Thévenin’s and Norton’s Theorems

Section 2.6 Thévenin’s and Norton’s Theorems

Circuits controlling devices that you use every day are often very complex with complicated scematic diagrams. This complexity can make it difficult to analytically determine the behavior of the output voltage and current when this complex circuit is connected to an external load. DEFINE LOAD. NOTE SOMEWHERE THAT \(\Delta v_\text{ext}=V_A-V_B\) .

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In this section, we introduce two theorems that allow us to replace a complicated circuit with a simplified circuit. For the moment, we will limit ourselves to resistor-based DC circuits and will generalize these theorems next chapter.

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Thévenin’s theorem states that any complicated circuit containing comprised of only ideal voltage sources, ideal current sources, and resistors will demonstrate a relationship between \(i_\text{ext}\) and \(\Delta v_\text{ext}\) that is identical to the relationship demonstrated by a Thévenin equivalent circuit containing a single voltage source (the Thévenin voltage \(V_\text{th}\)) in series with a single resistance (the Thévenin resistance \(R_\text{th}\)) for any load (DEFINE!) connected across the circuit output. Similarly, Norton’s theorem states that any complicated circuit containing comprised of only ideal voltage sources, ideal current sources, and resistors will demonstrate a relationship between \(i_\text{ext}\) and \(\Delta v_\text{ext}\) that is identical to the relationship demonstrated by a Norton equivalent circuit containing a single current source (the Norton current \(I_n\)) in parallel with a single resistance (the Norton resistance \(R_n\)) for any load (DEFINE!) connected across the circuit output.

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Below, we will start by analyzing the Thévenin and Norton equivalent circuits and determine the relationship between \(i_\text{ext}\) and \(v_\text{ext}\text{.}\) We will then show that any complicated circuit containing only ideal sources and resistors must demonstrate this same behavior, thus proving these two theorems.

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Subsection 2.6.1 Behavior of Thévenin/Norton circuits

Applying Kirchhoff’s voltage law to the Thévenin equivalent circuit in Figure 2.6.1, we find that

\begin{equation*} V_\text{th}-i_\text{ext}R_\text{th}-\Delta v_\text{ext}\text{.} \end{equation*}

Rearrangement of this equation shows

\begin{equation} i_\text{ext}=-\frac{1}{R_\text{th}}\Delta v_\text{ext} + \frac{1}{R_\text{th}} V_\text{th}\text{,}\tag{2.6.1} \end{equation}

a linear relationship between \(i_\text{ext}\) and \(v_\text{ext}\text{.}\)

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Application of Kirchhoff’s current law to the Norton equivalent circuit in Figure 2.6.1 shows

\begin{equation*} I_n - i_\text{ext} - \frac{1}{R_n}\Delta v_\text{ext} \end{equation*}

where Ohm’s law was used to express the current through \(R_n\) in terms of \(\Delta v_\text{ext}\text{.}\) Rearrangement of this equation shows that

\begin{equation} i_\text{ext} = -\frac{1}{R_n}\Delta v_\text{ext} + I_n\tag{2.6.2} \end{equation}

which again demonstrates a linear relationship between \(i_\text{ext}\) and \(\Delta v_\text{ext}\text{.}\)

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If both Thévenin’s theorem and Norton’s theorem are to be true, then Equations (2.6.1) and (2.6.2) must be equivalent. This can only be true if

\begin{equation*} I_n = \frac{V_\text{th}}{R_\text{th}} \end{equation*}

and

\begin{equation*} R_n = R_\text{th}\text{.} \end{equation*}

The linear relationship between \(i_\text{ext}\) and \(\Delta v_\text{ext}\) for these circuits is shown in Figure 2.6.2. Here we see that the slope is the negative inverse of \(R_\text{th}\) and \(R_n\text{.}\) The Thévenin voltage \(V_\text{th}\) is equal to \(\Delta v_\text{ext}\) when \(i_\text{ext}=0\) (an open-circuit where nothing is connected across the output terminals), and the Norton current \(I_n\) equals \(i_\text{ext}\) when the outputs are shorted by a wire (making \(\Delta v_\text{ext}=0\)).

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Figure 2.6.2. I-V plot for linear circuits. The norton current \(I_n\) is equal to \(i_\text{ext}\) when the output is shorted (\(v_\text{ext}=0\)). We call this the short-circuit current \(I_\text{sc}\text{.}\) The Thevenin voltage \(V_\text{th}\) is given by \(v_\text{ext}\) when \(i_\text{ext}=0\text{.}\) This occurs when there is no load placed across the output terminals, so we call it the open-circuit voltage \(V_\text{oc}\text{.}\)
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In the next section, we show that Figure 2.6.2 must represent the behavior of any circuit comprised of ideal voltage sources, ideal current sources, and resistors.

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Subsection 2.6.2 Proof of Thevenin’s and Norton’s Theorems

Let’s assume that inside of a box we have a circuit with two leads exiting the box (Figure 2.6.3). The circuit in the box is comprised of a complicated network of resistors \(R\text{,}\) \(\ell\) DC voltage sources \(V\text{,}\) and \(k\) ideal current sources \(I\text{.}\) We will also assume \(j\) unknown currents \(i\) that are dependent on the size of the voltage difference \(\Delta v_\text{ext}\) across the output terminals of the circuit.

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Let’s assume that if we apply the branch circuit analysis method, we end up with \(N_1\) independent junction law equations and \(N_2\) loop law equations. Each junction law equation will have the form

\begin{equation} a_{r,\text{ext}} i_{\text{ext}} + \sum\limits_{n=1}^j a_{rn} i_n + \sum\limits_{n=1}^k b_{rn} I_n=0\tag{2.6.3} \end{equation}

where \(a_{rn}\) and \(b_{rn}\) have values of +1, -1, or 0 depending on whether the currents \(i_n\) and \(I_n\) are entering, exiting, or not present at a junction and \(r\) is an index counting over KCL equations. Here, \(i_\text{ext}\) is the current that passes out of our circuit, through some external load, and back into our circuit.

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We will also end up with \(N_2\) independent loop law equations. All loops chosen for KVL analysis must be absent known current sources \(I_n\text{.}\) Since the current is already known on circuit legs containing circuit sources, so this restriction will not cost us anything. Each of these loop law equations will have the form

\begin{equation} m_{r,\text{ext}} \Delta v_\text{ext} + \sum\limits_{n=1}^\ell m_{rn} V_n + c_{r,\text{ext}} i_\text{ext} + \sum\limits_{n=1}^j c_{rn} i_n + \sum\limits_{n=1}^k d_{rn} I_n = 0\text{.}\tag{2.6.4} \end{equation}

The variables \(m_{rn}\) have values of either +1, or -1 depending on the battery orientation relative to the direction travelled around the loop, or a value of 0 for batteries not present in a given loop. Variables \(c_{rn}\) and \(d_{rn}\) have units of resistance (since each nonzero term in these sums represents a voltage change described by Ohm’s Law). Here, the index \(r\) counts over KVL equations.

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These \(N_1\) instances of (2.6.3) and \(N_2\) instances of (2.6.4) can be re-expressed as a matrix equation \(\mathbf{A}\mathbf{x}=\mathbf{v}\text{.}\) First, let’s rearrange these equations so that all unknowns are on the left side of the equation and all of the source terms are on the right

\begin{align*} a_{r,\text{ext}} i_{\text{ext}} + \sum\limits_{n=1}^j a_{rn} i_n \amp = - \sum\limits_{n=1}^k b_{rn} I_n\\ c_{r,\text{ext}} i_\text{ext} + \sum\limits_{n=1}^j c_{rn} i_n \amp = - m_{r,\text{ext}} \Delta v_\text{ext} - \sum\limits_{n=1}^\ell m_{rn} V_n - \sum\limits_{n=1}^k d_{rn} I_n \text{.} \end{align*}

We can express these equations in the form \(\mathbf{A}\mathbf{x}=\mathbf{v}\) if we let

\begin{equation*} \mathbf{A} = \begin{pmatrix} a_{1,\text{ext}} \amp a_{11} \amp \cdots \amp a_{1j}\\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ a_{N_1,\text{ext}} \amp a_{N_1 1}\amp \cdots \amp a_{N_1 j} \\ c_{1,\text{ext}} \amp c_{11} \amp \cdots \amp c_{1j} \\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ c_{N_2,\text{ext}} \amp c_{N_2 1} \amp \cdots \amp c_{N_2 j} \\ \end{pmatrix}\text{,} \end{equation*}

\begin{equation*} \mathbf{x} = \begin{pmatrix} i_\text{ext} \\ i_1 \\ \vdots \\ i_j \\ \end{pmatrix} \quad , \quad \mathbf{v} = \begin{pmatrix} -\sum\limits_{n=1}^{k} b_{1n} I_n \\ \vdots \\ -\sum\limits_{n=1}^{k} b_{N_1 n} I_n \\ -m_{1,\text{ext}}\Delta v_\text{ext} - \sum\limits_{n=1}^\ell m_{1n} V_n - \sum\limits_{n=1}^k d_{1n} I_n\\ \vdots \\ -m_{N_{2},\text{ext}}\Delta v_\text{ext} - \sum\limits_{n=1}^\ell m_{N_2 n} V_n - \sum\limits_{n=1}^k d_{N_2 n} I_n \end{pmatrix} \text{.} \end{equation*}

Now, every element in matrix \(\mathbf{A}\) is either a unitless number or a function of resistances only. More to the point, there is no voltage or current dependence in \(\mathbf{A}\text{.}\) If we multiply both sides of our matrix equation by \(\mathbf{A}^{-1}\text{,}\) we get

\begin{equation*} \mathbf{x}=\mathbf{A}^{-1}\mathbf{v}\text{.} \end{equation*}

The solution that is most important to us is \(i_\text{ext}\) as we wish to examine the relationship between output current and voltage. If we actually perform the calculations necessary to invert \(\mathbf{A}\text{,}\) use matrix multiplication to find an expression for \(i_\text{ext}\text{,}\) and then group source terms and rename constants multiplying each source (\(v_\text{ext},V,\ I\)), the resulting solution is of the form

\begin{equation*} i_\text{ext} = \alpha_0 v_\text{ext} + \sum\limits_{n=1}^\ell \alpha_n V_n + \sum\limits_{n=1}^k \beta_n I_n \end{equation*}

where all \(\alpha_n\) and \(\beta_n\) terms are functions of resistances only and are thus constants. This result demonstrates that \(i_\text{ext}\) is a linear function of \(v_\text{ext}\) with slope \(\alpha_0\) and a y-intercept that equals the two summations on the right side of the equation. So, we have now shown that any circuit comprised of solely linear circuit elements is itself a linear circuit.

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At this point, it is important to discuss that the internal behavior of the complicated circuit and simplified circuit are different. EXPAND ON THIS THOUGHT.

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Subsection 2.6.3 Finding Thevenin Equivalent Circuits

Given a complicated circuit, we can use the procedure provided in the proof above to find Thevenin and Norton equivalent circuit parameters.

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Example 2.6.4. Thevenin Example - Matrix Approach.

Coming soon.

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Here, we develop (2.6.3) and (2.6.4), convert them into a matrix equation, and use Python to solve for the Thevenin and Norton equivalent circuit parameters.

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Solution.

Coming soon

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If one wishes to find Thevenin and Norton equivalent parameters analytically, it is often easier to follow this procedure:

Use circuit analysis to find the open-circuit voltage \(V_{oc}\) which is equivalent to \(V_{oc}=V_A-V_B\) when nothing is connected externally between output terminals A and B. Then, \(V_{th}=V_{oc}\text{.}\)
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Find \(R_{th}=R_n\) which is equivalent to \(R_{AB}\) when all internal source voltages and currents are `turned off’.
- Turning a source `off’ means replacing voltage sources with wires and replacing current sources with breaks.
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- This method works as a consequence of application of the superposition theorem to the complicated circuit.
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Use your values for \(V_{th}\) and \(R_{th}\) to find \(I_n = V_{th}/R_{th}\)
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Let’s look at an example.

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Example 2.6.5. Thevenin Example - No Linear Algebra.

Find \(V_{th}\text{,}\) \(R_{th}\text{,}\) and \(I_n\) for the circuit in Figure 2.6.6.

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Answer.

\begin{align*} V_{th} \amp = 7.5\text{V} \\ I_n \amp = 1.36\text{mA} \\ R_{th} \amp = 5.5\text{k}\Omega \end{align*}

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Solution.

Find \(V_{th}:\) Assume an open circuit between terminals A and B such that \(I_2=0\text{.}\) Using mesh analysis, we find

\begin{equation*} 0 = V_1 - I_1 R_1 - I_1 R_3 - I_1 R_2 \end{equation*}

\begin{equation*} V_1 - I_1 (R_1 + R_2 + R_3) = 0\text{.} \end{equation*}

Then

\begin{align*} I_1 \amp = \frac{V_1}{R_1 + R_2 + R_3} \\ \amp = 1.5\text{mA} \text{.} \end{align*}

Thus, the voltage difference across \(R_2\) is \(I_1 R_3 = 4.5\text{V}\text{.}\) Since \(I_2=0\text{,}\) there is no voltage change across \(R_4\text{,}\) but there is an additional voltage change across \(V_2\) so that the open-circuit voltage between the output terminals is

\begin{equation*} V_{th}=V_{oc} = 7.5\text{V}\text{.} \end{equation*}

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Find \(R_{th}\text{:}\) If we turn off all sources in our circuit, we are left to find the resistance between terminals A and B in Figure 2.6.8. Looking for series and parallel resistances, we find that

\begin{equation*} R_{th}=\left(\left(R_1 + R_2\right)||R_3\right)+R_4\text{.} \end{equation*}

Plugging in resistance values, we find that

\begin{align*} R_{th} \amp = R_4 + \left(\left(R_1 + R_2\right)^{-1}+R_3^{-1}\right)^{-1}\\ \amp = 5.5\text{k}\Omega \text{.} \end{align*}

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Find \(I_{n}\text{:}\) The Norton current \(I_n=V_{th}/R_{th}=1.36\)mA.

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Sometimes, one does not have a circuit schematic but only has the physical circuit. In this case, a digital multimeter can be used to physically perform the following procedure:

Find the open-circuit voltage \(V_{oc}\) which is equivalent to \(V_{oc}=V_A-V_B\) when nothing is connected externally between output terminals A and B. Then, \(V_{th}=V_{oc}\text{.}\)
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Find the short-circuit current \(I_{sc}\) which is the current from terminal A to terminal B when an external wire connects the two outputs. Then \(I_n = I_{sc}\text{.}\)
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The Thevenin resistance (which is equivalent to the Norton resistance) is then \(R_{th} = R_n =V_{th}/I_n\text{.}\)
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Look at section 41.6 to add color to example blocks. also section 28.2 and similar. xsl stylesheets

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