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Block 06 — Real sources and source equivalents

Learning objectives

90-minute plan

1. Warm-up (8–10 min):
   1. Spot the difference: ideal vs. real source (show $U$–$I$ lines).
   2. Quick quiz: “How to deactivate an ideal voltage/current source?”
2. Core concepts & derivations (60–65 min):
   1. Linear source model; $U_{\rm OC}$, $I_{\rm SC}$; load line and operating point.
   2. Thevenin/Norton duality & conversion; internal resistance by deactivation.
   3. Efficiency $\eta$ and utilization $\varepsilon$; maximum power transfer.
   4. Two-terminal equivalents; loaded voltage divider as a Thevenin source.
3. Guided practice (10–15 min): 2 short numericals + 1 sim task (see “Exercises”).
4. Wrap-up (5 min): Summary + pitfalls.

Conceptual overview

Core content

It is known from everyday life that battery voltages drop under heavy loads. This can be seen, for example, when turning the ignition key in winter: The load from the starter motor is sometimes so large that the car lights or radio briefly cuts out.
Another example is a $1.5~\rm V$ battery: If such a battery is short-circuited by a piece of wire, the current flows does not let the wire glow. It is noticeably less cutrrent.

So it makes sense here to develop the ideal voltage source concept further. In addition, we will see that this also opens up the possibility of converting and simplifying more complicated circuits.

First, the concept of the two-terminal from the chapter basics and basic concepts is to be expanded (Abbildung 1).

1. As passive two-terminal network which acts exclusively as a consumer. Thus it is valid for the passive two-terminal network that the current-voltage-characteristic always runs through the origin (see also chapter simple circuits).
2. Active two-terminal networks, on the other hand, also act as generators of electrical energy. Thus, the current-voltage characteristic there does not pass through the origin. Active two-terminal networks always contain at least one source (i.e. at least one current or voltage source).

From ideal to linear sources: load line, $U_{\rm OC}$ and $I_{\rm SC}$

This realization shall now be described with some technical terms:

• It is called open circuit when no current is drawn from an active two-terminal network: $I_{\rm SC}=0$.
  The voltage corresponds to the open circuit voltage $U=U_{\rm OC}$ (German: Leerlaufspannung $U_{\rm LL}$).
  The open circuit power is $P_{\rm OC}=U_{\rm OC} \cdot I_{\rm OC} = 0$.
• The term short circuit is used when the terminals of the two-terminal network are bridged without resistance. The current then flowing is called the short-circuit current $I=I_{\rm SC}$ (German: Kurzschlussstrom $I_{\rm KS}$).
  The short-circuit voltage is $U_{\rm SC}=0~\rm V$.
  Also, the short-circuit power is $P_{\rm SC}=U_{\rm SC} \cdot I_{\rm SC} = 0$.
• The active two-terminal network outputs power to a connected load in the region between no-load and short-circuit.

Important: As seen in the following, the short-circuit current can cause considerable power loss inside the two-terminal network and thus a lot of waste heat. Not every real two-terminal network is designed for this.

What is interesting now is the current-voltage characteristic of the circuit in Abbildung 2. This can be seen in the simulation below. The result is a linear curve (see Abbildung 3).

From a purely mathematical point of view, the course can be represented by the basic equation of linear graphs with the y-axis intercept $I_{\rm SC}$ and a slope of $-{{I_{\rm SC}}\over{U_{\rm OC}}}$:

\begin{align*} I = I_{\rm SC} - {{I_{\rm SC}}\over{U_{\rm OC}}}\cdot U \tag{3.1.1} \end{align*}

On the other hand, the formula can also be resolved to $U$:

\begin{align*} U = U_{\rm OC} - {{U_{\rm OC}}\over{I_{\rm SC}}}\cdot I \tag{3.1.2} \end{align*}

So what does the inside of the linear source look like? In Abbildung 4 two possible linear sources are shown, which will be considered in the following.

Linear Voltage Source

The linear voltage source consists of a series connection of an ideal voltage source with the source voltage $U_0$ (English: EMF for ElectroMotive Force) and the internal resistance $R_\rm i$. To determine the voltage outside the active two-terminal network, the system can be considered as a voltage divider. The following applies:

\begin{align*} U = U_0 - R_{\rm i} \cdot I \end{align*}

The source voltage $U_0$ of the ideal voltage source will be measured at the terminals of the two-terminal network if this is unloaded. Then no current flows through the internal resistor $R_i$ and there is no voltage drop there. Therefore: The source voltage is equal to the open circuit voltage $U_0 = U_{\rm OC}$.

\begin{align*} U = U_{\rm OC} - R_{\rm i} \cdot I \end{align*}

When the external voltage $U=0$, it is the short circuit case. In this case, $0 = U_{\rm OC} - R_{\rm i} \cdot I_{\rm SC}$ and transform $R_{\rm i} = {{U_{\rm OC}}\over{I_{\rm SC}}}$. Thus, equation $(3.1.2)$ is obtained: \begin{align*} U = U_{\rm OC} - {{U_{\rm OC}}\over{I_{\rm SC}}} \cdot I \end{align*}

Is this the structure of the linear source we are looking for? To verify this, we will now look at the second linear source.

Linear Current Source

The linear current source now consists of a parallel circuit of an ideal current source with source current $I_0$ and internal resistance $R_{\rm i}$, or internal conductance $G_{\rm i} = {{1}\over{R_{\rm i}}}$. To determine the voltage outside the active two-terminal, the system can be considered as a current divider. Here, the following holds:

\begin{align*} I = I_0 - G_{\rm i} \cdot U \end{align*}

Here, the source current can be measured at the terminals in the event of a short circuit. The following therefore applies: $I_{\rm SC}= I_0$

\begin{align*} I = I_{SC} - G_{\rm i} \cdot U \end{align*}

When the external current $I=0$, it is the no-load case. In this case, $0 = I_{\rm SC} - G_{\rm i} \cdot U_{\rm OC}$ and transform $G_{\rm i} = {{I_{\rm SC}}\over{U_{\rm OC}}}$.
Thus, equation $(3.1.1)$ is obtained: \begin{align*} I = I_{\rm SC} - {{I_{\rm SC}}\over{U_{\rm OC}}} \cdot U \end{align*}

So it seems that the two linear sources describe the same thing.

Duality of Linear Sources

Through the previous calculations, we came to the interesting realization that both the linear voltage source and the linear current source provide the same result. It is true: For a linear source, both a linear voltage source and a linear current source can be specified as an equivalent circuit! As already in the case of the star-delta transformation, this not only provides two explanations for a black box. Also, here linear voltage sources can be transformed into linear current sources and vice versa.

The Abbildung 5 compares again the two linear sources and their characteristics:

1. The linear voltage source is given by the source voltage $U_0$, or the open circuit voltage $U_{\rm OC}$ and the internal resistance $R_{\rm i}$.
2. The linear current source is given by the source current $I_0$, or the short-circuit current $I_{\rm SC}$ and the internal conductance $G_{\rm i}$.

The conversion is now done in such a way that the same characteristic curve is obtained:

Operating Point of a real Voltage Source

Abbildung 6 shows the characteristics of the linear voltage source (left) and a resistive resistor (right). For this purpose, both are connected to a test system in the simulation: In the case of the source with a variable ohmic resistor, and in the case of the load with a variable source. The characteristic curves formed in this way were described in the previous chapter.

The operating point can be determined from both characteristic curves. This is assumed when both the linear voltage source is connected to the ohmic resistor (without the respective test systems). In Abbildung 7 both characteristic curves are drawn in a current-voltage diagram. The point of intersection is just the operating point that sets in. If the load resistance is varied, the slope changes in inverse proportion, and a new operating point is established (light grey in the figure).

The derivation of the working point is also here explained again in a video.

Conversion of any linear two-terminal Network

In Abbildung 8, it can be seen that the internal resistance of the linear current source measured by the ohmmeter (resistance meter) is exactly equal to that of the linear voltage source.

In the simulation, a measuring current $I_\Omega$ is used to determine the resistance value. ¹⁾ Let us have a look at the properties of the ohmmeter in the simulation by double-clicking on the ohmmeter. Here, a very large measuring current of $I_\Omega=1~\rm A$ is used. This could lead to high voltages or the destruction of components in real setups.

In order to understand why is this nevertheless chosen so high in the simulation, do the following: Set the measuring current for both linear sources to (more realistic) $1~\rm mA$. What do you notice?

The circuit in Abbildung 9 shows this circuit again. The ohmmeter is replaced by a current source and a voltmeter since only the electrical properties are important in the following. In this setup, it can be seen that the current through $G_{\rm i}$ is just given by $I_{\rm i} = I_0 + I_\Omega$ (node theorem). Thus, the two sources in the circuit can be reduced.

This should make the situation clear with a measuring current of $1~\rm mA$. The voltage at the resistor is now given by $U_\Omega = R \cdot (I_0 + I_\Omega)$. Only when $I_\Omega$ is very large does $I_0$ become negligible. The current of a conventional ohmmeter cannot guarantee this for every measurement.

Any interconnection of linear voltage sources, current sources, and ohmic resistors can be.

• as a single, linear voltage source
  (Thévenin theorem ) or
• as a single, linear current source
  (Norton theorem )

In Abbildung 10 it can be seen that the three circuits give the same result (voltage/current) with the same load. This is also true when an (AC) source is used instead of the load.

Simplified Determination of the internal Resistance

Thus also the equivalent resistance of the complex circuit above can be derived quickly.
For the source current $I_0$ ideal equivalent current source resp. the source voltage $U_0$ ideal equivalent voltage source this derivation can not be used.

Exercises

Common pitfalls

• Wrong deactivation: do not set an ideal voltage source to open or an ideal current source to short; the rules are: $U$-source→short, $I$-source→open.
• Confusing goals: max power ($R_{\rm L}=R_{\rm i}$, $\eta=50\%$) vs. high efficiency ($R_{\rm L}\gg R_{\rm i}$). Don’t equate them.
• Ignoring ratings: not every real source is short-circuit-proof—$I_{\rm SC}$ is a model parameter, not a recommended experiment.
• Mixed conventions: keep the passive sign convention for loads; use conventional current ($+$ to $-$).

Exercises

Quick checks

Longer exercises

Embedded resources

Summary

• A linear (real) source is fully determined by $(U_{\rm OC},I_{\rm SC})$ or equivalently $(U_0,R_{\rm i})$ / $(I_0,G_{\rm i})$; both forms are equivalent.
• Thevenin ↔ Norton: $U_{\rm OC}=I_{\rm SC}R_{\rm i}$, $G_{\rm i}=1/R_{\rm i}$; deactivation rules let you find $R_{\rm i}$ quickly.
• Efficiency vs. maximum power: choose $R_{\rm L}\gg R_{\rm i}$ for high $\eta$, or $R_{\rm L}=R_{\rm i}$ for max $P_{\rm L}$.
• Two-terminal reductions (e.g., loaded divider) simplify analysis of larger networks.