Ai R

R ZS

S i+

= 104.8 1 101 10 2.61 10

3

3 3

+

= 29.03

Example 3.4: Determine Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.35.

VS

10 k

10 kRE

3 k

CE

5 k

CC1

Vo

+VCC

CC2100 k

Zin Zo

hfe = 50hie = 1.1 k

hre = 2.5 104

hoe = 24


220 2.5 102.7 10 (2.7 10 18 10 220 1.5 10 )(2.5 10 )

3

3 3 6 4 3

+



Solution:


RS RCR2R1

RL

VS+



hoeRC RL

Ic

hf eIb

Io

1RS

RB

VS

Ii Ib hie

hreVo

Zi Z iZin Z o Zo

Vo

+

Vi

+

B C

E


RB = R1 || R2 = (100 103) || (10 103) = 9.09 k

RL = RC || RL = (5 103) || (3 103) = 1.875 k


h R1fe re L

oe L

+

= 1.1 103 50 2.5 10 1.875 101 24 10 1.875 10

4 3

6 3

+

= 1.077 k



Zi = Zi || RB = (1.077 103) || (9.09 103) = 0.963 k

Zin = Zi + RS = 0.963 103 + 10 103 = 10.963 k


h hR R h

1

( || )oefe re

B S ie

+

= 1

24 10 50 2.5 10[(9.09 10 ) || (10 10 )] 1.1 10

64

3 3 3

+

= 45.73 k

Zo = Zo || RC || RL = (45.73 103) || (5 103) || (3 103) = 1.8 k


h h h h h R( )fe L

ie ie oe fe re L

+

=

= 83.25

Avs = AvZ

Z Ri

i S+

= 83.25 0.963 100.963 10 10 10

3

3 3

+

= 7.313


i

L

= (83.25) 0.963 103 10

3

3

= 26.72

Ais = Ai R

R ZS

S i+

= 26.72 10 1010 10 0.963 10

3

3 3

+ = 24.75

50 1.875 101.1 10 (1.1 10 24 10 50 2.5 10 )(1.875 10 )

3

3 3 6 4 3

+


3.7Exact Analysis of Common-base Amplifi er3.27

3.7EXACT ANALYSIS OF COMMON-BASE AMPLIFIERFigure 3.38 shows a common-base amplifi er.

RERS

R1 RC

CC1Vo

+VCC

CC3

CC2RL

R2

VS

Fig. 3.38Common-base Amplifi er

The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replac-ing DC source by ground as shown in Fig. 3.39. Two resistors R1 and R2 are shorted to ground.

RERS

Vo

RC RL

VS


Figure 3.39 can be redrawn by rotating the BJT as shown in Fig. 3.40.

RS

Vo

RCRE RL

VS




The h-parameter equivalent circuit is obtained by replacing the BJT by its h-parameter model as shown in Fig. 3.41.

hob

Ic

hf bIe

Io

1ViRS

+

Vo

+

IeIi hib

hrbVoRE RC RL

E C

B

ZiZin Zi Zo Zo

VS


RL = RC || RLInput impedance

Zi = hib h h R

h R1fb rb L

ob L

+

Zi = Zi || RE Zin = Zi + RS

Output impedance

Zo = h

h hR R h

1

( || )obfb rb

E S ib

+

Zo = Zo || RC ||RL

Voltage gain

Av = h R

h h h h h R( )fb L

ib ib ob fb rb L

+

Considering source resistance,

Avs = Av Z

Z Ri

i S+Current gain




Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L


Ais = Ai R

R ZS

S i+

Example 3.5: Calculate Zi, Zo, Av and Ai for the circuit shown in Fig. 3.42.

3 k1 k

50 k 2.2 k

CC1Vo

+VCC

CC2

CC3

50 k

Zi Zo

VS

hrb = 2 104hib = 14.41

hfb = 0.991hob = 0.18 106


Solution:AC equivalent circuit (Fig. 3.43)

RERS

Vo

RC

VS




Figure 3.43 can be redrawn by rotating the BJT as shown in Fig. 3.44.Vo

RCRE

RS

VS



hob

Ic

hf bIe1Vi

RS

+

+

Vo

+

IeIi hib

hrbVoRE RC

E C

B

Zi Zi Zo Zo

VS


(i) Input impedance Zi = hib h h R

h R1f b rb C

ob C+

= 14.41 ( 0.991)(2 10 )(2.2 10 )1 0.18 10 2.2 10

4 3

6 3

+ = 14.84

Zi = Zi || RE = 14.84 || (3 103) = 14.76


h hR R h

1

( || )obf b rb

E S ib

+

= 1

0.18 10 ( 0.991)(2 10 )[(3 10 ) || (1 10 )] 14.41

64

3 3

+

= 2.276 M Zo = Zo || RC = (2.276 10

6) || (2.2 103) = 2.197 k



(iii) Voltage gain Av = +h R

h h h h h R( )fb C

ib ib ob fb rb C

= ( 0.991) 2.2 10

14.41 [14.41 0.18 10 2 10 ( 0.991)](2.2 10 )

3

6 4 3

+

= 146.7


i

C

= 146.7 14.762.2 103

= 0.9847

Example 3.6: Calculate Zi, Zo, Avs and Ais for the circuit shown in Fig. 3.46.

Vo

10 k 47 k7.5 k

6 V +12 V

100

CC1 CC2

Zi Zo

hfb = 0.99hob = 2.47 107

hib = 12

hrb = 3 105VS



Vo

RC RLRE

RS

VS





hob

Ic

hf bIe

Io

1ViRS

+

Vo

+

IeIi hib

hrbVoRE RC RL

E C

B

Zi Zi Zo Zo

VS


RL = RC || RL = (10 103) || (47 103) = 8.25 k

(i) Input impedance Zi = hib

+

h h R

h R1fb rb L

ob L

= 12 +

( 0.99)(3 10 )(8.25 10 )1 2.47 10 8.25 10

5 3

7 3

= 12.24 Zi = Zi || RE = 12.24 || (7.5 10

3) = 12.22


h hR R h

1

( || )obfb rb

E S ib

+

= 1

2.47 10 ( 0.99)(3 10 )[(7.5 10 ) ||100] 12

75

3

+

= 1.94 M

Zo = Zo || RC || RL = (1.94 106) || (10 103) || (47 103) = 8.209 k

(iii) Voltage gain Av =

+

h R

h h h h h R( )fb L

ib ib ob fb rb L

= ( 0.99)(8.25 10 )

12 [12 2.47 10 3 10 ( 0.99)](8.25 10 )

3

7 5 3

+

= 665.64


3.8Exact Analysis of Common-collector Amplifi er3.33

Avs = Av Z

Z Ri

i S+

= 665.64 +

12.2212.22 100

= 72.6


i

L

= 665.64 12.2247 103

= 0.173

Ais = Ai R

R ZS

S i+

= 0.173 100100 12.22+

= 0.154

3.8EXACT ANALYSIS OF COMMON-COLLECTOR AMPLIFIERFigure 3.49 shows a common-collector amplifi er.

RE

R1

CC1

+VCC

VoCC2

RLR2

RS

VS

Fig. 3.49Common-collector Amplifi er



The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.50.

RE

Vo

RL

R2R1

RS

VS


The h-parameter equivalent circuit is obtained by replacing the BJT by its h-parameter model as shown in Fig. 3.51.

hoc

Ie

hf cIb

Io

1ViRS

+

Vo

+

IbIi hic

hrcVoRB RE RL

B E

C

VS


RB = R1 || R2RL = RE || RL

Input impedance

Zi = hic

+

h h R

h R1fc rc L

oc L



Zi = Zi || RB

Zin = Zi + RS

Output impedance

Zo = h

h hR R h

1

( || )ocfc rc

B S ic

+

= Zo || RE || RL

Voltage gain

Av = h R

h h h h h R( )fc L

ic ic oc fc rc L

+


Avs = AvZ

Z Ri

i S+

Current gain


Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L


Ais = AiR

R ZS

S i+



Example 3.7: Calculate Zi, Zo, Av and Ai, for the circuit shown in Fig. 3.52.

5 k

10 k

1 k

CC1

+VCC

VoCC2

20 k10 k

Zi Zo

hic = 1.2 khrc = 1hfc = 101

hoc = 25 106

VS



RE

Vo

RL

R2R1

RS

VS



hoc

Ie

hf cIb

Io

1ViRS

+

Vo

+

IbIi hic

hrcVoRB RE RL

B E

C

VS


RB = R1 || R2 = (10 103) || (10 103) = 5 k

RL = RE || RL = (5 103) || (20 103) = 4 k



(i) Input impedance Zi = hic h h R

h R1fc rc L

oc L

+

= 1.2 103 ( 101)(1)(4 10 )

1 25 10 4 10

3

6 3

+

= 368.47 kZi = Zi || RB = (368.47 10

3) || (5 103) = 4.93 k


h hR R h

1

( || )ocfc rc

B S ic

+

= 1

25 10 ( 101)(1)[(5 10 ) || (1 10 )] 1.2 10

63 3 3

+

= 20.12 Zo = Zo || RE || RL = 20.12 || (5 10

3) || (20 103) = 20.02


h h h h h R( )fc L

ic ic oc fc rc L

+

= ( 101)(4 10 )

1.2 10 [1.2 10 25 10 ( 101)(1)](4 10 )

3

3 3 6 3

+

= 0.996

Avs = Av Z

Z Ri

i S+

= 0.996 4.93 104.93 10 1 10

3

3 3

+ = 0.828


i

L

= 0.996 4.93 1020 10

3

3

= 0.246

Ais = Ai R

R ZS

S i+

= 0.246 1 101 10 4.93 10

3

3 3

+ = 0.041



3.9APPROXIMATE ANALYSIS OF TRANSISTOR AMPLIFIERFor the CE and CB confi gurations, the magnitude of hr and ho is often such that the parameters Zi, Zo, Av and Ai are only slightly affected if hr and ho are not included in the model. Since the value of hr is very small, controlled source hrVo becomes very small. It is approximated by hrVo = 0 and is replaced by a short circuit. Similarly, the value of ho is very small, and the impedance

h1

o

becomes very large and can be ignored in comparison to a parallel load. It is approxi-

mated by h1

o

= and is replaced by an open circuit. Thus, the h-parameter model is simpli-

fi ed by short circuiting controlled source hrVo and open circuiting impedance h1

o

. Generally, if

hoeRL < 0.1, the approximate model can be used. The simplifi ed h-parameter model is shown in the Fig. 3.55.

RL

Io

hi

VS

RS

Ii

hfIi Vo

+

Vi

+

Fig. 3.55Simplifi ed h-parameter Model

Input impedance Zi = hi

Output impedance Zo = h1

o

Voltage gain Av = h R

hf L

i

Current gain Ai = hf

3.10APPROXIMATE ANALYSIS OF COMMON-EMITTER AMPLIFIER (FIXED-BIAS CONFIGURATION)

Fig. 3.56 shows a fi xed-bias common-emitter amplifi er.

Case I: When RE is bypassed


3.10Approximate Analysis of Common-emi er Amplifi er3.39

RE

RS

RB RC

CC1Vo

+VCC

CC2

RL

CEVS

Fig. 3.56Common-emi er Amplifi er with Bypassed RE


Vo

RC RLRB

RS

VS


The BJT is replaced by the approximate h-parameter model as shown in Fig. 3.58.

Ic

hf eIb

Io

ViRS

+

Vo

+

IbIi

hieRB RC RL

B C

E

ZiZin Zi Zo Zo

VS




Input impedance Zi = hieZi = Zi || RBZin = Zi + RS

Output impedance The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.

Zo =

Zo = Zo || RC || RL = RC || RL

Voltage gain

RL = RC || RLVo = hfe Ib RLVi = Ib hie

Av = VV

o

i

= h I R

I hfe b L

b ie

= h R

hfe L

ie


Avs = Av Z

Z Ri

i S+

Current gainFor transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.

Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L


Ais = Ai R

R ZS

S i+


Case II: When RE is not bypassed Figure 3.59 shows a CE amplifi er with unbypassed RE.

RE

R1 RC

CC1

+VCC

VoCC2

RLR2

RS

VS

Fig. 3.59Common-emi er Amplifi er with Unbypassed RE


RE RC RL

Vo

R2R1

RS

VS



Input impedance Vi = Ibhie + IeRE

= Ibhie + (1 + hfe) Ib RE = [hie + (1 + hfe) RE] Ib

Zi = VI

i

b




= h h R I

I[ (1 ) ]ie fe E b

b

+ +

= hie + (1 + hfe) RE

Due to unbypassed RE, input impedance Zi increases by a value (1 + hfe) RE. When RE is transferred to base side, it gets multiplied by 1 + hfe. Hence, input impedance looking from the base side, i.e. Zi is the sum of hie and (1 + hfe) RE.

Zi = Zi || RB= [hie + (1 + hfe) RE] || RB

Zin = Zi + RSOutput impedance The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.

Zo = Zo = Zo || RC || RL = RC || RL

Voltage gain

R R R||L C L =

Vo = hfe Ib RLVi = [hie + (1 + hfe) RE] Ib

Av = VV

o

i

RC

RE

RL

Ic

hfeIb

Io

RSRB

VS

Ii Ib

Ie

hie

Zi Z iZin Z o Zo

Vo

+

Vi

+

B C

E



=

+ +

h I R

h h R I[ (1 ) ]fe b L

ie fe E b

= h R

h h R(1 )fe L

ie fe E

+ +

The negative sign indicates a phase shift of 180 between input and output voltages. Due to unbypassed RE, voltage gain decreases.

Considering voltage gain with source resistance RS,

Avs = Av Z

Z Ri

i S+Current gain For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.

Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L


Ais = Ai R

R ZS

S i+

Example 3.8: Calculate Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.62.

VS

1 k

RE

10 k

CE

4.7 k

CC1

Vo

+VCC

CC2470 k

Zin Zo

hfe = 110hie = 1.6 k


Solution: AC equivalent circuit (Fig. 3.63)




VS

RS

Vo

RCRB

RL

Fig. 3.63AC Equivalent Circuith-parameter equivalent circuit (Fig. 3.64)

RC RL

Ic

hfeIb

Io

RSRB

VS

Ii Ib

hie

Zi Z iZin Z o Zo

Vo

+

Vi

+

B C

E


RL = RC || RL = (4.7 103) || (10 103) = 3.197 k

(i) Input impedance Zi = hie = 1.6 k Zi = Zi || RB = (1.6 10

3) || (470 103) = 1.595 k Zin = Zi + RS = 1.595 10

3 + 1 103 = 2.595 k(ii) Output impedance Zo =

Zo = RC || RL = (4.7 103) || (10 103) = 3.197 k


hfe L

ie

=

110 3.197 101.6 10

3

3

= 219.8

Avs = Av Z

Z Ri

i S+

= 219.8 +

1.595 101.595 10 1 10

3

3 3

= 134.94



i

L

= (219.8)

1.595 1010 10

3

3

= 35.05

Ais = Ai R

R ZS

S i+

= 35.05 +

1 101 10 1.595 10

3

3 3

= 13.53


VS

1 k

4.7 k

1.2 k

10 k

4.7 k

CC1

Vo

+VCC

CC2470 k

Zin Zo

hfe = 110hie = 1.6 k

Fig. 3.65Example 3.9Solution:AC equivalent circuit (Fig. 3.66)

RE RC RL

Vo

RB

RS

VS






RC

RE

RL

Ic

hfeIb

Io

RSRB

VS

Ii Ib

Ie

hie

Zi Z i Z o Zo

Vo

+

Vi

+

B C

E


RL = RC || RL = (4.7 103) || (10 103) = 3.197 k

(i) Input impedance Zi = hie + (1 + hfe) RE= 1.6 103 + (1 + 110) (1.2 103)= 134.8 k

Zi = Zi || R B = (134.8 103) || (470 103) = 104.76 k

(ii) Output impedance Zo = Zo = RC = 4.7 k


h h R(1 ) fe L

ie fe E

+ +

= 110 3.197 10

1.6 10 (1 110)(1.2 10 )

3

3 3

+ + = 2.61

Avs = Av Z

Z Ri

i S+

= 2.61 +

104.76 10104.76 10 1 10

3

3 3

= 2.56


i

L

= (2.61)

104.76 1010 10

3

3


= 27.34

Ais = Ai R

R ZS

S i+

= 27.34 +

1 101 10 104.76 10

3

3 3

= 0.26

3.11 APPROXIMATE ANALYSIS OF COMMON-EMITTER AMPLIFIER (VOLTAGE DIVIDER BIAS CONFIGURATION)

Figure 3.68 shows a common-emitter amplifi er with voltage divider bias circuit.

Case I: When RE is bypassed

RE

RS

R1 RC

CC1Vo

+VCC

CC2

R2

RL

CEVS

Fig. 3.68Common-emi er Amplifi er with Bypassed REThe AC equivalent circuit is obtained by replacing all capacitors by short circuits and

replacing DC source by ground as shown in Fig. 3.69.

RC RL

Vo

R2R1

RS

VS






Ic

hf eIb

Io

ViRS

+

Vo

+

IbIi

hieRB RC RL

B C

E

ZiZin Zi Zo Zo

VS


RB = R1 || R2Input impedance

Zi = hieZi = Zi || RB

Zin = Zi + RS

Output impedance

The output impedance Zo of the circuit is the impedance looking from collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.

Zo =

Zo = Zo || RC || RL = RC || RL

Voltage gain

RL = RC || RLVo = hfe Ib RLVi = Ib hie

Av = VV

o

i

= h I R

I hfe b L

b ie

= h R

hfe L

ie


Avs = Av Z

Z Ri

i S+


Current gain


Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L


Ais = Ai R

R ZS

S i+

Case II: When RE is not bypassed Figure 3.71 shows a CE amplifi er with unbypassed RE.

RE

R1 RC

CC1

+VCC

VoCC2

RLR2

RS

VS

Fig. 3.71Common-emi er Amplifi er with Unbypassed RE





RCRE RL

Vo

R2R1

RS

VS



RC

RE

RL

Ic

hfeIb

Io

RSRB

VS

Ii Ib

Ie

hie

ZiZin Z i Z o Zo

Vo

+

Vi

+

B C

E


RB = R1 || R2Input impedance

Vi = Ibhie + IeRE = Ibhie + (1 + hfe) Ib RE = [hie + (1 + hfe) RE] Ib

Zi = VI

i

b

= + +h h R I

I[ (1 ) ]ie fe E b

b

= hie + (1 + hfe) REDue to unbypassed RE, input impedance Zi increases by (1 + hfe) RE.

Zi = Zi || RB= [hie + (1 + hfe) RE] || RB

Zin = Zi + RS


Output impedance The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.

Zo =

Zo = Zo || RC || RL = RC || RLVoltage gain

R R R||L C L =

Vo = hfe Ib RLVi = [hie + (1 + hfe) RE] Ib

Av = VV

o

i

=

+ +

h I R

h h R I[ (1 ) ]fe b L

ie fe E b

=

+ +

h Rh h R(1 )

fe L

ie fe E

The negative sign indicates a phase shift of 180 between input and output voltages. Due to unbypassed RE, voltage gain decreases.

Considering voltage gain with source resistance RS,

Avs = Av Z

Z Ri


Ai = II

o

i

=

VRVZ

o

L

i

i

= VV

o

i

ZR

i

L

= AvZR

i

L


Ais = Ai R

R ZS

S i+

Example 3.10: Calculate Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.74.Solution:AC equivalent circuit (Fig. 3.75)




RE

500

120 k 3.3 k

CC1Vo

+VCC

CC2

39 k

10 k

CE

Zin Zo

hie = 2 khfe = 120

VS


RC RL

Vo

R2R1

RS

VS



RC RL

Ic

hfeIb

Io

RSRB

VS

Ii Ib

hie

ZiZin Z i Z o Zo

Vo

+

Vi

+

B

E

C


RB = R1 || R2 = (120 103) || (39 103) = 29.43 k

RL = RC || RL = (3.3 103) ||(10 103) = 2.48 k


(i) Input impedance Zi = hie = 2 kZi = Zi || RB = (2 10

3) || (29.43 103) = 1.87 kZin = Zi + RS = 1.87 10

3 + 500 = 2.37 k

(ii) Output impedance Zo = Zo = RC || RL = (3.3 10

3) || (10 103) = 2.48 k


hfe L

ie

=

120 2.48 102 10

3

3

= 148.8

Avs = Av Z

Z Ri

i S+

= 148.8 1.87 101.87 10 500

3

3

+

= 117.4


i

L

= ( 148.8)

1.87 1010 10

3

3

= 27.83

Ais = Ai R

R ZS

S i+

= 27.83 +

500500 1.87 103

= 5.87






1.2 k

540 k5.6 k

CC1

+VCC

VoCC2

540 k

Zi Zo

hie = 555 hfe = 120

VS


RE RC

Vo

R2R1

RS

VS



RE

RChfeIb

Io = Ic

RBVS

Ii Ib

Ie

hie

Zi Z i Z o Zo

Vo

+

Vi

+

B C

E


R B = R1 || R2 = (540 103) || (540 103) = 270 k


(i) Input impedance Zi = hie + (1 + hfe) RE

= 555 + (1 + 120) (1.2 103)

= 145.76 k

Zi = Zi || R B = (145.76 103) || (270 103) = 94.66 k

(ii) Output impedance Zo =

Zo = RC = 5.6 k

(iii) Voltage gain Av = + +h R

h h R(1 )fe C

ie fe E

= 120 5.6 10

555 (1 120)(1.2 10 )

3

3

+ +

= 4.61


i

C

= ( 4.61)

94.66 105.6 10

3

3

= 77.93


0.9 k

0.1 k

33 k 3.3 k

CC1Vo

+VCC

CC2

10 k

CE

Zi Zo

hie = 2 khfe = 100VS






RE1 RC

Vo

R2R1

RS

VS



RC

RE1

hfeIbRS

RB

VS

Ii Ib

Ie

hie

Zi Z i Z o Zo

Vo

+

Vi

+

B

E

Io = Ic C


R B = R1 || R2 = (33 103) || (10 103) = 7.6 k

(i) Input impedance Zi = hie + (1 + hfe) RE= 2 103 + (1 + 100)(0.1 103)= 12.1 k

Zi = Zi || RB = (12.1 103) || (7.67 103) = 4.694 k


Zo = RC = 3.3 k

(iii) Voltage gain Av = + +h R

h h R(1 )fe C

ie fe E


= 100 3.3 102 10 (1 100)(0.1 10 )

3

3 3

+ + = 27.272


i

C

= ( 27.272)

4.694 103.3 10

3

3

= 38.79

3.12 APPROXIMATE ANALYSIS OF COMMON-EMITTER AMPLIFIER (COLLECTOR TO BASE BIAS CONFIGURATION)

Figure 3.83 shows a common-emitter amplifi er with the collector to base bias circuit. The collector feedback circuit uses a feedback path from collector to base to increase the stability of the system. The resistor RB is connected between input and output.

RE

RS

RB

RC

CC1

Vo

+VCC

CC2

C

EB

RL

CEVS


This circuit provides negative feedback which reduces the gain of the amplifi er. In the CE amplifi er, output is 180o out of phase with input. If output is fed back to input, the net gain decreases. This process is known as AC degeneration. If AC signal voltage increases, base current increases and collector current IC increases (Fig. 3.84). Hence, the net base current through the transistor decreases as If opposes Ib. Due to this the gain of the amplifi er decreases.

AC degeneration can be avoided by splitting RB into two parts and connecting a capacitor at the centre of both resistors (Fig. 3.85). For DC operation, there is no effect of the capacitor




(XC = ). For AC operation, as Ib increases, Ic increases. Hence, the component of base current through RB is bypassed by C. Hence, there is no change in Ib and there is no degeneration of input signal.

RS

RB

RC

CC1

Vo

+VCC

CC2

RL

If Ic

Ib

VS

Fig. 3.84AC Degeneration

RS

RC

CC1

+VCC

CC2RB2RB1

RL

C

VS

Fig. 3.85Method to Avoid AC Degeneration



RB2 RC

Vo

RLRB1

RS

VS


The h-parameter equivalent circuit is obtained by replacing BJT by its h-parameter model as shown in Fig. 3.87.

Ic

hf eIb

Io

ViRS

+

Vo

+

IbIi

hieRB1 RB2 RC RL

B C

E

ZiZin Zi Zo Zo

VS


Input impedance

Zi = hieZi = Zi || RB1Zin = Zi + RS

Output impedance

The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.

Zo = Zo = Zo || RB2 || RC || RL = RB2 || RC || RL




Voltage gain

RL = RC || RL

Vo = hfe Ib RLVi = Ib hie

Av = VV

o

i

= h I R

I hfe b L

b ie

= h R

hfe L

ie


Avs = Av Z

Z Ri

i S+

Current gain


Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L


Ais = Ai R

R ZS

S i+

If a capacitor is not connected at the centre of the feedback circuit, Millers theorem is used for AC analysis. Resistor RB is splitted into two equivalent resistors at the input and output using Millers theorem.

Millers theorem

If an impedance Z is connected between the input and output of a circuit having voltage gain

Av = VV

2

1

, it can be replaced with two impedances Z1 and Z2 at the input and output terminals,

respectively as shown in Fig. 3.88.


AmplifierAV

ZI1

I1 I2

I2

V1

+

V2

+

AmplifierAV

V1

+

V2

+

Z1 Z2

Fig. 3.88Millers Theorem

IV V

Z

VVV

Z

1

11 2

12

1=

=

=

V AZ

VZ

A

VZ

(1 )

(1 )

v

v

1 1 1

1

=

=

where ZZ

A1 v1 =

IV V

Z

VVV

Z

1

22 1

21

2=

=

=

VA

ZV

AA

Z

VZ

1 1

1

v

v

v

22 2

2

=

=

where ZA

AZ

1.v

v2 =

The resistor RB, which is connected between input and output, is splitted into resistors RB1 and RB2 at the input and output using Millers theorem.


RB2 RC

Vo

RLRB1

RS

VS





The h-parameter equivalent circuit is obtained by replacing BJT by its h-parameter model as shown in Fig. 3.90.

Ic

hf eIb

Io

ViRS

+

Vo

+

IbIi

hieRB1 RB2 RCRC RL

B C

E

ZiZin Zi Zo Zo

VS


Voltage gain

RA

AR R

1B

v

vB B2

=

RL = RB2 || RC || RL = RB || RC || RLVo = hfe Ib RLVi = Ib hie

Av = VV

o

i

= h I R

I hfe b L

b ie

= h R

hfe L

ie


Avs = Av Z

Z Ri

i S+

Input impedance

RB1 = R

AB

v1

Zi = hieZi = Zi || RB1

Zin = Zi + RS

Output impedance

The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0.


When VS = 0, Ib = 0, hfeIb = 0. The current source hfeIb is replaced by an open circuit.

Zo =

Zo = Zo || RB2 || RC || RL= RB2 || RC || RL

Current gain


Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L


Ais = Ai R

R ZS

S i+


VS

3 k

CC1

Vo

CC2120 k 68 k

Zi Zo

+VCC

hfe = 140hie = 1.4 k






VS

Vo

RB2RB1RC


h-parameter equivalent circuit (Fig. 3.93)Ic

hf eIb

Io

Vi

+

Vo

+

IbIi

hieRB1 RB2 RCRC

B C

E

Zi Z i Zo Zo

VS


R R R|| (68 10 ) || (3 10 ) 2.87 kL B C3 3

2 = = =

(i) Input impedance Zi = hie = 1.4 k

Zi = Zi || RB1= (1.4 103) || (120 103) = 1.38 k


Zo = RB2 || RC = (68 103) || (3 103) = 2.87 k


hfe L

ie

=

140 2.87 101.4 10

3

3

= 287


i

C

= ( 287)

1.38 103 10

3

3

= 132.02


Example 3.14: Calculate Av, Zi, Zo and Ai for the circuit shown in Fig. 3.94.

VS

10 k

CC1

Vo

CC2200 k

Zi Zo

+VCC

hfe = 50hie = 1.1 k



Ic

hf eIb

Io

Vi

+

Vo

+

IbIi

hieRB1 RB2 RCRC

B C

E

Zi Z i Zo Zo

VS



Ic

hf eIb

Io

Vi

+

Vo

+

IbIi

hieRB1 RB2 RCRC

B C

E

Zi Z i Zo Zo

VS


RA

AR R

1B

v

vB B2

=

RL = RB2 || RC = (200 103) || (10 103) = 9.52 k




(i) Voltage gain Av = h R

hfe L

ie

= 50 9.52 101.1 10

3

3

= 432.73

(ii) Input impedance RB1 = R

A1B

v

= 200 101 ( 432.73)

3

= 461.12 Zi = hie = 1.1 k

Zi = Zi || RB1 = (1.1 103) || (461.12) = 324.92

(iii) Output impedance Zo = Zo = RB2 || RC = (200 10

3) || (10 103) = 9.52 k


i

C

= (432.73) 324.9210 103

= 14.06

Example 3.15: Calculate Av, Zi, Zo and Ai for the circuit shown in Fig. 3.97.

VS

10 k

CC1

Vo

CC2100 k

Zi Zo

+VCC

hfe = 100hie = 1 k

50

10 k




VS

Vo

RB2RB1 RE

RC RL



RB2

RE

RC

Ic

hfeIb

Io

RLRB1VS

Ii Ib

hie

Zi Z i Z o Zo

Vo

+

Vi

+

B C

E


RA

AR R

1B

v

vB B2

=

RL = RB2 || RC || RL = (100 103) || (10 103) || (10 103) = 4.76 k

(i) Voltage gain Av =

+ +

h Rh h R(1 )

fe L

ie fe E

= 100 4.76 10

1 10 (1 100)(50)

3

3

+ +

= 78.68

(ii) Input impedance RB1 = R

A1B

v

= 100 10

1 ( 78.68)

3

= 1.26 k




Zi = hie + (1 + hfe) RE

= 1 103 + (1 + 100) (50)

= 6.05 k

Zi = Zi || RB1 = (6.05 103) || (1.26 103) = 1.04 k

(iii) Output impedance Zo =

Zo = RB2 || RC || RL = (100 103) || (10 103) || (10 103) = 4.76 k


i

L

= (78.68) 1.04 1010 10

3

3

= 8.18

3.13 APPROXIMATE ANALYSIS OF COMMON-BASE AMPLIFIERFigure 3.100 shows a common-base amplifi er.

RERS

VS

R1 RC

CC1Vo

+VCC

CC3

CC2RL

R2

Fig. 3.100Common-base Amplifi er

The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.101. Two resistors R1 and R2 are shorted to ground.


3.13Approximate Analysis of Common-base Amplifi er3.69

RE VS

Vo

RC RL


Figure 3.102 can be redrawn by rotating BJT as shown in Fig. 3.101.

RS RCRE RL

VS

Vo


The h-parameter equivalent circuit is obtained by replacing the BJT by its h-parameter model (Fig. 3.103).

Ic

hf bIe

Io

ViRS

+

Vo

+

IeIi

hibRE RCRC RL

E C

B

ZiZin Zi Zo Zo

VS


Input impedance

Zi = hibZi = Zi || RE

Zin = Zi + RSOutput impedanceThe output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfb Ie = 0. The current source hfb Ie is replaced by an open circuit.



Zo = Zo = Zo || RC || RL = RC || RL

Voltage gainRL = RC || RL Vo = hfb Ie RLVi = Ie hib

Av = VV

o

i

= h I R

I hfb e L

e ib

= h R

hfb L

ibConsidering source resistance,

Avs = Av Z

Z Ri


Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L


Ais = Ai R

R ZS

S i+


Solution:

1 k

33 k 3.3 k

CC1Vo

+VCC

CC2

CC3

10 k

Zi Zo

hfe = 99hie = 2 k

VS





RE

Vo

RC

VS



hf bIe

Io = Ic

Vi

+

Vo

+

IeIi

hibRE RC

E C

B

Zi Z i Zo Zo

VS


hib = h

h1ie

fe+ = 2 10

1 99

3

+= 0.02 k

hfb = +h

h1fe

fe

= 991 99+

= 0.99

(i) Input impedance Zi = hib = 0.02 k Zi = Zi || RE = (0.02 10

3) || (1 103) = 19.6

(ii) Output impedance Zo = Zo = RC = 3.3 k


hfb C

ib

= ( 0.99)(3.3 10 )0.02 10

3

3

= 163.35




i

C

= 163.35 19.63.3 103

= 0.97


2.7 k

100 k 3.9 k

1 k

CC1Vo

+VCC

CC2

CC3

33 k10 k

Zi Zo

hfe = 99hie = 2 k

VS



RE

RS

Vo

RC RL

VS





Ic

hf bIe

Io

ViRS

+

Vo

+

IeIi

hibRE RCRC RL

E C

B

ZiZin Zi Zo Zo

VS


hib = h

h1ie

fe+= 2 10

99 1

3

+= 0.02 k

hfb = h

h1fe

fe

+

991 99+

= 0.99

R R R|| (3.9 10 ) || (10 10 ) 2.81 kL C L3 3

= = =

(i) Input impedance Zi = hib = 0.02 k

Zi = Zi || RE = (0.02 103) || (2.7 103) = 19.85

Zin = Zi + RS = 19.85 + 1 103 = 1.02 k


Zo = RC || RL = (10 103) || (3.9 103) = 2.81 k


hfb L

ib

= ( 0.99) 2.81 100.02 10

3

3

= 139.1

Avs = Av Z

Z Ri

i S+



= 139.1 19.8519.85 1 103+

= 2.71


i

L

= 139.1 19.8510 103

= 0.28

Ais = Ai R

R ZS

S i+

= 0.28 1 101 10 19.85

3

3

+= 0.27

3.14 APPROXIMATE ANALYSIS OF COMMON-COLLECTOR AMPLIFIER

Figure 3.110 shows a common-collector amplifi er.

RE RL

RS

R1

CC1

+VCC

Vo

CC2

R2

VS

Fig. 3.110Common-collector Amplifi er


hic = hiehfc = (1 + hfe) hfe


3.14Approximate Analysis of Common-collector Amplifi er3.75

RE RL

RSVo

R1 R2VS

Fig. 3.111AC equivalent Circuit

Hence, the h-parameter model of the CE confi guration is used for the AC analysis of the CC amplifi er (Fig. 3.112).

RE RL

Ic

hfeIb

Io

RS

RB

VS

Ii Ib

Ie

hie

Zi Z iZin Z o Zo

Vo

Vi

+

B C

E


RB = R1 || R2 RL = RE || RLInput impedance

Vi = Ib hie + Ie RL = Ib hie + (1 + hfe) Ib RL = [hie + (1 + hfe) RL]Ib

Zi = VI

i

b

= hie + (1 + hfe) RL Zi = Zi || RB = [hie + (1 + hfe) RL] || RB Zin = Zi + RS



Output impedance The base circuit can be converted into Thevenins equivalent circuit (Fig. 3.113).

VTh = V = R

R RB

S B+VS

RTh = (RB || RS) + hie

Io

Ib

Ie

Z o

Vo

V+

B E(RB||RS) + hie

Fig. 3.113Thevenin's Equivalent Circuit

The output impedance Zo of the circuit is the impedance looking from the emitter side when input voltage V = 0.

Zo = VI

oo V 0=

Applying Kirchhoffs voltage law to the base-emitter circuit,

V Ib [(RB || RS)+ hie] Vo = 0

When V = 0, Vo = Ib [(RB || RS) + hie] Io = Ie = (1 + hfe) Ib

VI

o

o

= R R h

h( || )

(1 )B S ie

fe

+

+

Zo = R R h

h( || )

(1 )B S ie

fe

+

+

When base resistance [(RB || RS) + hie] is transferred to the emitter side, it gets divided by

(1 + hfe). Hence, output impedance looking from the emitter side, i.e. Zo is R R h

h( || )

1B S ie

fe

+

+

.

Zo = Zo || RL

= R R h

h( || )

1B S ie

fe

+

+

|| RL



When RS is not present, i.e. RS = 0,

Zo = h

h1ie

fe+

|| RL

Generally output impedance of the common-collector confi guration is very low. Hence, it is used as buffer for impedance matching.

Voltage gain

Vo = Ie RL

= (1 + hfe) Ib RL Vi = [hie + (1 + hfe) RL ] Ib

Av = VV

o

i

= h I R

h R h I(1 )

[ (1 )]fe b L

ie L fe b

+

+ +

= h R

h h R(1 )

(1 )fe L

ie fe L

+

+ +

If (1 + hfe) RL >> hieAv 1

Thus, Vo Vi and output at the emitter follows the input at the base. Hence, the common-collector amplifi er is called as emitter follower.

Current gain


Ai = II

o

i

=

VR

VZ

o

L

i

i

= VV

o

i

ZR

i

L

= Av ZR

i

L

Example 3.18: Determine Zi, Zo, Av and Ai for the emi er follower circuit shown in Fig. 3.114.




3.3 k

220 k

CC1

Vo

Vi

+VCC

CC2

ZoZi

hfe = 98hie = 1.275 k


RBRE

Vo

Vi

Fig. 3.115AC Equivalent Circuith-parameter equivalent circuit (Fig. 3.116)

RE

Ic

hfeIb

IoRB

Ii Ib

Ie

hie

Zi Z i Zo

Vi

+

Vo

+

B C

E


(i) Input impedance Zi = hie + (1 + hfe) RE= 1.275 103 + (1 + 98) (3.3 103)= 327.98 k

Zi = Zi || RB = (327.98 103) || (220 103) = 131.67 k




h1ie

fe+

|| RE

= 1.275 101 98

3

+

|| (3.3 10

3)

= 12.83


h h R(1 )

(1 )fe E

ie fe E

+

+ +

= (1 98)(3.3 10 )

1.275 10 (1 98)(3.3 10 )

3

3 3

+

+ +

= 0.996


i

E

= 0.996 131.67 103.3 10

3

3

= 39.74

Example 3.19: Find Zin, Zo, Av and Ai for the circuit shown in Fig. 3.117.

20 k

91 k

10 k

100 k

CC1

Vo

+VCC

ZoZin

hfe = 50hie = 1.1 k

VS





R2RE

Vo

R1

RS

VS



RE

Ic

hfeIb

Io

RS

RB

VS

Ii Ib

Ie

hie

Zi Z iZin Zo

Vi

+

B C

E

Vo

+


RB = R1 || R2 = (91 103) || (100 103) = 47.64 k

(i) Input impedance Zi = hie + (1 + hfe) RE= 1.1 103 + (1 + 50) (20 103)= 1021.1 k

Zi = Zi || RB = (1021.1 103) || (47.64 103) = 45.52 k

Zin = Zi + RS = 45.52 103 + 10 103 = 55.52 k

(ii) Output impedance Zo = R R h

h( || )

1B S ie

fe

+

+

|| RE

= {(47.64 10 ) || (10 10 )} 1.1 10

1 50

3 3 3 +

+

|| (20 10

3)

= 181.954




h h R(1 )

(1 )fe E

ie fe E

+

+ +

= (1 50)(20 10 )

1.1 10 (1 50)(20 10 )

3

3 3

+

+ +

= 0.998


i

E

= 0.998 45.52 1020 10

3

3

= 2.27

Example 3.20: Find Zi, Zo and Av for the circuit shown in Fig. 3.120.

5 k 2 k

1 k

50 k

CC1

+VCC

Vo

CC2

25 k

Zi Zo

hfe = 100hie = 2 k

VS



R2RE

Vo

RL

R1

RS

VS





RE RL

Ic

hfeIb

Io

RS

RB

VS

Ii Ib

Ie

hie

Zi Z i Zo

Vo

Vi

+

B C

E


RB = R1 || R2 = (50 103) || (25 103) = 16.67 k

RL = RE || RL = (5 103) || (2 103) = 1.43 k

(i) Input impedance Zi = hie + (1 + hfe) RL

= 2 103 + (1 + 100) (1.43 103)

= 146.28 k

Zi = Zi || RB = (146.28 103) || (16.67 103) = 14.96 k

(ii) Output impedance Zo = R R h

h( || )

1B S ie

fe

+

+

|| RE

= {(16.67 10 ) || (1 10 )} 2 10

1 100

3 3 3 +

+

|| (5 103)

= 28.97


h h R(1 )

(1 )fe L

ie fe L

+

+ +

= (1 100)(1.43 10 )

2 10 (1 100)(1.43 10 )

3

3 3

+

+ +

= 0.9863


3.15Comparison of CE, CB and CC Amplifi ers3.83

3.15COMPARISON OF CE, CB AND CC AMPLIFIERS

Parameter CE CB CC

(1) Voltage gain (Av) Very high Very high Less than 1

(2) Current gain (Ai) Very high Less than 1 High

(3) Input impedance (Zi) Medium Low High

(4) Output impedance (Zo) High High Very low

(5) Phase of output signal Out of phase with input In phase with input In phase with input

(6) Application Voltage amplifi er Non-inverting amplifi er Bu er between high impedance source and low impedance load

SUMMARY OF BJT AMPLIFIERS

Confi guration Circuit Equation

CE amplifi er with bypassed RE

VS

RS

R1 RC

CC1Vo

+VCC

CC2

RL

RE CE

R2

Zi Z i Z o Zo

Zi = hieZi = Zi R1 R2

Zin = Zi + RSZo =

Zo = Zo RC RL

Av = h R R

h( || )fe C L

ie

Avs = Av Z

Z Ri

i S+

Ai = Av ZR

i

L

Ais = Ai R

R ZS

S i+

(Continued)




CE amplifi er with unbypassed RE

VS

RS

R1 RC

CC1Vo

+VCC

CC2

RL

RE

R2

Zi Z i Z o Zo


Zi = Zi R1 R2

Zin = Zi + RSZo =

Zo = Zo RC RL

Av = h R R

h h R( || )(1 )

fe C L

ie fe E+ +

Avs = Av Z

Z Ri

i S+

Ai = Av ZR

i

L

Ais = Ai R

R ZS

S i+

CB amplifi er

VS

R1 RC

CC1Vo

+VCC

CC3

CC2RL

RERS

R2

Zi Z i Z o Zo

Zi = hib

Zi = Zi REZin = Zi + RS

Zo =

Zo = Zo RC RL

Av = h R R

h( || )fb C L

ib

Avs = Av Z

Z Ri

i S+

Ai = Av ZR

i

L

Ais = Ai R

R ZS

S i+

SUMMARY OF BJT AMPLIFIERS (Continued)

(Continued)



CC amplifi er

VS

RS

R1

CC1

Vo

+VCC

CC2

RLRE

R2

Zi Z i Z o


Zi = Zi R1 R2

Zin = Zi + RS

Zo = RE R R h

h[( || ) ]

1S B ie

fe

+

+

Zo = Zo RL

Av = h R R

h h R R(1 )( || )

(1 )( || )fe E L

ie fe E L

+

+ +

Avs = Av Z

Z Ri

i S+

Ai = Av ZR

i

L

Ais = Ai R

R ZS

S i+


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Bjt Small Signal Amplifier