Electrical Engineering BMT & UFC › Real Sources

Real Current Source

A real current source has an output current that depends on the output voltage. This is modelled by a high-resistance resistor placed in parallel with an ideal internal current source. Part of the source current flows into this resistor, namely the error current I_F.

\[ \begin{gathered} I_F = \frac{U_{\mathrm{Aus}}}{R_i} \\[4pt] I_{\mathrm{Aus}} = I_0 - I_F \end{gathered} \]

Only a reduced output current is available at the terminals of the real current source. The higher the output voltage, the higher the voltage across the internal resistance. As a result, the error current increases and the output current decreases.

We again determine the parameters of the equivalent circuit in open circuit and short circuit. Short-circuiting a current source is not a problem. It then simply pushes the output current through an ideal short-circuit conductor. In the short circuit, the following applies:

\[ \begin{gathered} U_{\mathrm{Aus}} = U_{Ri} = 0\,\mathrm{V} \rightarrow I_F = \frac{U_{Ri}}{R_i} = 0\,\mathrm{A} \\[4pt] I_{\mathrm{Aus}} = I_0 = I_{\mathrm{Kurzschluss}} \end{gathered} \]

At this operating point, we measure the short-circuit current. The short-circuit current equals the current of the internal ideal current source. In open circuit, no load is connected to the output terminals. The following applies:

\[ \begin{gathered} I_{\mathrm{Aus}} = 0\,\mathrm{A} \\[4pt] I_F = I_0 \\[4pt] U_{\mathrm{Aus}} = R_i \cdot I_F \\[4pt] R_i = \frac{U_{\mathrm{Aus}}}{I_F} = \frac{U_{\mathrm{Aus}}}{I_0} = \frac{U_{\mathrm{Aus}}}{I_{\mathrm{Kurzschluss}}} \end{gathered} \]

In open circuit, we measure the output voltage. From the open-circuit voltage and the short-circuit current, we again determine the internal resistance of the real current source.

Let us look at a numerical example:

\[ \begin{gathered} I_0 = 100\,\mathrm{mA} \\[4pt] R_i = 10\,\mathrm{k\Omega} \\[4pt] U_{\mathrm{Aus}} = 10\,\mathrm{V} \\[4pt] I_{\mathrm{Aus}} = I_0 - \frac{U_{\mathrm{Aus}}}{R_i} = 100\,\mathrm{mA} - \frac{10\,\mathrm{V}}{10\,\mathrm{k\Omega}} = 100\,\mathrm{mA} - 1\,\mathrm{mA} = 99\,\mathrm{mA} \end{gathered} \]

Simulation

The deviation in this example is 1 % of the nominal current.

The idea of a power source

It is not possible to specify both the voltage and the current of a source at the same time. One of the two parameters can be fixed; then the other must be flexible. Otherwise it would be mathematically impossible to operate different resistors on the source.

Problem: Check whether a source can specify voltage and current at the same time. To do this, model an ideal combined voltage/current source that specifies both the output voltage U = 5 V and the output current I = 1 A and is operated with a resistor R = 1 Ω as the load. Do you see why this cannot work?