On the Selection of the Capacitance of a Capacitor Supplying an Electromagnet

T. Yu. Mogilevskaya

The purpose of the present article is to consider the selection of the capacitance of a capacitor for a pulse generator supplying a special electromagnet, under the condition of a specified drop in the magnetic flux.

1. Determination of the Magnetic Flux

It may often be assumed that the magnetic induction in the profiled part of the pole tip of an electromagnet, as the distance from the center increases, varies according to the law:

\[ B_r = B_0 \left( \frac{r_0}{r} \right)^n, \tag{1} \]

where \(n < 1\), and, in addition, the magnitude of the induction at points of a circle of specified radius \(r_0\) is half the mean value of the induction over the area bounded by this circle, i.e.

\[ B_{\mathrm{cp}} = 2B_0. \tag{2} \]

Under these conditions the magnetic flux in the air gap of the electromagnet will be

\[ \Phi = 2\pi r_0^2 B_0 + \int_{r_0}^{r_n} 2\pi r B_r\,dr, \tag{3} \]

where \(r_n\) is the outer radius of the pole tip.

Substituting \(B_r\) under the integral sign, we obtain:

\[ \Phi = 2\pi r_0^2 B_0 \left[ \frac{1-n}{2-n} + \frac{1}{2-n} \left( \frac{r_n}{r_0} \right)^{2-n} \right]. \tag{4} \]

Taking into account the leakage coefficient \(\sigma\), the magnitude of the total magnetic flux produced by the winding of the electromagnet must be

\[ \Phi = 2\pi r_0^2 B_0 \sigma \left[ \frac{1-n}{2-n} + \frac{1}{2-n} \left( \frac{r_n}{r_0} \right)^{2-n} \right]. \tag{5} \]

2. Determination of the Profile of the Pole Tips

In calculations of the profile of the pole tips required for satisfying the above conditions, the magnitude of the air gap at points of a circle of specified radius \(r_0\) is determined by the power of the apparatus and must be calculated in advance or specified, as are the principal dimen-

measures of the magnetic circuit.

The calculation of the air gap at any other radius in the profiled part of the poles is carried out from the condition:

\[ \frac{\delta_r}{\delta_0}=\frac{B_0}{B_r}=\left(\frac{r}{r_0}\right)^n . \tag{6} \]

Hence

\[ \delta_r=\delta_0\left(\frac{r}{r_0}\right)^n . \tag{7} \]

After the profile of the pole tips obtained as a result of the calculation has been plotted, it is necessary to check whether relation (2) is satisfied.

To calculate \(B_{\mathrm{cp}}\), we divide the curve \(B=f(r)\), constructed according to equation (1), into tubes and calculate the magnetic flux inside each of them by the formula:

\[ \Phi_n=2\pi r_n a_n B_n, \tag{8} \]

where \(r_n\) is the mean radius of the tube, \(a_n\) is its width, and \(B_n\) is the mean value of the induction inside the tube.

It is obvious that

\[ B_{\mathrm{cp}}=\frac{\sum \Phi_n}{\pi r_0^2}. \tag{9} \]

3. Determination of the parameters of the electromagnet winding

The inductance of the winding is

\[ L=\frac{w\Phi_m}{I_m}=\frac{w^2\Phi_m}{AW}, \tag{10} \]

where \(AW\) is the total required ampere-turns, calculated in the usual manner.

The wire cross section is selected according to the effective value of the current, which can be determined as

\[ I=\frac{I_m}{K_A}=\frac{AW}{K_A w}, \tag{11} \]

where \(K_A\) is the amplitude coefficient, depending on the shape of the curve.

The active resistance of the coil, without taking into account the skin effect, is

\[ R_k=4\rho\,\frac{lw}{\pi d_{\mathrm{pr}}^2}, \tag{12} \]

where \(l\) is the mean length of a turn.

If the winding is divided into two coils (one around each pole), \(b\) is the width of each of them, and \(h\) is the height of the pole at the location of the coils, then the number of layers in each coil (without allowance for insulation) is

\[ n=\frac{w d_{\mathrm{pr}}}{2h}, \qquad b=nd_{\mathrm{pr}}=\frac{w d_{\mathrm{pr}}^2}{2h} \]

and

\[ l=\pi(2r_{\mathrm{н}}+b)=\pi\left(2r_{\mathrm{н}}+\frac{w d_{\mathrm{pr}}^2}{2h}\right). \]

After substitution into (12), we obtain

\[ R_k=4\rho w\left(\frac{2r_{\mathrm{н}}}{d_{\mathrm{pr}}^2}+\frac{w}{2h}\right). \tag{13} \]

4. Selection of the capacitance of the capacitor

Let us calculate the aperiodic discharge of the capacitor into a coil with the parameters specified above (Fig. 1).

The voltage on the coil and the discharge current are obtained as equal to:

\[ u=\frac{U_0}{p_1-p_2}\left(p_1 e^{p_2 t}-p_2 e^{p_1 t}\right), \tag{14} \]

\[ i=\frac{C U_0 p_1 p_2}{p_1-p_2}\left(e^{p_1 t}-e^{p_2 t}\right), \tag{15} \]

where

\[ p_1=-\frac{R}{2L}+\sqrt{\frac{R^2}{4L^2}-\frac{1}{LC}}, \]

\[ p_2=-\frac{R}{2L}-\sqrt{\frac{R^2}{4L^2}-\frac{1}{LC}}, \]

\(U_0\) is the initial value of the voltage on the capacitor,

\(R=R_k+R_{\mathrm{add}}\) is calculated from the condition:

\[ R>2\sqrt{\frac{L}{C}}. \tag{16} \]

Fig. 1

From the condition \(\dfrac{di}{dt}=0\), the time for the current to rise to its maximum will be

\[ \tau_{\phi}=\frac{1}{p_1-p_2}\ln\frac{p_2}{p_1}. \tag{17} \]

Substituting the values of the roots \(p_1\) and \(p_2\), we obtain:

\[ \tau_{\phi}= \frac{1}{2\sqrt{\dfrac{R^2}{4L^2}-\dfrac{1}{LC}}} \ln\left[ RC\left( \frac{R}{2L}-\frac{1}{RC} +\sqrt{\frac{R^2}{4L^2}-\frac{1}{LC}} \right) \right]. \tag{18} \]

This time must be no greater than a quarter of the period. We take the limiting case:

\[ \tau_{\phi}=\frac{1}{4}T=\frac{1}{4f}. \]

Then the relation between the frequency and the parameters of the circuit is obtained as:

\[ f= \frac{ \sqrt{\dfrac{R^2}{4L^2}-\dfrac{1}{LC}} }{ 2\ln\left[ RC\left( \dfrac{R}{2L}-\dfrac{1}{RC} +\sqrt{\dfrac{R^2}{4L^2}-\dfrac{1}{LC}} \right) \right] }. \tag{19} \]

The further calculation is carried out as follows: we assign various values of \(w\), compute the winding parameters, and then, for several arbitrary values of the capacitance, having selected the value \(R\) from condition (16), we calculate \(\ln N\), where

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

The mean value is

\[ (\ln N)_{\mathrm{av}} = \frac{\displaystyle \sum_{i=1}^{i=k} \left(\ln N_i\right)}{k}. \]

Substituting this value into equation (19), we obtain:

\[ f = \frac{\displaystyle \sqrt{\frac{R^2}{4L^2} - \frac{1}{LC}}} {2(\ln N)_{\mathrm{av}}}. \]

Hence the value of the capacitance is

\[ C = \frac{1} {\displaystyle \frac{R^2}{4L} - 4Lf^2(\ln N)^2_{\mathrm{av}}}. \tag{20} \]

5. Example

As an example, let us consider an electromagnet whose basic dimensions are given in Fig. 2.

In addition, let us take \(r_0 = 7.5\ \mathrm{cm}\), \(\delta_0 = 4.6\ \mathrm{cm}\), \(B_0 = 1.25 \cdot 10^{-5}\ \mathrm{Wb/cm^2}\); \(n = 2/3\), \(\sigma = 1.6\).

From equation (5) we find

\[ \Phi_m = 2\pi \cdot 7.5^2 \cdot 1.25 \cdot 10^{-5} \cdot 1.6 \left[ \frac{1 - 2/3}{2 - 2/3} + \frac{1}{2 - 2/3} \left(\frac{10}{7.5}\right)^{2-2/3} \right] = 9.72 \cdot 10^{-3}\ \mathrm{Wb}. \]

It is more convenient to calculate the profile of the pole pieces in tabular form, using the existing dimensions of the electromagnet. In Tables 1 and 2 are given, respectively, the calculation of the magnitude of the air gap and the induction at various radii, and the calculation of the flux inside the elementary tubes.

Fig. 2

Table 1

We divide the curve $B=f(r)$ into tubes, as shown in Fig. 3, and calculate the magnitude of the magnetic flux inside each tube.

Table 2

From Table 2 the total magnetic flux inside the circle of radius $r_0$ is

$$ \sum_{1}^{7} \Phi_n = 4{,}386 \cdot 10^{-3}\ \text{Wb}, $$

$$ B_{\mathrm{av}} = \frac{\sum_{1}^{7}\Phi_n}{\pi r_0^2} = \frac{4{,}386 \cdot 10^{-3}}{\pi \cdot 7{,}5^2} = 2{,}48 \cdot 10^{-5}\ \text{Wb/cm}^2, $$

$$ 2B_0 = 2{,}5 \cdot 10^{-5}\ \text{Wb/cm}^2. $$

The difference obtained may be attributed to inaccuracy in the construction. The profile of the pole pieces has been drawn quite correctly.

The total required ampere-turns, taken with a certain margin, are

$$ AW = 5500. $$

The inductance of the winding is

$$ L = \frac{w^2 \Phi_m}{AW} = \frac{w^2 \cdot 9{,}72 \cdot 10^{-3}}{5500} = 1{,}77 \cdot 10^{-6} w^2\ \text{H}. $$

For the obtained shape of the discharge-current curve we take the amplitude coefficient $K_A=5$. The subsequent check confirmed the correctness of this choice. The effective value of the current is obtained as

$$ I = \frac{AW}{K_A \cdot w} = \frac{1100}{w}\ \text{A}. $$

The height of the pole at the location of the coils is $h=5$ cm, $r_n=10$ cm. For copper $\rho = 0{,}0175 \cdot 10^{-4}$ ohm·cm.

Substituting these quantities into (12), we obtain

\[ R=0.07\cdot 10^{-5}w\left(\frac{200}{d_{pr}^{2}}+w\right). \]

Calculation of \((\ln N)_{av}\):

\[ w=100. \]

\[ I=\frac{1100}{100}=11\ a. \]

The cross section of the insulated copper wire for this current magnitude is

\[ S_{pr}=1.5\ \text{mm}^{2}. \]

\[ d_{pr}=0.2\sqrt{\frac{S_{pr}}{\pi}}=0.14\ \text{cm}. \]

\[ R_k=0.07\cdot 10^{-5}\cdot 100\left(\frac{200}{0.14^{2}}+100\right)=721\cdot 10^{-3}\simeq 0.72\ \text{ohm}. \]

\[ L=1.77\cdot 10^{-6}\cdot 10^{1}=17.7\cdot 10^{-3}\ \text{henry}. \]

Fig. 3

We set the capacitance value at \(C=10\ \mu\text{F}=10^{-5}\ \text{F}\).

\[ 2\sqrt{\frac{L}{C}}=84.2 \]

We take \(R=85\) ohms, i.e., an additional resistance is needed, \(R_{\text{add}}=84.28\) ohms.

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

\[ =85\cdot 10^{-5}\left[\frac{85\cdot 10^3}{2\cdot 17.7}-\frac{10^5}{85} +\sqrt{\left(\frac{85\cdot 10^3}{2\cdot 17.7}\right)^2-\frac{10^8}{17.7}}\right]=1.32. \]

\[ \ln N=2.3\cdot 0.12=0.276. \]

Similarly, we carry out the calculation for other arbitrary values of \(w\) and \(C\), and, for convenience, collect the obtained data in Table 3.

Table 3

The mean value of the logarithm is

\[ (\ln N)_{\text{av}}=\frac{\displaystyle\sum_{1}^{9}\ln N_i}{9} =\frac{2.604}{9}=0.293. \]

Substituting this value into (20), we obtain the expression for the capacitance in the form

\[ C=\frac{1}{\dfrac{R^2}{4L}-4Lf^2\cdot 0.293^2} =\frac{1}{\dfrac{R^2}{4L}-0.344Lf^2}. \]