Correction of Inductive Dividers

A. I. Kramnyuk

(Presented by the scientific seminar of the Department of Radio Engineering)

Recently, inductive dividers (IDs) have been developed using a new type of winding, called the ladder type [1]. Owing to their uniform structure, achieved purely by technological means, such dividers can in principle ensure high accuracy of the division ratio up to very high frequencies. The finite value of the output resistance, which is mainly inductive in character, and the presence of parasitic capacitances \((C_{\mathrm{par}})\) lead, as frequency increases, to the appearance of frequency error.

Among the indicated capacitances that have the most substantial influence should be included the load capacitance \((C_{\mathrm{n}})\), the residual parasitic capacitance between individual windings within the decade itself \((C_{0g})\), and the input capacitance of the decade \((C_{\mathrm{in}})\).

The influence of \(C_{0g}\) is examined in detail in [2]. Ways of reducing its influence are also outlined there, and the necessary recommendations are given.

When designing wideband IDs, to reduce direct feedthrough, individual decades must be placed in shields, which leads to the appearance of capacitance with respect to the shield \((C_{\mathrm{e}})\), which also has a substantial effect on the magnitude of the frequency error.

In the absence of parasitic capacitances, the voltage ratio is strictly equal to the ratio of the corresponding number of turns in the windings. This ratio will also hold in the presence of \(C_{\mathrm{par}}\) if the values of the equivalent capacitances \((C_{\mathrm{eq}})\) referred to each individual winding are equal to one another. Equality of \(C_{\mathrm{eq}}\) can be ensured by connecting additional correcting capacitances.

Achieving complete correction for all values of the division ratio is, in practice, rather difficult, since it is necessary to use a large number of correcting elements and fairly complicated switching systems. If one limits oneself to some entirely permissible value of the frequency error, then the number of correcting elements can be sharply reduced. Let us take into account the circumstance that the share contributed to the total frequency error by each subsequent decade, compared with the preceding one, will be smaller by as many times as the voltage applied to it is smaller than the voltage applied to the preceding decade. Therefore it is sufficient to ensure complete frequency independence for those decades in which one terminal is connected to the common conductor, and the error of the entire divider will be significantly reduced.

Figure 1 shows the circuit of the \(i\)-th decade operating into the capacitive load \(C_{n i}\), where

\[ C_1 \div C_{10} \]

are the correcting capacitances;

\[ C_{\mathrm{in},\,i+1} \]

is the input capacitance of the \(i+1\)-st decade;

\[ C_{11} \]

is the capacitance for correcting \(C_{\mathrm{in},\,i+1}\).

We shall determine the values of the correcting capacitances on the basis of equality of the equivalent capacitances referred to the individual windings.

Figure labels: \(i\)-th decade; to the next decade; to the battery; \(U_{\mathrm{out}}\); \(C_1,\ldots,C_{10}\); \(C_{11}\); \(C_{n i}\); taps \(0,\ldots,10\).

Fig. 1.

Taking into account that there are ten unknowns, while the number of equations that can be written is nine, we shall prescribe the relation

\[ \frac{C_1}{C_{w i}} = a_1 . \tag{1} \]

The quantity \(a_1\) will be determined below, proceeding from the attainment of a minimum value of the correcting capacitances.

When \(C_{n i}\) is connected to terminal “1” of the \(i\)-th decade, the condition of frequency independence is written in the form

\[ \frac{\displaystyle\sum_{k=2}^{10} C_k}{9} = C_1 + C_{n i}. \tag{2} \]

To derive “2,” the indicated condition takes the form

\[ \frac{\displaystyle\sum_{k=3}^{10} C_k}{8} = \frac{\displaystyle\sum_{k=1}^{2} C_k + 4 C_{n i}}{2}, \tag{3} \]

Similarly, we obtain equations for the other positions of the switch:

$$ \frac{\displaystyle\sum_{\kappa=4}^{10} C_\kappa}{7} = \frac{\displaystyle\sum_{\kappa=1}^{3} C_\kappa + 9C_{H_i}}{3}; \tag{4} $$

$$ \frac{\displaystyle\sum_{\kappa=5}^{10} C_\kappa}{6} = \frac{\displaystyle\sum_{\kappa=1}^{4} C_\kappa + 16C_{H_i}}{4}; \tag{5} $$

$$ \frac{\displaystyle\sum_{\kappa=6}^{10} C_\kappa}{5} = \frac{\displaystyle\sum_{\kappa=1}^{5} C_\kappa + 25C_{H_i}}{5}; \tag{6} $$

$$ \frac{\displaystyle\sum_{\kappa=7}^{10} C_\kappa}{4} = \frac{\displaystyle\sum_{\kappa=1}^{6} C_\kappa + 36C_{H_i}}{6}; \tag{7} $$

$$ \frac{\displaystyle\sum_{\kappa=8}^{10} C_\kappa}{3} = \frac{\displaystyle\sum_{\kappa=1}^{7} C_\kappa + 49C_{H_i}}{7}; \tag{8} $$

$$ \frac{\displaystyle\sum_{\kappa=9}^{10} C_\kappa}{2} = \frac{\displaystyle\sum_{\kappa=1}^{8} C_\kappa + 64C_{H_i}}{8}; \tag{9} $$

$$ \frac{\displaystyle\sum_{\kappa=10}^{10} C_\kappa}{1} = \frac{\displaystyle\sum_{\kappa=1}^{9} C_\kappa + 81C_{H_i}}{9}. \tag{10} $$

Solving (2) and (3) with (1) taken into account, we find

$$ C_2 = C_{H_i}(a_1 - 1.4). \tag{11} $$

Next, solving (3) and (4) with (1) and (11) taken into account, we find

$$ C_3 = C_{H_i}(a_1 - 2.2). \tag{12} $$

For the remaining capacitances,

$$ \begin{aligned} C_4 &= C_{H_i}(a_1 - 2.4); &\quad C_5 &= C_{H_i}(a_1 - 2);\\ C_6 &= C_{H_i}(a_1 - 1); &\quad C_7 &= C_{H_i}(a_1 + 0.6);\\ C_8 &= C_{H_i}(a_1 + 1.8); &\quad C_9 &= C_{H_i}(a_1 + 5.6);\\ C_{10} &= C_{H_i}(a_1 + 9). \end{aligned} \tag{13} $$

The condition of minimality will be satisfied if the value of \(a_1\) is set equal to 2.4 (from the expression for determining \(C_4\)). In this case,

$$ \begin{aligned} C_1 &= 2.4C_{H_i}; &\quad C_2 &= C_{H_i}; &\quad C_3 &= 0.2C_{H_i};\\ C_4 &= 0; &\quad C_5 &= 0.4C_{H_i}; &\quad C_6 &= 1.4C_{H_i};\\ C_7 &= 3C_{H_i}; &\quad C_8 &= 5.2C_{H_i}; &\quad C_9 &= 8C_{H_i};\\ C_{10} &= 11.4C_{H_i}. \end{aligned} $$

When correcting an inductive divider, in addition to the values found, it is necessary to connect the corresponding capacitances that eliminate the influence of the decade’s self-capacitances, the values of which are given in [2].

To eliminate the effect of the input capacitance of the following decade, \(C_{11}\) is connected. We shall determine its value for the worst case, when the \(i+1\) decade is connected to terminals \(0\)—\(1\) of the \(i\)-th decade. Directly from Fig. 1 we determine

\[ C_{11}=\frac{C_{\mathrm{in}\, i+1}}{9}. \tag{14} \]

To determine the value of \(C_{\mathrm{l}i}\), one may use the relation

\[ C_{\mathrm{l}i}=\sum_{m=1}^{i-1} C_{\mathrm{e}m}+C_{\mathrm{l}}, \tag{15} \]

where \(C_{\mathrm{e}m}\) is the capacitance of all subsequent decades with respect to the shield;

\(C_{\mathrm{l}}\) is the capacitance of the load of the inductive divider.

The experiment carried out confirmed the results of the analysis. Thus, with the aid of the correction considered, the magnitude of the frequency error of a three-decade divider at a frequency of \(100\ \mathrm{kHz}\) was reduced from \(0.2\%\) to \(0.01 \div 0.02\%\).

REFERENCES

1. Standard sources of alternating voltages. Report on the research project “Vereteno,” “Vitamin,” deposited manuscript, state registration No. 68021571, No. 68021579, VNIITEIR, Moscow, 1969.

2. A. I. Kramnyuk, M. S. Roitman. Expansion of the operating range of inductive dividers. This collection.