CONVERSION OF A SINUSOIDAL VOLTAGE GENERATOR INTO AN EQUIVALENT CURRENT GENERATOR

V. P. SHERSTOBITOV

(Presented by the scientific seminar of the Faculty of Automation and Computer Engineering)

In practice, there sometimes arises a need for current generators, sources with a large internal resistance and a sufficiently large emf. For this purpose, voltage generators are converted into current generators, since natural current generators almost do not exist.

There is a method for converting a voltage source into a current source, the essence of which consists in the use of feedback. However, such a conversion requires summing amplifiers. Below we consider a simple resonant four-terminal network which, within limited ranges of load variation, provides conversion of a sinusoidal-voltage generator into an equivalent current generator.

Let there be a T-shaped four-terminal network, shown in Fig. 1. Here \(Z_1, Z_2, Z_3\) are the elements of the four-terminal network, \(U\) is the sinusoidal-voltage generator, and \(Z_{\mathrm{н}}\) is the total load resistance.

Fig. 1

For a current generator it is necessary that the current \(J_2\) not depend on the magnitude of the load. By known methods of analysis it is easy to find the current in the load for the given circuit; it will have the expression:

\[ I_2=\frac{U\cdot Z_3}{Z_{\mathrm{н}}(Z_1+Z_3)+Z_1\cdot Z_2+Z_1\cdot Z_3+Z_2\cdot Z_3}. \tag{1} \]

For ideal conversion to be performed, it is sufficient that the condition

\[ Z_1+Z_3=0 \tag{1a} \]

be satisfied.

The input current of the four-terminal network is determined by:

\[ I_1=\frac{U(Z_2+Z_3+Z_{\mathrm{н}})}{Z_{\mathrm{н}}(Z_1+Z_3)+Z_1\cdot Z_2+Z_1\cdot Z_3+Z_2\cdot Z_3}. \tag{2} \]

With an active load \(Z_{\mathrm{n}}=R_{\mathrm{n}}\), in order to ensure an active input resistance of the four-terminal network it is necessary that the condition be satisfied:

\[ Z_2+Z_3=0. \tag{2a} \]

Conditions 1a and 2a can be satisfied if \(Z_1, Z_2, Z_3\) are reactive resistances; in this case \(Z_3\) must have a reactance opposite in character to \(Z_1\) and \(Z_2\). The indicated reactances at the generator frequency must have resistances equal in magnitude. Thus, the four-terminal network becomes a resonant network of type \(a\) or \(b\) (Fig. 2).

Fig. 2

The general expressions 1 and 2 for these circuits are simplified:

\[ I_2=\frac{U}{Z_{1,2}}, \tag{3} \]

\[ I_1=\frac{U\cdot Z_{\mathrm{n}}}{Z_{1,2}\cdot Z_3}. \tag{4} \]

For the circuit of Fig. 2a:

\[ Z_{1,2}=-j\rho,\quad I_2=j\frac{U}{\rho},\quad I_1=\frac{U\cdot R_{\mathrm{n}}}{\rho^2}. \]

For the circuit of Fig. 2b:

\[ Z_{1,2}=j\rho,\quad I_2=-j\frac{U}{\rho},\quad I_1=\frac{U\cdot R_{\mathrm{n}}}{\rho^2}. \]

Here \(\rho\) is the modulus of the reactive resistances of the four-terminal network at the frequency \(f\) of the supply generator \(U\).

As can be seen, the conversion of the source \(U\) into a current generator is carried out with the corresponding phase shift of \(\pm 90^\circ\). Let us analyze the accuracy of the conversion of a sinusoidal-voltage generator into an equivalent current generator by these four-terminal networks.

For the analysis we introduce errors of the elements into \(Z_1, Z_2, Z_3\). We take \(Z_{\mathrm{n}}=R_{\mathrm{n}}\).

Let us first consider the circuit of Fig. 2b. For this circuit we have:

\[ \left. \begin{aligned} Z_1&=j\rho\left[1-\frac{j}{Q_1}+\frac{\Delta L_1}{L}+\frac{\Delta f}{f}\right],\\ Z_2&=j\rho\left[1-\frac{j}{Q_2}+\frac{\Delta L_2}{L}+\frac{\Delta f}{f}\right],\\ Z_3&=-j\rho\left[1+\frac{j}{Q_3}-\frac{\Delta C_3}{C}-\frac{\Delta f}{f}\right]. \end{aligned} \right\} \tag{5} \]

\(Q_1, Q_2, Q_3\) are, respectively, the quality factors of the reactive elements of the four-terminal network,

\(\Delta L_1,\ \Delta L_2,\ \Delta C_3\) are the absolute errors in setting the reactive elements relative to \(L\) and \(C\);

\(\Delta f\) is the deviation of the frequency of generator \(U\) from the nominal value

\[ f=\frac{1}{2\pi\sqrt{LC}}. \]

The internal active resistance of the voltage generator is included in the quality factor \(Q_1\). In expressions 5, quantities of small order have been discarded, which is valid for small deviations of the parameters.

We substitute the values \(Z_1, Z_2, Z_3\), and \(Z_{\mathrm{H}}\) into formula (1) and carry out the necessary algebraic transformations. Here we also neglect quantities of small order, determined by the product of small quantities.

Let us introduce the load coefficient \(\alpha\), defined by the expression:

\[ \alpha=\frac{R_{\mathrm{H}}}{\rho}. \tag{6} \]

Taking the above remarks into account, we obtain:

\[ I_2=I_{2m}\cdot e^{j\varphi}. \]

\[ I_{2m}=-j\frac{U}{\rho}\left[1-\frac{\alpha}{2Q_1}-\frac{\alpha}{2Q_3}-\frac{\alpha^2}{2}\left(\frac{\Delta L_1}{L}+\frac{\Delta C_3}{C}+\frac{2\Delta f}{f}+\frac{2}{\alpha Q_3}\right)^2\right] \tag{7} \]

\[ \varphi\simeq-\alpha\left(\frac{\Delta L_1}{L}+\frac{\Delta C_3}{C}+\frac{2\Delta f}{f}+\frac{2}{\alpha Q_3}\right). \tag{8} \]

Let us determine the coefficient \(K\), equal to the ratio of the current of the real four-terminal network to the current of the ideal four-terminal network:

\[ K=1-\frac{\alpha}{2Q_1}-\frac{\alpha}{2Q_3}-\frac{\alpha^2}{2}\left(\frac{\Delta L_1}{L}+\frac{\Delta C_3}{C}+\frac{2\Delta f}{f}+\frac{2}{\alpha Q_3}\right)^2 \tag{7a} \]

For \(\alpha=1,\ Q_1=100,\ Q_3=\infty\) (the capacitors have a very high quality factor),

\[ \frac{\Delta L_1}{L}=\frac{\Delta C_3}{C}=\frac{\Delta f}{f}=10^{-2}, \]

\(K=1-5.8\cdot10^{-3}\), i.e., the error does not exceed \(0.58\%\).

The phase shift \(\varphi\) according to formula (8):

\[ \varphi=-2^\circ18'. \]

For the circuit of Fig. 2a, analogously we have:

\[ K=1-\frac{\alpha}{2Q_1}-\frac{\alpha}{2Q_3}-\frac{\alpha^2}{2}\left(\frac{\Delta L_3}{L}+\frac{\Delta C_1}{C}+\frac{2\Delta f}{f}-\frac{2}{\alpha Q_3}\right)^2, \tag{9} \]

\[ \varphi=-\alpha\left(\frac{\Delta L_3}{L}+\frac{\Delta C_1}{C}+\frac{2\Delta f}{f}-\frac{2}{\alpha Q_3}\right). \tag{10} \]

For \(Q_1=100;\ Q_3=100;\ \alpha=1,\)

\[ \frac{\Delta L_3}{L}=\frac{\Delta C_1}{C}=\frac{\Delta f}{f}=10^{-2}, \]

\[ K=1-10^{-2}\quad \varphi=-0.02=-1^\circ9', \]

i.e., the error does not exceed \(1\%\) when the load varies from \(0\) to \(\rho\).

Thus, the circuits of Fig. 2 provide a sufficiently accurate conversion of sinusoidal voltage generators into equivalent current generators with the corresponding phase shift close to \(\pm 90^\circ\), depending on the nature of the reactive elements \(Z_1, Z_2, Z_3\).