Calculation of Current Transformers with an Air Gap

I. D. Kutyavin

At present, very high requirements are imposed on the operation of relay protection of electrical systems with respect to its sensitivity, selectivity, and speed of action. In connection with this, the quality of relay devices is being continuously improved, and protection schemes are also being improved. However, in developing modern protection schemes, difficulties often arise because of the unsatisfactory operation of current transformers during transient processes in the primary circuit.

The point is that the operating condition of relay protection is the transient process in the primary circuit of the installation. In this case the protection must accurately determine the nature and location of the fault and carry out the corresponding operation. This is possible only if the protection scheme is supplied from instrument transformers which, under these conditions, possess sufficient accuracy or constancy of errors that can be determined in advance.

As is known, the current transformers used at present have quite sufficient accuracy only during normal operation of the installation. When short-circuit currents containing an aperiodic component appear in the primary circuit, the errors of these transformers in many cases become inadmissibly large owing to strong saturation of the core steel and a sharp decrease in the transformer mutual-induction coefficient.

In connection with this, there follows the necessity of developing current transformers possessing sufficient accuracy under all operating conditions of the installation. These requirements are met by transformers with a rectilinear characteristic

\[ I_2 = f(I_1). \]

The simplest way to obtain a rectilinear transformer characteristic would be to assume that the maximum induction in the core, for any overcurrents, will not exceed \(8\text{–}10\ \text{kGs}\). Then, under all operating conditions of the installation, the current transformer will operate on the ascending rectilinear part of the magnetization curve. However, such transformers, because of the appearance of residual magnetism in the core, will have a very small working induction limit, since the residual induction may reach \(6\text{–}7\ \text{kGs}\). Then the working induction limit is \(B_r = B_m - B_0 = 10 - 7 = 3\ \text{kGs}\).

Current transformers calculated for such an induction limit will turn out to be of impractical dimensions.

However, the residual induction can be reduced to any value by introducing an air gap into the magnetic circuit; the magnitude of this gap can be chosen on the basis of the following considerations. Let us assume that the greatest residual magnetism is characterized by the residual induction \(B_0\), produced by the internal magnetizing (retarding)

the force \(H_c\), which we shall express in certain residual specific ampere-turns \(aw_0\). Then the total magnetizing ampere-turns that create the residual magnetism will be equal to \(AW_0=aw_0l\), where \(l\) is the length of the mean line of force. If an air gap is introduced into the magnetic circuit, the magnetic resistance of the circuit will increase; therefore the residual induction will decrease and will be determined by the following expression:

\[ AW_0=aw_0l=aw'_0l+0.8B'_0\delta_p, \tag{1} \]

where \(B'_0\) is the specified reduced residual induction,
\(\quad aw'_0\) are the specific ampere-turns expended in producing the induction \(B'_0\) in the steel of the core, and
\(\quad \delta_p\) is the length of the calculated air gap in centimeters, taking into account the bulging of the lines of force in it.

From (1) one can determine the magnitude of the calculated air gap at which the residual induction decreases to \(B'_0\):

\[ \delta_p=\frac{0.4\pi l}{B'_0}(aw_0-aw'_0)\simeq \frac{0.4\pi aw_0l}{B'_0}. \tag{2} \]

In expression (2), the ampere-turns \(aw'_0\) may be neglected in the case where the length of the mean line of force \(l\) in the steel is small.

The actual structural length of the air gap \(\delta\) will be greater than the calculated one because of the bulging of the lines of force in the air gap and the inaccuracy of its stacking, and may be determined from the following formula:

\[ \delta=k_\delta \frac{ a\cdot b-(a+b)\lambda\delta_p+ \sqrt{[ab-(a+b)\lambda\delta_p]^2-4ab\lambda^2\delta_p^2} }{ 2\lambda\delta_p }, \tag{3} \]

where \(a\) and \(b\) are the sides of the core cross-section in centimeters,
\(\quad k_\delta=1.1\) is a coefficient allowing for the inaccuracy of stacking of the air gap, and
\(\quad \lambda\) is a coefficient allowing for the bulging of the lines of force in the air gap (see A. Ya. Builov, Fundamentals of Electrical Apparatus Engineering, 1946, p. 249).

If the induction \(B'_0\) is taken equal (or close) to the maximum induction in the transformer core at rated primary current and rated load in the secondary circuit, then this induction may be disregarded in subsequent calculations; in that case the upper limit of the working induction is

\[ B_p=B_m=8\text{--}10\ \text{kG}. \]

The maximum permissible magnetizing current corresponding to \(B_m\) and referred to the turns of the secondary winding of the transformer is:

\[ I_{om}=\frac{aw_ml+0.8B_m\delta_p}{w_2}, \tag{4} \]

where \(aw_m\) are the specific magnetizing ampere-turns for the core steel at an induction \(B_m=8\text{--}10\ \text{kG}\),
\(\quad w_2\) is the number of turns of the secondary winding.

Having established the basic principles for determining the air gap of transformers, let us consider their parameters.

Induction in the core at rated transformer load

As was indicated above, the condition for linearity of the transformer characteristic will be satisfied if the induction in the core at

with a large asymmetric overcurrent will not exceed 8–10 kg. Then the induction under the transformer’s rated load, \(B_n\), should be no more than 100–300 gauss, since the ratio of the magnetizing current at large overcurrents to that in the rated mode may be high.

Rated Secondary Current of the Transformers

With a rated induction \(B_n = 100\text{–}300\) gauss, a rated primary current up to 1000 A, and a rated secondary current of 5 A, the transformer core becomes structurally large. Therefore current transformers with an air gap should be manufactured with a rated secondary current of 1 A. In this case the total resistance of the secondary circuit \(Z_2\) (including the resistance of the transformer secondary winding) can be brought up to 5–10 ohms, which corresponds to a rated transformer burden of 5–10 VA.

Core Cross Section

The cross section of the transformer core can be determined from the equation for the secondary e.m.f. at its rated load:

\[ E_2 = I_{n2} Z_2 = 222 W_2 q B_n \cdot 10^{-8}, \tag{5} \]

Then the core cross section from (5) is:

\[ q = \frac{I_{n2} Z_2 10^8}{222 W_2 B_n} = \frac{S_n 10^8}{222 B_n (I_n W_2)}, \tag{6} \]

where \(S_n\) is the rated secondary power of the transformer in VA,
\(I_{n2}\) is the rated secondary current in amperes,
\(B_n = 100\text{–}300\) gauss is the induction in the core under the rated load of the transformer,
\(Z_2\) is the total resistance of the transformer secondary winding and load.

The rated secondary ampere-turns entering the denominator of expression (6) are constant; therefore the core cross section proves to be directly proportional to the rated power of the transformer \(S_n\) and inversely proportional to the rated induction \(B_n\).

Number of Winding Turns

The number of winding turns is determined on the basis of the required transformation ratio, the resistance of the secondary circuit \(Z_2\), and the core cross section of the current transformer. Table 1 gives the principal relationships of these quantities, calculated by formula (6), for \(Z_2 = 5\) ohms. The data of this table are not the only solution to the problem, but they show the order of magnitude of the individual quantities.

Equivalent Circuit of a Loaded Transformer

Fig. 1 shows the equivalent circuit of the current transformer under consideration, with the resistances indicated on it.

The active resistance \(r_{T1}\) and the leakage reactance \(x_{T1}\) of the primary winding do not affect the operation of the current transformer and therefore are not considered here.

Table 1

The leakage inductance of the secondary winding, in view of the complexity of the question, can be determined only experimentally under the conditions of factory tests of current transformers.

The active resistance of the secondary winding \(r_T\) is easily determined from the length and cross section of the conductor of this winding.

The resistances \(r_0\) and \(x_0\) of the no-load branch for current transformers with an air gap, at inductions up to 8–10 kgs, remain practically constant (see Fig. 1), since almost all the magnetizing ampere-turns are expended in overcoming the magnetic resistance of the air gap. In addition, the magnetization curve at the indicated inductions has a form close to rectilinear.

Fig. 1

The inductance coefficient of the no-load branch can be determined from the following well-known expression:

\[ L_0=\frac{4\pi q W_2^{\,2}10^{-9}}{\dfrac{l}{\mu}+\delta_p}, \tag{7} \]

where \(l\) is the length of the mean line of force in the steel of the core, in centimeters,
\(\mu\) is the magnetic permeability of the steel.

When calculating \(L_0\) by expression (7), it is more convenient, instead of \(\mu\), to substitute the induction \(B\), lying within the limits of 0–10 kgs, and the corresponding specific magnetizing ampere-turns \(aw\):

\[ L_0=\frac{qBW_2^{\,2}10^{-8}}{awl+0.8B\delta_p}. \tag{8} \]

The active resistance of the no-load branch \(r_0\) can be determined from the specific losses in the steel at a given induction \(B\) by the following formula:

\[ r_0=\frac{P_{\mathrm{st}}ql\gamma}{1000I_0^2}, \]

where \(P_{\text{st}}\) is the specific loss in the steel, in watts per kilogram, at the specified induction,

\(I_0\) is the rms value of the no-load current under the same conditions, and

\(\gamma\) is the specific weight of the steel.

The current \(I_0\) is determined from expression (4) by substituting into it the specified induction \(B\) and the ampere-turns corresponding to it:

\[ I_0=\frac{awl+0.8B\delta_p}{\sqrt{2}W_2}. \tag{10} \]

Expression (4) gives the maximum value of the current; therefore \(\sqrt{2}\) has been introduced into the denominator of (10).

As was already indicated above, when calculating \(r_0\) and \(L_0\) from expressions (8)—(10), one may use any induction up to \(10\ \text{kGs}\), taking for it the corresponding values of \(P_{\text{st}}\) and \(aw\).

To determine \(P_{\text{st}}\), it is necessary to use data relating to current transformers, since the insulation between the sheets of steel in them is weaker, and the losses in the steel are greater than in power transformers.

Finally, when calculating \(E_0\) and \(I_0\) from expressions (8) and (10), it is necessary to use the magnetization curve taken on alternating current, as a function of the amplitude specific ampere-turns.

Fig. 1 gives curves of variation of the resistances of the no-load branch \(r_0\) and \(x_0(\beta=0)\) as functions of the maximum specific magnetizing ampere-turns for one experimental specimen of a current transformer with an air gap, having a core cross section of \(25\ \text{cm}^2\), a secondary winding of 400 turns, and an air gap of \(3\ \text{mm}\). Its core was made of steel from a TP-type current transformer.

As can be seen from this figure, the inductive resistance \(x_0\) \((\beta=0)\), within the range of induction up to \(10\ \text{kGs}\) (up to \(38\ \frac{\text{A-turns}}{\text{cm}}\)), changes by only \(5\ \Omega\), or by 6.25%, reaching a maximum at \(5\ \text{kGs}\) (or \(18\ \frac{\text{A-turns}}{\text{cm}}\)). The active resistance of the magnetizing circuit decreases, with increasing magnetizing ampere-turns, by approximately 25%; however, this resistance is small and cannot exert a significant influence on the operation of the transformer in the steady-state process.

The influence of the aperiodic component of the magnetizing current on \(x_0\) is shown in Fig. 1 by means of curves for \(\beta=0\) and \(\beta=1.2\). The upper curve is plotted for the absence of an aperiodic current in the magnetizing current, and the lower one for the case when its aperiodic component is equal to 120% of the amplitude of the periodic component. As can be seen from this figure, the appearance of an aperiodic component of the indicated magnitude in the magnetizing current leads to a reduction of \(x_0\) by 7—10%.

However, the indicated variations of \(x_0\) can be taken into account in calculating protection, since these variations are unambiguous and cannot depend on any other factors, such as, for example, residual magnetism, saturation of the steel, etc.

The impedance of the secondary circuit of the transformer is

\[ z_2=\sqrt{(r_{T2}+r_n)^2+(x_{T2}+x_n)^2}=\sqrt{r_2^2+x_2^2}, \tag{11} \]

where \(r_n\) and \(x_n\) are the impedances of the transformer secondary load.

The impedance of the secondary loop of the transformer equivalent circuit is

\[ z_{02}=\sqrt{(r_0+r_2)^2+(x_0+x_2)^2}. \tag{12} \]

Transformation Ratio of the Transformer

Figure 2 shows the vector diagram of the ampere-turns of the transformer under consideration.

The primary ampere-turns on the basis of this diagram are:

\[ I_1 W_1=\sqrt{(I_2 W_2)^2+(I_0 W_2)^2+2(I_2 W_2)(I_0 W_2)\sin(\varphi_2+\psi_0)}, \tag{13} \]

where \(\varphi_2=\operatorname{arctg}\dfrac{x_2}{r_2}\) is the phase-shift angle of the current and e.m.f. in the secondary circuit,

\[ \psi_0=\operatorname{arctg}\frac{r_0}{x_0} \]

is the angle of losses in the iron.

The transformation ratio of the current transformer on the basis of (13) is:

\[ n_T=\frac{I_1}{I_2}=\frac{W_2}{W_1} \sqrt{1+\left(\frac{I_0}{I_2}\right)^2+2\frac{I_0}{I_2}\sin(\varphi_2+\psi_0)}, \tag{14} \]

The right-hand side of (14) may be expressed through resistances if it is taken into account that

\[ \frac{I_0}{I_2}=\frac{z_2}{z_0}, \]

\[ \sin(\varphi_2+\psi_0)=\frac{x_2x_0+r_2r_0}{z_2z_0}, \tag{15} \]

\[ \cos(\varphi_2+\psi_0)=\frac{r_2x_0-x_2r_0}{z_2+z_0}, \]

then expression (14) assumes the following form:

\[ n_T=\frac{W_2}{z_0 W_1} \sqrt{z_0^2+z_2^2+2(x_2x_0+r_2r_0)}. \tag{16} \]

Fig. 2

In order that the transformer have no current error, its secondary number of turns should be made equal to:

\[ W_2=\frac{n_T W_1 z_0} {\sqrt{z_0^2+z_2^2+2(x_2x_0+r_2r_0)}}. \tag{17} \]

Determination of the number of turns of the secondary winding from expression (17) can be carried out only by the method of successive approximation, since the resistances entering into the right-hand side themselves depend on the number of turns.

However, it is possible to obtain an approximate expression for \(W_2\) which will not have this drawback. To this end, in expression (17) we neglect the resistance \(z_2\), and the resistances \(r_2\) and \(x_2\) will be considered independent of \(W_2\).

In addition, we shall use (8) and (9) to transform the expressions for the resistances \(r_0\) and \(x_0\).

The resistance \(r_0\), on the basis of (9) and (10), is:

\[ r_0= \frac{P_{\text{ж}} d l r} {1000\left(\dfrac{\alpha \omega l+0.8B\delta r}{\sqrt{2}\,w_э}\right)^2} = C W_2^2, \tag{18} \]

where the constant

\[ C=\frac{P_{\mathrm{sc}} q' l_T}{500(awl+0.8B\delta_p)^2}. \tag{19} \]

The resistance \(x_0\), on the basis of (8):

\[ x_0=\omega L_0=\frac{\omega q B\cdot W_2^2 10^{-8}}{awl+0.8B\delta_p}=DW_2^2, \tag{20} \]

where the constant

\[ D=\frac{\omega q B\,10^{-8}}{awl+0.8B\delta_p}. \tag{21} \]

The total resistance of the no-load branch is

\[ z_0=W_2^2\sqrt{C^2+D^2}=AW_2^2, \tag{22} \]

then the secondary number of turns, on the basis of (17)--(22), is

\[ W_2=\frac{1}{A}\sqrt{n_T^2 W_1^2 A^2-2(Cr_2+Dx_2)}. \tag{23} \]

When the transformer is made with the number of turns \(W_2\) determined by expression (23), its transformation ratio with respect to currents will be equal to \(n_T\).

If in expression (17) one retains \(z_2\) and also considers it independent of \(W_2\), then an exact expression for \(W_2\) can be obtained.

Transformer errors

If compensation of the current error by selecting the secondary number of turns \(W_2\) according to expression (17) or (23) has not been carried out, and the transformation ratio \(n_T\) is determined from the ratio of the numbers of turns of the transformer, this error in percent can be determined from the following well-known expression:

\[ K_{i\%}=\frac{I_2W_2-I_1W_1}{I_1W_1}\,100. \tag{24} \]

Eliminating the ampere-turns from (24) on the basis of (13) and (15), we obtain:

\[ K_{i\%}=\frac{z_0-\sqrt{z_0^2+z_2^2+2(x_2x_0+r_2r_0)}}{\sqrt{z_0^2+z_2^2+2(x_2x_0+r_2r_0)}}\,100. \tag{25} \]

The phase-angle error of the transformer can also be determined on the basis of the diagram in Fig. 2:

\[ \tan \delta=\frac{I_0W_2\cos(\varphi_2+\psi_0)}{I_2W_2+I_0W_2\sin(\varphi_2+\psi_0)} =\frac{r_2x_0-r_0x_2}{z_0^2+(x_2x_0+r_2r_0)}, \tag{26} \]

then the phase-angle error in degrees is

\[ \delta^\circ=\arctan \frac{r_2x_0-r_0x_2}{z_0^2+(x_2x_0+r_2r_0)}. \tag{27} \]

Example. For the above-mentioned prototype of a transformer with an air gap, determine the errors and the current transformation ratio for alternating-current transformation and at inductions

in the core up to 10 kGs. Data: \(W_1=1;\ W_2=400;\ r_2=2\ \text{ohm};\ x_2=3.2\ \text{ohm};\ z_2=3.8\ \text{ohm}\). The values of the resistances \(r_0\) and \(x_0\) are to be taken from Fig. 1.

For \(B=1\) kGs, according to Fig. 1, \(r_0=4.3\) ohm and \(x_0=74\) ohm (at \(\beta=0\)). The resistance \(z_0=74.2\) ohm.

The current transformation ratio, on the basis of (16), is:

\[ n_T=\frac{400}{74.2\cdot 1}\sqrt{74.2^2+3.8^2+2(3.2\cdot 74+2\cdot 4.3)}=417.8. \]

The current error, on the basis of (25), is:

\[ K_{\%}= \frac{ 74.2-\sqrt{74.2^2+3.8^2+(3.2\cdot 74+2\cdot 4.3)2} }{ \sqrt{74.2^2+3.8^2+2(3.2\cdot 74+2\cdot 4.3)} }\cdot 100=-4.26\%. \]

The phase-angle error, on the basis of (27), is:

\[ \delta=\operatorname{arctg} \frac{2.74-4.3\cdot 3.2}{74.2^2+(3.2\cdot 74+2\cdot 4.3)} =1^\circ 20'. \]

The results for other values of the induction are calculated similarly to the preceding case and are given in Table 2.

Table 2

As is seen from Table 2, when setting the protection, the transformation ratio should be taken as equal to 416, and not 400. Then the current error at inductions from 4 to 10 kGs will be allowed for with sufficient accuracy. Or, in order to obtain a sufficiently accurate transformer with a transformation ratio of 400, the secondary number of turns \(W_2'\) must be wound according to formula (23).

The phase-angle error proved to be small and, when such transformers are used to supply wattmetric and distance protections, will not be of importance.

Table 2 also gives the no-load currents and the primary currents corresponding to the inductions. The no-load current was determined from the expression

\[ I_0=\frac{awl}{\sqrt{2}\cdot w_2}, \tag{28} \]

where \(l=45.5\) cm is the length of the mean line of force for the experimental transformer, and

\(aw\) is the specific magnetizing ampere-turns, which were determined from the magnetization curve of Fig. 1.

The expression for determining the current \(I_{1\mathrm{dop}}\) is obtained from (13) and (15) in the following form:

\[ I_{1\mathrm{dop}}=\frac{I_0 W_2}{z_2 W_1} \sqrt{z_0^2+z_2^2+2(x_2x_0+r_2r_0)} . \tag{29} \]

The last line of Table 2 gives the multiple of the primary current with respect to the primary rated current of \(400\,a\), at which the corresponding induction is reached. As is evident from this line, the transformer may be overloaded by the steady-state alternating current by a factor of 61; in this case the error in current will also be \(3.9\%\), and in angle, \(-122'\). But if the transformation ratio is taken as 416 rather than 400, or, with a transformation ratio of 400, the number of turns is wound in accordance with formula (23), then this error too will be eliminated.

Above, the operation of a current transformer with an air gap under steady-state conditions was investigated. It was shown that the magnitude of the air gap should be selected on the basis of the need to eliminate the influence of residual induction on the operation of the transformer. The resulting gap of several millimeters is quite sufficient for straightening the characteristics of the transformer. This is illustrated by Fig. 1, which shows the magnetization curve \(B\) and the resistance curves \(r_0\) and \(x_0\), as functions of the maximum specific magnetizing ampere-turns within the range from 0 to \(10\,\mathrm{kGs}\).

It has also been shown that the errors of such a transformer are considerably greater than the errors of normal transformers, but remain almost constant at inductions from 0 to \(10\,\mathrm{kGs}\). This makes it possible to eliminate the current error by adjusting the number of turns of the secondary winding or by taking into account the current transformation ratio from expression (16).

\({}^{1)}\) On the operation of the current transformers under consideration during short circuits in the primary circuit, see the separate article printed in this volume of the Institute’s Proceedings.