Calculation of Reverse-Voltage Suppression in a Pulse Circuit

I. I. Chuchalin, V. M. Razin

(Presented by the scientific seminar of the Faculty of Physics and Technology)

Introduction

In power-supply circuits for pulsed magnetic-field generators, in order to create a current pulse in the windings of an electromagnet, the discharge of a previously charged capacitor bank is usually used. Gas-discharge valves—ignitrons and thyratrons—are used as the switching elements. The negative voltage that appears on the capacitor bank toward the end of its recharge is suddenly applied to the electrodes of the switching valve at the moment of its deionization and may cause reverse ignition.

The probability of reverse ignition can be substantially reduced if, by some means, the rate of rise of the reverse voltage is changed so that, by the time a voltage sufficient for breakdown appears between the anode and cathode, the process of ion recombination has practically ended.

The occurrence of reverse ignition can be eliminated if the appearance of a high reverse voltage is prevented or its magnitude is limited.

In a pulsed operating mode with a sufficiently large duty factor, the reverse voltage appearing on the main valve can be reduced or completely prevented by means of an additional valve. The additional valve, connected in series with an active resistance, is connected in parallel with the capacitor bank.

In the present work, a mathematical analysis is carried out of the transient processes occurring in the reverse-voltage suppression circuit under consideration, and experimental data obtained in studying the operation of the circuit are presented.

On the basis of the analysis performed, recommendations are given for selecting the parameters of the suppression circuit.

Reverse-Voltage Suppression Circuit

In pulsed power-supply circuits for electromagnets, the discharge circuit usually consists of a capacitance \(C\), an inductance \(L\) (the electromagnet winding), and a switching valve (ignitron, thyratron—Fig. 1).

At the moment the switching valve is opened, the previously charged capacitance \(C\) begins to discharge, creating a current pulse in the turns of the winding \(L\).

During the time while the switching valve conducts current, the voltage between its electrodes is equal to the voltage drop in the arc. The voltage drop in the arc, or the internal voltage drop in thyratrons and tyratrons, reaches a value of no more than 20–25 V.

Fig. 1. Pulse circuit for supplying an electromagnet.

At the moment when the current \(i_l\) passes through zero, i.e., at the moment the switching valve is locked, the latter is subjected to a reverse voltage, which rises abruptly. In practice the reverse voltage rises at a high rate, but not instantaneously, since the distributed capacitances of the inductance coil and the interelectrode capacitances of the valve delay its rise [1].

Fig. 2. Oscillogram of currents and voltages in a pulse circuit

1 — voltage across the capacitance, 2 — current in the electromagnet winding, 3 — current in the charging circuit, 4 — voltage on the valve, 5 — calibration voltage 50 V.

In Fig. 2 an oscillogram is presented, on which are shown the curves of the voltage across the capacitance, the current in the inductance coil, the current in the charging circuit, and the voltage between the electrodes of the valve.

If an additional valve 2 with active resistance \(R\) is connected into the discharge circuit (Fig. 3) and fired at the moment when the current \(i_L\) passes through its maximum value, or somewhat later, then, depending on the magnitude of the resistance \(R\), the reverse voltage on valve 1 will be completely or partially suppressed as a result of dissipation, in the active resistance \(R\), of the energy of the oscillatory circuit.

Fig. 3. Circuit for suppressing the reverse voltage.

By varying the magnitude of the resistance \(R\), it is possible to vary the magnitude of the current \(i\) (Fig. 3) and, consequently, the degree of recharging of the capacitance \(C\), i.e., the magnitude of the negative voltage \(U_C\).

Fig. 4. Oscillogram of currents and voltages in the case of complete suppression of the reverse voltage.

1 — voltage on the capacitance, 2 — current in the electromagnet winding,
3 — current in the charging circuit, 4 — voltage on the main valve 1,
5 — voltage on the additional valve 2.

Figure 4 shows oscillograms for the case in which no reverse voltage arises on the valves. This is evident from the fact that the oscillograms of the voltage on the electrodes of the valves after they are blocked have no voltage jump. On the additional valve, reverse voltage exists when the sign of the voltage on the capacitance is positive. This

the reverse voltage is not dangerous for the additional rectifier, since at this time there are no ions in its interelectrode space.

Figure 5 presents oscillograms for partial suppression of the reverse voltage on the main rectifier.

Fig. 5. Oscillogram of currents and voltages with partial suppression of the reverse voltage.

Theory and calculation of the suppression circuit

For mathematical analysis, the circuit shown in Fig. 3 can be replaced by the equivalent circuit of Fig. 6.

Fig. 6. Equivalent calculation circuit.

The capacitor $C$ is charged to the voltage $U_{p0}$. At the time instant $t = 0$ the switch $P_1$ is closed and the capacitor begins to discharge through $r$ and $L$; in this case the variations of the current $i_L$ and of the voltage $U_c$ are determined by the expressions [2]:

\[ i_L=i_c=\frac{U_{m0}}{\omega^1 L}\cdot \sin \omega^1 t \cdot e^{-\delta t}, \tag{1} \]

\[ U_c=\frac{U_{m0}}{\omega^1\sqrt{LC}}\cdot \sin\left(\omega^1 t-\frac{\pi}{2}-\xi\right)e^{-\delta t}, \tag{2} \]

where

\[ \delta=\frac{r}{2L}, \]

\[ \omega_0=\frac{1}{\sqrt{LC}} \]
is the angular frequency of the natural undamped oscillations of the discharge circuit,

\[ \omega^1=\sqrt{\omega_0^2-\delta^2} \]
is the angular frequency of the natural damped oscillations of the discharge circuit,

\[ \xi=\operatorname{arctg}\frac{\delta}{\omega^1}. \]

At the instant when the voltage \(U_c\) is equal to zero, the switch \(P_2\) is closed. Let us find the magnitude of the current \(I_L^1\) at the instant of closing the switch \(P_2\). To do this, we set expression (2) equal to zero and determine the time \(t\) after which \(U_c\) decreases to zero:

\[ t=\frac{\xi+\frac{\pi}{2}}{\omega^1}. \tag{3} \]

Substituting the obtained expression into (1), we find the value of the current \(I_L^1\) at the instant of closing the switch \(P_2\):

\[ I_L^1=\frac{U_{m0}}{\omega^1 L}\cdot \sin\left(\xi+\frac{\pi}{2}\right)\cdot e^{-\frac{\delta}{\omega^1}\left(\xi+\frac{\pi}{2}\right)}. \tag{4} \]

We shall next be interested in the processes occurring in the circuit after the switch \(P_2\) is closed, i.e., after valve 2 is switched on. The initial equations in this case will have the form:

\[ \left\{ \begin{aligned} -\frac{1}{C}\int i_c\,dt+L\frac{di_L}{dt}+ri_L&=0,\\ i_rR+L\frac{di_L}{dt}+ri_L&=0,\\ i_L&=i_r+i_c. \end{aligned} \right. \tag{5} \]

From the last equations of system (5) we have:

\[ i_c=\left(1+\frac{r}{R}\right)i_L+\frac{L}{R}\frac{di_L}{dt}. \]

Substituting the obtained expression into the first equation of system (5) and, after simple transformations, we obtain the differential equation

\[ \frac{d^2 i_L}{dt^2} +\left(\frac{1}{RC}+\frac{r}{L}\right)\frac{di_L}{dt} +\frac{1}{LC}\left(1+\frac{r}{R}\right)i_L=0. \tag{6} \]

The nature of the processes occurring in the circuit depends on the ratio of the parameters \(r\), \(R\), \(L\), and \(C\) and, ultimately, is determined by whether ...

the roots of the characteristic equation are real or complex. In the case of an aperiodic discharge,

\[ \frac{1}{4}\left(\frac{1}{RC}+\frac{r}{L}\right)^2 \geq \frac{1}{LC}\left(1+\frac{r}{R}\right). \]

Let us consider the case when

\[ \frac{1}{4}\left(\frac{1}{RC}+\frac{r}{L}\right)^2 > \frac{1}{LC}\left(1+\frac{r}{R}\right). \]

The solution of differential equation (6) in this case will have the form:

\[ i_L=C_1 e^{p_1 t}+C_2 e^{p_2 t}. \tag{7} \]

Using the initial conditions

\[ \left. \begin{aligned} i_L(t)_{t=0}&=C_1+C_2=I_L,\\ u_c(t)_{t=0}&=0 \end{aligned} \right\} \tag{8} \]

and finding and substituting the constants of integration into (7), we obtain

\[ i_L=\frac{I_L}{L(p_1-p_2)} \left[(Lp_1+r)e^{p_2 t}-(Lp_2+r)e^{p_1 t}\right]. \tag{9} \]

The voltage across the capacitor:

\[ U_c=\frac{I_L(Lp_1+r)(Lp_2+r)}{L(p_1-p_2)} \left(e^{p_1 t}-e^{p_2 t}\right). \tag{10} \]

Let us proceed to the consideration of the processes occurring with the discharge circuit when

\[ \frac{1}{4}\left(\frac{1}{RC}+\frac{r}{L}\right)^2 < \frac{1}{LC}\left(1+\frac{r}{R}\right). \]

In this case the roots of the characteristic equation will be complex:

\[ p_1=\alpha+\beta i, \]

\[ p_2=\alpha-\beta i, \]

where

\[ \alpha=-\frac{1}{2}\left(\frac{1}{RC}+\frac{r}{L}\right), \]

\[ \beta=\sqrt{\frac{1}{LC}\left(1+\frac{r}{R}\right)-\frac{1}{4}\left(\frac{1}{RC}+\frac{r}{L}\right)^2}. \]

The solution of differential equation (6) will have the form:

\[ i_L=(C_3\cos \beta t+C_4\sin \beta t)e^{\alpha t}. \tag{11} \]

Determining the constants \(C_3\) and \(C_4\) from the initial conditions, we obtain:

\[ i_L=I_L\left[\cos \beta t-\left(\frac{r+\alpha L}{\beta L}\right)\sin \beta t\right]e^{\alpha t}. \tag{12} \]

Voltage on the plates of the capacitor

\[ U_c = I_L L \beta \left[ 1 + \left( \frac{r+\alpha L}{\beta L} \right)^2 \right]\sin \beta t \cdot e^{\alpha t}. \tag{13} \]

In the case of an aperiodic discharge of the capacitor, the current approaches zero asymptotically. No reverse voltage appears on the main valve 1 (Fig. 4).

In the case of an oscillatory discharge, by the time the current passes through zero the capacitor has not had time to discharge completely, and the negative voltage remaining on it is applied to valve 1 (Fig. 5).

Let us determine the magnitude of the reverse voltage. To do this, we set expression (12) equal to zero and find the time \(t\) at which the current \(i_L\) passes through zero:

\[ t=\frac{1}{\beta}\left(\pi-\operatorname{arctg}\left|\frac{\beta L}{r+\alpha L}\right|\right).^{1)} \]

Substituting the value found into expression (13), we determine the magnitude of the reverse voltage:

\[ U_{\mathrm{rev}}=I_L\sqrt{(r+\alpha L)^2+\beta^2L^2}\cdot e^{\frac{\alpha}{\beta}\left(\pi-\operatorname{arctg}\left|\frac{\beta L}{r+\alpha L}\right|\right)}. \tag{14} \]

Substituting into (14) the value \(I_L\) from (4), we obtain:

\[ U_{\mathrm{rev}}=U_{m0}\, e^{\frac{\alpha}{\beta}\left(\pi-\operatorname{arctg}\left|\frac{\beta L}{r+\alpha L}\right|\right) -\frac{\delta}{\omega'}\left(\operatorname{arctg}\frac{\delta}{\omega'}+\frac{\pi}{2}\right)}. \tag{15} \]

The ratio of the reverse voltage to the initial voltage on the electrodes of valve 1 is

\[ \frac{U_{\mathrm{rev}}}{U_{m0}}= e^{\frac{\alpha}{\beta}\left(\pi-\operatorname{arctg}\left|\frac{\beta L}{r+\alpha L}\right|\right) -\frac{\delta}{\omega'}\left(\operatorname{arctg}\frac{\delta}{\omega'}+\frac{\pi}{2}\right)}. \tag{16} \]

If we take into account that \(\delta=\dfrac{r}{2L}\) and \(\omega'=\sqrt{\omega_0^2-\delta^2}\), then the ratio \(\dfrac{\delta}{\omega'}\) can be reduced to the form

\[ \frac{\delta}{\omega'}=\frac{1}{\sqrt{4Q^2-1}}, \qquad \text{where } Q=\frac{\omega_0 L}{r} \]

is the quality factor of the discharge circuit. Then formula (16) may be written as

\[ \frac{U_{\mathrm{rev}}}{U_{m0}}= e^{\frac{\alpha}{\beta}\left(\pi-\operatorname{arctg}\left|\frac{\beta L}{r+\alpha L}\right|\right) -\frac{1}{\sqrt{4Q^2-1}}\left(\operatorname{arctg}\frac{1}{\sqrt{4Q^2-1}}+\frac{\pi}{2}\right)}. \tag{17} \]

The term

\[ -\frac{\beta L}{r+\alpha L} \]

may be written approximately as

\[ -\frac{\beta L}{r+\alpha L}\simeq -\frac{\beta}{\alpha}. \tag{18} \]

\(^{1)}\) Vertical bars denote the absolute value of the fraction.

The indicated approximation is justified, since in practically used circuits

\[ r \ll aL . \]

For a quality factor of the discharge circuit \(Q > 5\), one may approximately, with an accuracy of up to 1%, write:

\[ \frac{\delta}{\omega'} \simeq \frac{1}{2Q}. \tag{19} \]

Taking into account the approximations (18) and (19), formula (17) may be represented in the following form:

\[ \frac{U_{\mathrm{rev}}}{U_{m0}} \simeq e^{-\frac{\gamma}{\sqrt{1-\gamma^{2}}}\left(\pi-\operatorname{arctg}\frac{\beta}{L}\right) -\frac{1}{2Q}\left(\operatorname{arctg}\frac{1}{2Q}+\frac{\pi}{2}\right)} . \]

Let us introduce the notation:

\[ \gamma = \sqrt{ \frac{ \frac{1}{4}\left(\frac{1}{RC}+\frac{r}{L}\right)^{2} }{ \frac{1}{LC}\left(1+\frac{r}{R}\right) } } = \sqrt{\frac{\alpha^{2}}{\alpha^{2}+\beta^{2}}}. \tag{20} \]

Finally, expression (17) may be written as

\[ \frac{U_{\mathrm{rev}}}{U_{m0}} \simeq K_{1}\cdot K_{2}. \tag{21} \]

where

\[ K_{1} = e^{-\frac{\gamma}{\sqrt{1-\gamma^{2}}}\left(\pi-\arccos \gamma\right)}, \tag{22} \]

\[ K_{2} = e^{-\frac{1}{2Q}\left(\operatorname{arctg}\frac{1}{2Q}+\frac{\pi}{2}\right)} . \tag{23} \]

The dependence \(K_{1}=f_{1}(\gamma)\) is shown graphically in Fig. 7. In Fig. 8 the dependence \(K_{2}=f_{2}(Q)\) is presented. The values of \(K_{2}\) for \(Q=1, 2, 3, 4, 5\) were calculated by the formula

\[ K_{2} = e^{-\frac{1}{\sqrt{4Q^{2}-1}} \left( \operatorname{arctg}\frac{1}{\sqrt{4Q^{2}-1}}+\frac{\pi}{2} \right)}, \]

and \(K_{2}\) for \(Q>5\) by the formula

\[ K_{2}\simeq e^{-\frac{1}{2Q}\left(\operatorname{arctg}\frac{1}{2Q}+\frac{\pi}{2}\right)} . \]

For an oscillatory discharge of the capacitor, specifying the value of the resistance \(R\) within the limits imposed by the condition for an oscillatory discharge, i.e.

\[ \frac{1}{4}\left(\frac{1}{RC}+\frac{r}{L}\right)^{2} < \frac{1}{LC}\left(1+\frac{r}{R}\right), \]

one can, using the graphs in Figs. 7 and 8, determine, for a circuit with a known quality factor, the ratio

\[ \frac{U_{\mathrm{rev}}}{U_{m0}} . \]

In practice it is often necessary to solve the inverse problem, namely: having chosen, for a given gas-discharge valve, the ratio \(\dfrac{U_{\text{rev.}}}{U_{m0}}\), it is required to find the value of the resistance \(R\) connected in series with the additional valve. The procedure for solving such a problem is indicated below.

Fig. 7. Plot of the dependence \(K_1 = f_1(\gamma)\).

1. Knowing the quality factor of the discharge circuit, we determine the coefficient \(K_2\) from the plot \(K_2 = f_2(Q)\) (Fig. 8).

2. Having chosen the ratio \(\dfrac{U_{\text{rev.}}}{U_{m0}}\), we determine the coefficient \(K_1\) from the formula

\[ K_1 = \frac{U_{\text{rev.}}}{U_{m0}} \cdot \frac{1}{K_2}. \]

1. Knowing \(K_1\), from the plot \(K_1 = f_1(\gamma)\) (Fig. 7) we determine \(\gamma\).

Fig. 8. Plot of the dependence \(K_2 = f_2(Q)\).

1. And finally, knowing \(\gamma\), we calculate the value of the resistance \(R\) from the formula:

\[ R = \frac{1}{4\gamma^2 - r^2 \dfrac{C}{L}} \left\{ r(1 - 2\gamma^2) + r \sqrt{ 2 \left[ (2\gamma^2 - 1) r^2 + 2 \frac{L}{C} \right] } \right\}. \tag{24} \]

Formula (24) is obtained from expression (20). In solving the quadratic equation with respect to \(R\), the root with the minus sign is not taken into account, since a negative value of resistance has no physical meaning.

1. The power dissipated in the resistance \(R\) may be calculated, with some design margin, from the approximate expression

\[ P \simeq f \frac{C U_{mo}^{2}}{2}, \tag{25} \]

where \(f\) is the pulse repetition frequency.

The last equation is written on the assumption that all the energy is dissipated only in the resistance \(R\). In reality, part of the energy is dissipated in other elements of the discharge circuit, and the actual power dissipated in the resistance \(R\) will be somewhat less than the value calculated from expression (25).

Results of Experimental Investigations

Investigations of reverse-voltage suppression were carried out in a circuit using ignitrons И-100/1000 as valves. Comparison of the calculated transient-process curves for particular cases with those obtained experimentally on the oscillograms of Fig. 9 (aperiodic process) and Fig. 10 (oscillatory process) indicates a satisfactory accuracy of the theoretical calculations. The increase in the discrepancy between the curves as the current decreases is explained by the influence of the nonlinearity of the ignitron resistance. As the current decreases, the equivalent resistance of the ignitron to direct current increases, and the actual transient process ends earlier than the calculated one (turning at the very end from aperiodic into oscillatory, as is seen from the oscillograms of Figs. 4 and 9).

Fig. 9. Oscillogram of the current in the electromagnet winding during an aperiodic process.

Fig. 10. Oscillogram of the current in the electromagnet winding during an oscillatory process.

In Fig. 11 are presented the theoretical and experimental dependences

\[ \frac{U_{\mathrm{rev}}}{U_{mo}} = f_{3}(\gamma) \]

for the particular case \(Q = 11\). The good agreement of the curves confirms the validity of the calculation relationships obtained above.

In conclusion, we note that the voltage-suppression method considered can find successful application in such pulse circuits where the considerable energy losses inherent in this method are not of substantial importance, and where there is a need to raise the voltage above the rated data of the valves used.

Fig. 11. Theoretical and experimental data for determining the relative magnitude of the reverse voltage.

LITERATURE

1. Tolstov Yu. G. Electromagnetic processes in rectifier installations with nonlinear anode chokes, Izvestiya AN SSSR, Department of Technical Sciences, No. 2, 1950.

2. P. L. Kalantarov, L. R. Neiman. Theoretical Foundations of Electrical Engineering, GEI, 1951.