Comparison of low-frequency non-sinusoidal measurement for transformer no-load characteristics based on different interpolation methods

12 Nov.,2022

Low Frequency Transformer

Appendix A

For sine wave excitation, it is assumed that the induced electromotive force at both ends of the core is:

$$e(t) = E_{m} \sin (\omega t)$$

(A1)

where Em is the peak value of electromotive force.

From formula (1), the magnetic flux density B(t) is:

$$B(t) = B_{m} \cos (\omega t)$$

(A2)

$$B_{m} = \frac{{E_{m} }}{NA \cdot 2\pi f} = \frac{E}{NA \cdot \sqrt 2 \pi f}$$

(A3)

In formulas (A2) and (A3), Bm is the magnitude of the magnetic flux density, and E is the effective value of the electromotive force.

Similarly, for square wave excitation, it is assumed that the induced electromotive force and magnetic flux density at both ends of the core are:

$$e(t) = \left\{ {\begin{array}{*{20}c} {E_{m} } &\quad {0 < t < T/2} \\ { - E_{m} } &\quad {T/2 < t < T} \\ \end{array} } \right.$$

(A4)

$$B(t) = \left\{ {\begin{array}{*{20}l} {B_{m} (4t/T - 1)} \hfill &\quad {0 < t < T/2} \hfill \\ {B_{m} (3 - 4t/T)} \hfill & \quad{T/2 < t < T} \hfill \\ \end{array} } \right.$$

(A5)

$$B_{m} = \frac{{E_{m} }}{4NAf} = \frac{E}{4NAf}$$

(A6)

From equations (A3) and (A6), it can be seen that to keep the amplitude of the magnetic flux density Bm generated by the excitation voltages of square wave and sine wave at the same frequency equal, and it is necessary to make:

$$\frac{{E_{\sin } }}{{E_{{{\text{squ}}}} }} = \frac{\pi }{2\sqrt 2 } \approx 1.11$$

(A7)

where Esin is the effective value of sine wave, and Esqu is the effective value of square wave.

Similarly, for triangular wave excitation, it is assumed that the induced electromotive force and magnetic flux density at both ends of the core are:

$$e(t) = \left\{ {\begin{array}{*{20}c} {E_{m} (4t/T - 1)} & \quad{0 < t < T/2} \\ {E_{m} (3 - 4t/T)} &\quad {T/2 < t < T} \\ \end{array} } \right.$$

(A8)

$$B(t) = \left\{ {\begin{array}{*{20}l} {\frac{{E_{m} }}{NA}(2t^{2} /T - t)} \hfill & \quad{0 < t < T/2} \hfill \\ {\frac{{E_{m} }}{NA}( - 2t^{2} /T + 3t - T)} \hfill & \quad{T/2 < t < T} \hfill \\ \end{array} } \right.$$

(A9)

$$B_{m} = \frac{{E_{m} }}{8NAf} = \frac{\sqrt 3 E}{{8NAf}}$$

(A10)

It can be seen from formulas (A3) and (A10) that to keep the amplitude of the magnetic flux density Bm generated by the excitation voltages of triangular wave and sine wave at the same frequency equal, and it is necessary to make:

$$\frac{{E_{\sin } }}{{E_{{{\text{tri}}}} }} = \frac{\sqrt 6 \pi }{8} \approx 0.962$$

(A11)

where Esin is the effective value of sine wave, and Etri is the effective value of triangular wave.

For the transformer, ignoring the voltages of the leakage inductance and DC resistance, so that the effective value of the sine wave excitation voltage is 1.11 times that of the square wave, then the amplitude of magnetic flux density Bm generated under the two waveform excitations can be equal. This is also the basis of the volt-ampere characteristic curve and core loss under square wave excitation that can be converted to a sine wave. At the same time, it can be known from Eqs. (5) and (8) that, for voltage excitations at different frequencies of the same waveform, as long as the ratio of the effective value of the applied voltage is equal to the ratio of the frequencies, the generated Bm is equal.

Appendix B

In reference [33], the core loss was calculated and derived, and the equivalent formula for eddy current loss was obtained:

$$P_{e} = \frac{1}{{8\pi \rho N^{2} A}}\frac{1}{T}\int\limits_{0}^{T} {e(t)^{2} {\text{d}}t}$$

(B1)

where N is the number of turns, A is the area of the cross section and ρ is the resistivity of the core material.

Considering the effective value of electromotive force:

$$E = \sqrt {\frac{1}{T}\int\limits_{0}^{T} {e(t)^{2} {\text{d}}t} }$$

(B2)

Combining equations (B1) and (B2), it can be concluded that:

$$P_{e} = \frac{{E^{2} }}{{8\pi \rho N^{2} A}}$$

(B3)

Combining equations (A3) and (B3), the eddy current loss Pe-sin under sine wave excitation can be obtained as follows:

$$P_{e - \sin } = \frac{\pi A}{{4\rho }}B_{m}^{2} f^{2}$$

(B4)

Combining equations (A6) and (B3), it can be concluded that the eddy current loss under square wave excitation Pe-squ is:

$$P_{{e - {\text{squ}}}} = \frac{2A}{{\pi \rho }}B_{m}^{2} f^{2}$$

(B5)

According to formulas (A10) and (B3), the eddy current loss Pe-tri under triangular wave excitation is as follows:

$$P_{{e - {\text{tri}}}} = \frac{8A}{{3\pi \rho }}B_{m}^{2} f^{2}$$

(B6)

From equations (B4), (B5) and (B6), it can be seen that the eddy current loss is still proportional to the square of Bm and f, no matter for sine wave, square wave or triangular wave excitation, only the difference of proportion coefficient.

Appendix C

See Tables 2, 3 and 4.

Table 2 The effective values of voltage and current of the test sine wave with different frequencies (10 Hz, 15 Hz, 20 Hz, 50 Hz)

Full size table

Table 3 The effective values of voltage and current of the test triangular wave with different frequencies (10 Hz, 15 Hz, 20 Hz)

Full size table

Table 4 The effective values of voltage and current of the test square wave with different frequencies (10 Hz, 15 Hz, 20 Hz)

Full size table