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Learn It!

Almost all electronic devices we use are "black boxes": we do not have to know what they contain or how exactly they work, but we do have to know how they will behave when we use them.

Figure 1

A black box subcircuit is an unknown subcircuit that can be connected to another subcircuit. As long as we know how the unknown subcircuit behaves when it is connected to any other subcircuit, we can determine how the completed circuit operates. The information we need about the unknown subcircuit is its I-V characteristic because we can use it to find the completed circuit's operating point using the load line method.

Figure 2

If two subcircuits have identical I-V characteristics, they behave exactly the same from the point of view of a connected subcircuit, so we say they are equivalent. The Thévenin theorem states that any black box subcircuit containing exclusively sources and resistors is equivalent to a subcircuit containing a voltage source in series with a resistor.

Figure 3

The Norton theorem states that any black box subcircuit containing exclusively sources and resistors is equivalent to a subcircuit containing a current source in parallel with a resistor.

Figure 4

The Thévenin and Norton equivalent subcircuits are also equivalent to each other for appropriate values of $V_T$, $I_N$ and $R_{\text{eff}}$. Note that the effective resistance $R_{\text{eff}}$ is the same value for both of these subcircuits.

The following table describes how to find the values of $V_T$, $I_N$ and $R_{\text{eff}}$ from the unknown subcircuit.

Value | Diagram/Equation | Method |
---|---|---|

$V_T$ | - Keep the terminals of the unknown subcircuit open (i.e. disconnected) so that the open-circuit current $I_{\text{oc} }=0$.
- Find the open-circuit voltage $V_{\text{oc} }$ across the terminals. Note that $V_{\text{oc} }$ can be measured in lab using a voltmeter across the terminals since an ideal voltmeter's infinite resistance keeps the circuit open.
- Set $V_T=V_{\text{oc} }$.
| |

$I_N$ | - Short the terminals of the unknown subcircuit so that the short-circuit voltage $V_{\text{sc} }=0$.
- Find the short-circuit current $I_{\text{sc} }$ between the terminals. Note that $I_{\text{sc} }$ can be measured in lab using an ammeter across the terminals since an ideal ammeter's zero resistance shorts the circuit.
- Set $I_N=I_{\text{sc} }$.
| |

$R_{\text{eff} }$ | - Turn off (i.e. zero) all the sources in the unknown subcircuit. Specifically, zero voltage sources behave as short circuits and zero current sources behave as open circuits. Remember these facts by noticing that there is always zero voltage drop across a short circuit and always zero current through an open circuit.
- Set $R_{\text{eff} }$ equal to the equivalent resistance between the terminals.
| |

$V_T$, $I_N$ or $R_{\text{eff} }$ | $R_{\text{eff}}=\frac{V_T}{I_N}$ | Given any two of $V_T$, $I_N$ and $R_{\text{eff} }$, solve $R_{\text{eff}}=V_T/I_N$ to find the other one. |

Most of the results in Table 1 can be derived by comparing the I-V characteristic of an unknown subcircuit consisting of sources and resistors with those of the Thévenin and Norton subcircuits.

Subcircuit | Schematic | I-V Characteristic | Explanation |
---|---|---|---|

Unknown subcircuit | Any subcircuit containing exclusively sources and resistors has a linear I-V characteristic. This line has $V$-intercept equal to $V_{\text{oc} }$ since $I_{\text{oc} }=0$, and it has $I$-intercept equal to $I_{\text{sc} }$ since $V_{\text{sc} }=0$. Therefore, the slope is $-I_{\text{sc} }/V_{\text{oc} }$. | ||

Thévenin subcircuit | This line is the I-V characteristic of a voltage source and resistor in series. | ||

Norton subcircuit | This line is the I-V characteristic of a current source and resistor in parallel. |

For the three subcircuits in Table 2 to be equivalent, their I-V characteristics must be identical. Therefore, we equate the intercepts and slopes to obtain three of the results in Table 1.

\begin{align}

V_T&=V_{\text{oc} } & & \text{by equating }V\text{-intercepts} \label{TNE:EQ1}\\

I_N&=I_{\text{sc} } & & \text{by equating }I\text{-intercepts} \label{TNE:EQ2} \\

R_{\text{eff}}&=\frac{V_{\text{oc}}}{I_{\text{sc}}}=\frac{V_T}{I_N} & & \text{by equating slopes} \label{TNE:EQ3}

\end{align}

Note that, in Table 2, we chose to label the voltage drop $V$ across the terminals with $+$ label at the top and $-$ label at the bottom and we chose to label the current $I$ with current arrow label pointing out of the terminal at the top. These choices are consistent with the labeling in Fig. 2 to Fig. 4 and in Table 1. If $V$ and $I$ are labeled differently, then the I-V characteristics appear horizontally and/or vertically flipped and some of the equations $\eqref{TNE:EQ1}$ to $\eqref{TNE:EQ3}$ change sign. It is better to reason about the labeling and signs than to try to memorize formulas and that is why we do not list all the cases.

We have seen that Thévenin and Norton equivalents can represent unknown subcircuits, but they can also be used to simplify known subcircuits. Consider two subcircuits connected together as shown below.

Figure 5

The table below gives example procedures for determining the Thévenin and Norton equivalents of the two subcircuits in Fig. 5.

Step | Subcircuit 1 | Subcircuit 2 |
---|---|---|

Draw known subcircuit | ||

Find one of the values $V_T$, $I_N$ or $R_{\text{eff} }$ | ||

Find another value out of $V_T$, $I_N$ or $R_{\text{eff} }$ | ||

Find the remaining value $V_T$, $I_N$ or $R_{\text{eff} }$ | $I_N=\frac{V_T }{R_{\text{eff} } }=\frac{3 \text{ V}}{2\text{ k}\Omega} =\frac{3}{2} \text{ mA}$ | $V_T=I_N R_{\text{eff} }=\left( 2 \text{ mA} \right)\left( 5\text{ k}\Omega \right)= 10\text{ V}$ |

Draw Thévenin equivalent | ||

Draw Norton equivalent |

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