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802 ⏐⏐⏐ NETWORK THEOREMS (ac)
+
ETh –
5.08 V ∠ –77.09°
Z Th
4.64 � + j2.94 �
a
a′
R 3
7 �
E 2
+
–
Step 5: The Thévenin equivalent circuit is shown in Fig. 18.32.
30 V ∠ 15°
+
ETh –
4.64 � 2.94 � 7 �
+
5.08 V ∠ –77.09°
FIG. 18.32
The Thévenin equivalent circuit for the network of Fig. 18.28.
R
The next example demonstrates how superposition is applied to electronic
circuits to permit a separation of the dc and ac analyses. The fact
that the controlling variable in this analysis is not in the portion of the network
connected directly to the terminals of interest permits an analysis of
the network in the same manner as applied above for independent sources.
EXAMPLE 18.9 Determine the Thévenin equivalent circuit for the
transistor network external to the resistor R L in the network of Fig.
18.33. Then determine V L.
+
Ei –
R s
0.5 k�
C 1
10 �
R B
12 V
1 M�
R C
X L
2 k�
FIG. 18.33
Example 18.9.
Solution: Applying superposition.
a
a′
C 2
10 �
Transistor
R 3
E 2
–
Thévenin
Th
30 V ∠ 15°
+
R L = 1 k� V L
–
dc Conditions Substituting the open-circuit equivalent for the coupling
capacitor C 2 will isolate the dc source and the resulting currents
from the load resistor. The result is that for dc conditions, V L � 0 V.
Although the output dc voltage is zero, the application of the dc voltage
is important to the basic operation of the transistor in a number of
important ways, one of which is to determine the parameters of the
“equivalent circuit” to appear in the ac analysis to follow.
ac Conditions For the ac analysis, an equivalent circuit is substituted
for the transistor, as established by the dc conditions above, that