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1

The Transistor At Low frequencies Graphical Analysis of CE configuration Consider a CE amplifier as shown below

• VCC and VBB provide the biasing for the transistor • RL is the load resistor • RS is the source resistance

The input and output characteristics are shown below

B

Q

A

iC

vBE V

iB vCE

RL

VCC

RB B

Collector voltage vCE, V

Collector current iC, mA IB3

IB2

IB

IB1

IB0



2

The intersection of input characteristics, corresponding to quiescent base current IB , and the load line results in the operating point Q. corresponding to this operating point, the base current, collector current, base voltage and collector voltage are indicated by the symbols IB, IC, VB and VC respectively

Due to the application of ac signal, the base current is subjected to variation around the q point. The instantaneous value of current and voltage are indicated by lower case letter with an upper case subscript ( for ex: iB, iC, vb, vc).

The instantaneous value of the varying component from the quiescent value is indicated by lower case letter with a lower case subscript (for ex: ib, ic, vb, vc). The following relations can be established ic=ic-Ic = Δ ic Ib = iB-IB = Δ iB Vc =vc-Vc= Δ Vc Vb = vB - VB = Δ vB

Thus, variation of collector current and voltage can be seen as alternating current and voltages. The waveforms When an AC signal is applied, the instantaneous base current will fluctuate between two extreme values iB1 and iB3 as indicated in the output characteristics . The corresponding Q point will fluctuate between two points A and B on the DC load line the corresponding collector current and voltage variations are as indicated in the output characteristics. Due to the nonlinearity of the output characteristics, the output characteristics are not parallel lines.

iB

B

A

Q

VBE

Input characteristics



3

Also these characteristics are not at equally spaced for equal increments in the base current. This results in the collector current and voltage waveforms being non sinusoidal. This called output linear distortion. The input characteristics is non linear, therefore the base voltage Vb is not identical to the base current ib, which is sinusoidal. This change in waveform is known as input nonlinear distortion. Due to the non-linearity for large voltage , the base current swing is not symmetrical around the Q point. The base current swing and hence collector current swing is larger in BQ region then in QA region. Thus if the operating point Q is situated in linear region, then for small changes , the distortion would be less. Under this condition of small amplitude of the input signal, linear circuit model can be used. NOTE: Refer to PPT slides for waveforms and diagrams

SMALL SIGNAL MID FREQUENCY ANALYSIS OF AMPLIFIER

MODELING OF TRANSISTOR AMPLIFIER

To analyze and assess the performance of transistor amplifiers, equivalent circuits are used. The behavior and performance of the transistor can be expressed by a set of mathematical equations. These equations are called a mathematical model. These equations are based on suitable theory. For example, considering the transistor to be operating in linear region it can be modeled with help of two-port network theory. The solution of these mathematical equations results in the response of the amplifier for a given excitation.

These mathematical equations can be synthesized in to an electrical network called equivalent circuit. This equivalent circuit along with the external components conned to the network represents the transistor amplifier. The solution of the equivalent circuit provides to the response of the system.

The elements of the equivalent circuit are called the parameters of the transistor. Various parameters can be defined based on the requirements of the designer. For example h-parameters are used for mid and low frequency analysis and hybrid-pi parameters are used for high frequency analysis.

Besides the above two models, Z-parameters, Y-parameters or A-B-C-_D parameters can also be used. Identification of proper independent and dependent variables results in suitable model.

THE TRANSISTOR HYBRID MODEL:

Consider the amplifier circuit as shown in the fig. A common emitter configuration is considered. The transistor can be considered as a linear device in the active region for small signal amplitude. Then, the two-port network theory can be applied to the transistor. According to this theory, the four variables of the transistor can be related by the following set of equations. For defining h-parameters, the quantities i b



4

(input current) and vce (output voltage) are taken as independent variables and the remaining two quantities are represented in terms of independent variables Vobe= f1(ib , v ce ) ic = f2(ib ,vce )

ΔvBE = δf1 Δ iB + δ f1 ΔvCE δ iB δ vCE ΔiC = δf2 Δ iB + δ f2 ΔvCE δ iB δ vCE ΔvBE, Δ iC, Δ iB , ΔvCE represent small change in the base and collector voltage and currents. Rewriting the above equation vbe = δf1 ib + δ f1 vce δ iB δ vCE iC = δf1 iB + δ f2 vce δ iB δ vCE



5

All the terms shown in the parenthesis which are the co-efficients of the equations also called transistor parameter. The above equations are rewritten as vbe = hie ib + hre vce --------------------------------(1) ic = hfe ib + hoe vce---------------------------------- (2) Put vce =0 in the equation (1), vbe = ib * hieor hie = vbe/ib Ohm = ΔvBE = δvBE ΔiB δ iB Thus, hie has the unit of resistance. The terms vbe and ib represents input voltage and input current. Therefore, hie is defined as the input resistance of the transistor when the output terminals are short-circuited. Put ib = 0 in equation (2), ic = vce * hoe or hoe = ic/vce mho ΔiC = with ΔvCE =0 ΔvCE δ iC = with ΔvCE =0 or at Vc δvCE Thus, hoe has the unit of conductance. The terms ic and vce represents the output current and voltage of the transistor. Therefore, hoe is called output conductance of the transistor when the input terminals are open circuited. Put vce = 0 in equation (2), ic = ib * hfe or hoe = ic/ib no-unit ΔiC = with vce = 0 Δ ib Thus, the parameter hfe is dimensionless and is equal to ratio output current to input current. Therefore, hfe is called current gain of the transistor when the output terminals are short-circuited. It is also called simply the short-circuit current gain. Put ib =0 in equation (1), vbe= vce * hre or hoe=vbe/vce no-unit. ΔvBE = with iB = constant Δ vCE



6

δ vBE = at IB = constant δ vCE Thus, hoe is also dimensionless. The above ratio represents the reciprocal of voltage gain (voltage gain is the ratio of output voltage to input voltage). Therefore, hoe is called reverse voltage ratio of reciprocal of voltage gain when the input terminals are open-circuited. Since the units of all parameters are different, they are said to be hybrid in nature or simply h-parameters. To obtain the equivalent circuit, the equation (1) and (2) can be used to write the following circuit. The equation (1) is a KVL, hence it can be realized with a series circuit consisting of a voltage source vs , a resistance hie and a voltage source hre* vce. The equation (2) represents a KCL, hence it can be realized by a parallel circuit with three branches. One branch has a dependent current source hfe* ib , the second branch is resistor of value 1/hoe and the third branch is the total current ic.

Analysis of CE amplifier

h- parameters equivalent circuit To find current gain

hre vce -

+

iL

+

- hre vce

V vce 1/hoe

hfe ib

Ri

ib hie

+ VS -

RL

E

C

-



7

)1(−−−−=

=

ibiLAi

currentinputcurrentoutputAi

Apply current divider rule to the output circuit

Loe

fei

Loe

feL

oeL

oebfe

L

Rhh

A

eqnfromRh

hi

hR

hih

i

+−

=

−+−

=

+−=

1

)1(1

1

1

To find input resistance

)2(−−−−−=b

si i

vR

Applying KVL to input circuit Vs = hie ib + hre vce Vs = ib hie + hre iL RL from equation (1) Vs = ib hie + hre Ai ib RL ( iL =Ai ib ) Substituting in equation (2)



8

Ri = hie + hre Ai RL To find voltage gain

Av = vsvce

geinputvoltavoltageoutput

=

ib

LLv Ri

RiA = since

b

si i

vR =

i

Liv R

RAA =

To find output resistance NOTE:Replace RL by a voltage source. Replace independent sources by internal impedance of the source

)3(−−−−=C

ceO i

vR

Applying KC L to the output circuit. iC = hfe ib + i1 iC = hfe ib + vce hoe-----(4)

RL

+

vce

-

1/hoe hfe ib hre vce

ib hie

RO1

RO



9

Applying KVL to input circuit - ( hie ib + hre vce) =0

ie

cereb h

vhi −=

substituting for ib in equation (4)

oeceie

cerefeC hv

hvhhi +

−= )(

substituting in equation (3)

oeie

erfeO

hh

hhR+

−=

1

To find output resistance with RL

RO1 = RO||RL

Since RL is in parallel with the voltage source, total output resistance is the parallel combination of RL and RO



10

PR0BLEMS: 1. A common emitter amplifier has the following h- parameters. hie =1KΩ, hre = 10-

4, hfe =100, hoe = 12µmho. Find current gain, Voltage gain, Ri, Ro, power gain. Take RL = 2KΩ. Also find output power take vS = 500 mV ( rms).

2.199468.980102656.97

468.980)102656.9710()101(

656.971021012(1

1001

3

343

36

−=

××−=

=

Ω=××−×+×=

+=

−=×××+

−=

+

−=

−

−

v

v

i

Liv

i

i

Liereii

i

i

Loe

efi

A

A

RRAA

RR

RAhhR

A

A

Rhh

A



11

3

1

1

1

10453.19

656.972.199

992.125002500

||

5005.0

1

×=

−×−=×=

Ω=+×

=

+==

Ω=Ω=

+−

=

P

P

iVP

O

O

LO

LOLOO

O

oeie

refO

AA

AAAKR

R

RRRR

RRR

KMR

hh

hhR

e

WattP

P

RVAP

RvP

O

O

L

SvO

L

ceO

96.4102

)10500()2.99.1(

).(

3

232

2

2

=×

×−=

=

=

−



12

2.Repeat the above problem if the internal resistance of the source is RS = 1KΩ

Without RS (The derivation is same as the previous problem but use vbe instead of vS )

b

bei

i

Loe

ef

b

Li

ivR

ARh

hiiA

=

−=+

−==

656.971

Ω=+=

KRRhAhR

i

Lreiiei

98.0

2.199−=

=

==

=

v

b

bei

i

Li

ib

LLv

be

cev

AivR

RRA

RiRiA

vvA

iL

+ hrevce

-

vce RL 1/hoe hfeib

Ri Ris

+ vs -

RS

hie

vbe

iC B

E



13

With RS

b

sis i

vR =

By inspection, RS is in series with the entire circuit whose resistance on the input side is Ri Ris = RS + Ri Ris = 1.98KΩ To find current gain Ais Converting the voltage source in to current source

S

Lis i

iA =

Multiplying and dividing by ib

)1(

.

−−−−×=

=

s

biis

s

bLis

iiAA

ii

ibiA

iS RS Ri

ib



14

Applying current divider rule to the modified circuit.

iS

SSb RR

Rii+×

=

substituting in (1)

is

Si

iS

Siis R

RARRRAA ×

=+×

=)(

32.4998.1

1656.97

−=

×−=

is

is

A

A

To find voltage gain

64.9898.12656.97

.

−=

×−=

=

===

vs

vs

b

Sis

is

Li

isb

LL

s

cevs

A

A

ivRSince

RRA

RiRi

vvA

To find output resistance.



15

C

ceO i

vR =

Applying KCL to output circuit iC = hfe ib + vce hoe Applying KVL to input circuit -(RS + hie )ib –hre vce =0

ieS

cereb hR

vhi+

−=

Substituting in the equation for iC

oeceieS

cerefeC hv

hRvhh

i ++

−=

hfeib vce 1/hoe hrevce RS

hie

RL

iC

RO

RO1

ib



16

)1012(

10)11()10(100

1

1

6

3

4−

− ×+

+−

=

++

−=

O

oeieS

refeO

R

hhRhhR

Ω=+

=

Ω=

KRR

RRR

RLwithdecreasesRRwithKR

LO

LOO

OSO

9724.1

),(857.142

1

Question:In a transistor amplifier, explain the effect of internal resistance of ac voltage source on the performance of the amplifier. Derive the necessary equations in support of the answer. CKT DIAGRAM Let Ri is the input resistance of the amplifier without Rs- internal resistance.

voi

Amp.2 source

-

+

i

Ri

Vs

Rs



17

b

bei i

vR =

Let Ris is the input resistance of the amplifier with RS

From the above circuit b

sis i

vR =

Ris = RS + Ri ( Since Rs is in series with input circuit) Thus due to the internal resistance Rs, the input resistance of the amplifier increases. It is desirable since the input resistance of the amplifier should be high to avoid loading effect on the previous stage of the amplifier, the input resistance of the amplifier should be high. Then the amplifier will draw less input current from the previous stage. Loading effect is the drop in the input voltage of the amplifier due to the large current drawn by the amplifier. The drop in the voltage is due to the voltage drop in the internal resistance of the previous amplifier. Current gain:

Let Ai is the current gain without Rs. b

Li i

iA =

Let Ais be the current gain with Rs

iS

Liis RR

RAA+

=

From the above equation, the current gain decreases with Rs. This is not desirable Voltage gain Let AV is the voltage gain without Rs.



18

)1(−−−−−−−−−=

=

i

Liv

be

cev

RRAA

vvA

Let Avs is the voltage gain with RS

)2(−−−=

=

is

Livs

s

cevs

RRAA

vvA

Comparing equations (1) and (2), voltage reduces with Rs. Output resistance

ieS

fereoe

O

hRhh

hR

+−

=1

Due to Rs present in the denominator, the output resistance decreases. This is desirable to reduce the loading effect. 3.For the following circuit, find ac performance quantities. hie =1KΩ,hre=10-4, hfe =100, hoe =12µmho.

RE

4K

CC

RC =4K

CC

RB 10K

VS

VCC

CE



19

Considering only Ac signal and taking the frequency of ac signal to be high

fcX C ∏

=2

1

becomes very low. It is approximated to zero, or all capacitors are replaced by short circuit. AC equivalent circuit. h- parameters equivalent circuit without RB

hre vce

B

RL 2K

1/hoe hfe ib

-

+

ib hie iS

+

VS -

Rib Ri

E iL

vce

+

-

C

+

4K

RC RB VS

C

-

RL=2K



20

2.199

98.0

656.971

−===

Ω=+==

−=+−

==

i

Li

s

cev

Leriieb

si

Loe

fe

b

Li

RRA

vvA

KRhAhivR

Rhh

iiA

With RB Looking from the input circuit ,RB is in parallel with rest of the circuit.

Ω=

×=

==

KR

R

RRRR

ib

ib

ib

Biib

89.098.10

1098.010||98.0

||

Current gain:

RB

VS

iS

ib1

Ri



21

)1(−−−=

×=

=

s

biib

b

b

S

Lib

S

Lib

iiAA

ii

iiA

iiA

Applying current divider rule

iB

BSb RR

Rii

+×

=

substituting in equation 1

94.88−=+

=iB

Biib RR

RAA

Aib is less than Ai. That is due to biasing resistance RB, The current gain reduces To find voltage gain

s

cevb v

vA =



22

Since RB is in parallel with ideal voltage source the resistance RB will not affect the input voltage.∴ voltage gain remains the same. To find output resistance

848.1771694.882.199

97.1||

5001

1

=−×−==

Ω==

Ω=−

=

gainPowerAAgainpower

KRRR

K

hhh

hR

ibv

LOO

ie

efreoe

O

4.With the help of suitable circuit and mathematical proof, explain the effect of biasing resistance on the performance of the amplifier.

RL

+

vce

-


ib hie

RO1

RO

vS RB

RL 2K

1/hoe hfe ib

-

+

ib hie iS

+ VS -

Rib Ri

E iL

vce

+

-

C



23

Let Ai be the current gain without RB b

Li i

iA =

Let Aib be the current gain with RB Ri is the input resistance of the amplifier without RS

S

Lib i

iA =

S

bi

b

b

S

Lib i

iA

ii

ii

A ×=×=

Applying current divider rule,

iB

Biib

iB

BSb

RRRA

A

RRRi

i

+=

+=

From the above equation current gain decreases with RB. This is undesirable. The voltage gain, without RS ( AV) and with RB ( Avb) are equal since the biasing resistor RB is in parallel with the ideal voltage source.

RB

VS

iS

ib1

Ri



24

Output resistance.

ceoeie

refeC

ie

cereb

ceoebefC

ie

refeoe

O

c

ceO

vhh

vcehhi

hvh

i

vhihih

hhh

R

iv

R

+−=

−=

+=

−=

=

1

The equation for RO is independent of RB. Hence the output resistance of the amplifier is unaffected due to the redundancy of RB Input resistance Let Ri and Rib be the input resistance of the amplifier without and with RB respectively.

RL

+

vce

-


ib hie

RO1

RO



25

Looking from the input circuit, RB is parallel with the rest of the circuit.

iib

iBbi

RRRRR

<∴

=∴ ||

Thus the input resistance of the amplifier decreases with RB. This is undesirable. 5.

4K

150K 100K

VCC

4.2K

VS RE

Output



26

Hybrid equivalent circuit

Without RB

b

Li i

iA =

RL = 4 || 4.2 ||100 RL = 2KΩ 4K

RC 4.2K

100K 150K RB

VS

Output

Ac equivalent circuit

hrevce vS RB RL 2K

1/hoe hfe ib

-

+

ib hie iS

+ VS -

Rib Ri

E iL

vce

+

-

C



27

Applying current divider rule for the output circuit.

b

si

i

Loe

fi

Loe

bef

Loe

oebfe

L

iv

R

A

Rheh

A

Rhih

Rh

hih

i

=

−=

+

−=∴

+−=

+−=

656.97

1

11

1

Applying KVL to input circuit Vs = hie ib + hre vce vce =( iL) RL Vs = hie ib + hre Ai ib RL vce=(Ai ib) RL ∴ Ri = hie + hre Ai RL Ri =0.98 KΩ



28

Ω=×

=

=Ω=

−=

===

KR

RRRKRwith

ARR

ARiRi

vsv

A

ib

iBib

B

v

i

Li

ib

LLcev

9736.098.150

98.0150||

1502.199

To find current gain Aib

S

bi

b

b

S

Lib i

iA

ii

ii

A ×=×=

Applying current divider rule to the above circuit

RB

VS

iS

ib

Ri



29

022.9798.150

150656.97−=

×−=

+=∴

+=

ib

iB

Biib

iB

BSb

A

RRRAA

RRRii



30

PROBLEMS CONTINUED 6. +VCC R1 RC 100K 2K CC Output CC CE R2 RE 10K

VS

+

ib

VS

output

RC

R1 R2

RB = R1 || R2 = 9.09 K



31

Ri 1 Ri

is ib hie + iL + hfe ib + 1/hoe RC Vs RB hre vce vce - - - WITH OUT R B To find current gain

b

Li i

iA =

Applying KCL,

oeC

oebfe

L

hR

hih

i 1

1

+

−= =

oe

bfe

hRih

+−

1

=+−

=Rh

hA

oe

fei 1

-97.656

To find input resistance.

b

si i

VR =

Applying KVL to input circuit,

KRRAhhR

RAhhVhihV

i

cireiei

cireie

cerebies

98.0=+=+=+=



32


i

Ci

ib

cL

s

cev R

RARiRi

VVA ===

vA = -199.29 WITH R B

0.98||9.09R|| R iB

| ==iR

|iR = 0.8846KΩ

To find current gain

152.88|

|

−=

+=

=

=

=

i

iB

Bi

S

bi

b

b

S

L

S

Li

A

RRRAiiA

iiX

iiiiA

iB

BSb RR

Rii

+=

VS

iS ib

RB Ri



33

To find voltage gain. Voltage gain does not alter with BR as it is connected across an ideal voltage source.

29.199| −=vA To find out put resistance.

ieh

VhhVhi

ihVhicircuitoutputtoKCL

VhihcircuitinputtoKVL

cerefeceoeC

bfeceoeC

cerebie

−+=

−+=

−=

1/

hoe

ib hie

+

-

hre vce

iC

hfeib

RC

RO

RO1

Vce

+

-

Ω=

=

Ω=

−+

=

=

KR

RRR

KRh

hhh

R

iVR

C

ie

fereoe

C

ce

992.1

||

500

1

|0

0|

0

0

0

0



34

7. Repeat the above problem with internal resistance of the ac signal Rs = 1K. Without Rs and RB Without Rs and RB

R2 R1

RS

VS

RC

OUTPUT

RB = R1 || R2 = 9.09K

ib

R

iL

1/hoe

hrevce

-

+

vbe VS

B

C

E

hie

hfeib

RC vce

+

-

CKT NO 1



35

2.199

98.0

656.971

−====

=+==

−=+−

==

i

Ci

ib

CL

be

CeV

Cireieb

bei

Coe

fe

b

Li

RRA

RiRi

VVA

KRAhhi

VR

Rhh

iiA

All above equations to be derived With Rs and RB

S

Si i

VR =|

By inspection, RB is in parallel with the rest of the circuit (Ri ) and this parallel combination is in series with RS.

K

RRRRR

RRR

iB

iBS

iBS

8846.1R

)||(R

|i

|i

=

++=

+=

To find the current gain Rewriting

Ri RS RB RS + RB

iS1

ib

RS

is

is1

ib

RB

Ri

CKT NO 2



36

11

|

S

bi

b

b

S

Li i

iAii

iiA ×=×=

Applying current divider rule to circuit 2,

77.46

98.09.09.0656.97

)||()||(

)||()||(

)||()||(

)||(

|

|

1

−=

+×−

=

+=

+=

+=

+=

i

iBS

BSii

iSB

BSS

iSB

BSS

iBS

SSb

A

RRRRRAA

RRRRRi

RRRRRi

RRRRii

To find voltage gain Use circuit 1,

BiS

iSSbS

iBS

BSS

si

bCi

b

b

iS

L

S

Ce

V

RRRRRiii

RRRRRi

iRiRA

ii

RiRi

V

VA

+==

+=

−−−−−−−=×==

)||()||()(

)||()||(i

2,circuit From

)1(

11b

||

|

Applying current divider rule to circuit 2,



37

55.93A

)98.009.9(09.9

8846.12656.97

A

(1)in thisusing

|V

||

V

−=

+××−=

+=

+=

iB

B

i

Ci

iB

BSb

RRR

RRA

RRRii

To find output resistance

KRR

RRRR

RBS

BS

9.009.9||1

==

+=

hrevce

hie

RB

RS

ib

+ vce

-

1/hoe

hfeib

RC

ib

hie

hrevce

R

vce

1/hoe

hfeib



38

Ω=

==

Ω=+

−×=

+−

=

=

−−

KR

RRR

KR

hRhh

hR

iVR

C

ie

fereoe

C

ce

973.1

2||437.148||

1401900.0)10(1001012

1

1

|0

0|

0

0

46

0

0

Approximate h-parameter equivalent circuit. hre is the reverse voltage gain or reciprocal of voltage gain. For an amplifier, voltage gain is high. There fore hre is very low. For approximation, take hre Vce because zero or the voltage source is replaced by short circuit.

In the h-parameter equivalent circuit, hoe

1 is the internal resistance of the current

source hfe ib. For an ideal condition, the internal resistance must be infinity. Because hoe

1 =

infinity or hoe =0Ω or the corresponding branch is replaced by open circuit. With the above approximations, the circuit is called approximate h-parameter equilent circuit .

8. Repeat the previous problem using approximate h-parameter equilent circuit .



39

Without Rs and RB

100−=

−=

=

i

fei

b

Li

AhA

iiA

b

si i

vR =

Since hre =0, the input circuit contains only one resistance hie. Therefore Ri = hie = 1KΩ

200−====i

Ci

ib

CL

s

ceV R

RARiRi

vvA

With RS and RB

Ri

RB

+ VS

-

RS iS

ib hie

Ri 1

iL

RC

vce

+

-

RB =9.09K RC = 2K RS = 1K



40

)||(1iBSi RRRR +=

Ω= KRi 9.11

1

1

1

1

1

1

*

s

bii

b

b

S

Li

s

li

iiAA

ii

iiA

iiA

=

=

=

iBS

Bsii RRR

RRAA

+=

)||()||(1

iBS

BSsb RRR

RRii

+=

)||()||(1

200

4.4719.0

)9.0(100

1

!

!

−==

−=+

−=

s

cev

i

i

vvA

A

A


iS1

Ri

RB

RS

iS1

iS ib

RB

hie

ib

RS

vce

hfeib

RO

RO1

iC



41

C

ceO i

vR =

Since hoe = 0Ω , ∞=oeh1

Therefore, the output resistance RO = ∞ Ω .

RO

11 = RO || RC RO

1 ≈2KΩ 9. For the following circuit, find ac performance quantities.

RE

Rc2

R4

R3

CE

RE

RC1

R2

R1

CE

Vcc



42

RC1 = RC2 =5K R1 =100K R2 =10K R3 =80K R4 =8K hie =1K

9999.0==

=

efhβα

Since hre, hoe values are not given for approximation. Assume the data as zero that is use approximate h-parameter equivalent circuit Ac equivalent circuit,

Rc2

R4

R3

Rc1

R2

R1

O/P



43

RB1 = 9.097K R= RC1 || R3 || R4 R= 2.96 K Application h-parameter equivalent circuit.

hie

RB1

ib1

is

Ri1 Ri

B1

R

hfeib1

ib2

hie C1 B2

RC2 hfeib2

C2

iL

E2 E1

RC2

R4 R3 RC1

R2

R1

RB1

Output



44

Without RB1 To find current gain

)1(1

−−−−−=b

Li i

iA

iL = - hfe ib2

ie

bfeb hR

XRihi

+−= 1

2 Applying current divider rule between collector of first transistor and base of second transistor. Substituting in equation (1)

7326196.2)96.2(992

2

=+

=

+=

i

i

ie

efi

A

A

hRRh

A


1b

si i

vR =

Since hre is zero, the input circuit contains only one resistance hie

Ω==∴ KhR iei 1 To find voltage gain

i

Ci

ib

CL

s

cev R

RARi

Riv

vA 2

1

22 ===



45

36630157326

=

∗=

V

V

AKKA

With RB1

KR

RRR

iVR

i

iBi

s

si

9.0

||1

11

1

=

=

=

To find current gain Current divider rule

iB

Bsb RR

Rii+

=1

11

44.6600

)1097.9(097.97326

11

1

1

1

11

11

1

=

+×

=

+=

=

×=

i

i

iB

Bii

S

bii

i

A

A

RRRAA

ii

AA

ibib

iSiLA



46


s

ceV v

vA 21 =

Voltage gain remains same as the resistor RB1 is connected across the ideal voltage source.

Av1= 36630


hie

RB

ib1 iB2

hie

R

hfeib1 RC2

+

vce2 -

hfeib2

iC2

RO1

RO

RB1

VS

iS

ib1

Ri



47

2

2

c

ceO i

vR =

ib1, hfe ib1, ib2, hfe ib2 =0 Since there is ideal current source in the output circuit, output resistance Ro =∞

KRRRR

RRR

RRR

Co

CO

COO

COO

5

111

111||

2

2

2

21

==

+∞

=

+=

=

10. hie = 1K hfe =100 RE is not bypassed.



48

Output

VS

RC 2K

R2 10K

100K

R1

RE 1K

VCC

+

-

RE RC

R1||R2

RB

VS

ib

VO

hfe ib

RC

ib + hfe ib

RE

RB

+ -

ib hie iS

Ri1

vS

Ri

iL

vO



49

RB = 9.09K Without RB To find current gain

100−=−=−

==ef

b

bef

b

Li h

iih

iiA


b

si i

vR =

Applying KVL, vs = hie ib + ( ib + hfe ib ) RE

Ri = hie + ( 1 + hfe ) RE Ri = 102KΩ TO find voltage gain

96.1102

2100 −=×−=

===

v

i

Ci

ib

CL

S

Ov

A

RRA

RiRi

vvA

with RB

iBi RRR ||1 =

Ω= KRi 3462.81

iB

Bi RR

AiRibibAi

ibib

iSiLA

+==×=1

RB

V

iS

ib1

Ri



50

1825.810209.9

)09.9(1001 −=+

−=iA

96.11 −==S

OV V

VA iB

Bsb RR

Rii+

=


Since there is no source on the input side , ib =0.

∞==

==∴

C

OO

Cbef

ivR

iih 0

In the above problem, what is the effect of connecting a bypass capacitor across RE ?? With emitter bypass capacitor, RE becomes redundant or mathematically RE = 0. Substituting RE = 0 in the above equations.

Ai= -hfe not affected since it does not depend on the RE value or emitter bypass capacitor does not affect current gain.

+

vo - RE

hie

ib hfe ib



51

Ri = hie + ( 1 + hfe ) RE Ri = hie Ri = 1KΩ With the emitter bypass capacitor, input resistance decreases. An amplifier should have high input resistance. But with bypass capacitor, Ri decreases. ∴Bypass capacitor should not be connected to get high input resistance.

2001

)2(100

−=

−=

=

V

i

CiV

A

RR

AA

The voltage gain increases due to reduction in the input resistance. ∴ Emitter bypass capacitor should be connected to get high voltage gain. ∞=0R not affected. ∴Bypass capacitor will not alter the output resistance. Thus the function of RE and CE is to provide high dc stability ( -||- behaves like open circuit for dc) and to get high voltage gain. For ac signal of mid frequency, reactance of the -||- becomes very low and capacitor behaves like short circuit.



52

High Input Resistance Amplifiers.

1. EMITTER FOLLOWER Draw the circuit of emitter follower. Find ac performance quantities. What are the limitations of emitter follower circuit?? Take R1=100K R2 = 10K RE = 1K hie=1K hfe=100

RE

R1

R2 Output

VCC



53

Ac equivalent circuit.

App h-parameter eq.ckt

RE R1||R2

VS vO

Ri Ri1

RB RE

+ -

iS

ib hie

ib + hfe ib

hfe ib

vS iL +

vO -



54

RB = 9.09K WITHOUT RB BUFFER High current gain impedance matching Logic circuit

Current gain

10110011 =+=+=+

=

=

feb

bfebi

b

Li

hi

ihiA

iiA

Input resistance

b

Si i

VR =

Applying KVL to input

Ω=++=

++=

++=

KR

RhhRRihihV

i

Efeiei

EbfebieS

1021)1001(1

)1(

)(

Voltage gain

( ) 99.0102

11010 =====i

Ei

ib

EL

SV R

RARiRi

VV

A

WITH RB

Ω=

==

KR

RRR

i

iBi

3462.8

102||09.9||

|

|

To find current gain

RB

Vs

iS

ib

Ri



55

264.810209.9

)09.9(101|

|

=+

=

+=

=×=

i

iB

Bi

S

bi

b

b

S

Li

A

RRRA

ii

Aii

iiA

iB

BSb RR

Rii

+=


99.0| ==S

oi V

VA


o

oo i

VR =

Applying KCL to output terminal (E) io+ hfe ib + ib = 0 Apliying KVL to input circuit -ib hie -Vo= 0 Vo = -ib hie

Ω=+

=+−

−= 9.9

1)1( ie

ie

feb

iebo h

hhi

hiR

Limitations of Emitter Follower

1. Input resistance Ri (102 KΩ) is very high for an emitter follower circuit. But when biasing network is considered (RE), input resistance reduces. .|||

ii RRR B= Thus the input resistance is

limited by BR , for any value of .iR Therefore the biasing network defeats the purpose of using emitter follower.

2. Current gain of emitter follower is very high [Ai = 1+ hfe]. But due to biasing network, the overall current gain |

iA is reduced.

iB

Bii RR

RAA

+=|

The above 2 limitations can be over come by using Boot strapping Technique.

iO +

vO -

ib

hie

hfe ib



56

BOOT STRAPPED EMITTER FOLLOWER For dc condition. Capacitor behaves like open circuit. DC equivalent circuit.

IB

RE

R

R2

R1

VCC

VS

R

CC

RE R2

R1

CC

Output



57

This is a voltage divider circuit. With an additional resistor R. assuming IB and R to be very small, the voltage drop across R can be neglected and circuit resembles standard voltage divider bias. Hence the biasing condition is unaltered. For ac condition

The capacitor behave like short circuit. fc

X C π21

= , if the frequency is in mid frequency range.

The boot strap resistor R is directly connected to output terminal E. therefore, the input VS is connected to output terminal through the resistance R. This method of connecting input and output terminals only under ac condition is called boot strapping STATE AND PROVE MILLER’S THEOREM

Consider an amplifier with a resistance ( or capacitance ) connected across input and output terminals. Statement: Miller’s theorem states that when a resistance ( or capacitance ) is connected across input and output terminals, the same can be replaced by two independent resistances ( or capacitances ) connected one across the input terminals and the other across output terminals. These are called Miller equivalent resistances ( or capacitances).

R

VS

-

RL +

vS1

-

RS

Amplifier A

VO

+

i



58

PROOF: Applying Ohm’s law

SOS

O

OS

AVVThenVV

RVV

i

==

−−−−−−−−−−−

=

ALet

)1(

Substituting in (1)

)2(0

1

01

)1(

−−−−−−−−−−−

=

−

−=

−

=

−=

−=

MI

S

S

S

S

SS

RV

AR

VA

RV

RAV

RAVV

i

comparing equations (1) and (2), the same current i will flow through a resistance MIR when it is between input terminal and ground. Therefore, the KCL at the input terminal is not affected by replacing the resistance R by MIR . MIR is called Miller equivalent resistance on the input side. To find Miller equivalent resistance on the output side.

R

-

RL +

vS1

-

RS

Amplifier A

VO

+

i



59

1(4);-----------

0V1

RAV

1V11V

RA

VV

AVVthen VV

ALet

)3(

O

O

OOO

O

SOS

O

−=

−=

−

=

−

=

−

=−

=

==

−−−−−−−−−−−−−−−

=

ARAR

Ri

A

RA

A

RAi

RVV

i

MOMO

SS

MOR is the Miller equivalent resistance on the output side.

The same current I will flow through the resistance MOR if R is replaced by MOR which is connected

across output terminal and ground. ∴Replacing R by MOR will not affect KCL at the output terminal. ∴Replacing R by Miller equivalent resistances, the circuit is as follows.

i

RMI

+

-

RMO RL +

vS1

-

RS

Amplifier A

i

+

-

Vo

+

-



60

Draw the circuit of Boot strapped Darlington Emitter Follower and find ac performance quantities.

R1=100K R2 = 10K RE = 1K hie=1K hfe=100

Dual of Miller’s Theorem Consider a general amplifier as shown below. In this amplifier, the resistance R is common to the input and output circuits do not have a common resistance. This resistance R can be removed so that the input and output circuits do not have a common resistance. Removing of the resistance R affects the KVL equation on input and output side. The circuit must be modified as explained follows so that the removal of resistance R should not affect the KVL equations. The purpose of the following analysis is to remove the inter dependence of input and output circuits. So that either input circuits or output circuits can be solved independently.

C

-

+

i1

Vi

-

+

i2

vO1

+

RL

-

+

R

RS

Amplifier

vO

iL

VS

-



61

Applying KVL to input circuit

0)( 211 =+−−− RiiVRiV iS S

But i2 = -iL

RARiRVRiV

RAiVRiVRiAiVRiV

RiiVRiV

iMI

MIiSS

iiSS

iiSS

LiSS

)1(0

0)1(0)(

0)(

11

11

111

11

−==−−−

=−−−−=−−−−

=+−−−

From the above equation, RAi )1( − is the resistance which when connected in series with the input circuit instead of R will not affect the input circuit. Repeating the above analysis for the output circuit.

RA

AR

RiVV

RA

iVV

RA

iVV

RiAiVV

RiiVV

iiButRiiVV

i

iMO

MOLOO

iLOO

iLOO

Li

LOO

LOO

L

OO

−=

=+−

=−+−

=−−−

=−−−

=−−−

−==+−−

1

0

0)11(

0)11(

0)(

0)(

0)(

|

|

|

|

1|

2

21|

We know that

i

LL

Li

Aii

iiA

=

=1

From the above equation, the resistance MOR when connected in the output circuit alone does not alter the KVL equation. Thus, without affecting the electrical nature of the circuit, the circuit can be redrawn a



62

1. Solve the following circuit using Dual of Millers Theorem.

Ac equivalent circuit

i1 RL

RMO = R( Ai – 1) Ai

RMI = R( 1-Ai) RS

Amplifier iL

VO

+

-

VS

RC = 2K

VCC

RE = 1K

VO

hie = 1K hfe =100

RC

RE

iL

vO

-

+



63

Current gain

100−=−=

−=

=

fei

bfeL

b

Li

hA

ihiiiA

To find MIR and MOR

( )( ) K

AA

RR

KAR

i

iEMO

iE

01.11001011)

1(

10110011)1(R MI

=−−

=−

=

=+=−=

iL hfe ib

VS

hie ib

RE

RC VO

hfe ib

Ri VS

RMI hie

RL = RC + RMO = 3.01K

iL ib



64


b

Si i

VR =

Applying KVL to input circuit

KRhRR

hRiV

i

ieMIi

ieMIbS

1021101

)(

=+=+=

+=


95.2102

01.3100

−=

−====

V

i

Li

ib

LL

S

OV

ARRA

RiRi

VV

A


hfe ib +

VO -

iC hie RMI

ib



65

C

OO i

VR =

Since there is no source on the input side, Ib ic = hfe ib ∞=∴ OR 2. Repeat the above problem if hie = 1K hre = 10-4 hoe = 12 μ hfe= 100

b

Li i

iA =

Appling current divider rule in the output circuit

VS vce (RMO + RC) = RL 1/hoe

hie ib

hre vce

hie

RMI

ib iL



66

( )

( )

( )

( )

5135.96)103(10121

100101012

)(1

1

]1

[

]1

[1

][11

11

1

36

36

−=××+−××

=

++

−=

−=−++

−=−

++

−++

−=

++

−=

+

−=

+

−=

+

−=

−

−

i

i

ECOe

feEOei

feiEOeCOeii

fei

iECOeii

i

iECOe

fei

MOCOe

fe

OeL

fei

OeL

bfe

OeL

Oebfe

L

A

A

RRhhRh

A

hARhRhAA

hA

ARRhAA

AA

RRh

hA

RRhh

hRh

A

hRih

hR

hih

i


b

Si i

VR =

Ω=Ω=

−=

=−=

KRKRAAR

R

KRARR

L

MO

i

iEMO

IM

iEMI

01.301.1

)1(5135.97

)1(

Applying KVL to input loop

Ω=−++=

++=++=++=

−

KR

RAhhRRRiAhhRi

VhhRiV

i

LireieMIi

LbireieMIb

CereieMIbS

4844.98)01.3()5135.96(1015135.97

)()(

4



67

Lbice

LLce

RiAvRiv

==


.9497.2484.9801.35135.96

−=

×−====

V

i

Li

ib

LL

S

CeV

ARRA

RiRi

VV

A


KCL to output

ceoebfeC Vhihi += KVL to input

( ) 0=−+− CereieMIb VhhRi

( )

( )

Ω=

+−

=

++

−=

=

K

hRhh

hR

VhhRVhh

i

iV

R

ieMI

refeoe

O

ceoeieMI

cerefeC

C

CeO

044.841

+

vce -

1/hoe

hie ib

hre vce

hie

RMI

ib

iC



68

HYBRID-π EQUIVALENT CIRCUIT To obtain Hybrid-π Equivalent circuit Consider a PNP transistor as shown above. The emitter current IE is divided in to base current IB and a component αIE of the collector current. This division of current takes place in the entire base layer at infinite number of points. For mathematical convenience, it is assumed that the division of current takes place at an imaginary terminal B1. rb

1e: It is the resistance of forward biased base to emitter junction and it is the resistance offered to the flow

of the current IE. rb

1c: It is the resistance of reverse biased collector to base junction. The flow of current in this resistance

represents the reverse saturation current Ico due to flow of minority charge carriers. rbb

1: It is the resistance of the base layer for the flow of the current IB. This is called base spreading resistance because the division of emitter current is spread across the entire region.

IC

Cb1e

rb1e

B1

Cb1C

rb1c

ICO

rbb1

IB

IE

C

E

P

N

P

B



69

αIE: This is the current in the collector due to transistor action. When charge carriers reach the base layer from emitter, the potential gradient at the collector junction will result in the movement of the charge carriers in to the collector. This forms the current. αIE depends on the emitter current IE which inturn depends upon the voltage across base to emitter junction.

]1[1

−=T

EVB

OE VeIIη

Therefore, the voltage VB1

E controls αIE. VB1

E is the independent variable. This depends on charge carrier concentration and temperature. cb

1e and cb

1c: This is the stray capacitance across the two P-N junction. The reactance of the capacitor is

very high at mid-frequency. Hence approximately, capacitors are replaced by open circuit (not considered).

But for high frequency, fc

X C ∏=

21

the reactance becomes finite. Hence considered in the analysis.

All the above terms are called Hybrid-π parameters. These parameters can be represented by the following circuit and it is called Hybrid-π equivalent circuit or Giacollette equivalent circuit. gm vb

1e is the component of collector current(αie) expressed as a function of independent variable vb

1e. gm

is the Transconductance of the transistor. This represents ability of the transistor in transforming the input voltage vb

1e in to output current.

rce: rce is the internal resistance of the current source.

Cb1e

ic

gmvb1e

rb1C

rbb1 B1

Cb1C

E

C B

ib

rb1e

rce



70

To find Hybrid-π parameters. Hybrid -π equivalent circuit: Let the output terminals be short circuited . Considering mid- frequency, reactance of all capacitors becomes infinite. Therefore, all capacitors can be replaced by open circuit. rb

1c is the resistance of reverse biased collector junction whose value is very high. Therefore it can be

approximated to open circuit. rce is short circuited, becomes redundant. Hence can be removed.

vs

Cb

1e

ic

gmvb1e

rb1

C

rbb1

B1

Cb1C

E

C B

ib

rb1e

rce

B11

rbb

1

rb1e

Vb1

e

vs

B ib

E

gmvb1e

C

Vce=0



71

To find gm

EB

Cm

eb

cm

ebmc

VI

g

vi

g

vgi

1

1

1

∆∆

=

=

=

where ΔIC and ΔVB

1E are the changes in the currents and voltages around quiescent condition.

)1(1

−−−=EB

Cm dV

dIg

We know that IC = αIE + ICO Since ICO is very small and α is very close to unity,

)2(][

]1[

1

1

−−−≈

−==

≅

T

EBOC

T

EBOEC

EC

VV

eII

VV

eIII

II

η

η

Differentiating with respect to VB1

E

)2()3(

1][

1

1

1

fromVI

dVdIC

VVV

eIddI

T

C

EB

TT

EBO

EVB

C

−−−=

=

η

ηη



72

600,11273+

=CtV

o

T

If t = 27oC

126

026.011600

300

==

==

ηmVV

VV

T

T

(3) mvI

dVdI C

EB

C

261

=

substituting in (1)

mVmAI

g Cm 26=

In general T

Cm V

Ig

η=

In the above equation, IC represents the dc collector current or quiescent current. Its value can be found graphically by drawing the dc load line, locating the Q point on the load line and then measuring IC. OR if know the biasing arrangement of the transistor, then the circuit can be solved using biasing technique and then IC can be calculated. To find rb

1e

From the two port network theory, we know that

vse = ib hie + hre vce -----(4) ic = ibhfe + hoe vce -----------(5) OR

-

ib

hie

hrevce oeh1

hfeib

vce vs

+

-

+

+

-



73

From equation –(5)

)6(0 −−−== ceb

cef whenv

ii

h

In the hybrid π equation circuit, VCE is already 0. Therefore obtain the ratiob

c

ii

From hybrid π equation circuit and equate it to equation (6).

ebmb

c

cmebb

cmeb

rgii

igri

igv

1

1

1

=

=

=

Equating to equation (6).

hfe =gm rb1

e

To find rbb1 From equation (4)

fe

meb

m

feeb

hg

g

gh

r

=

=

1

1



74

)7(0 −−−== ceb

sie whenv

iv

h

From the hybrid-π equivalent circuit, applying KVL to input circuit. Vs = ib(rbb

1 + rb1

e)

ebiebb

ebbbie

ebbbb

s

rhr

fromrrh

rriv

11

11

1

)7(

1

−=

+=

+=

ebiebb rh

g1

1

1−

=

To find rb1

c Rewriting the hybrid π equivalent circuit by neglecting all capacitances( open circuit) From equation (4)

)8(0 −−−−== bce

sre iwhen

vv

h

Taking ib = 0 in the hybrid-π equivalent circuit, since there is no voltage drop across rbb1, vs = vb

1e.

+

iC

vce rce

gmvb1e

B +

ib=0 i2

rb1e

rbb1

B1 rb

1c

Vb1

e

vs

- -

CKT (2)



75

Substituting in equation (8).

)9(01

−−== bce

ebre i

vv

h

From the hybrid -π equivalent circuit. Applying voltage divider rule to circuit(2).

ebre

ebcb

cbeb

eb

cbeb

eb

ce

eb

cbeb

ebceeb

rhr

r

rrr

hreinngsubstituti

rrr

vv

rrrv

v

1

1

1

11

1

11

11

11

1

1

),9(

−=

+=

+=

+=

rb

1e is the resistance of the forward biased junction and rb

1c is the resistance of the reverse biased junction.

Therefore rb1

e can be neglected in the denominator.

re

ebcb

cb

ebre

cb

eb

ce

eb

hr

r

rr

h

rr

vv

1

1

1

1

1

11

=

=

=



76

To find rce From equation—(5)

)10(0 −−−== bce

coe i

vi

h

Applying KCL at the output terminal

ic = i1 + gm vb1

e + i2

cbeb

ceebm

ce

cec rr

vvg

rv

i11

1+

++= substituting in the above equation

eb

cecerem

ce

cec r

vvhg

rv

i1

++=

Since rb1

e << rb1

c, rb1e + rb1

c ≅ rb1

c

remcb

oe

ce

ceermcb

oece

remcbce

oe

remcbcece

c

remebce

cec

hgr

hr

ghgr

hr

hgrr

h

equationingsubsitutin

hgrrv

i

hgrr

ivi

−−=

=−−=

++=

++=

++=

1

1

1

1

1

11

11

11)10(

11

]1[

To find Cb

1C



77

Cb1

C is the junction capacitance of reverse biased collector to base junction. When a PN junction is reverse biased, the width of the depletion layer increases and capacitance decreases. Therefore Cb

1C is very low of

the order of few pico farads. To find Cb

1e

This is the capacitance of forward biased PN junction. When a PN junction is forward biased, width of the depletion layer decreases and capacitance increases.

Cb1

e + Cb1

C =T

m

fg∏2

Where fT is called the transition frequency.

fT = hfe fβ

fβ is called upper cutoff frequency.

fβ = ][21

111 cbebeb ccr +∏

Problem: A transistor amplifier is operating with a dc condition of (10V,10mA). The operating temperature is 300C. The H-parameters of the transistor are hie =1Ko, hre =2.5X10-4, hfe=100, hoe=25X10-5mho. Calculate hybrid-π parameters given that CC=3PF. Take fT=1MHz.

mhomgmhog

mVmAg

VIg

m

m

m

T

Cm

1.3833831.0

1.2610

).1(

==

=

=η



78

mVVVV

V

CtV

SforGfor

T

T

T

T

i

e

1.260261.0

1160027330

11600273

21

0

==

+=

+=

==

ηη

Ω=

Ω=

=

=

Kr

r

r

gh

r

eb

eb

eb

m

feeb

261.0

2613831.0100

).2(

1

1

1

1

Ω=

−=

−=

Kr

r

rhr

bb

bb

ebiebb

739.0

261.01

).3(

1

1

11

Ω=

=

=

−

KrX

r

hr

r

cb

cb

re

ebcb

1044105.2

261.0

).4(

1

1

1

1

4



79

Ω=

−−=

−−=

−−−

−

Kr

XXXXr

hgr

hr

ce

ce

remcb

oe

ce

524.6

)105.2101.83.3(1044101025

1

11).5(

433

5

1

pFc

ppc

pXc

XX

c

cf

gmc

fgmcc

eb

eb

eb

eb

cbT

eb

Tcbeb

60967

360970

310097.6

)103(102

3831.02

2)6(

1

1

1

1

11

11

8

126

=

−=

−=

−∏

=

−∏

=

∏=+

−

−

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