JPH0329865A - Exciting current detector of transformer - Google Patents

Abstract

PURPOSE:To accurately detect an exciting current by mounting a current detector for collectively detecting the difference between the value (m) times the primary current of a transformer whose primary/secondary turn ratio is m/n and the value (n) times the secondary current thereof and a means increasing the output of said detector by 1/m or 1/n times. CONSTITUTION:A transformer TR whose primary/secondary turn ratio is m/n (m, n: integer), a current detector CT collectively detecting the difference between the value (m) times the promary current I1 of the transformer TR and the value (n) times the secondary current I2 thereof and a means increasing the output of said detector CT by 1/m or 1/n times are mounted. The detector CT detects the difference m.I1-n.I2 between the value (m) times the current I1 and the value (n) times the current I2 and said difference is increased by 1/m times by a proportional amplifier to detect an exciting current Io converted on a primary side. Further, the current Io required actually is one obtained by increasing the output of the detector CT by 1/m times and, by this constitution, a detection error can be markedly reduced.

Description

【発明の詳細な説明】 〔発明の目的〕 （産業上の利用分野） 本発明はインバータ等の電力変換器の交流側端子に接続
される変圧器の励磁電流検出装置に関する。DETAILED DESCRIPTION OF THE INVENTION [Object of the Invention] (Industrial Application Field) The present invention relates to an excitation current detection device for a transformer connected to an AC side terminal of a power converter such as an inverter.

（従来の技術） 近年、大容量の自己消弧素子（例えばゲートターンオフ
サイリスタ：ＧＴ○等）の開発が盛んに行われ、インバ
ータ等の電力変換装置に用いられるようになってきた。(Prior Art) In recent years, large-capacity self-extinguishing elements (for example, gate turn-off thyristors: GT○, etc.) have been actively developed and are being used in power conversion devices such as inverters.

特にパルス幅変調制御インバータは、可変電圧可変周波
数の正弦波出力が得られることから交流電動機の駆動電
源として盛んに用いられるようになってきた。In particular, pulse width modulation controlled inverters have come to be widely used as driving power sources for AC motors because they can provide a sine wave output with variable voltage and variable frequency.

当該インバータ等の電力変換器の交流側端子には、絶縁
の目的で、あるいは交流電圧を昇圧したり降圧したりす
る目的で、さらには、複数台のインバータと多重接続す
る目的で、変圧器が設置される。この変圧器は上記電力
変換器を構戊する素子のスイッチング特性のバラツキに
より、あるいは外部的要因により、直流バイアス電圧が
印加し、励磁電流が増太し、やがては片側方向に磁気飽
和（偏磁）することがある。A transformer is connected to the AC side terminals of the power converter such as the inverter for the purpose of insulation, to step up or step down the AC voltage, or for the purpose of multiple connection with multiple inverters. will be installed. Due to variations in the switching characteristics of the elements that make up the power converter, or due to external factors, this transformer receives a DC bias voltage, increases the excitation current, and eventually reaches magnetic saturation (biased magnetization) in one direction. ) There are things to do.

このような変圧器の直流偏磁を防ぐために、当該変圧器
の励磁電流を検出し、当該励磁電流が過大にならないよ
うに電力変換器の交流側出力電圧を調整する必要がある
。In order to prevent such DC bias of the transformer, it is necessary to detect the excitation current of the transformer and adjust the AC side output voltage of the power converter so that the excitation current does not become excessive.

第７図は、従来の変圧器の励磁電流検出装置を示す構！
戊図である。Figure 7 shows the structure of a conventional transformer exciting current detection device!
It is a diagram.

図中、ＴＲは変圧器、”Ｉ．ｌＷ２は変圧器Ｔ　Ｒの↓
？、２次巻線、ＣＴ，　，　ＣＴ２は電流検出器、Ｋは
比例増幅器、Ａは加減算器である。In the figure, TR is a transformer, and I.W2 is a transformer TR.
? , secondary winding, CT, , CT2 is a current detector, K is a proportional amplifier, and A is an adder/subtractor.

変圧器ＴＲの工次／２次巻数比を８１：８２とした場合
、比例増幅器Ｋは、ａ　２　／　ａ　１倍する増幅器を
用意する。When the transformer TR has a turn/secondary turns ratio of 81:82, the proportional amplifier K is an amplifier that multiplies a 2 / a by 1.

電流検出器ＣＴ１及びＣＴ２により、各々工次電流Ｔ■
及び２次電流■２を検出し比例増幅Ｋにより、２次電流
■２を（ａ２／ａ■）倍する。この値Ｉ２　＝　Ｉ２　
・（ａ２　／　ａ１）は２次電流のｌ次側換算値となる
。さらに１次電流検出値■，と上記２次電流の１次側換
算値■２を減算器Ａに入力し、変圧器ＴＲの励磁電流■
。−■、（ａ２／ａ．）・丁，を求めている。The current detectors CT1 and CT2 each detect a working current T■
The secondary current (2) is detected and the secondary current (2) is multiplied by (a2/a2) by proportional amplification K. This value I2 = I2
・(a2/a1) is the primary side conversion value of the secondary current. Furthermore, input the detected primary current value ■, and the primary side conversion value ■2 of the above secondary current into the subtracter A, and input the excitation current of the transformer TR ■
. -■, (a2/a.)・Ding, is sought.

（発明が解決しようとする課題） このような従来の変圧器の励磁電流検出装置は次のよう
な問題点がある。(Problems to be Solved by the Invention) Such conventional excitation current detection devices for transformers have the following problems.

すなわち、変圧器の励磁電流■。の値は、工次電流Ｉ，
や２次電流（１次側換器）Ｒに比例するとごくわずかな
値であり、電流検出器の精度が悪いと、上記のようにし
て検出された励磁電流は信頼性のない値となってしまう
。In other words, the excitation current of the transformer■. The value of is the industrial current I,
The excitation current detected as described above will be an unreliable value if the current detector is not accurate. Put it away.

＝３ ４ ？えば、大容量の変圧器では、励磁電流Ｔ．は↓次電流
■１の２〜３％となるので、■■の定格を１００ＯＡと
した場合、Ｉ．の定格は３０Ａ程度となる。ここで、電
流検出器ＣＴ，，　ＣＴ２の精度誤差を工％とした場合
、ＩＯＡの検出誤差が出て従来の手法により変圧器の励
磁電流Ｔ。を求めると、単純に考えても約３０％の誤差
を含むことになる。実際には，上記励磁電流丁。の計算
は、１次電流Ｔ，と２次電流Ｉｒのベクトルの差となる
ので、上記励磁電流検出値の誤差はさらに大きくなる。=3 4? For example, in a large capacity transformer, the exciting current T. is 2 to 3% of the secondary current ■1, so if the rating of ■■ is 100OA, the I. The rating is about 30A. Here, if the accuracy error of the current detectors CT, CT2 is expressed as %, the detection error of IOA will appear, and the excitation current T of the transformer will be calculated using the conventional method. Calculating this will include an error of approximately 30%, even if considered simply. In fact, the above excitation current D. Since the calculation of is the difference between the vectors of the primary current T and the secondary current Ir, the error in the excitation current detection value becomes even larger.

この励磁電流の検出誤差は当該励磁電流を制御したとき
指令値と実電流との相違をもたらし、従って変圧器から
必要な出力電圧が得られなくなる可能性もある。また変
圧器の直流偏磁対策等を行う場合でも正確な励磁電流が
検出できないため荊記電力変換器側から適正な補償電圧
を発生できなくなるという欠点があった。This detection error of the excitation current causes a difference between the command value and the actual current when the excitation current is controlled, and there is a possibility that the required output voltage cannot be obtained from the transformer. Further, even when taking measures against DC bias in the transformer, etc., there is a drawback that an appropriate compensation voltage cannot be generated from the power converter side because accurate excitation current cannot be detected.

本発明は、以上の問題点に鑑みてなされたもので変圧器
の励磁電流を正確に検出できる励磁電流検出装置を提供
することを目的とする。The present invention has been made in view of the above problems, and an object of the present invention is to provide an excitation current detection device that can accurately detect the excitation current of a transformer.

〔発明の構威〕[Structure of invention]

（課題を解決するための手段） 以」二の目的を達戒するために、本発明は、工次７２次
の巻数比がｍ対ｎ＋ｋ　（ｍ，ｎは整数、黍は−０．５
≦β≦０．５の実数）の変圧器と、当該変圧器の１次電
流のｍ倍と２次電流のｎ倍との差を一括して検出する第
１の電流検出器と、２次電流を検出する第２の電流検出
器と、当該第２の電流検出器の出力を沼倍する手段と、
前記第１の電流検出器の出力と」二記β倍する手段の出
力との差を求め、その値を（　１　／　ｍ　）倍又は１
／（ｎ十濾）倍する手段とを具備している。(Means for Solving the Problems) In order to achieve the second object, the present invention provides a method in which the turn ratio of the 72nd order of work is m to n+k (m, n are integers, millet is -0.5
≦β≦0.5); a first current detector that collectively detects the difference between m times the primary current and n times the secondary current of the transformer; a second current detector for detecting current; a means for multiplying the output of the second current detector;
The difference between the output of the first current detector and the output of the means for multiplying by 2 β is calculated, and the value is multiplied by (1/m) or 1.
/(n ten filters).

また、同目的を達或するために、本発明は、１次７２次
の巻数比がｍ対ｎ　（ｍ，ｎは整数）の変圧器と、当該
変圧器のｌ次電流のｍ倍と２次電流のｎ倍との差を一括
して検出する電流検出器と、当該電流検出器の出力を（
　１　／　ｍ　）倍又は（１／ｎ）倍する手段とを具備
している。Furthermore, in order to achieve the same object, the present invention provides a transformer in which the primary and 72nd turns ratio is m to n (m and n are integers), and m times the primary current of the transformer and 2 A current detector that collectively detects the difference between n times the next current and the output of the current detector (
1/m) or (1/n).

（作用） 変圧器のｔ次７２次巻数比がｍ対ｎ＋ｋの場合、工次電
流Ｔ１はｍターンの電流を、また２次電流■２は電流方
向を逆にしてｎターンの電流を貫通形ホールＣＴ等の電
流検出器（第１の電流検出器）により一括して検出する
。その結果、ｍ・■ユーｎ・■２が検出される。また、
第２の電流検出器により、２次電流工２を検出し，演算
増幅器等の手段によりＡ倍する。眉は−０．５≦ｋ≦０
．５の実数である。それをさらに減算器により、上記第
工の電流検出器の出力から差し引く、その値を（　１　
／　ｍ　）倍することにより、次式のように励磁電流（
１次側換算値）Ｔｏが求められる。(Function) When the t-order 72nd turns ratio of the transformer is m to n+k, the primary current T1 is the current of m turns, and the secondary current 2 is the current of n turns with the current direction reversed. The current is detected all at once by a current detector (first current detector) such as a Hall CT. As a result, m・■yu n・■2 is detected. Also,
The secondary current 2 is detected by the second current detector and multiplied by A by means such as an operational amplifier. Eyebrows -0.5≦k≦0
．． It is a real number of 5. This is further subtracted from the output of the above-mentioned first current detector using a subtracter, and the value is (1
/ m ), the excitation current (
The primary side conversion value) To is calculated.

？。＝　　（ｍ４．−ｎ４２−７＆４２）／ｍ二■■一
（ｎ十沼）・■２／ｍ Ｉ１−Ｉ． このとき、第１の電流検出器はｍ・Ｉ１−ｎ−Ｉ２を検
出すればよいので、■１を直接検出する場合に比較する
と、検出電流の最大値は格段に小さくなり、その分検出
誤差も小さくなる。また、第２の電流検出器はβ■２を
検出すればよく、　沼は−０．５≦ｋ≦０．５の値をと
るので、やはり■２を直接検出する場合に比較すると、
ボの値に比例して検出誤差も小さくなる。? . = (m4.-n42-7&42)/m2 ■■1 (n Tonuma)・■2/m I1-I. At this time, the first current detector only needs to detect m・I1-n-I2, so compared to the case of directly detecting 1, the maximum value of the detected current becomes much smaller, and the detection error increases accordingly. will also become smaller. Also, the second current detector only needs to detect β■2, and since the swamp takes a value of -0.5≦k≦0.5, compared to the case where ■2 is directly detected,
The detection error also decreases in proportion to the value of .

また、変圧器の１次／２次巻数比がｍ対ｎの場合、↓つ
の電流検出器により、１次電流工，のｍ倍と２次電流■
２のｎ倍の差を一括して検出する。その結果、ｍ・Ｉ１
−ｎ・■２が検出され、さらに比例増幅器により（　１
　／　ｍ　）倍することにより１次側に換算した励磁電
流１。を検出できる。すなわち？ｏ＝（ｍ・Ｉ１−ｎ・
Ｉ２）／ｍ ＝Ｉ■一　（　ｎ　／　ｍ　）　　・■２＝■■−■２ となる。このときの電流検出器はｍ・Ｉ，−ｎ・工，の
電流を検出すればよく、さらに実際に必要とされる励磁
電流■。は当該検出器の出力を１　／　ｍ倍したもので
、全体の検出誤差は従来方式より格段に小さくなる。In addition, if the primary/secondary turns ratio of the transformer is m to n, ↓ two current detectors can be used to calculate m times the primary current and the secondary current
A difference of n times 2 is detected all at once. As a result, m・I1
-n・■2 is detected, and then ( 1
/ m ) excitation current 1 converted to the primary side by multiplying. can be detected. In other words? o=(m・I1−n・
I2)/m=I■1 (n/m) ・■2=■■−■2. The current detector at this time only needs to detect the current of m·I, -n·cm, and the excitation current (2) that is actually required. is the output of the detector multiplied by 1/m, and the overall detection error is much smaller than that of the conventional method.

このようにして変圧器の励磁電流を精度良く検出するこ
とが可能となり、正確な励磁電流制御を７ ８ー ？えるようになる。また変圧器の直流偏磁に対しても適
正な補償が可能となる。In this way, the excitation current of the transformer can be detected with high precision, allowing accurate excitation current control. You will be able to understand. It is also possible to appropriately compensate for DC bias in the transformer.

（実施例） 第１図乃至第６図は本発明の変圧器の励磁電流検出装置
の実施例を示す構或図である。(Embodiment) FIGS. 1 to 6 are configuration diagrams showing an embodiment of the excitation current detection device for a transformer according to the present invention.

図中．ＴＲは変圧器、り，は１次巻線、ｗ２は２次巻線
、ＣＴエ，　ＣＴ２，　ＣＴは電流検出器、Ｋ，Ｋ■ｌ
　Ｋ２は比例増幅器、Ａは加減算器である。In the figure. TR is the transformer, ri, is the primary winding, w2 is the secondary winding, CT, CT2, CT is the current detector, K, Kl
K2 is a proportional amplifier, and A is an adder/subtractor.

第１図は変圧器ＴＲの１次７２次の巻数比が１対ｌの場
合を示す。FIG. 1 shows a case where the turns ratio of the primary and 72nd orders of the transformer TR is 1:1.

電流検出器ＣＴは第１図（ｂ）に示すように、流れてい
る電流線を直接貫通させるタイプのホールＣＴ等を用い
る。当該ホールＣＴに工次電流■１が流れている電線Ｌ
エと２次電流工２が流れている電線Ｌ２をその電流方向
が逆になるように貫通させ、励磁電流■。＝　１：，　
一Ｉ２を直接検出している。説明の便宜上、位置差を無
視して考えると、例えば，｛次電流Ｉ，　＝　１０１０
Ａ、２次電流Ｉ２＝１００ＯＡとした場合、励磁電流工
。は■。＝工■−Ｉ２＝１ＯＡを電流検出器ＣＴにより
検出することになる。ＣＴの精度を１％？した場合、検
出誤差はＯ．ＬＡとなる。従来の電流検出器はＩＯＯＯ
Ａに対する１％の誤差すなわち１ｏＡの誤差があったの
に対し、格段の精度の向上が図れる。As the current detector CT, as shown in FIG. 1(b), a Hall CT or the like of a type through which a flowing current line passes directly is used. Electric wire L through which the electrical current ■1 is flowing through the relevant hole CT
The excitation current (2) is passed through the electric wire L2 through which the secondary current wire (d) and the secondary current wire (2) are flowing, so that the current direction is reversed. = 1:,
-I2 is directly detected. For convenience of explanation, ignoring the position difference, for example, {order current I, = 1010
A. When the secondary current I2 = 100OA, excitation current. ■. =Engine -I2=1OA will be detected by the current detector CT. CT accuracy 1%? In this case, the detection error is O. Becomes LA. Conventional current detector is IOOO
Whereas there was an error of 1% with respect to A, that is, an error of 1oA, the accuracy can be significantly improved.

変圧器ＴＲの↓次／２次巻数比がｍ対ｎの場合、ホール
ＣＴを貫通する電線数は１次電流線がｍ本、２次電流線
がｎ本になる。すなわち、電流検出器ＣＴは、ｍ・工■
−ｎ４２の電流を検出する。従ってこの値を（１／ｍ）
倍することにより、ｌ次側に換算した励磁電流■。が次
式のように求められる。When the ↓/secondary turns ratio of the transformer TR is m to n, the number of electric wires passing through the hole CT is m for the primary current line and n for the secondary current line. In other words, the current detector CT is
-Detect the current of n42. Therefore, this value is (1/m)
By multiplying, the excitation current is converted to the primary side. is calculated as follows.

？。＝　　（ｍ・■■−ｎ・Ｌ）／ｍ ＝Ｉ．−（ｎ／ｍ）　　・Ｔ■ Ｉ，　−　Ｉ２ ただし、Ｉ２は２次電流■２の１次側換算値、例えば、
ｍ＝２，ｎ＝３として、Ｉ．　＝１５１０Ａ、Ｉ２＝　
ＩＯＯＯＡの場合 ＝１０ ？なる。電流検出器ＣＴは、ｍ　４．　−　ｎ　４２＝
　２０Ａの電流を検出すればよく、その精度を１％とす
ると検出誤差は０．２Ａとなる。　しかも実際に使用す
る励磁電流工。はその（１／ｍ）＝０．５倍であるので
、実際の検出誤差は０．ＩＡとなる。従来の検出器では
、１５００Ａに対する１％すなわち、　１５Ａの検出誤
差があるのに対し、約１５０倍の精度の向上が図れる。? . = (m・■■−n・L)/m =I. -(n/m) ・T■ I, - I2 However, I2 is the primary side conversion value of the secondary current ■2, for example,
Assuming m=2 and n=3, I. =1510A, I2=
For IOOOA = 10? Become. Current detector CT is m4. −n42=
It is sufficient to detect a current of 20 A, and if the accuracy is 1%, the detection error will be 0.2 A. What's more, it's an exciting current machine that is actually used. is (1/m) = 0.5 times that, so the actual detection error is 0. Becomes an IA. While conventional detectors have a detection error of 1% for 1500A, or 15A, the accuracy can be improved by about 150 times.

第２図は、変圧器のｌ次／２次巻数比が１対１．２の場
合を示す。FIG. 2 shows a case where the transformer's primary/secondary turns ratio is 1:1.2.

第１の電流検出器ＣＴ■は１次電流■■と２次電流工２
の差を検出する。また、第２の電流検出器ＣＴ２は２次
電流工２を検出する。この第２の電流検出器ＣＴ２の出
力は比例増幅器Ｋを介して、ｄ＝０．２倍される。さら
に減算器Ａにより、前記第１の電流検出器ＣＴ■の出力
信号■■一１２と前記比例増幅器Ｋの出力信号０．２・
■２の差を求め、次のように励磁電流Ｔ。を算出してい
る。The first current detector CT■ detects the primary current ■■ and the secondary current
Detect the difference between Further, the second current detector CT2 detects the secondary current detector 2. The output of this second current detector CT2 is multiplied by d=0.2 via a proportional amplifier K. Further, a subtracter A converts the output signal 0.2 of the first current detector CT 12 and the output signal 0.2 of the proportional amplifier K.
■ Find the difference between 2 and calculate the excitation current T as follows. is being calculated.

？．＝Ｉ■−Ｉ２−０．２・Ｔ７． ？■■−１．２・Ｔ２ ？１■−Ｉ２ ？えば、Ｉ１＝　１２１ＯＡ、Ｉ２＝　ＩＯＯＯＡとし
て、位相差を無視して考えると、第工の電流検出器ＣＴ
■はＩ、−Ｉ■＝２１０Ａを検出することになる。その
精度を１％とすると検出誤差は２．ＩＡとなる。　また
、第２の電流検出器ＣＴ２はＩＯＯＯＡを検出し、精度
を１％とすると誤差はＩＯＡとなる。しかし次の比例増
幅器Ｋにより、０．２倍されるので、検出値は２００Ａ
で誤差は、２Ａとなる。すなわち、減算器Ａに入力され
る２つの信号の誤差はともに約２Ａということになる。? ．． =I■-I2-0.2・T7. ? ■■-1.2・T2? 1■-I2? For example, assuming I1 = 121OA and I2 = IOOOA, ignoring the phase difference, the current detector CT of the first part
(2) detects I, -I2 = 210A. If the accuracy is 1%, the detection error is 2. Becomes an IA. Further, the second current detector CT2 detects IOOOA, and assuming the accuracy is 1%, the error is IOA. However, the next proportional amplifier K multiplies it by 0.2, so the detected value is 200A.
The error is 2A. That is, the error between the two signals input to the subtracter A is approximately 2A.

従来の検出装置では１２００Ａの１％すなわち、１２Ａ
の誤差があったのに対し、約６倍の精度の向上が図れる
ことになる。Conventional detection equipment uses 1% of 1200A, or 12A.
This means that the accuracy can be improved by about 6 times, compared to the previous error.

第３図は、変圧器ＴＲの１次／２次巻数比が１対２．２
の場合を示す。Figure 3 shows that the primary/secondary turns ratio of transformer TR is 1:2.2.
The case is shown below.

第１の電流検出器ＣＴ１は、　１次電流■、と２次電流
■２の２倍との差を検出し、減算器Ａにその信号を入力
する。　また、第２の電流検出器ＣＴ２は２次電流■２
を検出し、比例増幅器Ｋにより０．２倍し、１１ １２ 減算器Ａにその信号を入力する。その結果、励磁電流■
。が次式のように求められる。The first current detector CT1 detects the difference between the primary current (2) and twice the secondary current (2), and inputs the signal to the subtracter A. In addition, the second current detector CT2 has a secondary current ■2
is detected, multiplied by 0.2 by proportional amplifier K, and the signal is input to 11 12 subtracter A. As a result, the excitation current
. is calculated as follows.

？。＝Ｉ１−２・Ｉ２−０．２・工２ ＝Ｉよ−２．２・■２ ＝■■−■２ ？えば、位相差を然視して工■＝２２１ＯＡ、工２１０
００Ａで考えると、第１の電流検出器ＣＴ１は、Ｔ１２
・Ｉ２＝２１０Ａを検出することになる。その精度を１
％とすると、検出誤差は２．１Ａとなる。また、第２の
電流検出器ＣＴ２は１０００　Ａを検出し精度を１％と
すると、誤差はＩＯＡとなる。しかし、比例増幅器Ｋを
介して０．２倍されるので、誤差は２Ａとなる。すなわ
ち、減算器Ａに入力される２つの信号の誤差はともに約
２Ａということになる。従来の検出装置では、２２００
Ａの１Ｉすなわち、２２Ａの誤差があったのに対し、約
１１倍の精度の向」二が図れることになる。? . =I1-2・I2-0.2・ENG2 =Iyo-2.2・■2 =■■-■2 ? For example, by looking directly at the phase difference, we have
Considering 00A, the first current detector CT1 is T12
・I2=210A will be detected. Its accuracy is 1
%, the detection error is 2.1A. Further, assuming that the second current detector CT2 detects 1000 A and has an accuracy of 1%, the error is IOA. However, since it is multiplied by 0.2 via the proportional amplifier K, the error becomes 2A. That is, the error between the two signals input to the subtracter A is approximately 2A. With conventional detection equipment, 2200
Compared to the error of 1I of A, that is, 22A, the accuracy can be improved by about 11 times.

第４図は、変圧器ＴＲの１次７２次巻数比がｍ対ｎ＋沼
の場合を示すもので、ｍ，ｎは整数、β？一０．５≦β
≦０．５の実数となる。第３図に対応するとｍ＝＝１，
ｎ＝２，β＝０．２である。Figure 4 shows the case where the primary 72nd turns ratio of the transformer TR is m to n + swamp, where m and n are integers and β? -0.5≦β
It is a real number of ≦0.5. Corresponding to Figure 3, m==1,
n=2, β=0.2.

第１の電流検出器ＣＴ１は、　１次電流■、のｍ倍と２
次電流１■のｎ倍との差ｍ４■一ｎ・１２を一括して検
出する。第２の電流検出器ＣＴ２は２次電流丁，を検出
しその出力を比例増幅器Ｋ１により｛倍する。The first current detector CT1 has m times the primary current ■, and 2
The difference m4■-n·12 between the next current 1■ and n times is detected at once. The second current detector CT2 detects the secondary current D, and its output is multiplied by the proportional amplifier K1.

減算器Ａにより、前記第１の電流検出器ＣＴ■の出力信
号ｍ・工■−ｎ−Ｉ２と前記比例増幅器Ｈ■の出力信号
沼・■２との差を演算し、さらに比例増幅器Ｋ２により
（１／ｍ）倍することにより，次式で示される励磁電流
（１次側換算値）■ｏが求められる。The subtracter A calculates the difference between the output signal m-n-I2 of the first current detector CT-1 and the output signal N-2 of the proportional amplifier H2, and the difference is calculated by the proportional amplifier K2. By multiplying by (1/m), the excitation current (primary side converted value) ■o is determined by the following formula.

？。＝　　（ｍ・Ｉ，−（ｎ十沼）Ｌ）／ｍ＝■■−（
ｎ十４）Ｔｉ／ｍ ＝Ｉ１−Ｉ２ ただし、工２は２次電流Ｉ２の１次側換算値である。? . = (m・I,−(ntonuma)L)/m=■■−(
n14) Ti/m = I1 - I2 However, 2 is the primary side converted value of the secondary current I2.

変圧器の１次７２次巻数比が１対１．４の場合、　２対
２．８として、ｍ＝２，ｎ＝３，　　ｔＡ＝−０．２と
して励磁電流■。を検出したほうが、検出誤差が小さく
なる利点がある。When the primary and 72nd turns ratio of the transformer is 1 to 1.4, 2 to 2.8, m = 2, n = 3, tA = -0.2, and the exciting current ■. Detecting this has the advantage of reducing the detection error.

？５図は、変圧器の１次／２次巻数比が１．２５対１の
場合を示す。降圧形の変圧器である。? FIG. 5 shows a case where the transformer has a primary/secondary turns ratio of 1.25:1. It is a step-down type transformer.

第１の電流検出器ＣＴ１は、　１次電流■１と２次電流
■２の差を一括して検出する。また、第２の電流検出器
ＣＴ２は工次電流■１を検出し、　比例増幅器Ｋ１によ
り、その信号を０．２５倍する。加減器Ａにより、第工
の電流検出器ＣＴ１の出力信号■■一Ｉ２　と比例増幅
器Ｋ■の出力信号０．２５・■１の和を求め、さらに第
２の比例増幅器Ｋ２により（１．／１．２５）倍するこ
とにより、励磁電流Ｔ。を検出する。The first current detector CT1 collectively detects the difference between the primary current (1) and the secondary current (2). Further, the second current detector CT2 detects the output current ■1, and the proportional amplifier K1 multiplies the signal by 0.25. The adder/subtractor A calculates the sum of the output signal of the first current detector CT1 (I2) and the output signal of the proportional amplifier K (0.25.■1), and then the second proportional amplifier K2 calculates (1./ 1.25) By multiplying the excitation current T. Detect.

Ｉｏ＝　（Ｉ１−Ｌ　＋０．２５Ｉ．）／］．．２５？
■、−　（１／１．２５）・工２＝工■一■２？こで、
第２の比例増幅器Ｋ２として、１倍を用いると、２次側
に換算した励磁電流■。が得られることは言うまでもな
い。Io=(I1-L+0.25I.)/]. ．． 25?
■、- (1/1.25)・Work 2 = Work■１■２? Here,
If 1x is used as the second proportional amplifier K2, the excitation current converted to the secondary side is ■. Needless to say, you can obtain

Ｉ１＝］．０１０Ａ、１■＝１２５ＯＡの場合を考える
と、第１の電流検出器ＣＴ■は■■一Ｉ．　＝　−　２
４０Ａを検出すればよく，精度を１％とすると、　検出
誤差は２．４Ａとなる。　また、第２の電流検出器ＣＴ
２はＩ１＝］Ｏ］．０？を検出し、精度を工％とじた場
合、誤差は１０．ｌＡとなるが、次の比例増幅器Ｋ■に
より０．２５倍されるので、誤差は２．５２５Ａとなる
。　さらに第２の比例増幅器Ｋ２により（　１　／１．
２５）倍されているので、上記２つの信号の誤差として
は、約２Ａということになる。従来の検出装置では、Ｉ
ＯＯＯＡの１％すなわちＩＯＡの誤差があったのに対し
、約５倍の精度の向上が図れたことになる。I1=]. 010A, 1■ = 125OA, the first current detector CT■ is ■■ - I. = − 2
It is sufficient to detect 40A, and if the accuracy is 1%, the detection error is 2.4A. In addition, a second current detector CT
2 is I1=]O]. 0? Detected and the accuracy is divided into %, the error is 10. 1A, but it is multiplied by 0.25 by the next proportional amplifier K■, so the error becomes 2.525A. Furthermore, by the second proportional amplifier K2 (1/1.
25) Since the signal is multiplied, the error between the above two signals is approximately 2A. In conventional detection devices, I
While there was an error of 1% of OOOA, that is, IOA, this means that the accuracy has been improved by about 5 times.

第６図は、変圧器ＴＲの１次／２次巻数比が、工対０．
８の場合を示す。変圧器としては第５図と全く同一のも
のであるが、第１及び第２の電流検出器ＣＴ１及びＣＴ
２の入れ方を変えている。FIG. 6 shows that the primary/secondary turns ratio of the transformer TR is 0.0.
8 is shown. The transformer is exactly the same as that shown in Fig. 5, but the first and second current detectors CT1 and CT
The way in which 2 is inserted is changed.

第１の電流検出器ＣＴ■は１次電流Ｉ１と２次電流■２
の差を一括して検出する。第２電流検出器ＣＴ２は、２
次電流■２を検出し、比例増幅器Ｋにより、０．２倍す
る。さらに加算器Ａにより、上記ＣＴ１の出力とＫの出
力の和を演算し、励磁電流■。を求めている。The first current detector CT■ has a primary current I1 and a secondary current ■2.
Detect differences in all at once. The second current detector CT2 is 2
The next current ■2 is detected and multiplied by 0.2 by the proportional amplifier K. Furthermore, the adder A calculates the sum of the output of CT1 and the output of K, and the excitation current (■) is calculated. I'm looking for.

：［ｏ＝Ｉ１−Ｉ２＋０．２Ｉ２ １５− １６− ＝Ｉよ−０．８・工２ ＝工、−■２ ？れは、第４図に対応させるとｍ＝１，ｎ＝１，沼＝−
０．２になる。:[o=I1-I2+0.2I2 15- 16- =Iyo-0.8・Work2 =Work, -■2? Corresponding to Fig. 4, m = 1, n = 1, swamp = -
It becomes 0.2.

■■＝１０１０Ａ，　Ｉ２＝１２５０Ａとした場合、第
１の電流検出器ＣＴ１は■、−Ｉ２＝−２４０Ａを検出
し、その精度を工％とすると、誤差は２．４Ａとなる。When ■■=1010A and I2=1250A, the first current detector CT1 detects ■, -I2=-240A, and if the accuracy is expressed as work%, the error is 2.4A.

　また、第２の電流検出器ＣＴ２は、Ｌ　＝　１２５０
Ａを検出し、その精度を１％とすると、誤差は１２．５
Ａとなるが、比例増幅器Ｋを介して０．２倍されるので
、誤差は２．５Ａとなる。第５図の手法に比較すると検
出誤差は少し大きくなる。しかし、従来の検出装置と比
較すれば、約４倍の精度の向上が期待できる。Further, the second current detector CT2 has L = 1250
If A is detected and its accuracy is 1%, the error is 12.5
A, but since it is multiplied by 0.2 via the proportional amplifier K, the error becomes 2.5A. Compared to the method shown in FIG. 5, the detection error is slightly larger. However, compared to conventional detection devices, it is expected that the accuracy will be improved by about four times.

一般に、変圧器ＴＲは、工次７２次巻数比をｍ対ｎとし
て整数比で表わすことができる。例えば、１次巻線数５
０ターン、２次巻線数１００ターンではｍ＝１＋　ｎ＝
２となって、簡単な数値となり、端数はでてこないで済
む。このような場合には、第１図あるいは、その応用例
で述べたように１つの電流検出器を用いて非常に精度良
く、励磁電流工。In general, the transformer TR can be expressed as an integer ratio, where the turn ratio of the 72nd order is m to n. For example, the number of primary windings is 5
For 0 turns and 100 turns of secondary winding, m=1+ n=
2, which is a simple number, and there are no fractions. In such a case, as shown in Figure 1 or its application example, one current detector can be used to perform the excitation current measurement with high accuracy.

を検出できる。can be detected.

しかし例えば、工次巻線数、５０ターンに対し、２次巻
線数１１０ターンの変圧器では、ｍ　＝　５、ｎ＝１１
となり、合計１６本の電線がホールＣＴの磁路を交差し
なければらならず、構威が複雑になる。However, for example, in a transformer with a primary winding of 50 turns and a secondary winding of 110 turns, m = 5, n = 11
Therefore, a total of 16 electric wires must cross the magnetic path of the Hall CT, making the structure complicated.

そこで、第４図で示したように、工次／２次巻数比とし
て、ｌ対２．２として表現し、　ｎ＝２，，４＝０．２
として励磁電流■。を検出する。検出精度はやや低下す
るが、構或が簡単になる利点があり、従来の方法に比較
すると大幅な精度の向上が期待できる。Therefore, as shown in Figure 4, the ratio of turns/secondary turns is expressed as l to 2.2, n=2,,4=0.2
As excitation current■. Detect. Although the detection accuracy decreases slightly, it has the advantage of simplifying the structure and can be expected to significantly improve accuracy compared to conventional methods.

降圧形の変圧器の励磁電流Ｉ。を検出する場合、１次／
２次巻数比をｎ＋ｋ対ｍに考え、第工の電流検出器ＣＴ
１で（ｎ−■．一ｍ−Ｉ２）を一括して検出し、第２の
電流検出器ＣＴ２で検出したＩエをす倍し、加算器によ
り上記ＣＴエの出力信号に加える。その結果を１／（ｎ
＋沼）倍すれば、１次側に換算した励磁電流■。が求め
られることは第５図の実施例で示した。Excitation current I of a step-down transformer. When detecting the primary/
Considering the secondary turns ratio as n+k to m, the current detector CT of the first stage
1, (n-■.1m-I2) is detected all at once, and the Ie detected by the second current detector CT2 is multiplied and added to the output signal of the CTe by an adder. The result is 1/(n
+ Swamp) If you multiply it, it will be the excitation current converted to the primary side■. It was shown in the example of FIG. 5 that this is required.

〔発明の効果〕〔Effect of the invention〕

以上のように、本発明の変圧器の励磁電流検出装置によ
れば、変圧器の王次７２次巻数比により、整数ｍ，ｎ及
び実数Ａを適当に選ぶことにより、検出誤差を格段に小
さくすることが可能となり、正確な励磁電流制御が達或
できる。又、変圧器の直流偏磁に対し、正確な検出が可
能となり、その補正も適正に行うことができるようにな
る。As described above, according to the excitation current detection device for a transformer of the present invention, the detection error can be significantly reduced by appropriately selecting the integers m, n, and the real number A according to the 72nd order turns ratio of the transformer. This makes it possible to achieve accurate excitation current control. Further, it becomes possible to accurately detect DC bias of the transformer, and to appropriately correct it.

【図面の簡単な説明】[Brief explanation of drawings]

第工図乃至第６図は本発明の変圧器の励磁電流検出装置
のそれぞれ異る実施例を示す構成図、第７図は従来の励
磁電流検出装置の構戒図である。 ＴＲ：変圧器 Ｗｌｌ臀２：変圧器の１次、２次巻線 ＣＴ．　，　ＣＴ２，　ＣＴ　：電流検出器Ｋ，　Ｋ１
，　Ｋ２：比例増幅器 Ａ：加減算器6 to 6 are configuration diagrams showing different embodiments of the excitation current detection device for a transformer according to the present invention, and FIG. 7 is a configuration diagram of a conventional excitation current detection device. TR: Transformer Wll Butt 2: Primary and secondary windings of transformer CT. , CT2, CT: Current detector K, K1
, K2: Proportional amplifier A: Adder/subtractor

Claims (1)

【特許請求の範囲】 （１）１次／２次の巻数比がｍ対ｎ（ｍ，ｎは整数）の
変圧器と、当該変圧器の１次電流のｍ倍と２次電流のｎ
倍との差を一括して検出する電流検出器と、当該電流検
出器の出力を（１／ｍ）倍又は（１／ｎ）倍する手段と
を具備してなる変圧器の励磁電流検出装置。（２）１次
／２次の巻数比がｍ対ｎ＋ｋ（ｍ，ｎは整数、ｋは−０
．５≦ｋ≦０．５の実数）の変圧器と、当該変圧器の１
次電流のｍ倍と２次電流のｎ倍との差を一括して検出す
る第１の電流検出器と、前記変圧器の２次電流を検出す
る第２の電流検出器と、当該第２の電流検出器の出力を
ｋ倍する手段と、前記第１の電流検出器の出力と上記ｋ
倍する手段の出力の差又は和を求め、その値を（１／ｍ
）倍又は１／（ｎ＋ｋ）倍をする手段とを具備してなる
変圧器の励磁電流検出装置。 （３）１次／２次の巻数比がｎ＋ｋ対ｍ（ｍ，ｎは整数
、ｋは−０．５≦ｋ≦０．５の実数）の変圧器と、当該
変圧器の１次電流のｎ倍と、２次電流のｍ倍との差を一
括して検出する第１の電流検出器と、前記変圧器の１次
電流を検出する第２の電流検出器と、当該第２の電流検
出器の出力をす倍する手段と、前記第１の電流検出器の
出力と上記ｋ倍する手段の出力の和又は差を求め、その
値を（１／ｍ）倍又は１／（ｎ＋ｋ）倍をする手段とを
具備してなる変圧器の励磁電流検出装置。[Claims] (1) A transformer with a primary/secondary turns ratio of m to n (m and n are integers), and m times the primary current and n times the secondary current of the transformer.
An excitation current detection device for a transformer, comprising a current detector that collectively detects the difference between the current detector and a means for multiplying the output of the current detector by (1/m) or (1/n). . (2) The primary/secondary turns ratio is m to n+k (m, n are integers, k is -0
．． 5≦k≦0.5 real number) and 1 of the transformer
a first current detector that collectively detects the difference between m times the secondary current and n times the secondary current; a second current detector that detects the secondary current of the transformer; means for multiplying the output of the first current detector by k; and the output of the first current detector and the k
Find the difference or sum of the outputs of the multiplying means and calculate the value (1/m
) or 1/(n+k) times. (3) A transformer with a primary/secondary turns ratio of n+k to m (m, n are integers, k is a real number of -0.5≦k≦0.5) and the primary current of the transformer. a first current detector that collectively detects the difference between n times the secondary current and m times the secondary current; a second current detector that detects the primary current of the transformer; and a second current detector that detects the primary current of the transformer; Find the sum or difference between the means for multiplying the output of the detector and the output of the first current detector and the means for multiplying by k, and multiply that value by (1/m) or 1/(n+k). An exciting current detection device for a transformer, comprising means for doubling the current.