ON THE COMBINED OPERATION OF A SHOCK GENERATOR AND A CAPACITOR BANK INTO AN INDUCTIVE LOAD

G. A. Sipailov, K. A. Khorkov

(Presented by the scientific seminar of the Department of Electrical Machines and General Electrical Engineering)

In modern experimental physics, in the construction of charged-particle accelerators for high energies, in installations for thermonuclear research, and in certain other cases, it becomes necessary to obtain pulsed magnetic fields with energies of tens of millions of joules and with pulse durations on the order of tens of milliseconds. To obtain such magnetic fields it is most expedient to accumulate energy in the rotating flywheel masses of electric machines. In particular, a shock-power generator can be used for this purpose [1, 2, 4].

When a shock generator is short-circuited, depending on its reactance and rotor diameter, from 5 to 50% of the kinetic energy stored by the rotating rotor can be converted during one pulse into electromagnetic energy of the stator.

When a shock generator is closed onto an inductive load, the fraction of kinetic energy converted into electromagnetic energy decreases; in this case, the part of the energy that is transferred into the energy of the magnetic field of the load, in the limiting case when the active and, correspondingly, inductive resistances of the generator and the load are equal, amounts to 25% of the electromagnetic energy of the stator under short-circuit conditions.

The fraction of kinetic energy transferred into the magnetic field of the load can be substantially increased if, in parallel with the generator, a capacitor bank is connected whose energy is approximately one quarter of the load energy.

The article presents the results of a theoretical study of the proposed scheme for increasing the energy delivered by a shock generator into an inductive load.

Scheme of Combined Operation of the Generator and the Capacitance

In the case of the usual scheme for switching a shock generator onto a load, that is, without an auxiliary capacitance, between the generator and the load there is a switching device \(K\), which must connect the generator at the moment when the voltage passes through zero (Fig. 1,a). In this case the current of the first half-wave, necessary for creating the magnetic field of the load, due to the aperiodic component—

…will have the maximum amplitude and duration, while the current of the second, negative half-wave will have the minimum amplitude and duration (Fig. 1, b). The rate of rise of the current and its amplitude are determined by the parameters of the generator and the load.

Fig. 1. a) Circuit for direct connection of a shock generator to a load; b) pattern of variation of voltage and current when the shock generator is switched on at the instant when the voltage passes through zero.

Disconnection of the generator is carried out in the same way or, when the operations of switching on and switching off are separated, by another switching device at the first or any other preassigned passage of the current through zero [3, 4].

The disadvantage of such a circuit is that, in the best case, only one fourth of the electromagnetic energy of the stator corresponding to a short circuit can be transferred to the load.

Fig. 2. Circuit for the combined operation of a shock generator and a capacitor bank into a load.

To increase the energy transferred by the shock generator to the load, it is possible to use the combined operation of the generator with a capacitor bank precharged to some voltage \(U_{c0}\). The circuit of combined operation is shown in Fig. 2, where

UG—shock-power generator; \(K\)-1, \(K\)-2—fast-acting arcless contact apparatuses [3, 4]; \(C\)—capacitor bank; \(r_{\mathrm{n}}, L_{\mathrm{n}}\)—load.

The initial state of the circuit is as follows: contact apparatuses \(K\)-1 and \(K\)-2 are open, and the capacitor bank \(C\) is charged to the voltage \(U_{c0}\). The generator is switched on by closing the contacts \(K\)-1 at the instant when the generator emf passes through zero; moreover, the polarity of the half-wave of the emf must be chosen so that the generator emf and the voltage on the capacitor bank act in the same direction. This leads to an increase in the steepness of the leading edge of the generator current \(I_{\mathrm{g}}\) by several times in comparison with the current \(I_1\) that would flow in the initial time interval if the generator were connected directly to the load, i.e., without capacitance.

The process of switching the shock generator onto the capacitor bank is described, in operational form, by the equation

\[ I(p)=\frac{e(p)}{pL+r+\dfrac{1}{pC}}+\frac{U_{c0}}{pL+r+\dfrac{1}{pC}}, \tag{1} \]

where, for

\[ e=E_m\sin\omega t, \]

\[ e(p)=E_m\frac{\omega p}{p^2+\omega^2}. \]

Solving equation (1), we obtain

\[ \begin{aligned} i_c=2I_m\Bigg\{& \frac{\delta\omega^3}{(\omega^2-\omega_{k0}^{\,2})^2+\delta^2\omega^2}\sin\omega t -\frac{\omega^2(\omega^2-\omega_{k0}^{\,2})}{(\omega^2-\omega_{k0}^{\,2})^2+\delta^2\omega^2}\cos\omega t \\ &-\left[ \frac{0.5\delta\omega^2(\omega^2+\omega_{k0}^{\,2})}{(\omega^2-\omega_{k0}^{\,2})^2+\delta^2\omega^2} -\omega\frac{U_{c0}}{E_m} \right] \frac{e^{-0.5\delta t}\sin\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}\,t} {\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}} \\ &+\frac{\omega^2(\omega^2+\omega_{k0}^{\,2})}{(\omega^2-\omega_{k0}^{\,2})^2+\delta^2\omega^2} e^{-0.5\delta t}\cos\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}\,t \Bigg\}. \end{aligned} \tag{2} \]

A second important equation is the variation of the voltage on the capacitor bank

\[ u_c=\frac{1}{C}\int i_c\,dt. \tag{3} \]

Since \(i_c=i_{\mathrm{g}}\), substituting the expression for the current (2) into equation (3), we obtain

\[ \begin{aligned} u_c=E_m\Bigg\{& \frac{\delta\omega^3}{(\omega^2-\omega_{k0}^{\,2})^2+\delta^2\omega^2}\cos\omega t -\left[ 1+\frac{\omega^2(\omega^2-\omega_{k0}^{\,2})}{(\omega^2-\omega_{k0}^{\,2})^2+\delta^2\omega^2} \right]\sin\omega t \\ &-\frac{(\omega^2-\omega_{k0}^{\,2})^2\omega\omega_{k0}^{\,2}-0.5\delta^2\omega^3} {(\omega^2-\omega_{k0}^{\,2})^2+\delta^2\omega^2} \cdot \frac{e^{-0.5\delta t}}{\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}} \sin\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}\,t \\ &-\frac{\delta\omega^3e^{-0.5\delta t}} {(\omega^2-\omega_{k0}^{\,2})^2+\delta^2\omega^2} \cos\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}\,t \Bigg\} +U_{c0}e^{-0.5\delta t}\cos\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}\,t \\ &-U_{c0}\frac{0.5\delta e^{-0.5\delta t}}{\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}} \sin\sqrt{\omega_{k0}^{\,2}-0.25\delta^2}\,t . \end{aligned} \tag{3,a} \]

In those cases in which the active resistance of the \(L_{\mathrm{g}}C\) circuit may be neglected, the expressions for the generator current and the voltage across the capacitor bank are substantially simplified:

\[ \delta=\frac{r}{L}=0; \]

\[ i_{\mathrm{g}}=2I_m\left[ \frac{\omega^2}{\omega_{\mathrm{k}0}^{\,2}-\omega^2}\cos\omega t -\frac{\omega^2}{\omega_{\mathrm{k}0}^{\,2}-\omega^2}\cos\omega_{\mathrm{k}0}t +\frac{\omega U_{c0}}{\omega_{\mathrm{k}0}E_m}\sin\omega_{\mathrm{k}0}t \right]; \tag{4} \]

\[ u_c=-E_m\left[ \left(1+\frac{\omega^2}{\omega^2-\omega_{\mathrm{k}0}^{\,2}}\right)\sin\omega t +\frac{\omega\omega_{\mathrm{k}0}}{\omega^2-\omega_{\mathrm{k}0}^{\,2}}\sin\omega_{\mathrm{k}0}t \right]+U_{c0}\cos\omega_{\mathrm{k}0}t, \tag{5} \]

where

\[ I_m=\frac{E_m}{2\omega L}, \]

\[ \omega_{\mathrm{k}0}^{\,2}=\frac{1}{L_{\mathrm{g}}C}; \]

\(\omega_{\mathrm{k}0}\) is the natural angular frequency of the \(L_{\mathrm{g}}C\) circuit;
\(\omega\) is the angular frequency of the generator emf;
\(E_m\) is the amplitude of the generator emf.

After the voltage of the capacitance charged to \(U_{c0}\), in the course of discharge in the \(L_{\mathrm{g}}C\) circuit, falls to zero (the instant \(t_1\) in Fig. 3), the load is connected to the generator and the capacitance by closing contacts \(K\)-2. There are no data in the literature on the solution of problems of this kind; therefore we present the calculation of the circuit (Fig. 2) in some detail. The initial conditions are: at \(t=t_1\)

\[ u_c=U_{c1}=0,\qquad i_{\mathrm{g}}=I_{\mathrm{g}1}. \]

In addition, for the purpose of simplification, in all subsequent derivations we shall assume

\[ L_{\mathrm{n}}=L_{\mathrm{g}}=L; \]

\[ r_{\mathrm{n}}=r_{\mathrm{g}}=r. \]

The process of switching the load onto the generator and the capacitance for

\[ e=E_m\sin(\omega t+\psi) \]

and \(U_c=U_{c1}\) is described, in operational form, by the equations

\[ i_{\mathrm{n}}(p)= \frac{\left(e_1(p)+pLI_{\mathrm{g}1}\right)\dfrac{1}{pc}+U_{c1}(pL+r)} {\left(pL+r+\dfrac{1}{pc}\right)^2-\left(\dfrac{1}{pc}\right)^2} = \]

\[ = \frac{e_1(p)}{cL^2(p+\delta)\left(p^2+p\delta+\omega_{\mathrm{k}}^2\right)} +\frac{pI_{\mathrm{g}1}}{cL(p+\delta)\left(p^2+p\delta+\omega_{\mathrm{k}}^2\right)} + \]

\[ +\frac{pU_{c1}}{L\left(p^2+p\delta+\omega_{\mathrm{k}}^2\right)}; \tag{6} \]

\[ i_{\mathrm{g}}(p)= \frac{(e_1(p)+I_{\mathrm{g}1}pL)(pL+r)+(e_1(p)+I_{\mathrm{g}1}pL)\dfrac{1}{pc}-U_{c1}(pL+r)} {\left(pL+r+\dfrac{1}{pc}\right)^2-\left(\dfrac{1}{pc}\right)^2} = \]

\[ = \frac{p e_1(p)}{L\left(p^2+p\delta+\omega_k^2\right)} + \frac{e_1(p)}{cL^2(p+\delta)\left(p^2+p\delta+\omega_k^2\right)} - \frac{pU_{c1}}{L\left(p^2+p\delta+\omega_k^2\right)} + \frac{p^2 I_{r1}}{p^2+p\delta+\omega_k^2} + \frac{p I_{r1}}{cL(p+\delta)\left(p^2+p\delta+\omega_k^2\right)}, \tag{7} \]

where, for

\[ e_1(t)=E_m\sin(\omega t+\psi); \]

\[ e_1(p)=E_m\frac{p^2\sin\psi+\omega p\cos\psi}{p^2+\omega^2}; \]

\[ \delta=\frac{r}{L}; \qquad \omega_k^2=\frac{2}{Lc}; \]

\(\omega_k\) is the natural angular frequency of the circuit \(L_r C L_n\).

As an example, let us show the solution of the first term in expression (6)

\[ \frac{e_1(p)}{cL^2(p+\delta)\left(p^2+p\delta+\omega_k^2\right)} = \frac{E_m}{2L}\,\omega_k^2 \frac{p^2\sin\psi+\omega p\cos\psi} {(p+\delta)(p^2+\omega^2)\left(p^2+p\delta+\omega_k^2\right)}; \]

\[ \frac{p^2\sin\psi+\omega p\cos\psi} {(p^2+\omega^2)\left(p^2+p\delta+\omega_k^2\right)(p+\delta)} = \frac{pA_1+A_2}{p^2+\omega^2} + \frac{pA_3+A_4}{p^2+p\delta+\omega_k^2} + \frac{A_5}{p+\delta}. \]

To determine the constants \(A_1—A_5\), we set up a system of five equations. As a result of solving the system, we find:

\[ A_1= \frac{ (\omega_k^2-\omega^2+\delta^2)\sin\psi + \frac{\delta}{\omega}(\omega_k^2-2\omega^2)\cos\psi } { (\omega_k^2-\omega^2+\delta^2)^2 + \frac{\delta^2}{\omega^2}(\omega_k^2-2\omega^2)^2 }; \]

\[ A_2= \frac{ -(\omega_k^2-2\omega^2)\delta\sin\psi + (\omega_k^2-\omega^2+\delta^2)\omega\cos\psi } { (\omega_k^2-\omega^2+\delta^2)^2 + \frac{\delta^2}{\omega^2}(\omega_k^2-2\omega^2)^2 }; \]

\[ A_3= \frac{ -\left[ (\omega_k^2-\omega^2) + \frac{\delta^2}{\omega^2}(\omega_k^2-\omega^2) + \frac{\delta^2}{\omega_k^2}(\omega^2+\delta^2) \right]\sin\psi + \frac{\omega\delta}{\omega_k^2}(\omega^2+\delta^2)\cos\psi } { (\omega_k^2-\omega^2+\delta^2)^2 + \frac{\delta^2}{\omega^2}(\omega_k^2-2\omega^2)^2 }; \]

\[ A_4= \frac{ (\omega^2+\delta^2)\delta\sin\psi + \left(\omega_k^2-\omega^2-\delta^2+\delta^2\frac{\omega_k^2}{\omega^2}\right)\omega\cos\psi } { (\omega_k^2-\omega^2+\delta^2)^2 + \frac{\delta^2}{\omega^2}(\omega_k^2-2\omega^2)^2 }; \]

\[ A_5= \frac{ \frac{\delta^2}{\omega^2}(\omega^2+\delta^2-\omega_k^2)\sin\psi + \frac{\delta}{\omega} \left[ \omega_k^2-2\omega^2+\frac{\omega^2}{\omega_k^2}(\omega^2+\delta^2) \right]\cos\psi } { (\omega_k^2-\omega^2+\delta^2)^2 + \frac{\delta^2}{\omega^2}(\omega_k^2-2\omega^2)^2 }. \]

Using the formulas for passing from the operational form to the time form, we ultimately obtain the expression of the 1st term in time form.

\[ \begin{aligned} C_1(t)={}&\frac{E_m}{2L}\cdot \frac{\omega_k^2} {\left(\omega_k^2-\omega^2+\delta^2\right)^2+ \dfrac{\delta^2}{\omega^2}\left(\omega_k^2-2\omega^2\right)^2} \times\\ &\times\Bigg\{ \left[\left(\omega_k^2-(\omega^2+\delta^2)\right)\sin\psi +\frac{\delta}{\omega}\left(\omega_k^2-2\omega^2\right)\cos\psi\right] \frac{1}{\omega}\sin\omega t+\\ &\quad+\left[-\left(\omega_k^2-2\omega^2\right)\delta\sin\psi +\omega\left(\omega_k^2-\omega^2+\delta^2\right)\cos\psi\right] \frac{1}{\omega^2}\left(1-\cos\omega t\right)+\\ &\quad+\left[-\left(\omega_k^2+\omega^2+\delta^2 -\delta^2\frac{\omega_k^2}{\omega^2} -\delta^2\frac{\omega^2}{\omega_k^2} -\delta^2\frac{\delta^2}{\omega_k^2}\right)\sin\psi\right.\\ &\qquad\left. +\frac{\delta\omega}{\omega_k^2}\left(\omega^2+\delta^2\right)\cos\psi\right]\cdot \frac{e^{-0{,}5\delta t}}{\sqrt{\omega_k^2-0{,}25\delta^2}}\, \sin\sqrt{\omega_k^2-0{,}25\delta^2}\,t-\\ &\quad-\left[\left(\omega^2+\delta^2\right)\delta\sin\psi+ \left(\omega_k^2-\omega^2-\delta^2+\delta^2\frac{\omega_k^2}{\omega^2}\right) \omega\cos\psi\right]\frac{1}{\omega_k^2}\times\\ &\qquad\times\left[1-e^{-0{,}5\delta t} \left(\cos\sqrt{\omega_k^2-0{,}25\delta^2}\,t+ \frac{0{,}5\delta\sin\sqrt{\omega_k^2-0{,}25\delta^2}\,t} {\sqrt{\omega_k^2-0{,}25\delta^2}}\right)\right]-\\ &\quad-\left[\frac{\delta^2}{\omega^2}\left(\omega^2+\delta^2-\omega_k^2\right)\sin\psi +\frac{\delta}{\omega}\left(\omega_k^2-2\omega^2+ \omega^2\frac{\omega^2}{\omega_k^2} +\delta^2\frac{\omega^2}{\omega_k^2}\right)\cos\psi\right]\times\\ &\qquad\times\frac{1}{\delta}\left(1-e^{-\delta t}\right) \Bigg\}. \end{aligned} \]

In an analogous manner we solve the remaining equations. The second term of equation (6) will be

\[ C_2(t)=0{,}5 I_{\Gamma 1} \left\{ \frac{0{,}5\delta e^{-0{,}5\delta t}} {\sqrt{\omega_k^2-0{,}25\delta^2}}\, \sin\sqrt{\omega_k^2-0{,}25\delta^2}\,t - e^{-0{,}5\delta t}\cos\sqrt{\omega_k^2-0{,}25\delta^2}\,t +e^{-\delta t} \right\}. \]

The third, last, term of equation (6)

\[ C_3(t)=\frac{U_{c1}}{L}\, \frac{e^{-0{,}5\delta t}} {\sqrt{\omega_k^2-0{,}25\delta^2}}\, \sin\sqrt{\omega_k^2-0{,}25\delta^2}\,t. \]

The current in the load will be equal to the sum of three components

\[ i_{\mathrm{n}}=C_1(t)+C_2(t)+C_3(t). \tag{8} \]

In the equation for the generator current (7), in addition to the components \(C_1(t)\), \(C_2(t)\), and \(C_3(t)\) indicated above, two more components enter, which in time form will have the form:

\[ \begin{aligned} C_4(t)={}&\frac{E_m}{2\omega L}\cdot \frac{2\omega} {\left(\omega^2-\omega_k^2\right)^2+\delta^2\omega^2} \Bigg\{ \left[\left(\omega_k^4-\omega^2\omega_k^2+0{,}5\delta^2\omega^2\right)\sin\psi\right.\\ &\qquad\left. -\left(\omega_k^2+\omega^2\right)0{,}5\delta\omega\cos\psi\right] \frac{e^{-0{,}5\delta t}} {\sqrt{\omega_k^2-0{,}25\delta^2}}\, \sin\sqrt{\omega_k^2-0{,}25\delta^2}\,t+ \end{aligned} \]

\[ +\left[(\omega^2-\omega_k^2)\omega\cos\psi-\omega^2\delta\sin\psi\right]e^{-0.5\delta t}\cos\sqrt{\omega_k^2-0.25\delta^2}\,t+ \]

\[ +\left[\delta\omega^2\cos\psi+(\omega^2-\omega_k^2)\omega\sin\psi\right]\sin\omega t-\left[(\omega^2-\omega_k^2)\omega\cos\psi-\omega^2\delta\sin\psi\right]\cos\omega t\}; \]

\[ C_5(t)=I_{r1}e^{-0.5\delta t}\left[\cos\sqrt{\omega_k^2-0.25\delta^2}\,t-\frac{0.5\delta}{\sqrt{\omega_k^2-0.25\delta^2}}\sin\sqrt{\omega_k^2-0.25\delta^2}\,t\right]. \]

The generator current will consist of five components

\[ i_g=C_1(t)+C_2(t)-C_3(t)+C_4(t)+C_5(t). \tag{9} \]

Obviously, it is practically impossible to use such cumbersome formulas in calculations; therefore, in order to simplify the expressions for \(i_n\) and \(i_g\), we neglect the active resistance of the circuit \(L_gCL_n\), that is, we assume \(\delta=0\); then:

\[ \begin{aligned} i_n={}&0.5I_{r1}+I_m\cos\psi+\frac{\nu^2}{\nu^2-1}\sin\psi I_m\sin\omega t -\frac{\nu^2}{\nu^2-1}\cos\psi I_m\cos\omega t \\ &-\left(\frac{\nu^2}{\nu^2-1}\sin\psi-\frac{2U_{c1}}{\nu E_m}\right)I_m\sin\nu\omega t \\ &+\left(\frac{1}{\nu^2-1}\cos\psi-\frac{I_{r1}}{2I_m}\right)I_m\cos\nu\omega t . \end{aligned} \tag{10} \]

\[ \begin{aligned} i_g={}&0.5I_{r1}+I_m\cos\psi+\frac{2-\nu^2}{\nu^2-1}\sin\psi I_m\sin\omega t \\ &-\frac{2-\nu^2}{\nu^2-1}\cos\psi I_m\cos\omega t +\left(\frac{2\nu}{\nu^2-1}\sin\psi-\frac{2U_{c1}}{\nu E_m}\right)I_m\sin\nu\omega t \\ &-\left(\frac{2}{\nu^2-1}\cos\psi-\frac{I_{r1}}{2I_m}\right)I_m\cos\nu\omega t . \end{aligned} \tag{11} \]

Since \(i_c=i_n-i_g\) and \(u_c=\dfrac{1}{c}\int i_cdt\), then

\[ \begin{aligned} u_c={}&-0.5E_m\left\{ \frac{\nu^2}{\nu^2-1}\sin\psi\cos\omega t +\frac{\nu^2}{\nu^2-1}\cos\psi\sin\omega t \right.\\ &\left. -\frac{\nu^2}{\nu^2-1}\sin\psi\cos\nu\omega t -\frac{\nu}{\nu^2-1}\cos\psi\sin\nu\omega t +\frac{\nu I_{r1}}{2I_m}\sin\nu\omega t \right\}\\ &-U_{c1}\cos\nu\omega t, \end{aligned} \tag{12} \]

where

\[ \nu^2=\frac{\omega_k^2}{\omega^2}. \]

Calculations show that, in the case of operation of the shock generator into an inductive load, when \(L_n=L_g\) and \(r_n=r_g\), setting \(\delta=0\) introduces an error in the results of current calculations of the order of \(5\div10\%\).

Figure 3 presents the curves of currents and voltages in the elements of the system (Fig. 2), calculated from equations (10), (11), and (12) for the following parameters:

\[ L_{\mathrm{n}}=L_{\mathrm{g}}=L=3.2\cdot 10^{-4}\ \mathrm{H}; \]

\[ E_m=10\ \mathrm{kV}; \]

when operating into the load

\[ I_m=\frac{E_m}{2\omega L}=50000\ \mathrm{A}; \]

under short circuit

\[ I_{mk}=\frac{E_m}{\omega L}=100000\ \mathrm{A}. \]

Fig. 3. Character of the variation of voltage and currents in the elements of the combined-operation circuit: \(i_{\mathrm{g}}\)—generator current; \(i_c\)—capacitor current; \(i_{\mathrm{n}}\)—load current; \(i_1\)—current when the generator is connected directly to the load; \(e_1\)—generator emf; \(u_c\)—voltage across the capacitor.

By correctly choosing the ratio between the inductances of the generator and the load and of the auxiliary capacitance, overvoltages in the circuit can be avoided by limiting them, for example, to the value \(1.25E_m\). In the example under consideration, from the condition of obtaining overvoltages on the load of no more than \(1.2E_m\) with a relatively small capacitor bank, we take \(\nu^2=3.5\) and \(U_{c0}=0.5E_m\).

The choice of the capacitance value is determined from the expression

\[ C=\frac{L_{\mathrm{n}}+L_{\mathrm{g}}}{L_{\mathrm{n}}L_{\mathrm{g}}}\cdot \frac{1}{\nu^2\omega^2} \]

for

\[ L_{\mathrm{n}}=L_{\mathrm{g}}, \]

\[ C=\frac{2}{\nu^2\omega^2 L}, \]

in the case under consideration \(C=0.018\ \mathrm{F}\).

As can be seen from Fig. 3, the current in the load when the shock generator operates jointly with the capacitor bank is considerably (in the case under consideration, by a factor of 1.7) greater than the current obtained when the generator operates according to the usual circuit.

Energy relations

Let us consider, for comparison, the energy indices of operation of the shock generator according to the circuits (Fig. 1) and (Fig. 2) for the various instants of time marked in Fig. 3.

The electromagnetic energy of the generator in the case of a sudden short circuit at its terminals at \(t=t_4\) is

\[ Q = 0.5 L_{\mathrm{g}} I_{\mathrm{ud}}^2; \]

\[ I_{\mathrm{ud}} = 2 I_{\mathrm{tk}} K_{\mathrm{z}}. \]

If current damping is neglected, i.e., if the damping coefficient is taken as \(K_{\mathrm{z}}=1\), then

\[ Q = 0.5 \cdot 3.2 \cdot 10^{-4} (2 \cdot 10^5)^2 = 6.4 \cdot 10^6 \text{ J}. \]

The energy transmitted to the load when the shock generator is switched on according to the usual circuit (Fig. 1) at \(t=t_4\) is

\[ Q_{1(4)} = 0.5 L_{\mathrm{n}} I_{1(4)}^2 = 0.5 \cdot 3.2 \cdot 10^{-4} (1 \cdot 10^5)^2 = 1.6 \cdot 10^6 \text{ J}. \]

The ratio \(Q_{1(4)}:Q=0.25\) shows the fraction of the energy transferred into the magnetic field of the load at \(L_{\mathrm{n}}=L_{\mathrm{g}}\), in comparison with the electromagnetic energy of the stator in a short circuit.

The nature of the change in energy in the individual elements of the circuit when the shock generator operates jointly with the capacitance is shown in Fig. 4; the same figure also gives the change in energy when the generator operates according to the circuit without capacitance.

The capacitor bank, charged to the voltage \(U_{c0}=0.5E_m\), has an initial store of energy

\[ Q_{c0} = 0.5 c U_{c0}^2 = 0.5 \cdot 0.018 (0.5 \cdot 10^4)^2 = 0.225 \cdot 10^6 \text{ J}. \]

At the instant \(t_1\), corresponding to the angle \(\psi=50^\circ\), when the voltage across the capacitance \(U_{c1}=0\) and the generator current \(I_{\mathrm{g}1}=1.3 I_m\), electromagnetic energy is stored in the leakage fields of the stator winding:

\[ Q_{\mathrm{g}1} = 0.5 L_{\mathrm{g}} I_{\mathrm{g}1}^2 = 0.5 \cdot 3.2 \cdot 10^{-4} (1.3 \cdot 0.5 \cdot 10^5)^2 = 0.675 \cdot 10^6 \text{ J}. \]

Consequently, by the moment the load is switched on, from the kinetic energy of the rotating rotor there has passed into the electromagnetic energy of the stator leakage fields

\[ Q_{\mathrm{g}1} - Q_{c0} = 0.675 \cdot 10^6 - 0.225 \cdot 10^6 = 0.45 \cdot 10^6 \text{ J}, \]

which, in comparison with the energy of the leakage fields when the generator operates according to the usual circuit (\(Q_{11}=0.016 \cdot 10^6\) J), is almost 30 times greater. Thus, as a result of connecting the capacitor bank, an acceleration of the process of converting the kinetic energy of the rotor into electromagnetic energy is achieved. When the load is connected, part of the generator current continues to flow through the capacitive circuit, recharging the capacitance, and part of the current branches into the load. Therefore, in the initial period of time the load current increases smoothly from its zero value.

Such a shape of the current curve facilitates the process of arcless switching-on of the load; moreover, in the case of the application of joint operation of the ge-

generator and the capacitance, for example, in power-supply circuits for charged-particle accelerators, such a change in the current can provide the required reduction in the rate of rise of the controlling magnetic field from zero to the value determined by the injection level.

Fig. 4. Character of the change in energy in the elements of the circuits for operation of a shock generator into a load: \(Q_{\mathrm{n}}\)—energy in the load during combined operation; \(Q_1\)—energy in the load when operating according to the circuit without capacitance; \(Q_{\mathrm{r}}\)—energy of the stray fields of the stator winding during combined operation; \(Q_{\mathrm{c}}\)—energy of the capacitor bank.

At the instant \(t=t_2\), the generator current reaches its greatest value \(I_{\mathrm{r}2}=I_{\mathrm{rmax}}\). In this case the maximum energy stored in the stray fields of the generator is

\[ Q_{\mathrm{r}2}=0.5 L_{\mathrm{r}} I_{\mathrm{r}2}^{2}=1.6\cdot 10^{6}\ \text{J}. \]

At the instant \(t=t_3\), the capacitor bank is charged to the greatest voltage \(U_{\mathrm{c}3}=U_{\mathrm{c\,max}}=1.2E_m\), and the current through the capacitance is zero; in this case the maximum energy stored in the bank is equal to

\[ Q_{\mathrm{c}3}=0.5cU_{\mathrm{c}3}^{2}=0.5\cdot 0.018(1.2\cdot 10^{4})^{2}=1.3\cdot 10^{6}\ \text{J}. \]

For \(t>t_3\), the entire generator current and the current of the discharging capacitance pass through the load. The load current reaches its maximum value \(I_{\mathrm{n}4}=I_{\mathrm{nmax}}=3.41I_m\) at the instant \(t_4\), corresponding to the angle \(\psi=180^\circ\). In this case the capacitor bank completely gives up the stored-

energy to the load, $\dot U_{c4}=0$ and $Q_{c4}=0$ (Figs. 3 and 4). Part of the energy stored in the leakage fields of the generator is also transferred to the load, and for a certain ratio $\nu=\dfrac{\omega_k}{\omega}$ it is transferred completely.

Thus, the maximum energy transferred to the load from the shock generator by the instant of time $t_4$ will be equal to

\[ Q_{n4}=0.5 L_n i_{n4}^{2}=0.5\cdot 3.2\cdot 10^{-4}(3.41\cdot 0.5\cdot 10^5)^2=4.65\cdot 10^6\ \text{J}, \]

which is 2.9 times greater than when the generator operates according to the usual circuit, and amounts to 73% of the energy of a sudden short circuit of the generator. In this case the capacitor bank must be designed for 28% of the energy transferred to the load.

An analysis of the circuits (Figs. 1 and 2) from the energy point of view shows that, in order to obtain magnetic fields with high energy, with the required duration of the field or current pulse creating this field on the order of 0.02 sec, in a number of cases it is expedient to use the combined operation of a shock generator with a capacitor bank. However, the question of choosing the energy source in each particular case requires additional comprehensive study.

The magnitude of the energy transferred to the load during the combined operation of the shock generator and the capacitance, $Q_n$, as well as the magnitude of the energy of the capacitor bank, $Q_c$, depends strongly on the value of the capacitance $C$, its initial charge $U_{c0}$, and the load parameters. Table 1 gives the results of calculations of $Q_n$ (for $t=t_4$) and $Q_c$ (for $t=t_3$) in comparison with the energy $Q_1$ transferred to the load when the shock generator operates according to the usual circuit; it also shows what part of the load energy is constituted by the energy entering the load from the capacitance. The calculations were performed for several values of $\nu$, $C$, and $U_{c0}$ with $L_n=L_g=3.2\cdot 10^{-4}\ \text{H}$ and $r_n=r_g=0$.

Table 1

A more detailed analysis of the influence of individual parameters on the combined operation of the generator and the capacitance, carried out with the aid of computers, will be given in a separate article.

Disconnecting the generator and the load

For $t>t_4$, the energy stored in the load is returned to the generator and is again converted into the kinetic energy of the rotating rotor. When the generator current passes through zero, or at the instant when the current $i_g$ is close to zero ($t_5$), the generator is dis-

by opening the contacts \(K-1\). In this case, part of the energy remains in the capacitor bank and in the load; for example, in the case considered:

\(U_{c5}=-0.245E_m\) and \(I_{n5}=0.51I_m\), then

\[ Q_{c5}=0.5\,cU_{c5}^{2}=0.06\cdot 10^{6}\ \text{J} \]

and

\[ Q_{n5}=0.5\,L_n I_{n5}^{2}=0.1\cdot 10^{6}\ \text{J}. \]

The energy remaining in the load then passes into the capacitor bank. At the time \(t=t_6\), \(I_{n6}=0\) and \(Q_{n6}=0\); the load circuit is disconnected by opening the contacts \(K-2\). All the energy remaining in the \(L_nC\) circuit after the generator is disconnected has been transferred to the capacitance. In this case, the sign of the capacitance voltage will be the same as it was at the beginning of the cycle at \(t=0\).

The process of energy transfer from the load to the bank from the instant \(t=t_5\) is described by the equations:

\[ I_n=I_m\frac{2U_{c5}}{\nu_k E_m}\sin \nu_k\omega t+I_{n5}\cos \nu_k\omega t; \tag{13} \]

\[ U_c=-U_{c5}\cos \nu_k\omega t+\frac{\nu_k}{2}E_m\sin \nu_k\omega t, \tag{14} \]

where

\[ \nu_k=\frac{\omega_{kk}}{\omega}; \qquad \omega_{kk}=\frac{1}{L_nC}; \]

\(\omega_{kk}\) is the natural angular frequency of the \(L_nC\) circuit. The magnitude of the energy remaining stored in the capacitor bank, \(Q_{c6}\), depends to a considerable degree on the relationship between \(L_n\) and \(C\). With a corresponding choice of \(L_n\) and \(C\), one can obtain \(Q_{c6}\) equal to \(Q_{c0}\). For the example considered, \(Q_{c6}=0.156\cdot 10^6\ \text{J}\), which corresponds to \(U_{c6}=0.42E_m\), i.e., \(U_{c6}=0.84U_{c0}\).

Thus, at the end of the cycle the contact devices \(K-1\) and \(K-2\) are open, the capacitance is charged to the initial voltage; that is, the circuit has been brought to its initial state.

Conclusions

1. The combined operation of a shock generator and a capacitor bank leads to a significant increase in the utilization of the electromagnetic energy of the shock generator. In the presence of an auxiliary capacitance, the energy transferred to the load increases by a factor of 2--4 compared with the energy transferred to the load in a circuit without capacitance.

2. The amount of energy for which the capacitor bank must be designed is 25--30% of the amount of energy transferred by the generator to the load. In this case, powerful energy sources are not required for charging the capacitor bank.

3. The combined operation of a shock generator and a capacitance may find application in accelerator technology, in installations for thermonuclear research, and in other cases where it is necessary to obtain magnetic fields with high energy at a field-pulse duration on the order of 0.02 s.

References

1. P. L. Kapitsa. Proceedings of the Royal Society, vol. 105, 1924; vol. 115, 1927.
2. G. M. Mitot et al. Atomic Technology Abroad, no. 10, 1958.
3. V. V. Ivashin. Switching device. Author’s certificate no. 155206. Bulletin of Inventions and Trademarks, no. 12, 1963.
4. V. V. Ivashin, G. A. Sipailov. Switching the current of a shock-power generator. In: Problems of the Theory and Design of Electrical Machines. Interuniversity collection, Novosibirsk, 1963.