circuitos electronicos

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Objective :
To understand the Hybrid Equivalent model of Transistor.
Hello! Students I hope now you are well versed with the
concepts of transistor re
-model and the different configuration
that we have studied in the last lectures. Now in today’s class we
are going to discuss the hybrid-model for the various configura-
tion.
To analyze the behavior of transistor amplifier with the help of
β and the values of resistances used in the circuit, is not so
accurate. It is because the input and output circuits of the
amplifier are considered to be completely independent and
some of the effects are ignored such as IC
is taken as constant
but actually its value depends upon load resistance.
Therefore, more accurate method to analyze a transistor
amplifier is hybrid parameters (h- parameters) method, we can
study h-parameter at low frequency and at high frequency. So
this chapter first introduces h- parameter at low frequency and at
high frequency. So this chapter first introduces h – parameter at
low 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 linear
circuit having input and output terminals are called hybrid or
h-parameters.
Meaning of hybrid is ‘mixed’. Since these parameters have
mixed dimensions, they are called hybrid-parameters.
Determination of h-parameters
Consider a linear circuit having two input and two output
terminals (See Fig. 2.29). Their input and output voltage and
currents are labeled with their positive directions. The conven-
tions used are the standard one which may not correspond to
the actual directions. However, while analyzing the circuit, if any
direction is opposite that may be considered as negative.
LINEAR CIRCUIT
Fig. 2.29
In Fig. 4.1, the voltages and currents can be related by the
following 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 as
hybrid-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 we
look at the eqn.. (ii), h2l
has no dimension but h22
, has the
dimension of mho. Hence, the four parameters are named as
hybrid (having mixed dimensions) parameters.
These parameters can be determined very easily as explained
below:
(i) Short-circuit the output terminals, as shown in Fig. 2.30(a),
the output voltage reduces to zero i.e. v2
, = 0. Substituting
this value in eqns. (i) and (ii). we get
v1
= h11
+ h12 × 0
Fig. 2.30 (a) and (b)
or h11 =
1
1
t
v
(output short circuited; v2 = 0) ….(iii)
and t2 = h21 i1 + h22
× 0
or h21 =
1
2
i
i
(output short circuited; v2 = 0) ….(iv)
Here, h11
is called input impedance (i.e. v1
/ ii
) with output
shorted and h2l
is called current gain (i.e. i2
./ i1
) with output
shorted.
(ii) Open circuit the input terminals, as shown in Fig. 2.30b, the
input current reduces to zero i.e. i1
= 0. Substituting this
value in eqns. (i) and (ii), we get,
v1 = h11× 0 + h12 v2
or h12 =
2
1
v
v
(input opened; i1 = 0) ….(v)
and i2 = h21
×
0 + h22 v2
or h22 =
2
2
v
i
(input opened; i1 = 0) ….(vi)
Here, h12
is called voltage feedback ratio (i.e. v1
/v2
) with input
terminals open and h2
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 MULTISTAGE
AMPLIFIER)

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Hybrid-Parameter Equivalent Circuit
Fig. 2.31 &2.32
A two port linear circuit is shown in Fig. 2.31. The voltages and
currents of the circuit can be expressed in terms of h-parameters
by 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. The
input circuit (or port) is derived from the exp. (i). Here, input
impedance (resistor) h is connected in series with a voltage
generator h12
v2
. The output circuit (or port) is derived from exp.
(ii) it involves current generator h2l
v2.
and shunt resistor h22
.
Example.1 Determine the h-parameters of the circuit
shown in Fig. 2.33.
Fig. 2.33 & 2.34
Solution. To determine h- parameters of the circuit proceed as
follows:
(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 Ω
Now, current i1
flows into the box and the same current flows
out of the box.
∴ i2
= i1
.
And h21 = 1
1
1
1
2
−=
−
=
i
i
i
i
(ii) Open the input terminals and make the arrangement as
shown in Fig. 2.35. It may be noted that output terminals
are driven by voltage v2
and no current flows through 20
Ωresistor. Hence, the voltage across 10 Ω resistor i.e. v2
reaches across the input terminals.
∴ v1
= v2
Figure 2.36
and 1
2
2
2
1
==
v
v
v
v
Output impedance = 10 Ω
∴ Output impedance,
h22
= 01.0
10
1
mh=
Hence the various h-parameters of the circuit are
h11
= 20 Ω; h21
= -1
h12
= 1; h12
= 0.1 mho
Example.2 To determine the h-parameter of the circuit shown
in Fig 2.36
Solution. To determine h- parameters of the circuit proceed as
follows:
(i) Short circuit the output terminals as shown in Fig. 4.10.
h11 = 6 + 8 | | 8 = 6 + =
+
×
88
88
10Ω
Now, the input current i1
is divided equally at the junction.
Fig.2.36
∴ i2 = 1
1
5.0
2
i
i
−=−
h21 = 5.0
5.0
1
1
1
2
−=−=−
i
i
i
i

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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 il
resistor as the output terminals are driven by voltage v2
.
v1 = 2
22
5.08
16
8
88
v
vv
=×=×
+
∴ h12 = 5.0
5.0
2
1
2
2
==
v
v
v
v
Impedance looking into the output terminals = 8 + 8 = 16 Ω
h22 =
16
1
= 0.0625 mho
Hence, the various h-parameters of the circuit are
h11
. = 10 Ω ; h21
= - 0.5 ; h12
= 0.5 ; h22
= 0.0625 mho
Performance of a Linear Circuit In
H- Parameters
It has already been seen that a linear circuit has a set of h-
parameters. Now, we shall study the performance of such a
circuit by developing expressions for input impedance, current
gain, voltage gain etc. in terms of h-parameters.
Figure 2.40
Consider a circuit having load resistance RL
across its output
terminal as shown in Fig. 2.39.
Input Impedance
The ratio of input voltage to input current is called the input
impedance Zin
.
∴ Zin.=
i
i
i
v
or Zin.=
1
212111
i
vhih +
(∴ v1= h11 i1 + h12 v2)
or Zin. = h11 +
1
212
i
vh
….(i)
Now, i2 = h21 i1 + h22 v2
or 






 −
=∴+=
−
LL i
v
ivhih
i
v 2*
2222121
2
or -h21 i1 = v2 = 





+
L
r
h
1
22
The ratio of output current to input current is called current
gain Ai
.
Zin = h11 -
Lr
h
hh
1
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
h
i
i
22
21
1
2
1+
=
Substituting the value of
1
2
i
i
in eqn.(iv), we get,
A1
=
Lrh
h
22
21
1+
…..(v)
If h22
rL
<< 1 then Ai ≅ h21
.
Voltage Gain
The ratio of output to input voltage is called voltage gain Av
.
Av
=
1
2
v
v
……(vi)
or Av
=
in
zi
v
1
2
( )11 inZiv =Θ

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Substituting the value of
1
2
i
v
from eqn. (ii), we get,
Av
=
in
L
Z
r
h
h






+
−
1
22
21
H- Parameters of a Transistor
A transistor in a three–terminal device. If any one of the
terminals is made common to the input and output, it will
have two ports. Thus it will have two input terminals and two
output terminals (See Fig.2.40), for a small ac signal transistor
behave 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 the
following 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
i
v
02=v = Input impedance (with
output shorted) = hi
(in ohms)
h21
= 02
1
2
=v
i
i
= Forward current ratio (with
output shorted) = hf
(no unit)
h12
= 01
2
1
=i
v
v
= Reverse voltage ratio (with
input open)= hr
(no unit)
h22
= 01
2
2
=i
v
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 in
different way. For instance, h11
is represented as **hie
, hib
and
hic
for CE, CB and CC connections.
ii) While checking the performance, ac output resistance is
considered as load resistance i.e.
LC
LC
LCACL
RR
RR
RRRr
+
===
The values of h-parameters depend upon the Q point. If Q
point changes, the values of h-parameters are also changed.
In transistor circuits the values of voltage and currents are taken
depending 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 transistor
depending upon its connections is given in the following table
4.1.
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 are
given below:
hie = 3.5 k Ω; hre =1.3 x 10-4
hfe = 120; hoe = 8.5 µ mho
*It may be noticed that subscript used is the first letter of the
description i.e. input, forward, reverse and output respectively.
** It may be noticed that the second letter of the subscript used
indicates the type of transistor connections.
Performance of (Ce Circuit) Transistor in
H-Parameters
While studying the performance of a transistor, we are inter-
ested in the following terms:
i) Input Impedance
The general expression for Zin is
L
in
r
h
hh
hZ
1
22
2112
11
+
−=
Substituting the values of h-parameters for transistor in CE
configuration. We get,

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L
e
fere
iein
r
h
hh
hZ
1
0 +
−=
(ii) Current gain
The general expression is
L
i
rh
h
A
22
21
1+
=
Substituting the values of h-parameters for transistor in CE
configuration, we get,
Le
fe
i
rh
n
A
01+
=
(iii) Voltage gain
The general expression is
in
L
v
Z
r
h
h
A






+
−
=
1
22
21
Substituting the values of h-parameters for transistor in CE
configuration, we get,
in
L
e
fe
v
Z
r
h
h
A






+
−
=
1
0
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 out
the following quantities for the circuit:
i) ac emitter current
ii) ac voltage at emitter, based and collector
iii) voltage gain.
Assume hie
or rin
=250 W
Figure 2.41
Solution. (i) Base current due to signal
ib =
)(20
250
5
valuepeakA
mV
r
v
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 earth
through CE
)
)(5 peakmVvb =
VkmARiRiv ccACcc 111 =Ω×=×==
(iv) Voltage gain, Av
= .200
5
1
===
mV
V
v
v
v
v
in
c
in
out
Example.4 A single stage amplifier circuit using transistor AC
126 is shown in Fig. 2.42. Draw its ac equivalent circuit and
calculate the voltage gain with an without RL
. Assume the
following transistor parameters:
Ω== krorhorh inieacfe 5.1;150β
Figure 2.42

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Solution. To draw the ac equivalent circuit of an amplifier the
dc voltages and capacitors are short circuited. Thus, the resultant
ac 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×=×= µ
Collector current,
mAAii bc 5.1150010150 ==×=×= µβ
Output voltage, Lcout riv ×=
(i) When load resistance RL
is considered
rL
= RAC
= RC
|| RL
= Ω=
+
×
k5.0
11
11
VkmAvout 75.05.05.1 =Ω×=∴
Voltage gain, 50
1015
75.0
15
75.0
3
=
×
=== −
mV
V
v
v
A
in
out
v
(ii) When load resistance RL
is not considered
rL
= RC
= 1kW
VkmAvout 5.115.1 =Ω×=∴
Voltage gain, .100
15
5.1
===
mV
V
v
v
A
in
out
v
Example. 5 A CE amplifier has the following h-parameters:
4
105.2,1100 −
×== reie hohmh
mhomicrohh oefe 25,50 ==
If the load and source resistance both are 1 Kilo-ohm, find
current and voltage gain.
Solution: Here Rs
=1kW and rL =
1kW= Ω× 3
101
Current gain, Ai =
Loe
fe
Rh
h
×+1
36
10110251
50
×××+
= − = 78.48
025.01
50
=
+
Voltage gain,
inoe
fe
v
Z
rL
h
h
A






+
−
=
1
Where Zin
=
36
4
1011025
50105.2
1100
1 −−
−
×+×
××
−=
+
−
L
oe
fere
ie
r
h
hh
h
=100-12.5=1087.5W
977.45
5.1087)1011025(
50
36
−=
××+×
−
=∴ −−vA
The negative sign shows that the output voltage is 1800
out of
phase 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.
hie
= 1500W ; hfe
= 100
hre
=4 mhohoe
44
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=
+
×
= 80008
4010
4010
kRAC
Input impedance, Zin
=
L
oe
fere
ie
r
h
hh
h
1
+
−
=
Ω=
+×
××
−
−
−
1424
8000
1
104
100104
1500
4
4
AC input resistance of the entire stage *(Rac
)
= Zin
| | R1
| | R2
=1424 | |
100 Ω=×× 1366100050||1000
Voltage gain, Av
=
134
1424
8000
1
104
100
1 4
−=






+×
−
=






+
−
−
in
L
oe
fe
Z
r
h
h
The magnitude of gain is 134 but the output is 1800
out of
phase to the input signal.
Experimental Determination of
Transistor H-parameters
For the determination of transistor h-parameters, consider the
circuit shown in Fig. 2.45. The standard equations for linear
circuit are
2121111 vhihv +=
2221212 vhihi +=
Taking rms values of voltages and currents and using standard
transistor 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 is
accomplished by making the capacitor C2
deliberately of large
value 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 dc
collection to emitter voltage VCE
Substituting this value in equa. (i) and (ii), we get,
0×+= rebiebe hIhV
or
b
be
ie
I
V
h = ….(iii)
0×+= oebfec hIhI
or
b
c
fe
I
I
h = …..(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. Measure
Vbe,
Vce
and Ic.
. The large reactance connected in the base circuit
does not allow the ac current to enter base resistor RB
. At the
same time reactor has a low resistance so that may not affect the
operating point. Under this condition Ib
=0

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Substituting this value in eqns. (i) and (ii), we get,
cereiebe VhhV ×+×= 0
or
ce
be
re
V
V
h = …(v)
ceoefec VhhI ×+×= 0
or
ce
c
oe
V
I
h = ….(vi)
Figure 2.47
Example.7 In a CE amplifier circuit, the following quantities
are 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., Ib
=0)
VVAImVV cecbe 5.1;90;1 === µ
Determine all the h-parameters of the circuit. Assuming that all
the values are ac rms.
Solution. The various h-parameters are calculated as under:
Ω=== 1000
15
15
A
mV
I
V
h
b
be
ie
µ
Ω=== 100
5.1
5.1
A
mV
I
I
h
b
c
fe
µ
3
1066.0
5.1
1 −
×===
A
mV
I
V
h
b
be
re
µ
mho
V
A
V
I
h
ce
c
oe µ
µ
60
15
90
===
Approximate Analysis
For a typical transistor RC
= 1K and hoe
=25mS
CC
oe
oe
RR
h
K
h
≈
=
||
1
40
1
Θ
hence
∴ hoe
may be neglected.
Also , hre
=2.5 4
10−
×
Figure 2.49a Common Emitter Configuration
Θ feedback voltage hre
vc
is very small and can be omitted.
Therefore, hybrid ac equivalent circuit becomes. Thus , the
comparison for hybrid versus re
model for (a) common emitter
and (b) common base configuration can be given, as shown in
fig. 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=feh
rehib =
1−≅−= αfbh
Example.9: Given IE
= 2.5mA, hfe
=140, hoe
=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=== 4.10
5.2
2626
mA
mV
I
mV
r
E
e
Ω=Ω== kh reie 456.1)4.10)(140(β
Ω=== k
Sh
r
oe
o 50
20
11
µ
(b) Ω===≅
Ω=
M
Sh
r
r
ob
o
e
2
5.0
11
1
4.10
µ
α
(a) Common-Emitter Configurations using hybrid model.
(i) Fixed bias configuration
For the fixed bias configuration of fig.2.53 a, the small signal ac
equivalent 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−=−=
and
ie
i
b
h
V
I =
with '0 R
h
V
hV
ie
i
fe−=
so that
ie
oeCfe
i
V
h
hRh
V
V
A
)/1||(0
−==
AI
: Assuming that RB
>>hie
and 1/hoe
³10R,
then Ib
@Ii and Io
=
Ic
= hfe
Ii
with
fe
i
o
h
I
I
Ai ≅=
Av
:
ie
oeCfe
h
hRh
Av
)/1||(
−=
Ai:
ie
fe
hR
Rh
Ai
+
=
'
'
(ii) Unbypassed Emitter-Bias Configuration.
For the CE unbypassed emitter-bias configuration of Fig. 2.54
the small-signal ac model will be the same as re - model , with
bre
replaced by hie
and bIb
by hfe
Ib
. The analysis will proceed in
exactly the same manner.
Zi
: Zb
@ hfe
RE
and ZI
= RB
||Zb
Zo
: Zo
=RC
Av
:
Efe
Cfe
b
Cfe
v
Rh
Rh
Z
Rh
A −≅−=
and Av
@
E
C
R
R−
AI
:
bB
Bfe
i
ZR
Rh
A
+
=
Example.10 For the network of Fig. 2.55, determine:
(a) Zi
.
(b) Zo
.
(c) Av
.
(d) Ai
.

10. © Copy Right: Rai University
4A.273 39
ELECTRONICDESIGNTECHNOLOGY
Solution:
Figure 2.55
(a) ZI
= RB
||hie
= 330kW||1.175 kW @ hie
=1.171kW
(b) Ω=== k
VAh
r
oe
o 50
/20
11
µ
CC
oe
o RkkkR
h
Z ≅Ω=ΩΩ== 56.27.2||50||
1
(c)
34.262
171.1
)50||7.2)(120()/1||(
−=
Ω
ΩΩ
−=−=
k
kk
h
hRh
A
ie
oeCfe
v
(d) 120=≅ fei hA
(iii) Voltage-divider Configuratioin
For the voltage-divider bias configuration of Fig. 2.56 the
resulting small-signal ac equivalent network will have the same
appearance as Fig. 2.53b with RB
replaced by R’=R1
||R2
ZI
: From Fig. 2.54 with RB
=R’
ZI
=R’|| hie
Zo
: From Fig. 2.54,
Co RZ ≅
Fig. 2.56
or
C
i
R
Z
AvA 1
−=
(b) Emitter-Follower configuration
For the emitter-follower of fig. 2.57a the small –signal ac model
will match Fig. of re model bre
= hie
and b = hfe
The resulting equations will therefore be quite similar.
Zi
:
Zb
@ hfe
RE
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 equations
will appear as shown in Fig. 4.32.
Zo
=RE
||
fe
ie
h
h
+1
or since 1+ fefe hh ≅
Zo
= RE
||
fe
ie
h
h

11. 40 4A.273
© Copy Right: Rai University
ELECTRONICDESIGNTECHNOLOGY
Av
: For the voltage gain the voltage-divider rule can be applied
to Fig. 2.57a as follows
)1/(
)(
feieE
iE
o
hhR
VR
V
++
=
since 1+hfe
@hfe’
feieE
E
i
o
v
hhR
R
V
V
A
/+
≅=
Ai:
bB
Bfe
i
ZR
Rh
A
+
=
or Ai
= -Av
E
i
R
Z
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 configuration
Fig. 2.58b Substituting the approx., hybrid equivalent circuit
into the ac equivalent network of fig. 2.58b