CN101071142A - Method for compensating secondary current amplitude and phase error for current transformer - Google Patents

Abstract

The invention relates to a method for offsetting secondary current amplitude error of current mutual inductor, comprising the steps of: a. deriving the amplitude error caused by transfer of the current mutual inductor, i.e. mu d is approximately equal to the sum of mu I and k I2, from an equivalent circuit diagram of the current mutual inductor; b. using test data to find values A and B, where A=4.83, B=51.66, and obtaining an amplitude error offsetting model of the current mutual inductor; and c. using the amplitude error of the current mutual inductor to offset the secondary-side current amplitude of the current mutual inductor.

Description

The method of compensation current transformer secondary current amplitude and phase error

Technical field

The present invention relates to electricity field, relate in particular to the electric power power transmission and transformation system and utilize Current Transformer Secondary to measure the method for its transmission error of current compensation.

Background technology

Along with the electric pressure of transmission line of electricity is more and more higher, the power of circuit transmission is also increasing, the electric current of the measure and control device of transformer station and the measuring accuracy of power are had higher requirement, and the electric current under the especially little electric current and the measuring accuracy of power have been to weigh an important index of the measure and control device (RTU) of voltage levels.Present most measure and control device can only descend electric current and power just to have than high measurement accuracy near rated current, and the electric current under the little electric current and the measuring accuracy of power can not satisfy the requirement of GB.

Summary of the invention

Task of the present invention is that it transmits phase place, the amplitude error that produces by utilizing Current Transformer Secondary to measure the electric current dynamic compensation, makes electric current and power have very high measuring accuracy in gamut, satisfies the measurement requirement of current with high accuracy and power.

To achieve the above object of the invention, technical program of the present invention lies in 1, a kind of method of compensation current transformer secondary current amplitude error, the step of this method is as follows:

A. the equivalent circuit diagram by current transformer draws the amplitude error that the current transformer transmission is produced:

$f\≈\frac{I_L{N}_1}{I_1{N}_1}\sin (\mathrm{\α}+\mathrm{\β})\≈\frac{{\mathrm{BL}}_{\mathrm{AV}}}{I_1{N}_1\mathrm{\μ}}\sin (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{(Z+Z_2)L_{\mathrm{AV}}}{4.44N_2^2\mathrm{FS\μ}}\sin (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{1}{A+{\mathrm{BI}}_2}$

In the formula: $A=\frac{4.44N_2^2\mathrm{FS}{\mathrm{\μ}}_i}{(Z+Z_2)L_{\mathrm{AV}}\sin (\mathrm{\α}+\mathrm{\β})}$

$B=\frac{4.44N_2^2\mathrm{FSk}}{(Z+Z_2)L_{\mathrm{AV}}\sin (\mathrm{\α}+\mathrm{\β})}$

μ _d≈μ _i+k?I ₂

B. utilize test figure to obtain A, B value, A=4.83, B=51.66 draws current transformer amplitude error compensation model:

$f\≈\frac{1}{4.83+51.66I_2}$

C. utilize current transformer amplitude error dynamic current compensation mutual inductor secondary to measure the amplitude of electric current.

Another technical scheme of the present invention is: a kind of method of dynamic current compensation mutual inductor secondary current amplitude error, and the step of this method is as follows:

A. the equivalent circuit diagram by current transformer draws the phase error that the current transformer transmission is produced:

$\mathrm{\δ}\≈\frac{I_L{N}_1}{I_1{N}_1}\cos (\mathrm{\α}+\mathrm{\β})\≈\frac{{\mathrm{BL}}_{\mathrm{AV}}}{I_1{N}_1\mathrm{\μ}}\cos (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{(Z+Z_2)L_{\mathrm{AV}}}{4.44N_2^2\mathrm{FS\μ}}\cos (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{1}{C+{\mathrm{DI}}_2}$

In the formula: $C=\frac{4.44N_2^2\mathrm{FS}{\mathrm{\μ}}_i}{(Z+Z_2)L_{\mathrm{AV}}\cos (\mathrm{\α}+\mathrm{\β})}$

$D=\frac{4.44N_2^2\mathrm{FSk}}{(Z+Z_2)L_{\mathrm{AV}}\cos (\mathrm{\α}+\mathrm{\β})}$

μ _d≈μ _i+k?I ₂

B. utilize test figure to obtain C, D value, C=1.06, D=1.79 draws current transformer phase error compensation model:

$\mathrm{\δ}\≈\frac{1}{1.06+1.79I_2}$

C. utilize current transformer phase error dynamic current compensation mutual inductor secondary to measure the phase place of electric current.

The present invention utilizes Current Transformer Secondary to measure amplitude and phase error that the transmission of electric current dynamic current compensation mutual inductor produces, makes electric current and power have very high measuring accuracy in gamut, can satisfy the measurement requirement of current with high accuracy and power.

Description of drawings

Fig. 1 is the equivalent circuit diagram of current transformer;

Fig. 2 is the magnetization curve of magnetic material;

Fig. 3 is the current amplitude measuring error curve that does not have after phase compensation and the phase compensation;

Fig. 4 is the wattful power messurement graph of errors that does not have after phase compensation and the phase compensation.

Embodiment

The first, the amplitude, the phase error that are produced of current transformer transmission is as follows as can be known by the equivalent circuit diagram of current transformer:

$f\≈\frac{I_L{N}_1}{I_1{N}_1}\sin (\mathrm{\α}+\mathrm{\β})\≈\frac{{\mathrm{BL}}_{\mathrm{AV}}}{I_1{N}_1\mathrm{\μ}}\sin (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{(Z+Z_2)L_{\mathrm{AV}}}{4.44N_2^2\mathrm{FS\μ}}\sin (\mathrm{\α}+\mathrm{\β})---\left(1\right)$

$\mathrm{\δ}\≈\frac{I_L{N}_1}{I_1{N}_1}\cos (\mathrm{\α}+\mathrm{\β})\≈\frac{{\mathrm{BL}}_{\mathrm{AV}}}{I_1{N}_1\mathrm{\μ}}\cos (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{(Z+Z_2)L_{\mathrm{AV}}}{4.44N_2^2\mathrm{FS\μ}}\cos (\mathrm{\α}+\mathrm{\β})---\left(2\right)$

Parameter declaration in the error function: L _AVBe the average length of magnetic path, F is a frequency, and S is the cross-sectional area of magnetic circuit, N ₂Be the number of turn of secondary coil, Z is secondary load, Z ₂Be the equivalent secondary impedance of current transformer, α is E ₂In advance

Angle, β is the angle (core loss angle) of exciting current and main flux.

By error function as can be known: the amplitude of the generation that current transformer transmits is relevant with the secondary load and the magnetic permeability of current transformer with phase error, under the certain situation of magnetic permeability, it transmits the amplitude and the phase error that are produced and increases along with the increase of secondary load, under the certain situation of secondary amplitude, its error is mainly determined by the magnetic permeability of current transformer, and when magnetic permeability reduced, the error of amplitude and phase place increased, and the error of the error ratio amplitude of phase place is bigger.

From last surface analysis as can be known: core material magnetization do not have saturated before, μ _dAlong with exciting current I _μIncrease and increase, along with exciting current I _μMinimizing and reduce.

The second, the amplitude that the transmission of dynamic current compensation mutual inductor produces and the model of phase error:

By the magnetization curve of (1) formula, (2) formula summation current transformer as can be known, along with the minimizing of primary current, amplitude that current transformer transmits and phase error increase, and want to have high-precision measurement in gamut, the amplitude that must correcting current mutual inductor magnetic permeability causes and the error of phase place, μ _dWith i _μNot linear relation, for the saturated measuring accuracy in the past of the only guaranteed current transformer of measurement TA,

Suppose μ _d≈ μ _i+ k ₁I _μ(3)

Exciting current I _μSize by E ₂Amplitude decision, and amplitude E ₂With I ₂Be directly proportional, so (3) formula can become:

μ _d≈μ _i+k?I ₂ (4)

By (1), (2), (4) Shi Kede

Amplitude error:

$f\≈\frac{1}{A+{\mathrm{BI}}_2}---\left(5\right)$

In the formula: $A=\frac{4.44N_2^2\mathrm{FS}{\mathrm{\μ}}_i}{(Z+Z_2)L_{\mathrm{AV}}\sin (\mathrm{\α}+\mathrm{\β})}$

$B=\frac{4.44N_2^2\mathrm{FSk}}{(Z+Z_2)L_{\mathrm{AV}}\sin (\mathrm{\α}+\mathrm{\β})}$

Phase error:

$\mathrm{\δ}\≈\frac{1}{C+{\mathrm{DI}}_2}---\left(6\right)$

In the formula: $C=\frac{4.44N_2^2\mathrm{FS}{\mathrm{\μ}}_i}{(Z+Z_2)L_{\mathrm{AV}}\cos (\mathrm{\α}+\mathrm{\β})}$

$D=\frac{4.44N_2^2\mathrm{FSk}}{(Z+Z_2)L_{\mathrm{AV}}\cos (\mathrm{\α}+\mathrm{\β})}$

Transmit loop, the L in the error function for given current transformer _AV, F, S, N ₂, Z, Z ₂, α is constant, β is the angle of exciting current and main flux, it is relevant with the size of exciting current, but its variation is little, can think a constant, the amplitude of current transformer transmission, the A in the phase error function, B, C, D are constant like this, can measure the size of electric current according to Current Transformer Secondary, dynamically compensate amplitude, phase error that its transmission brings.

Three, engineering is used

(5) formula has been described the current transformer transmission and has been produced the relative error of amplitude and the relation of measurement electric current, magnetic permeability, (6) formula has been described the absolute error and the relation of measuring electric current, magnetic permeability of the phase place of current transformer transmission generation, because measure A, B, C, D are very difficult, so utilize test figure to obtain A, B, C, D, be a current transformer amplitude and phase error compensation model below.

$f\≈\frac{1}{4.83+51.66I_2}$

$\mathrm{\δ}\≈\frac{1}{1.06+1.79I_2}$

Whether to adopting the current amplitude of dynamic compensation technology, the measuring error of active power to compare, the voltage that applies is rated voltage, and the angle of electric current falls behind 60 ° of voltages, and the computing formula of error is among the figure: $E=\frac{A_C{\∫A}_L}{A_L},$ In the formula: A _CBe measured value, A _LBe theoretical value.

Claims (2)

1, a kind of method of compensation current transformer secondary current amplitude error is characterized in that, the step of this method is as follows:

A. the equivalent circuit diagram by current transformer draws the amplitude error that the current transformer transmission is produced:

$f\≈\frac{I_L{N}_1}{I_1{N}_1}\sin (\mathrm{\α}+\mathrm{\β})\≈\frac{{\mathrm{BL}}_{\mathrm{AV}}}{I_1{N}_1\mathrm{\μ}}\sin (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{(Z+Z_2)L_{\mathrm{AV}}}{{4.44N}_2^2\mathrm{FS\μ}}\sin (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{1}{A+{\mathrm{BI}}_2}$

In the formula: $A=\frac{{4.44N}_2^2\mathrm{FS}{\mathrm{\μ}}_i}{(Z+Z_2)L_{\mathrm{AV}}\sin (\mathrm{\α}+\mathrm{\β})}$

$B=\frac{{4.44N}_2^2\mathrm{FSk}}{(Z+Z_2)L_{\mathrm{AV}}\sin (\mathrm{\α}+\mathrm{\β})}$

μ _d≈μ _i+k?I ₂

B. utilize test figure to obtain A, B value, A=4.83, B=51.66 draws current transformer amplitude error compensation model:

$f\≈\frac{1}{4.83+{51.66I}_2}$

C. utilize current transformer amplitude error dynamic current compensation mutual inductor secondary to measure the amplitude of electric current.

2, a kind of method of compensation current transformer secondary current amplitude error is characterized in that, the step of this method is as follows:

A. the equivalent circuit diagram by current transformer draws the phase error that the current transformer transmission is produced:

$\mathrm{\δ}\≈\frac{I_L{N}_1}{I_1{N}_1}\cos (\mathrm{\α}+\mathrm{\β})\≈\frac{{\mathrm{BL}}_{\mathrm{AV}}}{I_1{N}_1\mathrm{\μ}}\cos (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{(Z+Z_2)L_{\mathrm{AV}}}{{4.44N}_2^2\mathrm{FS\μ}}\cos (\mathrm{\α}+\mathrm{\β})$

$\≈\frac{1}{C+{\mathrm{DI}}_2}$

In the formula: $C=\frac{{4.44N}_2^2\mathrm{FS}{\mathrm{\μ}}_i}{(Z+Z_2)L_{\mathrm{AV}}\cos (\mathrm{\α}+\mathrm{\β})}$

$D=\frac{{4.44N}_2^2\mathrm{FSk}}{(Z+Z_2)L_{\mathrm{AV}}\cos (\mathrm{\α}+\mathrm{\β})}$

μ _d≈μ _i+k?I ₂

B. utilize test figure to obtain C, D value, C=1.06, D=1.79 draws current transformer phase error compensation model:

$\mathrm{\δ}\≈\frac{1}{1.06+{1.79I}_2}$

C. utilize current transformer phase error dynamic current compensation mutual inductor secondary to measure the phase place of electric current.

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