Thévenin and Mayer-Norton Circuits
Samantha R. Summerson
4 September, 2009
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Equivalent Resistance
𝑅1
𝑖
+ 𝑣1 −
𝑖2
𝑖1
+
+
+
𝑣𝑖𝑛 (𝑡) ±
←𝑣
𝑅2
−
−
−
Figure 1: Example: how do we find an equivalent circuit?
From the last lecture, we found that
𝑣=
𝑅2
𝑅1 𝑅2
𝑣𝑖𝑛 +
𝑖.
𝑅1 + 𝑅2
𝑅1 + 𝑅2
𝑖1
𝑖
𝑅𝑒𝑞
+ 𝑣𝑅 −
+
+
𝑣𝑖𝑛 (𝑡) ±
𝑣𝑜𝑢𝑡 (𝑡)
−
−
Figure 2: Thévenin equivalent circuit.
If we re-write this equation in the form
𝑣 = 𝑅𝑒𝑞 𝑖 + 𝑣𝑒𝑞 ,
the equation looks like a KVL equation for the circuit shown in Fig.2. It turns out that all circuits consisting
of sources and resistors can be equivalently representing in this manner; this is called the Thévenin equivalent
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circuit. From the two equations, we see that
𝑅𝑒𝑞
=
𝑣𝑒𝑞
=
(𝑅1 ∣∣𝑅1 ),
𝑅2
𝑣𝑖𝑛 .
𝑅1 + 𝑅2
In general, how do we find these equivalent circuit values?
1. To find 𝑅𝑒𝑞 , set all sources to zero and play the series/parallel ”game” (i.e. reduces resistors to a single
equivalent resistance).
2. To find 𝑣𝑒𝑞 , set 𝑖 = 0 (open circuit) and find 𝑣. The voltage divider rule is helpful here.
3. To find 𝑖𝑠𝑐 , set 𝑣 = 0 (short circuit) and solve for the current. This time the current divider rule is
useful.
Example 1. Find the Thévenin equivalent for the circuit in Fig.3.
𝑅2
𝑖3
𝑖
+
𝑖2
𝑖𝑖𝑛
↑
𝑅1
𝑅3
𝑣𝑜𝑢𝑡
−
Figure 3: Thévenin equivalent circuit.
First, we set 𝑖𝑖𝑛 = 0, since 𝑖𝑖𝑛 is the only current source. The equivalent resistance is
𝑅𝑒𝑞 = (𝑅1 + 𝑅2 )∣∣𝑅3 .
Second, we set 𝑖 = 0. We can use our current divider rule to find 𝑖3 .
𝑖3 =
𝑅1
𝑖𝑖𝑛 .
𝑅1 + 𝑅2 + 𝑅3
Now, using our 𝑣 − 𝑖 relation for resistor 𝑅3 , we have
𝑣
=
=
𝑅3 𝑖3 ,
𝑅1 𝑅3
𝑖𝑖𝑛 .
𝑅1 + 𝑅2 + 𝑅3
This is our equivalent voltage, 𝑣𝑒𝑞 . Using 𝑅𝑒𝑞 and 𝑣𝑒𝑞 , we can find 𝑖𝑠𝑐 . We proceed through the third step
for the sake of completion. Setting 𝑣 = 0, we use our current divider rule to find
𝑖𝑠𝑐 =
−𝑅1
𝑖𝑖𝑛 .
𝑅1 + 𝑅2
2
𝑖
+
↑
𝑖𝑠𝑐
𝑅𝑒𝑞
𝑣
−
Figure 4: Mayer-Norton equivalent circuit.
𝑅1
𝑖1
𝑖
+
+ 𝑣1 −
+
𝑣𝑖𝑛 (𝑡) ±
𝐶
←𝑣
−
−
Figure 5: Example 2: how do we find output voltage?
In addition to the Thévenin circuit representation, we can also represent any circuit with sources and
resistors using the Mayer-Norton model. Notice that 𝑣 = 𝑅𝑒𝑞 𝑖𝑠𝑐 .
Example 2. Find the output voltage for the circuit with a capacitor.
Using KVL we have
𝑣𝑖𝑛
= 𝑅𝑖 + 𝑣,
(
)
𝑑𝑣(𝑡)
= 𝑅 𝐶
+ 𝑣.
𝑑𝑡
This is a partial differential equation. We will solve this using a neat trick. Suppose
𝑣𝑖𝑛 = 𝑣 𝑗2𝜋𝑓 𝑡 .
Using this value for 𝑣𝑖𝑛 , we find the current through the capacitor.
(
)
𝑑𝑣𝑖𝑛
𝑖 = 𝐶
𝑑𝑡
=
𝐶𝑗2𝜋𝑓 𝑣 𝑗2𝜋𝑓 𝑡
=
𝐼 𝑗2𝜋𝑓 𝑡
In the above, we have defined 𝐼 = 𝐶𝑗2𝜋𝑓 𝑣. In the next class, we will finish this example.
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