Thévenin’s Equivalent Circuits For Circuits with Dependent sources Example problems Home Work problems

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Session 12D: Focus
 Thévenin’s Equivalent Circuits
◦ For Circuits with Dependent sources
 Example problems
 Home Work problems

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Thévenin’s Equivalent Circuits
for
Circuits with Dependent Sources

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When Dependent Sources are Present
 If network A contains a dependent source, then again
we must ensure that the controlling variable and its
associated element(s) cannot be in network B.
 Up to now, we have only considered circuits with
resistors and independent sources.
 Although technically speaking it is correct to leave a
dependent source in the “inactive” network when
creating a Thévenin or Norton equivalent
◦ In practice this does not result in any kind of simplification

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When Dependent Sources are Present …
contd.
 What we really want is an independent voltage
source in series with a single resistor
◦ Or an independent current source in parallel with a
single resistor
◦ In other words, a two-component equivalent

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Thévenin’s Equivalent for
a Circuit with Dependent Sources
Note: This is Example 5.9 on page 148, Figure 5.31 in the Ref 1 book (by Hayt)
Using KVL on the mesh:
+
-
vx
-4 + 2k (- vx/4000) + vx = 0
-4 - 0.5 vx + vx = 0 0.5 vx = 4
vx = Voc = 8 V
A possible, but rather useless, form of
Thévenin equivalent circuit is:
Since no current can flow through 3 kΩ,
the voltage across the dependent source is
also vx and Open Circuit Voltage is
also (Voc) vx
vx = 0 V Voc = 8V
Dead network (RTH)

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Example: With Dependent Sources
Note: This is Example 5.9 on page 148, Figure 5.31 in the Ref 1 book (by Hayt)
+
-
vx
A possible, but rather useless, form of
the Thévenin equivalent.
What we need is a two element equivalent
circuit with RTH and an independent
voltage source.
That means, we need to somehow get rid of
the dependent voltage source and replace it
with only two elements

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Example: With Dependent Sources contd…
Note: This is Example 5.9 on page 148, Figure 5.31 in the Ref 1 book (by Hayt)
+
-
vx
vx = Voc = 8 V
To compute RTH across the open terminals,
apart from Voc , we also need Isc
Then, RTH = Voc / Isc
When the terminals are short circuited,
vx = 0, which means the dependent current
Source will also be zero. If, no current is
Passing through the dependent source, it is
Equivalent to open. Then,
Thévenin’s
Equivalent
Circuit
Isc = 4 / (2k + 3k) = 4/5k = 0.8 mA
RTH = Voc / Isc = 8 / 0.8 mA = 10 kΩ

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Example1: Thévenin’s Equivalent Circuit
 Find Vo by using Thévenin’s theorem:
Note: This is Example 5.7 on page 201, Figure 5.9 in the Ref 2 book (by Irwin)
Vo = 8 V

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Example2: Thévenin’s Equivalent Circuit
 Find Vo by using Thévenin’s theorem:
Note: This is Example 5.8 on page 202, Figure 5.10 in the Ref 2 book (by Irwin)
Vo = 48/7 V

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Home Work Problem

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HW2: Find Thévenin Circuit for
the Circuit with a Dependent Source
Note: This is Practice 5.8 on page 148, Figure 5.32 in the Ref 1 book (by Hayt)

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Session 12D: Summary
 Thévenin’s Equivalent Circuits
◦ For Circuits with Dependent sources
 Example problems
 Home Work problems

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References
Ref 1 Ref 2