Lecture 08 hibridequivalentmodel

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Objective :To understand the Hybrid Equivalent model of Transistor.Hello! Students I hope now you are well versed with theconcepts of transistor re-model and the different configurationthat we have studied in the last lectures. Now in today’s class weare going to discuss the hybrid-model for the various configura-tion.To analyze the behavior of transistor amplifier with the help ofβ and the values of resistances used in the circuit, is not soaccurate. It is because the input and output circuits of theamplifier are considered to be completely independent andsome of the effects are ignored such as IC is taken as constantbut actually its value depends upon load resistance.Therefore, more accurate method to analyze a transistoramplifier is hybrid parameters (h- parameters) method, we canstudy h-parameter at low frequency and at high frequency. Sothis chapter first introduces h- parameter at low frequency and athigh frequency. So this chapter first introduces h – parameter atlow frequency and after that hybrid-π model (for high fre-quency).

Hybrid Parameters (For Low Frequency)The four * parameters which are used to analyze any linearcircuit having input and output terminals are called hybrid orh-parameters.Meaning of hybrid is ‘mixed’. Since these parameters havemixed dimensions, they are called hybrid-parameters.

Determination of h-parameters

Consider a linear circuit having two input and two outputterminals (See Fig. 2.29). Their input and output voltage andcurrents are labeled with their positive directions. The conven-tions used are the standard one which may not correspond tothe actual directions. However, while analyzing the circuit, if anydirection is opposite that may be considered as negative.

LINEAR CIRCUIT

Fig. 2.29In Fig. 4.1, the voltages and currents can be related by thefollowing set of equations: v1= h11 i1 + h12 v2.…(i) i2= h21 i1 + h22 v2.…(ii)Where h11, h21, h12 and h22 are fixed constants and are known ashybrid-parameters. These parameters relate the four variables i.e.i1, i2, v1 and v2 by the above-said two equations.

If we look at the equation. (i), it is clear that h11 has the dimen-sion of ohm whereas h12 , has no dimension. Similarly, if welook at the eqn.. (ii), h2l has no dimension but h 22, has thedimension of mho. Hence, the four parameters are named ashybrid (having mixed dimensions) parameters.These parameters can be determined very easily as explainedbelow:(i) Short-circuit the output terminals, as shown in Fig. 2.30(a),

the output voltage reduces to zero i.e. v2, = 0. Substitutingthis value in eqns. (i) and (ii). we get

v1 = h11 + h12 × 0

Fig. 2.30 (a) and (b)

or h11 = 1

1

tv

(output short circuited; v2 = 0) ….(iii)

and t2 = h21 i1 + h22 × 0

or h21 = 1

2

ii

(output short circuited; v 2 = 0) ….(iv)

Here, h11 is called input impedance (i.e. v1 / ii ) with outputshorted and h 2l is called current gain (i.e. i2 ./ i1) with outputshorted.(ii) Open circuit the input terminals, as shown in Fig. 2.30b, the

input current reduces to zero i.e. i1 = 0. Substituting thisvalue in eqns. (i) and (ii), we get,

v1 = h 11 × 0 + h 12 v2

or h12 = 2

1

vv

(input opened; i1 = 0) ….(v)

and i 2 = h21 × 0 + h22 v2

or h22 = 2

2

vi

(input opened; i1 = 0) ….(vi)

Here, h12 is called voltage feedback ratio (i.e. v1/v2) with inputterminals open and h 2 is called output admittance (i.e.i2 / v2)with input terminal open.

LESSON 8:(HYBRID EQUIVALENT MODEL)

UNIT - 2 (SMALL SIGNAL ANALYSIS FOR BJT :

SINGLE STAGE AND MULTISTAGEAMPLIFIER)

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YHybrid-Parameter Equivalent Circuit

Fig. 2.31 &2.32A two port linear circuit is shown in Fig. 2.31. The voltages andcurrents of the circuit can be expressed in terms of h-parametersby the expressions;

V1= h11 i1 + h12 v2

...(i)

i,2 = h21 i1 + h22 v2

...(ii)

The h-parameters’ equivalent circuit is shown in Fig. 2.32. Theinput circuit (or port) is derived from the exp. (i). Here, inputimpedance (resistor) h is connected in series with a voltagegenerator h12 v2. The output circuit (or port) is derived from exp.(ii) it involves current generator h 2l v2. and shunt resistor h 22.Example.1 Determine the h-parameters of the circuitshown in Fig. 2.33.

Fig. 2.33 & 2.34Solution. To determine h- parameters of the circuit proceed asfollows:(i) Short-circuit the output terminals as shown in fig 2.34 can

determine h11 and h21.

Since 10 Ω resistor is short circuited h11 = 2 ΩNow, current i1 flows into the box and the same current flowsout of the box. ∴ i2 = i1.

And h21 = 11

1

1

2 −=−

=ii

ii

(ii) Open the input terminals and make the arrangement asshown in Fig. 2.35. It may be noted that output terminalsare driven by voltage v2 and no current flows through 20Ω resistor. Hence, the voltage across 10 Ω resistor i.e. v2

reaches across the input terminals. ∴ v1 = v2

Figure 2.36

and 12

2

2

1 ==vv

vv

Output impedance = 10 Ω∴ Output impedance,

h22 = 01.0101

mh=Hence the various h-parameters of the circuit are h11 = 20 Ω ; h21 = -1 h12 = 1; h12 = 0.1 mhoExample.2 To determine the h-parameter of the circuit shownin Fig 2.36Solution. To determine h- parameters of the circuit proceed asfollows:(i) Short circuit the output terminals as shown in Fig. 4.10.

h11 = 6 + 8 | | 8 = 6 + =+×

8888

10Ω

Now, the input current i1 is divided equally at the junction.

Fig.2.36

∴ i2 = 11 5.02

ii

−=−

h21 = 5.05.0

1

1

1

2 −=−=−i

iii


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Fig.2.37&2.38(ii) Open the input terminals and make the arrangement as

shown in Fig. 2.38. Here, no current flows through 6 ilresistor as the output terminals are driven by voltage v2.

v1 = 222 5.08

168

88v

vv=×=×

+

∴ h12 = 5.05.0

21

2

2 ==v

vvv

Impedance looking into the output terminals = 8 + 8 = 16 Ω

h22 = 161

= 0.0625 mho

Hence, the various h-parameters of the circuit areh11. = 10 Ω ; h21 = - 0.5 ; h12 = 0.5 ; h22 = 0.0625 mho

Performance of a Linear Circuit In

H- ParametersIt has already been seen that a linear circuit has a set of h-parameters. Now, we shall study the performance of such acircuit by developing expressions for input impedance, currentgain, voltage gain etc. in terms of h-parameters.

Figure 2.40

Consider a circuit having load resistance RL across its outputterminal as shown in Fig. 2.39.

Input ImpedanceThe ratio of input voltage to input current is called the inputimpedance Z in.

∴ Z in.= i

i

iv

or Z in.= 1

212111

ivhih +

(∴ v1= h11 i1 + h12 v2)

or Zi n. = h11 +1

212

ivh

….(i)

Now, i2 = h21 i1 + h22 v2

or

−=∴+=

−

LL iv

ivhihi

v 2*2222121

2

or -h21 i1 = v2 =

+

Lrh

122

The ratio of output current to input current is called currentgain Ai.

Zin = h11 -

Lrh

hh1

22

2112

+ …..(iii)

We know, i2 = h21 i1 + h22 v2

and v2 = -i2 RL.(from output circuit) i2 = h21 i1 + h22 (-i2 RL) i2 = h21 i1 + h22 rL i2.

or i2 (1 + h22 RL) = h21 i1.

or Lrh

hii

22

21

1

2

1+=

Substituting the value of 1

2

ii

in eqn.(iv), we get,

A1 = Lrh

h

22

21

1+ …..(v)

If h22 rL << 1 then Ai ≅ h21.Voltage GainThe ratio of output to input voltage is called voltage gain Av.

Av = 1

2

vv

……(vi)

or Av = inzi

v

1

2

( )11 inZiv =Θ


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Substituting the value of 1

2

iv

from eqn. (ii), we get,

Av = in

L

Zr

h

h

+

−

122

21

H- Parameters of a TransistorA transistor in a three–terminal device. If any one of theterminals is made common to the input and output, it willhave two ports. Thus it will have two input terminals and twooutput terminals (See Fig.2.40), for a small ac signal transistorbehave as a linear device, hence it can be described in terms of h-parameters.The voltages and currents of the circuit can be related by thefollowing sets of equations:

Figure 2.40

V1 = h11 i1 + h12 v2 ….(i) i2 = h21 i1 + h22 v2 ….(ii)where the various h-parameters are;

h11 = 1

1

iv

02=v = Input impedance (with

output shorted) = h i (in ohms)

h21 = 021

2=vi

i= Forward current ratio (with

output shorted) = h f(no unit)

h12 = 012

1=iv

v= Reverse voltage ratio (with

input open)= hr(no unit)

h22 = 012

2=iv

i= Output admittance (with

input open=* h0(in mho)While considering the behaviour of ransistor in terms of h-parameters, the following points need attention:i) The value of h-parameters of transistor will depand upon the

transistor connection (i.e. CE, CB or CC). Therefore. h-

parameters for different connections are abbreviated indifferent way. For instance, h11 is represented as **h ie, h ib andh ic for CE, CB and CC connections.

ii) While checking the performance, ac output resistance isconsidered as load resistance i.e.

LC

LCLCACL RR

RRRRRr

+===

The values of h-parameters depend upon the Q point. If Qpoint changes, the values of h-parameters are also changed.In transistor circuits the values of voltage and currents are takendepending upon transistor configuration. For example, for CE

configuration cce

bbe

IiVv

IiVv

==

==

22

11

;

;

where Vbe, Ib, Vce and Ic are the rms values.The nomenclature used for the h-parameters of a transistordepending upon its connections is given in the following table4.1.Table 2.1

S. No. h-parameters CE

Configuration

CB

Configuration

CC

Configuration

1 h11 hie hib hie

2 h12 hre hrb hrc

3 h21 hfe hfb hfc

4 h22 h0e h0b h0c

The typical values of h-parameters of a 2N 3904 transistor aregiven below:hie = 3.5 k Ω ; hre =1.3 x 10-4hfe = 120; hoe = 8.5 µ mho*It may be noticed that subscript used is the first letter of thedescription i.e. input, forward, reverse and output respectively.** It may be noticed that the second letter of the subscript usedindicates the type of transistor connections.

Performance of (Ce Circuit) Transistor in

H-ParametersWhile studying the performance of a transistor, we are inter-ested in the following terms:

i) Input ImpedanceThe general expression for Zin is

L

in

rh

hhhZ

122

211211

+−=

Substituting the values of h-parameters for transistor in CEconfiguration. We get,


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Le

fereiein

rh

hhhZ

10 +

−=

(ii) Current gainThe general expression is

Li rh

hA

22

21

1+=

Substituting the values of h-parameters for transistor in CEconfiguration, we get,

Le

fei rh

nA

01+=

(iii) Voltage gainThe general expression is

inL

v

Zr

h

hA

+

−=

122

21

Substituting the values of h-parameters for transistor in CEconfiguration, we get,

inL

e

fev

Zr

h

hA

+

−=

10

The expressions for Zin Ai and Av for other transistor connec-tions (i.e.CB and CC) can be obtained similarly.Example.3 An amplifier circuit is shown in Fig. 2.41. Work outthe following quantities for the circuit:i) ac emitter currentii) ac voltage at emitter, based and collectoriii) voltage gain.Assume h ie or rin =250 W

Figure 2.41

Solution. (i) Base current due to signal

ib = )(202505

valuepeakAmV

rv

in

in µ=Ω

=

Collection current due to signal (peak value)

mAAAii bc 110002050 ==×== µµβ

Emitter current due to signal (peak value)

mAmAAiii bce 102.11020201000 ≅==+=+= µ(ii) ac voltage at emitter,

0=ev (since it is connected to earththrough CE)

)(5 peakmVvb =

VkmARiRiv ccACcc 111 =Ω×=×==

(iv) Voltage gain, Av = .20051

===mVV

vv

vv

in

c

in

out

Example.4 A single stage amplifier circuit using transistor AC126 is shown in Fig. 2.42. Draw its ac equivalent circuit andcalculate the voltage gain with an without RL. Assume thefollowing transistor parameters:

Ω== krorhorh inieacfe 5.1;150β

Figure 2.42


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YSolution. To draw the ac equivalent circuit of an amplifier thedc voltages and capacitors are short circuited. Thus, the resultantac equivalent circuit is shown in Fig. 2.43. Let the ac base current due to signal be 10 mA,i.e.

andAib µ10= signal voltage, mVkAriv inbin 155.110 =Ω×=×= µCollector current,

mAAii bc 5.1150010150 ==×=×= µβOutput voltage, Lcout riv ×=

(i) When load resistance RL is considered

rL = RAC = RC || RL= Ω=+×

k5.01111

VkmAvout 75.05.05.1 =Ω×=∴

Voltage gain, 50101575.0

1575.0

3=

×===

−mVV

vv

Ain

outv

(ii) When load resistance RL is not consideredrL= RC= 1kW

VkmAvout 5.115.1 =Ω×=∴

Voltage gain, .10015

5.1===

mVV

vv

Ain

outv

Example. 5 A CE amplifier has the following h-parameters:

4105.2,1100 −×== reie hohmh

mhomicrohh oefe 25,50 ==

If the load and source resistance both are 1 Kilo-ohm, findcurrent and voltage gain.

Solution: Here Rs=1kW and rL = 1kW= Ω× 3101

Current gain, Ai = Loe

fe

Rh

h

×+1

36 1011025150

×××+= − = 78.48

025.0150

=+

Voltage gain, inoe

fev

ZrL

h

hA

+

−=

1

Where Zin

=36

4

101102550105.2

11001 −−

−

×+×××

−=+

−

Loe

fereie

rh

hhh

=100-12.5=1087.5W

977.455.1087)1011025(

5036

−=××+×

−=∴

−−vA

The negative sign shows that the output voltage is 1800 out ofphase to the input signal.Example.6. A transistor amplifier circuit is shown in Fig. 4.17.The h-parameters of the transistor are as under.

h ie = 1500W ; h f e = 100

hre =4 mhohoe44 104;10 −− ×=×

Determine the ac input impedance of the amplifier and the voltage gain.

Figure 2.44


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Solution. At load resistance of the amplifier,

rL = Ω=Ω=+×

= 8000840104010

kRAC

Input impedance, Zin = L

oe

fereie

rh

hhh

1+

−

=Ω=

+×

××−

−

−

1424

80001

104

1001041500

4

4

AC input resistance of the entire stage *(Rac) = Z in | | R1 | | R2

=1424 | |

100 Ω=×× 1366100050||1000Voltage gain, Av

=

1341424

80001

104

1001 4

−=

+×

−=

+

−

−in

Loe

fe

Zr

h

h

The magnitude of gain is 134 but the output is 1800 out ofphase to the input signal.

Experimental Determination ofTransistor H-parametersFor the determination of transistor h-parameters, consider thecircuit shown in Fig. 2.45. The standard equations for linearcircuit are

2121111 vhihv +=2221212 vhihi +=

Taking rms values of voltages and currents and using standardtransistor nomenclature, the above equations cab ne written as

cerebiebe VhihV += … (i)

cecebfec VhihI += ….(ii)

Determination of hfe and hie

Short circuit the output as showing in Fig. 2.46. This isaccomplished by making the capacitor C2 deliberately of largevalue so that it can carry the short circuit current.This makes *Vce = 0

Fig.2.45*Vce = 0 means only ac output is zero. It does not effect the dccollection to emitter voltage VCE

Substituting this value in equa. (i) and (ii), we get,

0×+= rebiebe hIhV

orb

beie I

Vh = ….(iii)

0×+= oebfec hIhI

orb

cfe I

Ih = …..(iv)

Determination of hre and hoe

Open circuit the input ( no signal is applied) as shown in Fig.4.20 but a signal generator is applied across the output. MeasureVbe, Vce and Ic. . The large reactance connected in the base circuitdoes not allow the ac current to enter base resistor RB. At thesame time reactor has a low resistance so that may not affect theoperating point. Under this condition Ib=0


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YSubstituting this value in eqns. (i) and (ii), we get,

cereiebe VhhV ×+×= 0

orce

bere V

Vh = …(v)

ceoefec VhhI ×+×= 0

orce

coe V

Ih = ….(vi)

Figure 2.47

Example.7 In a CE amplifier circuit, the following quantitiesare measured:i) When ac output is short circuited (i.e Vce=0)

mVVmAIAI becb 15;5.1;15 === µ

ii) When ac input is opened (i.e., I b=0)

VVAImVV cecbe 5.1;90;1 === µ

Determine all the h-parameters of the circuit. Assuming that allthe values are ac rms.Solution. The various h-parameters are calculated as under:

Ω=== 10001515

AmV

IV

hb

beie µ

Ω=== 1005.15.1

AmV

II

hb

cfe µ

31066.05.1

1 −×===A

mVI

Vh

b

bere µ

mhoVA

VI

hce

coe µ

µ60

1590

===

Approximate Analysis

For a typical transistor RC= 1K and hoe =25mS

CCoe

oe

RRh

Kh

≈

=

||1

401

Θ

hence∴ hoe may be neglected.

Also , hre =2.5 410 −×

Figure 2.49a Common Emitter Configuration

Θ feedback voltage hrevc is very small and can be omitted.Therefore, hybrid ac equivalent circuit becomes. Thus , thecomparison for hybrid versus re model for (a) common emitterand (b) common base configuration can be given, as shown infig. 4.33

Figure 2.50 Common Base Configuration


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Thus, the re model can be converted to a hybrid model and vice-versa using the given relations

rehie β=β=fehrehib =

1−≅−= αfbhExample.9: Given IE= 2.5mA, h fe=140, h oe=20 mS (mmho)and hob =0.5mS, determine:a) The common-emitter hybrid equivalent circuit.b)The common-base re model.Solution:

Figure 2.51 Common Emitter Configuration

(a) Ω=== 4.105.2

2626mAmV

ImV

rE

e

Ω=Ω== kh reie 456.1)4.10)(140(β

Ω=== kSh

roe

o 5020

11µ

(b) Ω===≅

Ω=

MSh

r

r

obo

e

25.011

1

4.10

µα

(a) Common-Emitter Configurations using hybrid model.

(i) Fixed bias configuration

For the fixed bias configuration of fig.2.53 a, the small signal acequivalent model will appear as shown in Fig. 2.53b. Here,

Zi: ZI = RB||hie

Zo: Zo=RC||1/hoe

Av: Using R=1/hoe||RC

''00 RIhRIV bfe−=−=

andie

ib h

VI =

with '0 RhV

hVie

ife−=

so that

ie

oeCfe

iV h

hRh

VV

A)/1||(

0 −==

AI: Assuming that RB>>hie and 1/hoe ³10R, then Ib@Ii and I o =Ic = h feIi with

fei

o hII

Ai ≅=

Av:ie

oeCfe

h

hRhAv

)/1||(−=

Ai:ie

fe

hR

RhAi

+=

'

'

(ii) Unbypassed Emitter-Bias Configuration.For the CE unbypassed emitter-bias configuration of Fig. 2.54the small-signal ac model will be the same as re - model , withbre replaced by h ie and bIb by h f eIb . The analysis will proceed inexactly the same manner.Zi: Zb @ hfeRE

and ZI = RB||Zb

Zo: Zo =RC

Av:Efe

Cfe

b

Cfev Rh

Rh

Z

RhA −≅−=

and Av @E

C

RR−

AI:bB

Bfei ZR

RhA

+=

Example.10 For the network of Fig. 2.55, determine:(a) Zi.(b) Zo.(c) Av.(d) Ai.


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YSolution:

Figure 2.55(a) ZI = RB||h ie = 330kW||1.175 kW @ h ie

=1.171kW

(b) Ω=== kVAh

roe

o 50/20

11µ

CCoe

o RkkkRh

Z ≅Ω=ΩΩ== 56.27.2||50||1

(c)

34.262171.1

)50||7.2)(120()/1||(−=

ΩΩΩ

−=−=k

kkh

hRhA

ie

oeCfev

(d) 120=≅ fei hA

(iii) Voltage-divider ConfiguratioinFor the voltage-divider bias configuration of Fig. 2.56 theresulting small-signal ac equivalent network will have the sameappearance as Fig. 2.53b with RB replaced by R’=R1||R2

ZI : From Fig. 2.54 with RB=R’ZI=R’|| h ie

Zo : From Fig. 2.54,

Co RZ ≅Fig. 2.56

orC

i RZ

AvA 1−=

(b) Emitter-Follower configurationFor the emitter-follower of fig. 2.57a the small –signal ac modelwill match Fig. of re model bre= h ie and b = h f e

The resulting equations will therefore be quite similar.Zi :

Zb @ h f eRE

Zi =RB||Zb

Fig. 2.57a Emitter-follower configuration.

Fig. 2.57b Defining Zo for the emitter-follower configura-tion.Zo:

For Zo, the output network defined by the resulting equationswill appear as shown in Fig. 4.32.

Zo =RE||fe

ie

hh+1

or since 1+ fefe hh ≅

Zo = RE||fe

ie

hh


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Av: For the voltage gain the voltage-divider rule can be appliedto Fig. 2.57a as follows

)1/()(

feieE

iEo hhR

VRV

++=

since 1+h fe @hfe’

feieE

E

i

ov hhR

RVV

A/+

≅=

Ai:bB

Bfei ZR

RhA

+=

or Ai = -Av

E

i

RZ

Fig. 2.57a Emitter-follower configuration.Fig. 2.57b Defining Zo for the emitter-follower configuration.(c) Common-Base Configuration using hybrid model

Fig. 2.58a Common-base configurationFig. 2.58b Substituting the approx., hybrid equivalent circuitinto the ac equivalent network of fig. 2.58b

Lecture 08 hibridequivalentmodel

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Transcript of Lecture 08 hibridequivalentmodel