ON THE CALCULATION OF CIRCUITS WITH A VALVE

V. A. Lukutin, I. P. Guk

(Presented by the scientific seminar of the Department of Theoretical Foundations of Electrical Engineering)

At present there are many different methods for calculating currents and voltages in circuits with rectifiers. All of them have certain advantages and disadvantages and, as a rule, can be applied to a quite definite class of problems. With a piecewise-linear approximation of the volt-ampere characteristic of the valve, the most accurate solution is given by the method of fitting. It makes it possible to calculate not only steady-state but also transient processes in circuits with one or two valves.

The essence of the method of fitting consists in the fact that, for separate intervals of valve operation, the process in the electrical circuit can be described by linear differential equations with constant coefficients. In this case it becomes necessary to determine the boundaries of the intervals and to match the currents and voltages at these boundaries.

The application of the method of fitting is complicated by the fact that, in determining the boundaries of the intervals of valve operation, one has to encounter serious difficulties in solving transcendental equations. Nevertheless, the method of fitting proves to be very effective for a certain type of problems rather often encountered in engineering practice. Thus, the widely used circuit of a rectifier feeding an active load through a choke is calculated very easily, since in it there is essentially no transient process and, from the moment of switching on, a steady-state process immediately arises in the circuit. In this case it is necessary to calculate only one interval, first determining its boundaries.

In the circuit under consideration, with the adopted approximation of the volt-ampere characteristic, the valve can be replaced by a switch that closes at the moment when the supply voltage passes through zero and remains closed until the moment of time when the current of the transient process passes through zero from the region of positive values into the negative. The moment of opening of the switch will correspond to the beginning of the valve-blocking period, and the reserve of energy in the magnetic field of the inductance will be equal to zero. The next conduction period of the valve will again begin from the moment when the supply emf passes through zero, and the whole process begins to repeat itself.

For the closed state of the switch, the equation

\[ L\frac{di}{dt}+ri=E_m\sin\Theta, \]

will be valid, whose solution gives us the desired current through the valve:

\[ i(t)=\frac{E_m}{\sqrt{r^2+x^2}}\left[\sin(\Theta-\varphi)+\sin\varphi\cdot e^{-\frac{r}{x}\Theta}\right]. \tag{1} \]

The boundary of the interval in which this current exists is found from the following equality:

\[ \sin(\Theta-\varphi)+\sin\varphi e^{-\frac{r}{x}\Theta}=0. \tag{2} \]

This transcendental equation can be solved approximately if the corresponding functions are replaced by power series, after first carrying out certain transformations:

\[ p\sin\Theta-\cos\Theta+e^{-p\Theta}=0. \]

\[ p\left(\Theta-\frac{\Theta^3}{3!}+\frac{\Theta^5}{5!}-\ldots\right) -\left(1-\frac{\Theta^2}{2!}+\frac{\Theta^4}{4!}-\ldots\right)+ \]

\[ +\left(1-\frac{p\Theta}{1!}+\frac{p^2\Theta^2}{2!}-\frac{p^3\Theta^3}{3!}+\ldots\right)=0. \tag{3} \]

It is known that algebraic equations above the fourth degree cannot be solved in radicals; therefore we must restrict the series to the first two or three terms. It is quite obvious that the fewer terms of the series we take, the greater the error we shall allow in the solution. However, the convergence of the series of the functions \(e^{-p\Theta}\), \(\sin\Theta\), and \(\cos\Theta\), for a small number of terms, will be the better the smaller the values of \(p\Theta\) and \(\Theta\) are. It turns out that, if \(\Theta\) or \(p\Theta\) is less than unity, then for a sufficiently accurate solution of equation (3) it is necessary to take into account only the second degree of the corresponding series.

In the circuit under consideration, the cut-off angle may vary within rather wide limits:

\[ \Theta=\pi \quad \text{for} \quad p=\frac{r}{\omega L}=\infty, \]

\[ \Theta=2\pi \quad \text{for} \quad p=\frac{r}{\omega L}=0. \]

In any case the angle \(\Theta\) varies from 3.14 to 6.28, and therefore, for its accurate determination, many terms of the series must be taken, which makes an analytical solution impossible. A way out of this difficulty can be found if in equation (3) the corresponding substitution is made:

\[ \Theta=\psi+\Theta', \]

where the angle \(\psi\) is chosen on the basis of the specific values of the circuit parameters, and then \(\Theta'\) may be appreciably less than unity.

On the basis of the investigations, the following substitutions may be recommended:

a) for \(0 \leq p \leq 0.1\), substituting \(\Theta = 2\pi + \Theta'\) into equation (3), we obtain the simple formula

\[ p\sin\Theta' - \cos\Theta' + e^{-2\pi p}\cdot e^{-p\Theta'} = 0, \]

in which we replace the functions by series and obtain the computational equation

\[ \Theta'^2 + 2pA\Theta' - 2A = 0,\quad \text{where } A = 1 - e^{-2\pi p}. \]

The required cutoff angle will be

\[ \Theta' = -pA - \sqrt{p^2A^2 + 2A}; \tag{4} \]

b) for \(0.1 \leq p \leq 1\), substituting \(\Theta = \frac{3}{2}\pi + \Theta'\) into equation (3) and performing operations analogous to those in item \(a\), we find the required cutoff angle

\[ \Theta' = \left(B + \frac{1}{p}\right) - \sqrt{B^2 + \frac{1}{p^2} + 2}, \tag{5} \]

where \(B = e^{-\frac{3}{2}\pi p}\);

c) for \(1 < p \leq 5\), \(\Theta = 1\frac{1}{6}\pi + \Theta'\),

\[ \Theta' = \frac{(2Cp + p\sqrt{3} + 1) - \sqrt{4p^2C + 4\sqrt{3}C + 5p^2 + 7 - 2p\cdot \sqrt{3}}} {2Cp + p - \sqrt{3}}, \tag{6} \]

where \(C = e^{-\frac{7}{6}\pi p}\);

d) for \(p > 5\), \(\Theta = \pi + \Theta'\),

\[ \Theta' = -pD + \sqrt{p^2D^2 + 2D}, \tag{7} \]

where

\[ D = \frac{1 + e^{-\pi p}}{1 - p^2 e^{-\pi p}}. \]

Thus, knowing the parameters of the circuit, we find the cutoff angle; and from the found \(\Theta\) it is not particularly difficult to determine all harmonics of the rectified current and the voltage across the load.