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5.6 
 determInatIon of equIvaLentS for cIrcuItS wIth 
dependent SourceS
Method-1
(i) Find v
oc
by nodal analysis or mesh analysis or superposition principle.
(ii) Find i
sc
by nodal analysis or mesh analysis or superposition principle.
(iii) Obtain R
o
by R
v
i
o
oc
sc
=
Method-2
(i) Find v
oc
by nodal analysis or mesh analysis or superposition principle.
(ii) Assume that a current source of 1 A is applied to the output terminals such that the 
current flows into the network at the first terminal. Carry out a node or mesh analysis 

---

5.26
Circuit Theorems
and find out the voltage appearing at first terminal with respect to second terminal. The 
numerical value of this voltage gives the value of R
o
.
(iii) Determine i
sc
by i
v
R
sc
oc
o
=
Method-3
(i) Find i
sc
by nodal analysis or mesh analysis or superposition principle.
(ii) Assume that a voltage source of 1 V is applied to the output terminals with positive 
polarity at the first terminal. Carry out a node or mesh analysis and find out the current 
flowing into the first terminal. The numerical value of this current gives the value of 
G
o
=
1/R
o
.
(iii) Determine v
oc
by v
R i
oc
o sc
=
example: 5.6-1
Find the Thevenin’s equivalent of the circuit in Fig. 
5.6-1 with respect to the terminals a and b and 
thereby find the ratio of v
x
(t) to v
s
(t) when a resistor 
of 2k
W
is connected across the output. The circuit is 
the low-frequency signal model for an RC-coupled 
common emitter amplifier using a bipolar junction
transistor.
Solution
The first method is used in this example. Let b be the 
reference node and let the node voltage at top end of 
5k
W
be v
1
as marked in the figure. Then, 
i
v
v
v
i
v
v
x
x
x
x
x
= ×
−
= − ×
= −
−
= −
−
1 10
0 0005
2 10
200
0 0005
20
3
1
5
1
(
.
)
(
.
)
and
00
0 1
0 9
200
222 2
0 0005
0 11
1
1
1
1
v
v
v
v
v
v
v
v
x
x
x
x
+
∴
= −
⇒
= −
∴
= −
.
.
.
.
.
Now writing KCL at the node where v
1
is assigned,
0 2 10
1
0 11
1 10
0 02
0
0 021
3
1
3
1
1
.
(
(
. ))
. (
( ))
. ., .
×
+ − −
× ×
+
−
=
−
−
v
v
v
v t
i e
s
331
0 02
0 9385
1
1
v
v t
v
v t
s
s
=
∴ =
.
( )
.
( )
Since v
x
=
-
222.2 v
1
, v
x
=
-
208.5 v
s
(t) 
Therefore v
oc
=
-
208.5 v
s
(t)
When the terminals 
a
-
a´
 are shorted, v
x
=
0, and therefore the independent voltage source at the 
input side is zero-valued. The value of v
1
under this condition is given by,
v
k
k
k
k
v t
v t
s
s
1
5
1
50 5
1
0 9434
=
+
=
||
||
( )
.
( )
Fig. 5.6-1 
Circuit for Example 5.6-1
b
a
50 
Ω
1 k
Ω
5 k
Ω
2 k
Ω
+
–
+
–
+
–
i
x
i
x
v
x
v
S 
(
t
)
v
1
0.0005
100
v
x

---

Determination of Equivalents for Circuits with Dependent Sources 
5.27
Now the current i
x
is 0.9434
× 
10
-
3
v
s
(t), and hence the current in the dependent source at the 
output side is 0.09434v
s
(t). All this current flows out of 
a´
to a through the short-circuit. Hence 
i
sc
(t) 
=
-
0.09434v
s
(t).
∴
=
=
=
Ω
R
v
i
v t
v t
o
oc
sc
s
s
208 5
0 09434
2 21
.
( )
.
( )
. k
The two equivalent circuits are shown in Fig. 5.6-2.
(a)
2.21 k
Ω
208.5 
v
S 
(
t
)
b
a
+
–
2.21 k
Ω
(b)
b
a
0.09434 
v
S 
(
t
)
Fig. 5.6-2 
(a) Thevenin’s equivalent circuit for the amplifier in Example 5.6-1 (b) 
Norton’s equivalent circuit
If a load resistance of 2k
W
is connected at the output, then the output voltage will be 
−
×
+
= −
208 5
2
2 2 21
99 05
.
( )
.
.
( )
v t
v t
s
s
. Hence the ratio between output and input (i.e., the gain of the 
amplifier) is – 99.05.
example: 5.6-2
The equivalent circuit of DC current source realised 
using a transistor and few resistors is shown in Fig. 
5.6-3. The design is expected to deliver –2mA at a. 
Find the Norton’s equivalent circuit for this current 
source design.
Solution
We find the short-circuit current at output first. The 
circuit for this is shown in Fig. 5.6-4. 
0.0005
2.15 V
b a
100 k
Ω
1 k
Ω
1 k
Ω
5 k
Ω
100
+
–
+
+
–
–
i
x
i
x
i
SC
v
x
v
x
Fig. 5.6-4 
Circuit for determining short-circuit current in Example 5.6-2 
Fig. 5.6-3 
Circuit for Example 5.6-2
0.0005 
v
x
2.15 V
b
a
100 k
Ω
1 k
Ω
1 k
Ω
5 k
Ω
100
+
–
+
–
+
–
i
x
i
x
v
x

---

5.28
Circuit Theorems
We solve this circuit by mesh analysis. The current into 1k//100k (
=
0.99k) is 101i
x
.We will assume 
that the unit of i
x 
is in mA. Therefore the KVL in the first mesh is
−
+ × +
× −
×
+
×
=
=
2 15 6
0 0005
0 99 101
0 99 101
0
105 94
.
.
( .
)
.
. .,
.
i
i
i
i e
i
x
x
x
x
22 15
0 0203
.
.
⇒ =
i
x
mA
Therefore, i
i
sc
x
= −
= −
100
2 03
.
mA
To find R
o
We assume that we are injecting 1 mA into the network from terminal a after deactivating the circuit. 
We determine the voltage v
ab
. This voltage directly gives R
o 
in k
W
units. Refer to the circuit in Fig. 5.6-5 (a).
0.0005
100
1 mA 2.03 mA
(b)
(a)
a
a
b
e
b
5.13 M
Ω
1 k
Ω
1 k
Ω
5 k
Ω
100 k
Ω
+
–
+
–
i
x
i
x
v
x
v
x
Fig. 5.6-5 
(a) Circuit for determining Thevenin’s equivalent resistance in Example 5.6-2 
(b) The Norton’s equivalent circuit required
Let i
x
be in mA. The current in 100k resistor is 1
- 
100i
x
(by applying KCL at node a) and therefore 
v
x
=
100
- 
10
4
i
x
V. The current into 1k resistor at node e is (1
+ 
 i
x
) mA and hence voltage of node e 
with respect to node b is (1
+ 
 i
x
) V. Now applying KVL in the first mesh,
5
0 0005
100 10
1
0
0 5025
100 10
4
i
i
i
i
i
v
x
x
x
x
x
x
+ +
×
−
+ +
= ∴ = −
∴ =
−
.
(
) (
)
,
.
mA
44
5125
5125
1 0 5025
5125 5
i
v
v
v
R
x
ab
x
e
o
=
∴
= + =
+ −
=
⇒
volts and
volts
(
.
)
.
==
Ω
5125 5
. k
If we really apply 1 mA into the output of the transistor circuit, either the current source we are 
applying will fail to function as a current source or the transistor will fail on overvoltage! But then, 
this is only a ‘thought experiment’ aimed at evaluating R
o
. The Norton’s equivalent circuit for this 
current source design is shown in Fig. 5.6-5 (b). 
example: 5.6-3
Find the Thevenin’s equivalent of the circuit in 
Fig. 5.6-6 with respect to a and b.
Solution
We find the open-circuit voltage across a
-
b first. 
Assume two mesh currents i
1
(mA) and i
2
(mA) in 
the clockwise direction in the first and second mesh, 
respectively. The two mesh equations are
Fig. 5.6-6 
Circuit for Example 5.6-3
a
b
0.5 k
Ω
1 k
Ω
9 k
Ω
100 k
Ω
1000
+
+
+
–
–
–
v
x
v
x
v
S 
(
t
)

---

Reciprocity Theorem 
5.29
100
9
0 5
1000 100
0
1
1
2
2
1
2
2
1
i
i
i
v t
i
i
i
i
i
s
+ −
=
− +
+
+
×
=
(
)
( )
(
)
.
Solving these two equations we get i
1
=
v t
s
( )
.
9624 7
and i
2
=
-
0.9895 v
s
(t).
∴
=
+
×
=
v
i
i
v t
oc
s
0 5
1000 100
9 9
2
1
.
.
( )
To find R
o
We assume that 1 V is applied across a
-
b after deactivating the circuit and find out the current 
drawn by the circuit from this 1 V source. The value of current gives G
o
. The circuit required to solve 
for R
o
is shown as circuit in Fig. 5.6-7 (b).
(b)
(a)
9.9
b
a
b
a
9 k
Ω
1 k
Ω
1000
0.5 k
Ω
5 
Ω
100 k
Ω
1 V
+
+
+
–
–
–
+
–
v
OC
R
O
i
v
x
v
S 
(
t
)
v
x
Fig. 5.6-7 
(a) Circuit for determining 
R
o
in Example 5.6-3 (b) The Thevenin’s equivalent
The current drawn from 1 V has two components – one flowing into the 9k
W
resistor and the 
second flowing into 0.5k
W
path. The first component is 1/(9
+ 
1/100) 
=
1/9.99 
=
0.1 mA. The value of 
v
x
is given by applying voltage division principle as 
-
1
× 
(100/1)/(9
+ 
100//1) 
=
-
0.099 V. Therefore 
the voltage across 0.5k
W
resistor is 1
- 
(
-
0.099
× 
1000) 
=
100 V. Therefore the current flowing into 
0.5k
W
path is 100/0.5 
=
200 mA. Then, the total current drawn from 1 V source is 200.1 mA and the 
value of G
o
is 0.2 S. Therefore, the value of R
o
is 5 
W
.
The Thevenin’s equivalent for the circuit is shown in circuit of Fig. 5.6-7 (b).