On the Theory of Operation of a Loaded Cascade Generator

A. A. Vorob'ev and V. S. Melikhov

Principle of Operation of a Cascade Generator

A cascade generator, which has become widespread as a source of direct current at voltages up to several million volts, consists of an alternating-voltage source (a high-voltage transformer) and a large number of rectifiers and capacitors. In these circuits, at no load, the inverse voltage that the rectifiers must withstand and the direct voltage on the capacitors are equal to twice the maximum voltage of the transformer. The total direct voltage obtained from such a generator is the sum of the voltages across half of the capacitors, and may exceed the original alternating voltage many times over.

The physical principle of operation of a cascade generator can be explained with the aid of the electrical circuit shown in Fig. 1. Capacitances \(C_1\), \(C_2\), and \(C_3\), equal in magnitude, are connected in series. Capacitance \(C_3\) is connected to a source of direct voltage \(E\). Two other capacitances, \(C'_1\) and \(C'_2\), also equal in magnitude and likewise connected in series, are, by means of switches \(K_1\), \(K_2\), and \(K_3\), alternately connected either to capacitances \(C_2\) and \(C_3\), when the position of the switches is indicated by the symbols \(K'_1\), \(K'_2\), and \(K'_3\), or to capacitances \(C_1\) and \(C_2\). In the latter case the position of the switches is indicated by the letters \(K_1\), \(K_2\), and \(K_3\). In the first case the capacitances \(C'_1\) and \(C'_2\) are charged; in the second position of the switches the capacitances \(C_1\) and \(C_2\) are charged.

During operation of the circuit of Fig. 1, from the moment it is connected to the direct voltage, the following cycles and instants may be considered:

First cycle: 1) the position of the switches is indicated by dashed lines; the capacitances \(C'_1\) and \(C'_2\) are connected to the capacitances \(C_2\) and \(C_3\); the capacitance \(C'_2\), from the direct-current source, is charged to the voltage \(E\); 2) the position of the switches is indicated by solid lines; the capacitances \(C'_1\) and \(C'_2\) are connected to the capacitances \(C_1\) and \(C_2\). In this case the capacitance \(C_2\) is charged from the capacitance \(C'_2\). Under the condition that the capacitances are equal, as is the case under consideration, after recharging they will be at a voltage equal to

\[ \frac{E}{2}. \]

Second cycle: 1) with the position of the switches indicated by dashed lines, the capacitances \(C'_1\) and \(C'_2\) will be respectively connected to the capacitances \(C_2\) and \(C_3\); the capacitance \(C'_2\) is thereby again charged to the voltage \(E\), and the capacitance \(C'_1\) receives half the charge from the capacitance \(C_2\), after which they will be charged to the potential

\[ \frac{E}{4}; \]

2) the capacitances \(C'_1\) and \(C'_2\), when the position

the switches indicated by solid lines will be connected to the capacitances \(C_1\) and \(C_2\). The charges accumulated in the capacitances \(C_2\) and \(C'_2\), respectively, are equal to \(Q_2 = C \dfrac{E}{4}\) and \(Q'_2 = C \cdot E\).

Their total charge is equal to

\[ Q_{\Sigma} = C\left(\frac{E}{4} + E\right) = \frac{5}{4} C \cdot E, \]

and the voltage to which the capacitance \(C'_2\) recharges the capacitance \(C_2\) is determined by the formula:

\[ U = \frac{Q_{\Sigma}}{C_{\Sigma}} = \frac{5}{8}E. \]

The capacitance \(C'_1\) gives up half of its charge to the capacitance \(C_1\), after which they prove to be charged to the voltage \(\dfrac{E}{8}\). Thus, after

Fig. 1. Circuit for connecting capacitances, illustrating the principle of operation of a cascade generator.

two switching cycles of the circuit, the total voltage on all the capacitances of the left-hand column of the circuit in Fig. 1 proves to be equal to

\[ U_{\mathrm{out}} = E + \frac{5}{8}E + \frac{1}{8}E = \frac{14}{8}E. \]

With the further sequence of switching, charging of the capacitances of the left-hand column of the circuit in Fig. 1 and recharging of the capacitances of the right-hand column will occur, as a result of which all the capacitances will be charged to the voltage \(E\). The voltage at the output of the circuit in Fig. 1 will reach a steady-state value,

equal to \(3E\). If there are \(n\) stages in the circuit, the total output voltage in the steady state will be \(nE\) volts. The charging of the upper capacitance in the column to the full voltage will occur the faster, the more frequently the switching takes place. It is clear that mechanical high-voltage switches capable of carrying out these switchings at high frequency cannot be made. Only electronic and ionic devices can serve as such switches.

Figure 2 shows an electrical circuit of a high-voltage cascade generator using electronic or ionic devices. In this circuit the switches \(K_3\) and \(K'_3\) are replaced either by electronic or by ionic devices \(T_3\) and \(T'_3\) with a control grid. The remaining switches

Fig. 2. Variant of the circuit of a cascade generator with electronic or ionic devices.

are replaced by kenotrons or gas-filled rectifiers, depending on the power of the installation. The voltage to the control grids of the devices \(T_3\) and \(T'_3\) is applied with a phase shift of \(180^\circ\), so that in a given interval of time only one device (thyratron) will pass current.

The circuit shown in Fig. 2 operates as follows. At the moment when, under the action of the control voltage applied to the control grids, the device \(T'_3\) becomes conducting, the capacitance \(C'_2\) is charged from the direct-current source to the voltage \(E\) through the rectifier \(K'_2\) and the conducting device \(T'_3\). In the next half-period of the control voltage, when the device \(T_3\) passes current, the capacitance \(C'_2\) is connected in parallel with the capacitance \(C_2\), and through the rectifier \(K_2\) and the device \(T_3\) gives it half of its charge.

In the next half-period, when the device \(T'_3\) again opens, the capacitance \(C'_2\) is recharged through the rectifier \(K'_2\), and the thyratron \(T'_3\) from the capacitance

\(C_3'\) to the voltage of the source \(E\). The capacitance \(C_1'\), through the rectifiers \(K_1'\) and \(K_2'\), proves to be connected to the capacitance \(C_2\) and receives from it part of its charge.

The process of charging the capacitances \(C\) of the left column and of additionally charging the capacitances of the right column continues in the same way until all capacitances acquire a potential equal to \(E\).

From the above analysis of the operation of the circuit in Fig. 2 it follows that, by means of the devices with control grid \(T_3\) and \(T_3'\), the potential of point \(a\) of the circuit in Fig. 2 is forcibly varied from 0 to \(+E\) when it is connected either to point \(C\) or to \(b\).

It is known that in the ordinary circuit of half-wave rectification with one capacitor, the voltage on the kenotron varies from 0 to \(2U_m\), where \(U_m\) is the maximum value of the rectified voltage supplied by the transformer. This circumstance is also used in composing the electrical circuit of cascade generators, for example the one shown in Fig. 4. Each capacitor in this circuit, except \(C_n'\), is charged to a voltage equal to twice the voltage of the supply transformer.

2. Operation of the cascade generator under load

Let us consider the operation of a cascade generator assembled according to the circuit shown in Fig. 4, at the output of which some load is connected.

In practice it is of considerable interest to find the magnitude and the shape of the voltage curve at the output of the cascade generator.

The operation of the cascade-generator circuit, as was already shown above, consists in the charging and additional charging of the capacitor capacitances of both columns. At the maximum positive potential \(+E\) of point \(a\) in the circuit of Fig. 2, the capacitances of the left column of capacitors are charged. When the potential of point \(a\) reaches zero, the capacitances of the capacitors of the right column are charged. In this case the lowest capacitance of the right column is charged from the current source, while the others are additionally charged from the capacitances of the left column.

Fig. 3. Shape of the voltage curve at the output of the cascade-generator circuit when operating under load.

Charging and discharging of the capacitances of the left column proceed in the following manner: a) charging; b) discharge into the load during a half-period equal to \(\tau/2\); c) discharge of the capacitances of the left column after a time \(\tau/2\), elapsed from the time of charging, into the capacitances of the right column; and d) discharge into the load during the remaining part of the half-period \(\tau/2\). As a result, at the output of the cascade-generator circuit assembled according to the circuit of Fig. 4, a complex shape of the voltage curve shown in Fig. 3 will be obtained.

To determine the magnitude of the voltage pulsations at the output of the circuit, let us find the currents flowing in the circuit at the instants of charging of the capacitances of the left column and of recharging of the right one.

Figure 4 shows the distribution of the electric currents flowing in the circuit of a loaded cascade generator during the charging of the capacitances of the left column of capacitors and the recharging of the right one. The magnitudes of the currents are expressed in terms of the total charge that has passed through the corresponding elements of the circuit over the entire charging or recharging time. Solid arrows denote the currents flowing during charging of the left column, and dashed arrows—during recharging of the right one. This distribution of currents is given for the general case, when the generator consists of \(n\) stages.

During the discharge of the left column of capacitors into the load, a charge \(Q\) flows through the latter. Since the capacitances are connected in series, this charge flows through each capacitor of the left column. The upper first capacitor \(C_1\) thereby loses charge \(Q\). In order to maintain an unchanged output voltage, the charge delivered to the external circuit must be restored. This restoration is possible by the inflow of charges along the only path—through the kenotron \(K_1\) from the capacitor \(C'_1\). The latter, having no other load, operates only to charge the capacitance \(C_1\). In order to keep the voltage unchanged, it is necessary that the charge lost by the capacitance \(C'_1\) be replenished. This recharging will occur through the rectifier \(K'_1\) from the capacitance \(C_2\); therefore, over the period the capacitance \(C_2\) must deliver charge \(Q\) to the load and charge \(Q\) to the capacitance \(C_1\), in all \(2Q\), which must also be restored during charging. From the circuit shown in Fig. 4 it is seen that, during charging, the capacitance \(C_2\) will receive a charge \(Q\) through the capacitance \(C_1\), and it must receive still another charge \(Q\) from the capacitance \(C'_2\) through the rectifier \(K_2\). During charging of the left column, the capacitance \(C'_2\) delivers charge \(Q\) to the capacitance \(C_2\) and charge \(Q\) through the capacitance \(C'_1\), in all \(2Q\), which must also be restored during recharging. The capacitance \(C'_2\) will receive charge \(Q\) through the capacitance \(C'_1\), and still another charge \(Q\) from the capacitance \(C_3\) through the kenotron \(K_2\), and so on.

Fig. 4. Distribution of currents at the instants of charging and recharging of the capacitors of a cascade generator having \(n\) stages and connected to a load.

The greatest value of the voltage jump on each capacitance of the left column will occur at the instant of charging, when through the \(k\)-th capacitance of the left column, counting from the top, a charge equal to \(kQ\) passes. The magnitude of this voltage jump \(\delta U_k\) can be determined from the formula

\[ \delta U_k=\frac{kQ}{C_k}. \]

The total voltage jump on the entire column of capacitances, equal to the sum of the voltage pulsations on each capacitance, is written in the form

\[ \delta U_\Sigma=\sum_{k=1}^{n} k\,\frac{Q}{C_k}. \]

Taking the capacitances of all the capacitors to be equal, we obtain the formula for determining the voltage jump at the output:

\[ \delta U_{\Sigma}=\frac{Q}{C}(1+2+3+\ldots+n)=\frac{Q}{C}\cdot\frac{n(n+1)}{2}. \]

We shall call the deviation of the voltage \(\delta U\) at the output of the circuit from its mean value the voltage ripple. From Fig. 3 it is seen that the relation \(\delta U=\frac{1}{2}\delta U_{\Sigma}\) holds. Taking into account that \(Q=i\tau\), we may write:

\[ \frac{Q}{C}=\frac{i\tau}{C}=\frac{i}{fC}, \]

where \(f\) is the frequency of the applied voltage, and \(i\) is the load current. For determining the magnitude of the voltage ripple, we then obtain the following formula:

\[ \delta U=\frac{1}{4}n(n+1)\frac{i}{fC}. \tag{1} \]

Let us find the magnitude of the voltage at the output of the circuit when operating into a load.

If, with no load, the voltage on each capacitor reaches the value \(2U_m\), where \(U_m\) is the maximum transformer voltage, then under generator load the mean value of the voltage will be smaller. Consequently, the output voltage will be less than \(2nU_m\) by the amount of the internal voltage drop. Let us find the magnitude of this voltage drop; for this purpose we shall consider successively what components make up the voltage on the \(k\)-th capacitor under generator load.

During operation of the circuit in the steady state, the \(n\)-th capacitor of the right-hand column, \(C_n\), being recharged directly from the transformer through the rectifier \(K'_n\), will be charged to a certain voltage equal to \(U_m\). Therefore, we shall begin the consideration of the operation of the circuit under load from this section. After recharging, the capacitor \(C'_n\) is charged to the voltage \(U_m\). When the transformer voltage changes sign, the voltage of the capacitor \(C'_n\) and that of the transformer add, and at their terminals a voltage \(2U_m\) is obtained. In this case the capacitor \(C_n\) is charged both through the rectifier \(K_n\) and through all the remaining capacitors and rectifiers. The total charge delivered by the capacitor \(C'_n\) in charging the capacitors of the left-hand column will be \(nQ\) (see Fig. 4); therefore the voltage on it will decrease by an amount equal to

\[ \frac{nQ}{C}. \]

Since the rectifier \(K_n\) conducts current at this time, the voltage on the capacitor \(C_n\) will, by the end of charging, be equal to the sum of the voltages of the transformer and of the capacitor \(C'_n\), i.e. approximately

\[ U_m+U_m-\frac{nQ}{C}=2U_m-\frac{nQ}{C}. \]

After this, during the time \(\tau/2\), the capacitors of the left-hand column discharge into the load; therefore the voltage on the capacitor \(C_n\) decreases by

\[ \frac{i\cdot\tau/2}{C}=\frac{Q}{2C}. \]

When the transformer voltage reaches the value \(-U_m\) with respect to ground, recharging of the capacitors of the right-hand column begins; moreover, from Fig. 4 it is seen that the capacitor \(C_n\) then loses a charge \((n-1)Q\), and the voltage on it decreases by the amount

\[ (n-1)\frac{Q}{C}. \]

Since at this instant the rectifier \(K''_{n-1}\) conducts...

gives a current, then the voltage on capacitor \(C'_n\) will turn out to be equal to the voltage on capacitor \(C_n\) after the end of recharging, i.e.,

\[ 2U_m-\frac{nQ}{C}-\frac{(n-1)Q}{C}-\frac{Q}{2C} \]

and so on.

The process of algebraic summation of the voltages on the capacitors over the given time interval is conveniently represented in the form of Table 1.

This table has been compiled by following the distribution of the charging and recharging currents shown in Fig. 4.

Table 1

Distribution of voltages on the capacitors of a loaded cascade generator

From Table 1 let us take the expression for the magnitude of the voltage on the \(k\)-th capacitor, counting from the top of the column, which will correspond to the tabular index \(n-k'\), and sum the series obtained. We have:

\[ Uc_{n-k'\max} = 2U_m+\frac{nQ}{C} -\frac{2Q}{C} \left[ \overbrace{n+(n-1)+(n-2)+\cdots+(n-k')}^{k'+1\ \text{terms}} \right] -k'\frac{Q}{2C} = 2U_m+\frac{nQ}{C} -\frac{k'Q}{2C} -\frac{2Q}{C}\frac{[n+(n-k')](k'+1)}{2} = \]

\[ = 2U_m+\frac{(2n-k')Q}{2C} -\frac{2Q(2n-k')(k'+1)}{2C} = 2U_m-\frac{(2n-k')(2k'+1)}{2C}. \]

In order to pass to the ordinal number of the capacitor, counting from the top of the column, replace \(k'\) by \(n-k\). We obtain:

\[ Uc_{k\max}=2U_m-(n+k)(2n-2k+1)\frac{Q}{2C}, \tag{2} \]

hence

\[ U_{Ck\max}=2U_m-n(2n+1)\frac{Q}{2C}+k(2k-1)\frac{Q}{2C}. \tag{3} \]

Let us find the magnitude of the internal voltage drop \(\Delta U\) (Fig. 3). Denote

\[ \Delta U_{k\min}=2U_m-U_{Ck\max}=n(2n+1)\frac{Q}{2C}-k(2k-1)\frac{Q}{2C}. \tag{4} \]

It follows from formula (4) that, as the ordinal number of the capacitor increases, the magnitude of the voltage drop across the capacitor \(\Delta U\) decreases. It reaches its maximum value on the upper capacitors. Physically this is explained by the fact that the upper capacitors are farthest from the current source (the transformer) and are weakly coupled to it. The minimum value of the total internal voltage drop \(\Delta U_{\min}\) (Fig. 3) over the entire column of capacitors is expressed as the sum of the voltage drops across the individual capacitors, i.e.,

\[ \Delta U_{\min}=\sum_{k=1}^{n}\Delta U_k. \]

From formula (4) we have:

\[ \frac{\Delta U_{\min}}{\dfrac{Q}{2C}} = \sum_{k=1}^{n}\left[n(2n+1)-k(2k-1)\right] = n^2(2n+1)-2\sum_{k+1}^{n}k^2+\sum_{k=1}^{n}k = \]

\[ = n^2(2n+1)-\frac{2n(n+1)(2n+1)}{6} +\frac{n(n+1)}{2} = \frac{8n^3+3n^2+n}{6}. \]

Therefore

\[ \Delta U_{\min}=\frac{Q}{12C}\,(8n^3+3n^2+n). \]

From Fig. 3 it is seen that the mean value of \(\Delta U\) will exceed \(\Delta U_{\min}\) by the amount \(\delta U\), i.e.,

\[ \Delta U=\frac{Q}{12C}\,(8n^3+3n^2+n)+\frac{n(n+1)}{4}\frac{Q}{C}. \]

Hence we obtain the mean value of the total internal voltage drop in the form:

\[ \Delta U=\frac{Q}{6C}(4n^3+3n^2+2n)=\frac{i}{6fC}(4n^3+3n^2+2n). \tag{5} \]

As is seen from formulas (1) and (5), the voltage pulsations and the voltage drop increase very rapidly with increasing number of stages \(n\). To obtain ultrahigh voltages, the number of stages \(n\) has to be made large; therefore it is important to obtain the smallest values of \(\delta U\) and \(\Delta U\).

Since the capacitor \(C'_n\) operates at half voltage, its capacitance may be taken twice as large. If this change in capacitance is taken into account in deriving formula (5), then the term containing \(n^2\) disappears from it and the formula takes the form:

\[ \Delta U=\frac{i}{3fC}\,(2n^3+n). \]

It follows from this that doubling the capacitance of the capacitor \(C'_n\) will have a substantial effect only when the term \(3n^2\) in formula (5) is comparable in magnitude with the term \(4n^3\), i.e., for a small number of stages \(n\). The voltage at the output of the cascade generator, taking into account the internal drop, is expressed by the following formula:

\[ U_{\text{out}} = 2nU_m - \Delta U = 2nU_m - \frac{i}{6fC}\left(4n^3 + 3n^2 + 2n\right). \tag{6} \]

It follows from formula (6) that, as \(n\) increases, the output voltage at first increases almost proportionally to \(n\). Then, because of the rapid increase of the internal voltage drop with increasing \(n\), the rise of the voltage slows down. Let us determine the optimal number of generator stages \(n\) for which the highest average output voltage is obtained. Differentiating equation (6) with respect to \(n\) and equating the derivative to zero, we obtain:

\[ 2U_m - \frac{6n^2 + 3n + 1}{3}\,\frac{i}{fC} = 0. \]

Considering the terms \(3n\) and 1 small in comparison with \(6n^2\) and neglecting them, we obtain:

\[ n_{\text{opt}} = \sqrt{\frac{fC}{i}\,U_m}. \tag{7} \]

Let us consider an example. For a cascade generator with \(U_m = 110\,kV\); \(f = 200\,Hz\); \(C = 0.02\,\mu F\), at a load current \(i = 4\,mA\), we obtain:

\[ n_{\text{opt}} = \sqrt{\frac{200 \cdot 0.02 \cdot 10^{-6}}{4 \cdot 10^{-3}} \cdot 110 \cdot 10^3} = \sqrt{110} \approx 10. \]

In practice it is advisable to take the number of stages smaller than that calculated by formula (7), since the dimensions of the installation, as \(n\) approaches the optimum value, increase proportionally to \(n\), while the output voltage increases only insignificantly.

3. Cascade generator with unequal capacitances in the stages

It is evident from Fig. 4 that the charging and discharging currents flowing through the capacitors of both columns are proportional to their ordinal number, counting from the top. Therefore it may turn out that increasing the capacitance of the lower capacitors will be accompanied by a large reduction in the magnitude of the ripple and of the internal voltage drop.

Let us consider the operation of a cascade generator for the case in which the capacitance of the capacitors in the column varies proportionally to their ordinal numbers:

\[ C_k = kC \quad \text{and} \quad C'_k = kC', \]

i.e., it increases linearly from top to bottom.

The distribution of voltages on the capacitors for this case is given in Table 2, compiled similarly to Table 1.

Table 2

Distribution of voltage on the capacitors of a cascade generator under load
The capacitance of the capacitors varies proportionally to their ordinal numbers

The general term of the series by means of which the voltage on any \(k\)-th capacitor is computed can be represented in the form:

\[ \begin{aligned} U_{Ck\max}=U_{Cn-k'\max} &=2U_m-\frac{nQ}{nC}-\frac{(n-1)Q}{nC} -\frac{(n-1)Q}{(n-1)C}-\frac{(n-2)Q}{(n-1)C}\\ &\quad-\frac{(n-2)Q}{(n-2)C}-\ldots -\frac{n-k'}{\,n-k'+1\,}\frac{Q}{C} -\frac{n-k'}{\,n-k'\,}\frac{Q}{C} -\frac{Q}{2nC}-\frac{Q}{2(n-1)C}-\ldots\\ &\quad\ldots-\frac{Q}{2(n-k'+1)C} =2U_m-(k'+1)\frac{Q}{C}\\ &\quad-\frac{Q}{C}\left(\frac{n-1}{n}+\frac{n-2}{n-1}+\frac{n-3}{n-2}+\ldots+\frac{n-k'}{n-k'+1}\right)\\ &\quad-\frac{Q}{2C}\left(\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{n-k'+1}\right). \end{aligned} \]

Here also \(k'=n-k\), where \(k\) is the ordinal number of the capacitor, counting from the top of the column.

Since the table gives the values \(U_C=U_{C\max}\), we obtain the minimum value of the voltage drop, which is expressed by the following series:

\[ \Delta U_{k\min}=(k'+1)\frac{Q}{C} +\frac{Q}{C}\left(\frac{n-1}{n}+\frac{n-2}{n-1}+\ldots+\frac{n-k'}{n-k'+1}\right)+ \]

\[ +\frac{Q}{2C}\left(\frac{1}{n}+\frac{1}{n-1}+\ldots+\frac{1}{n-k'+1}\right)=(k'+1)\frac{Q}{C}+ \]
\[ +\frac{Q}{C}\left(\frac{n}{n}-\frac{1}{n}+\frac{n-1}{n-1}-\frac{1}{n-1}+\ldots+\frac{n-k'+1}{n-k'+1}-\frac{1}{n-k'+1}\right)+ \]
\[ +\frac{Q}{2C}\left(\frac{1}{n}+\frac{1}{n-1}+\ldots+\frac{1}{n-k'+1}\right) =(k'+1)\frac{Q}{C}+k'\frac{Q}{C}- \]
\[ -\frac{Q}{2C}\left(\frac{1}{n}+\frac{1}{n-1}+\ldots+\frac{1}{n-k'+1}\right). \]

Or, after replacing \(k'\) by \(n-k\),

\[ \Delta U_{k\min}=[2(n-k)+1]\frac{Q}{C} -\frac{Q}{2C}\left(\frac{1}{n}+\frac{1}{n-1}+\ldots+\frac{1}{k+1}\right). \tag{8} \]

Let us find the sum of the voltage drops \(\Delta U_{k\min}\) as \(k\) varies from 1 to \(n\), i.e. the complete minimum value of the voltage drop in the entire generator.

\[ \Delta U_{\min}=\sum_{k=1}^{n}\Delta U_{k\min} =\sum_{k=1}^{n}(2n+1-2k)\frac{Q}{C}- \]
\[ -\sum_{k=1}^{n}\left(\frac{1}{n}+\frac{1}{n-1}+\ldots+\frac{1}{k+1}\right)\frac{Q}{2C}. \tag{9} \]

Here:

\[ \sum_{k=1}^{n}[(2n+1)-2k]\frac{Q}{C} =n(2n+1)\frac{Q}{C}- \]
\[ -\sum_{k=1}^{n}2k\frac{Q}{C} =n(2n+1)\frac{Q}{C}-n(n+1)\frac{Q}{C}. \]

The second term of formula (9) is summed as follows:

For \(k=1\) we have the series
\[ \frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+\ldots+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}, \]

\[ k=2 \qquad \frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+\ldots+\frac{1}{4}+\frac{1}{3}, \]

\[ k=3 \qquad \frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+\ldots+\frac{1}{4}, \]

\[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \]

\[ k=n-2 \qquad \frac{1}{n}+\frac{1}{n-1}, \]

\[ k=n-1 \qquad \frac{1}{n}, \]

\[ k=n \qquad 0, \]

Adding by vertical rows, we obtain

\[ \sum_{k=1}^{n-1}\sum_{p=k}^{n-1}\frac{1}{p+1} =\frac{n-1}{n}+\frac{n-2}{n-1}+\frac{n-3}{n-2}+\ldots+\frac{3}{4}+\frac{2}{3}+\frac{1}{2}. \tag{10} \]

Series (10) can be expanded and reduced to the form:

\[ \left(\frac{n}{n}-\frac{1}{n}\right)+ \left(\frac{n-1}{n-1}-\frac{1}{n-1}\right)+\cdots+ \left(\frac{4}{4}-\frac{1}{4}\right)+ \left(\frac{3}{3}-\frac{1}{3}\right)+ \]

\[ +\left(\frac{2}{2}-\frac{1}{2}\right)=(n-1)-\sum_{k=1}^{n-1}\frac{1}{k+1}. \]

Substituting the values found into formula (9), we obtain:

\[ \Delta U_{\min}=\sum_{k=1}^{n}\Delta U_{k\min} =n(2n+1)\frac{Q}{C}-n(n+1)\frac{Q}{C}-(n-1)\frac{Q}{2C}+ \]

\[ +\frac{Q}{2C}\sum_{k=1}^{n-1}\frac{1}{k+1} =\frac{(2n^{2}-n+1)Q}{2C} +\frac{Q}{2C}\sum_{k=1}^{n-1}\frac{1}{k+1} \tag{11} \]

The average value of the voltage drop will be greater than the quantity given by formula (11) by the amount \(\delta U\) (Fig. 3). The voltage ripple \(\delta U\) for the \(k\)-th capacitor is expressed as

\[ \delta U_k=\pm \frac{1}{2}\frac{kQ}{kC}=\pm \frac{Q}{2C}. \]

The total value of the voltage ripple over the whole column will be determined by the condition:

\[ \delta U=\sum_{1}^{n}\delta U_k=\pm \frac{nQ}{2C}. \tag{12} \]

Consequently, the average magnitude of the total voltage drop is determined by the following formula:

\[ \Delta U= \frac{(2n^{2}-n+1)Q}{2C} +\frac{nQ}{2C} +\frac{Q}{2C}\sum_{k=1}^{n-1}\frac{1}{k+1} =n^{2}\frac{Q}{C}+ \]

\[ +\sum_{k=1}^{n}\frac{1}{k}\cdot\frac{Q}{2C}. \tag{13} \]

The sum \(\sum_{k=1}^{n}\frac{1}{k}\) increases with increasing \(n\) very slowly in comparison with the first term of expression (13). For \(n=10\), \(\sum_{1}^{n}\frac{1}{k}=2.93\). Therefore, for

\[ n \geq 10, \]

this term may be taken equal to \(\frac{3Q}{2C}\). Then formula (13) assumes the form:

\[ \Delta U \cong \frac{2n^{2}+3}{2}\frac{Q}{C} =\frac{i}{2fC}(2n^{2}+3). \tag{14} \]

Thus, the voltage at the output of the cascade generator in the case under consideration is determined by the formula

\[ U_{\text{out}}=2nU_m-\frac{i}{2fC}(2n^{2}+3). \tag{15} \]

Let us see by what factor the voltage ripple and the voltage drop have decreased in comparison with the first case considered, when the capacitances of all the capacitors of the cascade generator were identical. From formulas (1), (5), and (12), (14) we obtain the ratio of the quantities of interest to us in the first and second cases:

\[ \frac{\delta U_1}{\delta U_2}=\frac{n+1}{2} \quad \text{and} \quad \frac{\Delta U_1}{\Delta U_2}=\frac{4n^3+3n^2+2n}{3(2n^2+3)} . \]

At the same time, the total capacitance of the capacitors in the second case, when \(C_k=kC\), is equal to

\[ C_{2\Sigma}=\sum_1^n kC=\frac{n(n+1)}{2}\,C; \quad \text{but } nC=C_{1\Sigma}, \text{ therefore } \frac{C_{2\Sigma}}{C_{1\Sigma}}=\frac{n+1}{2}, \]

i.e., it has increased in comparison with the first case by a factor of \(\dfrac{n+1}{2}\).

Thus, the magnitude of the voltage ripple has decreased by the same factor by which the total capacitance has increased. The same result is also obtained with a uniform distribution of capacitance among the stages, if the capacitance of each capacitor is increased. Therefore, from the point of view of reducing voltage ripple, increasing the capacitances in proportion to their number offers no advantage. Consequently, changing the capacitance of the capacitors in proportion to their ordinal number will be advantageous only in the case where the magnitude of the internal voltage drop \(\Delta U\) decreases by a larger factor than the total capacitance increases. It follows from this that the inequality

\[ \frac{\Delta U_1}{\Delta U_2}>\frac{C_{2\Sigma}}{C_{1\Sigma}}, \text{ i.e. } \frac{4n^3+3n^2+2n}{3(2n^2+3)}>\frac{n+1}{2}. \tag{16} \]

must be satisfied.

This inequality is satisfied for \(n \geqslant 3\). Therefore, if the number of stages is greater than or equal to three, it is advantageous to change the capacitances of the capacitors in proportion to their ordinal number. In this case, for a given total capacitance of the capacitors, one can obtain a smaller value of the voltage drop \(\Delta U\) than with a uniform distribution of capacitances among the stages.

The value of the ratio of the left-hand side of inequality (16) to the right-hand side characterizes the degree of advantage of using the variant of the cascade-generator circuit under consideration. For large \(n\) this quantity approaches

\[ \frac{8}{6}, \]

i.e. \(\simeq 1.3\).

Thus, by distributing the capacitances among the stages according to the law \(C_k=kC\), one can obtain a reduction of the voltage drop by no more than 30%.

In practice, therefore, the question of the advantage of one or another method of distributing capacitances among the stages will be decided by design considerations.

4. Additional considerations on the cascade generator

From formulas (1) and (5) it follows that both \(\delta U\) and \(\Delta U\) are inversely proportional to the product \(fC\); therefore, in order to reduce the ripple and the voltage drop it is necessary either to increase the frequency of the charging current or to increase the capacitance of the capacitors.

In the general case it is more advantageous to increase the frequency, since the cost of high-voltage capacitors, as their capacitance is increased, rises more rapidly than the cost of the alternating-current source as the frequency is raised. Moreover, when using the cascade-generator circuit shown in Fig. 2, it is possible to feed it from a direct-current source of comparatively low voltage. The cost of such a source will also be small.

Let us note that an interesting feature of the circuit of Fig. 2 is its reversibility. If the triodes or thyratrons \(T_3\) and \(T'_3\) are interchanged with the rectifiers \(K_1\) and \(K'_1\), then the cascade generator is transformed into a transformer by means of which high-voltage direct current is converted into low-voltage direct current.

The frequency \(f\) can be increased up to such limits at which the transit time of the electrons over the entire length of the generator will already begin to have an effect, leading to a rapid increase in its internal resistance. For a two-stage generator, for example, the limiting frequency proved to be \(21\ \mathrm{MHz}\).

For a multistage generator this frequency will be lower. It is advantageous to increase the number of generator stages up to a certain limit, determined by formula (7), since the cost and dimensions of the capacitors, as their operating voltage is reduced, fall sharply; therefore, despite the increase in the number of stages \(n\), the cost and dimensions of the installation as a whole decrease.

Thus, by using a high frequency \(f\) and a large number of stages, one can obtain installations of sufficiently high power. With large \(n\), capacitors designed for a comparatively low operating voltage and having a large capacitance can be used, as a result of which the large value of the product \(fC\) compensates for the increase in the internal resistance of the installation as the number of stages is increased.

In low-power, small-size but multistage generators, in which capacitors of small capacitance are used, the parasitic capacitances between the columns of capacitors and between each of the columns and ground acquire great importance.

The parasitic capacitances smooth out the voltage jumps at the nodal points when the potential of point \(a\) in Fig. 2 changes from 0 to \(E\). This leads to a reduction of the output voltage, even in no-load operation.

In addition, large high-frequency currents will flow through the lower capacitors. These phenomena can for the most part be eliminated by the use of high-frequency chokes, connected through capacitors in parallel with each rectifier and forming tuned circuits with the parasitic capacitances. At the same time, the problem of supplying the heater filaments of the rectifiers is solved. They can be supplied with high-frequency voltage taken from taps on the choke.

The possibility of obtaining a comparatively high voltage from the installation with small dimensions (in the case of high-frequency supply and a large number of stages) opens broad prospects for the use of cascade generators. Thus, for example, a cascade generator was used to produce high voltage on the electrodes of a television tube. In this case it was supplied by periodic pulses of high-frequency currents. The simplest method of obtaining them is as follows. An inductance is inserted in the anode circuit of an electron tube blocked by a negative bias on the grid, and a cascade generator is connected in parallel with this inductance. The inductance, together with the parasitic capacitances, forms an oscillatory circuit. When positive pulses are applied to the control grid of the tube, current pulses pass in the anode circuit, and damped high-frequency oscillations are excited in the circuit, po-

fed to the cascade generator.

As a result, at the output of the last stage, with sufficiently frequent repetition of the high-frequency current pulses, a rectified high voltage is obtained.

The schematic diagram of such an installation is shown in Fig. 5, where \(L_a\) is the inductance included in the anode circuit of the tube, and \(C_a\) is the self-capacitance of the coil and other parasitic capacitances.

The cascade generator still provides one of the best technical solutions to the problem of obtaining very high direct voltages and even of transforming direct current. An essential advantage of the cascade generator is the possibility of using, in its construction, various standard parts whose production has been mastered, such as capacitors, kenotrons, insulators, resistors, etc.

By using gas-filled rectifiers in the high-voltage circuit, without reducing the overall efficiency of the installation, it is possible to obtain a considerable current (several amperes). For obtaining direct voltages above \(1000\ kV\), cascade generators at present constitute the most powerful current sources. The possibility of constructing powerful cascade generators is of interest for the practical solution of the problem of transmitting high-voltage direct current.

Fig. 5. Diagram of a cascade generator producing high-voltage pulses.

Fig. 6. Dependence of the efficiency of a ten-stage cascade generator at \(1500\ kV\) on the magnitude of the current drawn.

Figure 6 shows the dependence of the efficiency on the consumed current for a ten-stage cascade generator operating at a direct voltage of \(1500\ kV\) and a frequency of \(500\ Hz\). Here all the energy consumed is taken into account, including the filament supplies of the rectifiers, whose share will decrease as the power of the installation increases.

This circuit has good prospects from the standpoint of designing high-voltage installations. Let us point out, for example, that a ten-stage \(2000\ kV\) generator installed in the open air has an overall height of about \(6.5\ m\). The installation, operating at average working gradients along the insulation of \(3.1\ kV/cm\), has proved itself well in operation. An installation for \(3000\ kV\) relative to ground has a height of about \(10\ m\). Na-

the voltage at each stage is 300 kV. In a room with dimensions of 15 × 35 m² and a height of 16 m, a symmetrical voltage of 6000 kV was obtained.

By placing the cascade generator in a vessel filled with gas under pressure, it is possible, just as in the case of electrostatic generators with a moving belt, to reduce substantially the overall dimensions of the installation.

Fig. 7. Layout of a cascade generator for 5000 kV, current 4 mA. Generator height 10 m, vessel height 15 m.
\(G\)—generator, \(R\)—accelerating tube, \(R_1\)—voltage divider.

Fig. 8. Layout of a cascade generator for 3000 kV.

\(g\)—generator, \(T\)—100 kV transformer,

\(H\)—h.-f. generator, \(E\)—high-voltage electrode at 3000 kV, \(R\)—discharge tube, \(R_1\)—measuring resistance, \(R_2\)—potentiometer serving to equalize the potential distribution and to maintain a specified potential of the points of the shield.

A cascade generator for a voltage of 1000 kV and a current of 4 milliamperes, with a neutron or X-ray tube and auxiliary instruments such as a measuring resistance, etc., can be housed in a room measuring 4 × 8 × 4 m.

Figure 7 shows the layout of a cascade generator for 5 MV. The generator is placed in a pear-shaped vessel at an air pressure of 3 atm.

Figure 8 gives the layout of an installation designed in a tower at ordinary pressure. Buwers’ experiments showed that, with a diameter

with a tower diameter of 15 m and a height of 20 m, a cascade generator with a voltage of 3000 kV at negative polarity with respect to earth, and 2500 kV at positive polarity, can be accommodated in it.

Let us note that, both for an electrostatic generator with a moving belt and for a cascade generator, the same causes limit the volume of the room to definite dimensions. In both installations the electric charges are delivered to electrodes whose shape and distance from the wall make it possible to maintain them at a sufficiently high potential. Observations of the operation of such generators have made it possible to establish the following interesting circumstance. It turns out that corona discharge and the conduction current through a resistor placed between the wall and the installation, at the given distances, increase the permissible voltage of the generator. In this case it is important to have sufficient generator power in order to supply the required current. Owing to the more uniform distribution of the potential over the insulation, the height of a cascade generator for 1000 kV operating voltage may be taken smaller than that of an electrostatic one.