1、1The Transformer on loadIt has been shown that a primary input voltage can be transformed to any 1Vdesired open-circuit secondary voltage by a suitable choice of turns ratio. is 2E2Eavailable for circulating a load current impedance. For the moment, a lagging power factor will be considered. The sec
2、ondary current and the resulting ampere-turns will change the flux, tending to demagnetize the core, reduce and with it 2NI m. Because the primary leakage impedance drop is so low, a small alteration to 1E 1Ewill cause an appreciable increase of primary current from to a new value of 0IIequal to . T
3、he extra primary current and ampere-turns nearly ijXRV11/cancel the whole of the secondary ampere-turns. This being so, the mutual flux suffers only a slight modification and requires practically the same net ampere-turns as on no load. The total primary ampere-turns are increased by an amount 10NIn
4、ecessary to neutralize the same amount of secondary ampere-turns. In the 2vector equation, ; alternatively, . At full load, 1021NII 2101NIIthe current is only about 5% of the full-load current and so is nearly equal to0. Because in mind that , the input kVA which is 12/NI 211/Eapproximately is also
5、approximately equal to the output kVA, .1IE 2IEThe physical current has increased, and with in the primary leakage flux to which it is proportional. The total flux linking the primary, is 1mpshown unchanged because the total back e.m.f., （ ）is still equal dtNE/11and opposite to . However, there has
6、been a redistribution of flux and the mutual 1V2component has fallen due to the increase of with . Although the change is 11Ismall, the secondary demand could not be met without a mutual flux and e.m.f. alteration to permit primary current to change. The net flux linking the secondary swinding has b
7、een further reduced by the establishment of secondary leakage flux due to , and this opposes . Although and are indicated separately, they 2Imm2combine to one resultant in the core which will be downwards at the instant shown. Thus the secondary terminal voltage is reduced to which can be dtNVS/2con
8、sidered in two components, i.e. or vectorially tdm/2. As for the primary, is responsible for a substantially constant 22IjXEVsecondary leakage inductance . It will be noticed that the primary 22/Nileakage flux is responsible for part of the change in the secondary terminal voltage due to its effects
9、 on the mutual flux. The two leakage fluxes are closely related; , 2for example, by its demagnetizing action on has caused the changes on the mprimary side which led to the establishment of primary leakage flux.If a low enough leading power factor is considered, the total secondary flux and the mutu
10、al flux are increased causing the secondary terminal voltage to rise with load. is unchanged in magnitude from the no load condition since, neglecting presistance, it still has to provide a total back e.m.f. equal to . It is virtually the 1Vsame as , though now produced by the combined effect of pri
11、mary and secondary 1ampere-turns. The mutual flux must still change with load to give a change of and 1Epermit more primary current to flow. has increased this time but due to the vector 1E3combination with there is still an increase of primary current.1VTwo more points should be made about the figu
12、res. Firstly, a unity turns ratio has been assumed for convenience so that . Secondly, the physical picture is 21Edrawn for a different instant of time from the vector diagrams which show , 0mif the horizontal axis is taken as usual, to be the zero time reference. There are instants in the cycle whe
13、n primary leakage flux is zero, when the secondary leakage flux is zero, and when primary and secondary leakage flux is zero, and when primary and secondary leakage fluxes are in the same sense.The equivalent circuit already derived for the transformer with the secondary terminals open, can easily b
14、e extended to cover the loaded secondary by the addition of the secondary resistance and leakage reactance.Practically all transformers have a turns ratio different from unity although such an arrangement is sometimes employed for the purposes of electrically isolating one circuit from another opera
15、ting at the same voltage. To explain the case where the reaction of the secondary will be viewed from the primary winding. 21NThe reaction is experienced only in terms of the magnetizing force due to the secondary ampere-turns. There is no way of detecting from the primary side whether is large and
16、small or vice versa, it is the product of current and turns which 2I2causes the reaction. Consequently, a secondary winding can be replaced by any number of different equivalent windings and load circuits which will give rise to an identical reaction on the primary .It is clearly convenient to chang
17、e the secondary winding to an equivalent winding having the same number of turns as the 1Nprimary. With changes to , since the e.m.f.s are proportional to turns, 2N1N4which is the same as .212)/(EN1EFor current, since the reaction ampere turns must be unchanged 122NImust be equal to .i.e. .2I 212)/(
18、INIFor impedance, since any secondary voltage becomes , and VV)/(21secondary current becomes , then any secondary impedance, including II)/(12load impedance, must become . Consequently, IN/)/(/21and .212)/(RN212)/(XNXIf the primary turns are taken as reference turns, the process is called referring
19、to the primary side.There are a few checks which can be made to see if the procedure outlined is valid.For example, the copper loss in the referred secondary winding must be the same as in the original secondary otherwise the primary would have to supply a different loss power. Must be equal to . 2R
20、I 2RI ）21212/()/(NRNdoes in fact reduce to .Similarly the stored magnetic energy in the leakage field which is )/(2LIproportional to will be found to check as . The referred secondary 2XI 2XI.21212 )/()/( ENINEkVAThe argument is sound, though at first it may have seemed suspect. In fact, if the actu
21、al secondary winding was removed physically from the core and replaced by the equivalent winding and load circuit designed to give the parameters , , and , measurements from the primary terminals would be unable to 1N2RX2I5detect any difference in secondary ampere-turns, demand or copper loss, under
22、 kVAnormal power frequency operation.There is no point in choosing any basis other than equal turns on primary and referred secondary, but it is sometimes convenient to refer the primary to the secondary winding. In this case, if all the subscript 1s are interchanged for the subscript 2s, the necess
23、ary referring constants are easily found; e.g. ,21R; similarly and .21X12R12XThe equivalent circuit for the general case where except that has 21Nmrbeen added to allow for iron loss and an ideal lossless transformation has been included before the secondary terminals to return to .All calculations o
24、f 2V2internal voltage and power losses are made before this ideal transformation is applied. The behavior of a transformer as detected at both sets of terminals is the same as the behavior detected at the corresponding terminals of this circuit when the appropriate parameters are inserted. The sligh
25、tly different representation showing the coils and side by side with a core in between is only used for convenience. On the 1N2transformer itself, the coils are, of course, wound round the same core.Very little error is introduced if the magnetizing branch is transferred to the primary terminals, bu
26、t a few anomalies will arise. For example, the current shown flowing through the primary impedance is no longer the whole of the primary current. The error is quite small since is usually such a small fraction of . Slightly 0I 1Idifferent answers may be obtained to a particular problem depending on
27、whether or not allowance is made for this error. With this simplified circuit, the primary and referred secondary impedances can be added to give:And 2121)/(ReN 2121)/(NXe6It should be pointed out that the equivalent circuit as derived here is only valid for normal operation at power frequencies; ca
28、pacitance effects must be taken into account whenever the rate of change of voltage would give rise to appreciable capacitance currents, . They are important at high voltages and at dtCVIc/frequencies much beyond 100 cycles/sec. A further point is not the only possible equivalent circuit even for po
29、wer frequencies .An alternative , treating the transformer as a three-or four-terminal network, gives rise to a representation which is just as accurate and has some advantages for the circuit engineer who treats all devices as circuit elements with certain transfer properties. The circuit on this b
30、asis would have a turns ratio having a phase shift as well as a magnitude change, and the impedances would not be the same as those of the windings. The circuit would not explain the phenomena within the device like the effects of saturation, so for an understanding of internal behavior.There are tw
31、o ways of looking at the equivalent circuit:(a) viewed from the primary as a sink but the referred load impedance connected across ,or2V(b) Viewed from the secondary as a source of constant voltage with internal 1Vdrops due to and . The magnetizing branch is sometimes omitted in this 1Re1Xrepresenta
32、tion and so the circuit reduces to a generator producing a constant voltage (actually equal to ) and having an internal impedance (actually equal to 1E1VjXR).1RejXIn either case, the parameters could be referred to the secondary winding and this may save calculation time.The resistances and reactanc
33、es can be obtained from two simple light load tests.7负载运行的变压器通过选择合适的匝数比，一次侧输入电压 可任意转换成所希望的二次侧1V开路电压 。 可用于产生负载电流，该电流的幅值和功率因数将由而次侧2E电路的阻抗决定。现在，我们要讨论一种滞后功率因数。二次侧电流及其总安匝 将影响磁通，有一种对铁芯产生去磁、减小 和 的趋向。因为一次2NI m1E侧漏阻抗压降如此之小，所以 的微小变化都将导致一次侧电流增加很大，从1E增大至一个新值 。增加的一次侧电流和磁势近似平衡0IijXRVI1/了全部二次侧磁势。这样的话，互感磁通只经历很小的变化，
34、并且实际上只需要与空载时相同的净磁势 。一次侧总磁势增加了 ，它是平衡同量的10NI 2NI二次侧磁势所必需的。在向量方程中， ，上式也可变换成1021I。满载时，电流 只约占满载电流的 5%，因而 近似等于2101INI0 1I。记住 ，近似等于 的输入容量也就近似等于输出容2/ 211/NE1IE量 。IE一次侧电流已增大，随之与之成正比的一次侧漏磁通也增大。交链一次绕组的总磁通 没有变化，这是因为总反电动势1mp仍然与 相等且反向。然而此时却存在磁通的重新分配，由于dtNE/11V随 的增加而增加，互感磁通分量已经减小。尽管变化很小，但是如果没I有互感磁通和电动势的变化来允许一次侧电流变
35、化，那么二次侧的需求就无法满足。交链二次绕组的净磁通 由于 产生的二次侧漏磁通（其与 反相）s2I m的建立而被进一步削弱。尽管图中 和 是分开表示的，但它们在铁芯中是m一个合成量，该合成量在图示中的瞬时是向下的。这样，二次侧端电压降至8，它可被看成两个分量，即 ，或dtNVS/2 dtNdtVm/22者向量形式 。与一次侧漏磁通一样， 的作用也用一个大体为22IjXE常数的漏电感 来表征。要注意的是，由于它对互感磁通的作/Ni用，一次侧漏磁通对于二次侧端电压的变化产生部分影响。这两种漏磁通，紧密相关；例如， 对 的去磁作用引起了一次侧的变化，从而导致了一次侧2m漏磁通的产生。如果我们讨论一个
36、足够低的超前功率因数，二次侧总磁通和互感磁通都会增加，从而使得二次侧端电压随负载增加而升高。在空载情形下，如果忽略电阻， 幅值大小不变，因为它仍提供一个等于 的反总电动势。尽管现在p1V是一次侧和二次侧磁势的共同作用产生的，但它实际上与 相同。互感磁1通必须仍随负载变化而变化以改变 ，从而产生更大的一次侧电流。此时1E的幅值已经增大，但由于 与 是向量合成，因此一次侧电流仍然是增大1EV的。从上述图中，还应得出两点：首先，为方便起见已假设匝数比为 1，这样可使 。其次，如果横轴像通常取的话，那么向量图是以 为零21E 0m时间参数的，图中各物理量时间方向并不是该瞬时的。在周期性交变中，有一次侧
37、漏磁通为零的瞬时，也有二次侧漏磁通为零的瞬时，还有它们处于同一方向的瞬时。已经推出的变压器二次侧绕组端开路的等效电路，通过加上二次侧电阻和漏抗便可很容易扩展成二次侧负载时的等效电路。实际中所有的变压器的匝数比都不等于 1，尽管有时使其为 1 也是为了使一个电路与另一个在相同电压下运行的电路实现电气隔离。为了分析时的情况，二次侧的反应得从一次侧来看，这种反应只有通过由二次21N9侧的磁势产生磁场力来反应。我们从一次侧无法判断是 大， 小，还是2I2N小， 大，正是电流和匝数的乘积在产生作用。因此，二次侧绕组可用任2I2N意个在一次侧产生相同匝数 的等效绕组是方便的。1N当 变换成 ，由于电动势与
38、匝数成正比，所以 ，与21 212)/(EN相等。1E对于电流，由于对一次侧作用的安匝数必须保持不变，因此，即 。2122NII21)/(INI对于阻抗，由于二次侧电压 变成 ，电流 变为 ，因VV)/(IIN)/(12此阻抗值，包括负载阻抗必然变为 。因此，I/21， 。212)/(RN21)/(XNX如果将一次侧匝数作为参考匝数，那么这种过程称为往一次侧的折算。我们可以用一些方法来验证上述折算过程是否正确。例如，折算后的二次绕组的铜耗必须与原二次绕组铜耗相等，否则一次侧提供给其损耗的功率就变了。 必须等于 ，而2RI2I事实上确实简化成了 。）21212/()/(NRNI R类似地，与 成
39、比例的漏磁场的磁场储能 ，求出后验证与XI )/1(2LI成正比。折算后的二次侧 。2XI 212222 )/( IENNEIkVA尽管看起来似乎不可理解，事实上这种论点是可靠的。实际上，如果我们将实际的二次绕组当真从铁芯上移开，并用一个参数设计成 ， ， ，12RX的等效绕组和负载电路替换，在正常电网频率运行时，从一次侧两端无法2I10判断二次侧的磁势、所需容量及铜耗与前有何差别。在选择折算基准时，无非是将一次侧与折算后的二次侧匝数设为相等，除此之外再没有什么更要紧的了。但有时将一次侧折算到二次侧倒是方便的，在这种情况下，如果所有下标“1”的量都变换成了下标“2”的量，那么很容易得到必需的折
40、算系数，例如。值得注意的是，对于一台实际的变压器， ；同样地 ， 。21R21X12R12X的通常情形时的等效电路，它除了为了考虑铁耗而引入了 ，且N mr为了将 折算回 而在二次侧两端引入了一理想的无损耗转换外，其他方面2V2是一样的。在运用这种理想转换之前，内部电压和功率损耗已进行了计算。当在电路中选择了适当的参数时，在一、二次侧两端测得的变压器运行情况与在该电路相应端所测得的请况是完全一致的。将 线圈和 线圈并排放置在一1N2个铁芯的两边，这一点与实际情况之间的差别仅仅是为了方便。当然，就变压器本身来说，两线圈是绕在同一铁芯柱上的。如果将激磁支路移至一次绕组端口，引起的误差很小，但一些不
41、合理的现象又会发生。例如，流过一次侧阻抗的电流不再是整个一次侧电流。由于 通0I常只是 的很小一部分，所有误差相当小。对一个具体问题可否允许有细微差1I别的回答取决于是否允许这种误差的存在。对于这种简化电路，一次侧和折算后二次侧阻抗可相加，得 和2121)/(ReN 2121)/(NXe需要指出的是，在此得到的等效电路仅仅适用于电网频率下的正常运行；一旦电压变化率产生相当大的电容电流 时必须考虑电容效应。这dtCVIc/对于高电压和频率超过 100Hz 的情形是很重要的。其次，即使是对于电网频率也并非唯一可行的等效电路。另一种形式是将变压器看成一个三端或四端网络，这样便产生一个准确的表达，它对
42、于那些把所有装置看成是具有某种传递性能的电路元件的工程师来说是方便的。以此为分析基础的电路会拥有一个既产生11电压大小的变化，也产生相位移的匝比，其阻抗也会与绕组的阻抗不同。这种电路无法解释变压器内类似饱和效应等现象。等效电路有两个入端口形式：（a） 从一次侧看为一个 U 形电路，其折合后的负载阻抗的端电压为 ；2V（b） 从二次侧看为一其值为 ，且伴有由 和 引起内压降的恒压1V1ReX源。在这种电路中有时可省略激磁支路，这样电路简化为一台产生恒值电压（实际上等于 ）并带有阻抗 （实际上等于 ）的发电机。1E1VjXR1je在上述两种情况下，参数都可折算到二次绕组，这样可减小计算时间。其电阻和电抗值可通过两种简单的轻载试验获得。
