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).
Practical Example of a realistic Source: For the ideal voltage source, it was defined that it always supplies the same voltage independent of the load. In Abbildung 2, in contrast, an example of a „realistic“ voltage source is shown as an active two-terminal network.
This realization shall now be described with some technical terms:
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.
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.
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.
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:
The conversion is now done in such a way that the same characteristic curve is obtained:
From linear voltage source to linear current source:
Given: Source voltage $U_0$, resp. open circuit voltage $U_{\rm OC}$, internal resistance $R_\rm i$
in question: source current $I_0$, resp. short circuit current $I_{\rm SC}$, internal conductance $G_\rm i$
$\boxed{I_{\rm SC} = {{U_{\rm OC}}\over{R_\rm i}}}$ , $\boxed{G_\rm i = {{1}\over{R_\rm i}}}$
From linear current source to linear voltage source:
Given: Source current $I_0$, resp. short-circuit current $I_{\rm SC}$, internal resistance $G_\rm i$
in question: source voltage $U_0$, resp. open-circuit voltage $U_{\rm OC}$, internal resistance $R_\rm i$
$\boxed{U_{\rm OC} = {{I_{\rm SC}}\over{G_\rm i}}}$ , $\boxed{R_\rm i = {{1}\over{G_\rm i}}}$
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.
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 seen as
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.
This knowledge can now be used for more complicated circuits. In Abbildung 11 such a circuit is drawn. This is to be converted into a searched equivalent conductance $G_{\rm eq}$ and a searched equivalent current source with $I_{\rm eq}$.
Important here: Only two-terminal networks can be converted via source duality. This means that only 2 nodes may act as output terminals for selected sections of the circuit. If there are more nodes the conversion is not possible.
The equivalent circuits for the ideal sources can be seen via the circuit diagrams (see Abbildung 12).
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.
Given a Thevenin source $U_{\rm OC}=5.00~\rm V$ and $R_{\rm i}=250~\Omega$, find the Norton pair $(I_{\rm SC},G_{\rm i})$.
Simplify the following circuits by the Norton theorem to a linear current source (circuits marked with NT) or by Thevenin theorem to a linear voltage source (marked with TT).
Next, we figure out the current in the short circuit. In case of a short circuit, we have $2~V$ in a branch which in turn means there must be $−2~V$ on the resistor. The current through that branch is \begin{equation*} I_R=\frac{2~V}{8~\Omega} \end{equation*}
The current in question is the sum of both the other branches \begin{equation*} I_S= I_R + 1~A \end{equation*}
To substitute the circuit in $b)$ first we determine the inner resistance. Shutting down all sources leads to \begin{equation*} R_{\rm i}= 4 ~\Omega \end{equation*}
Next, we figure out the voltage at the open circuit. Thus we know the given current flows through the ideal current source as well as the resistor. The voltage drop on the resistor is \begin{equation*} R_{\rm i}= -4~\Omega \cdot 2~A \end{equation*}
The voltage at the open circuit is \begin{equation*} U_{\rm S}= 2~V + 1~V + U_R \end{equation*}
DC Voltage & Current Source Theory
Intro to superposition (method used when deactivating sources)
A more complex superposition example