2 TRANSFORMERS 2.1. Introduction Basic transformer principles were covered in Electromechanics [1], and the main results are given below. Figure 2-1 is a schematic representation of a single-phase transformer with two coils on a magnetic core, where the magnetic coupling is assumed to be perfect: the same flux ¢ passes through each turn of each coil, @ ¢ ‘N, tums: NN, turns Figure 2-1: Transformer with source and load Voltage relationships Kirchhoff’s voltage law applied to the two windings gives: a9 yet Rian, 2eR OL tf 21) 1p; yy) =@-Ri=N,% Ri, 22) =~ Ron = Na Roi (2-2) If the resistances Ry and R; are negligible, then equations 2-1 and 2-2 become: ds nent on do nent a dt en Dividing these equations gives the important result eM v2 Ne 5) Sinusoldal operation If the voltage source is sinusoidal, then the core flux will also be sinusoidal, so we may put: = Substituting this expression in equation 2-3 gives: SINE (2-6) ¥=N, B= No, coser=Vigcosor (2-7) Thus the maximum primary voltage is: Vig = NO, = 2 FN YDy, = 2 FNAB, (2-8) where A is the cross-sectional area of the core and Bw is the maximum flux density in the core. A typical value for By is 1.4 T for the silicon ste! characteristic in figure 2-2. 2 18 16 14 12 1 os 06 o4 oz 0 Flux density 8, T 0 2 4 6 8 10 12 14 16 18 20 Magnetic intensity H, kA‘ Figure 2-2: Silicon transformer steel. Current relationships ‘The relationship between the primary and secondary currents can be found by considering the ‘magnetic circuit of the transformer. From the basic ‘magnetic circuit equation, we have: F = Ni Noy = RO 29) In a well-designed transformer, the reluctance Nis small, so equation 2-9 becomes: yi Nai =0 (2-10) This gives the counterpart of equation 2-5 for voltage: = (11) 2 Electrical Machines and Systems Course NotesIf sinusoidal voltages and currents are represented by phasors, the corresponding forms for the basic voltage and current equations are: N, Me 212) V."N2 Lh (2-13) LN Transformer rating ‘The maximum voltage at the primary terminals of a transformer is determined by equation 2-8, and is. independent of the current. The maximum primary current is determined by the FR power loss in the resistance of the transformer windings, which ‘generates heat in the transformer. This power loss is independent of the applied voltage. Consequently, for a given design of transformer, there is a maximum value for the product Vil: at the primary terminals. To a first approximation, this is also equal to the product Vahs at the secondary terminals. This maximum value does not depend on the phase angle between the voltage and the current. Transformer ratings therefore specify the apparent power VI (volt-amperes, VA) rather than the real power VI cos ¢ (watts, W). 2.2 Types of power transformer In addition to the ordinary single-phase power transformer, two other types are in common use: auto-wound transformers, and 3-phase transformers, Auto-wound transformers ‘A transformer can have a single coil with an output taken from a portion of the coil, as shown in figure 2.3. This is known as an auto-wound transformer ‘or auto-transformer. Vs Figure 2-3: Auto-wound transformer. Unlike the normal transformer with two windings, known as a double-wound transformer, the auto- wound transformer does not provide electrical isolation between the primary and the secondary, However, an auto-wound transformer can have a much larger apparent power rating than a double- wound transformer of the same physical size. Let N be the number of turns on the upper part of the winding in figure 2-3, and NV, the number of turns on the lower part. The conventional transformer equations 2-12 and 2-13 apply to these parts of the winding, since they are equivalent to two separate windings with a common connection. Applying Kirchhoff’ law to this circuit gives: Vs =Vy4Vy (2414) 1-14, (2-15) As an example, suppose that Nj = No. Ifthe transformer is regarded as ideal (see section 2.3), then I = Ip and V; = V2, Equations 2-14 and 2-15, give: Vs-Vi+ y= 2V)=2V, (2-16) = 2, = 2s an ‘where Vz is the voltage across the load and Is is the ccurrent supplied by the source. This auto-wound ‘transformer behaves as a step-down transformer with a ratio of 2:1, and the current in each winding is equal to half of the load current. ‘An elegant application of the auto-wound transformer principle is the variable transformer, which has a single-layer coil wound on a toroidal core, The output is taken from a carbon brush that makes contact with the surface of the coil; the brush can be moved smoothly from one end of the coil to the other, thus varying the output voltage. Examples of variable transformers are shown in figure 2-4 Transformers 3Figure 2-4: Variable transformers. (RS Components Ltd) ‘phase transformers In 3-phase systems (see section 4.1), itis common practice to use sets of three single-phase transformers. It is also possible, however, to make 3.phase transformers with three sets of windings on three limbs of a core, as shown in figure 2-5 Figure 2-5: 3-phase transformer model. ‘The corresponding fluxes are shown in figure 2-6. Ga de Ge Figure 2-6: 3-phase transformer flux. In a balanced system with sinusoidal phase voltages, the fluxes will be given by: 6, =0,,cosot $5 = @y,cos(or—120") (218) 6, =, cos(@t 240°) = ©, cost +120°) Figure 2-7 shows flux plots for the transformer at the instants when cx = 0, 120° and 240°. {b) Figure 2-7: 3-phase transformer flux plots: (a) 0°, (b) 120°, (c) 240", There is no requirement for another limb to form a flux return path, because the fluxes fy, gy and J. sum to zero in a balanced 3-phase system. The proof is as follows. From equation 2-18, the sum is given by: ba* +e Om -0s01 + cos(@ot —120°) +cos(@t +120°) (2-19) = coset + 2cosertcos120° 0st coset = 0 Because the fluxes in the three limbs sum to zero at all instants of time, there is no leakage of flux from, the core, as the flux plots in figure 2-7 demonstrate 4 Electrical Machines and Systems Course Notes2,3 Ideal transformer properties If the primary and secondary windings have zero resistance, and the magnetic core has zero reluctance, then the approximate equalities in ‘equations 2-12 and 2-13 become exact equalities. This leads to the concept of an ideal transformer ‘element, to accompany the other ideal elements of circuit theory. Figure 2-8 shows a circuit symbol for the ideal transformer element. Figure 2-8: Ideal transformer element The voltage and current relationships in the time and frequency domains are given in table 2-1, Table 2-1: Ideal transformer relationships. ye domain Frequency domain V2 No 2-20) YOM 4M, bi, ew i, My LM ‘The following properties of the ideal transformer may be deduced from equations 2-16 and 2-17: * The voltage transformation is independent of the current, and vice versa. + Ifthe secondary is short-circuited, so that v2 = 0, the primary terminals appear to be short-circuited since v; = 0 + Ifthe secondary is open-circuited, so that in= 0, the primary terminals appear to be open- circuited since i; = 0. + The output power is equal to the input power, so there is no power loss in the element. Impedance transformation ‘The ideal transformer has the important property of transforming impedance values in a circuit. Consider an ideal transformer with an impedance Z, connected to its secondary terminals, as shown in figure 2-9, Nu Figure 2-9: Ideal transformer with a load, ‘The secondary impedance is given by: (2-22) At the primary terminals, the circuit presents an impedance given by: (2.23) ( Thus the combination of an ideal transformer of ration and an impedance Z;, can be replaced by an ‘equivalent impedance Z,,/ Referred impedances Figure 2-10(a) shows an ideal transformer with a load impedance Z, connected to the secondary. Another impedance Z is in series with Z,. The input impedance of this circuit is: ~hath, Es Z, (224) ‘The input impedance of the circuit in figure 2- 10(b) is: Z, ZL, - 0,4 in Lt (2.25) ‘The two expressions for Z,, will be identical if: 2 n (2-26) Transformers 5° Zin g Lh ° NNa @) Za Zin ; é | Z NaN (b) Figure 2-10: Referred impedance ~ 1. The impedance Z's is termed the secondary impedance Zp referred to the primary. Ina similar way, a primary impedance Z) can be referred to the secondary, as shown in figure 2-11. In this case, the referred impedance is given by: ZLi=wZ, (2-27) Zs ° Bn 3 hi ° NiNy @ ° NiNy ) Figure 2-11: Referred impedance ~ 2. ‘The concept of referred impedance is often a useful device for simplifying circuits containing transformers, as will be shown in the next section, It is conventional to use a single prime (’) to denote quantities referred to the primary side, and a double prime (") to denote quantities referred to the secondary side. 2.4 Circuit model of a transformer Ina practical transformer, the winding resistances and the core reluctance are not zero. In addition, there will be some power loss in the core because of eddy currents and hysteresis in the magnetic ‘material, All of these effects can be represented by the equivalent circuit (3, 4] shown in figure 2-12. MN Figure 2-12: Transformer equivalent circuit This circuit is based on the ideal transformer clement, with additional circuit elements to represent the imperfections. The resistances Ry and Ry represent the physical resistances of the windings, and R. represents the power lost in the core. The reactance X,, known as the magnetising reactance, allows for the current required to ‘magnetise the core when the reluctance is not zero. Reactances x, and x, known as leakage reactances, represent the leakage flux that exists ‘when the magnetic coupling between the primary and the secondary is not perfect. Figure 2-13 shows, the leakage flux when the core has an artificially ow relative permeability of 10, and one winding at atime is energised. In practice, the leakage is much less than this, so the leakage reactances are normally very much smaller than the magnetising, reactance Xr (a) 6 Electrical Machines and Systems Course NotesRit dR nn sen. | 4) x 3 b) Nas Figure 2-13: Transformer leakage flux: (a) eft coil energised, (b) right coll energised. In the circuit of figure 2-12, the current I's is ® the effective value of the secondary current as seen no om ies from the primary side of the transformer. Itis also nen. fren. known as the secondary current referred to the primary. The current Iy is the no-load current,whichis jx, [Ip 3 g the current taken by the primary when there is no load connected to the secondary. It has a ‘component Ion, known as the magnetising current, which represents the current required to set up the WisNy ‘magnetic flux in the core. The current Io is the ccore loss component of the no-load current. ) ite Bin Approximate equivalent circuit For power transformers with ratings above 100 VA, the values of the series elements Ry and x Rnd [IR are gencrally much smaller than the shunt elements R_ and X». Under normal working conditions, the voltage drop in Ry + x; is much smaller than the ° applied voltage Vj. Similarly, the no-load current /y NisNa is much smaller than the load current f. It follows that the shunt elements can be moved to the input terminals, as shown in figure 2-14(a), with very Figure 2-14: Approximate equivalent circuit litte loss of accuracy. The secondary elements Ry and jr; can be replaced by equivalent elements = R:/rand 2.5 Parameter determination = ™m/ 7 on the primary side (see section 2.3), giving the circuit shown in figure 2-14(b), Finally, The parameters of the approximate equivalent the series elements can be combined to give an circuit (igure 2-14) can be determined cfective esistance R, and leakage reactance x, as experimentally from two tests shown in figure 2-14(c), where the values are + An open-cireuit test, where the secondary is left unconnected and the normal rated voltage R=R, +h, x= +22 228) is applied to the primary. ” " + A short-circuit test, where the secondary terminals are short-circuited and a low voltage is applied to the primary, sufficient to circulate the normal full-load current, Transformers 7Open-circuit test With the secondary unconnected, I; = 0, so the equivalent circuit reduces to the form shown in figure 2-15. Troe oS + Vow Xe 3 []R o—_ Figure 2-15: Open-circult test, ‘Toa very close approximation, the current Ig. supplied to the primary is equal to the no-load. current Ip in figure 2-12. The values of the elements R, and X,,can be determined from ‘measurements of the input voltage Vj, current Ii and power Pcs follows. The input power is entirely dissipated in the resistance R,, giving: Va R (2:29) v, L Z, ae 2.30) a (230) Xm In terms of magnitudes, equation 2-30 becomes: ¥, 1 toe = ——= (231) Toe | 1 ’ 1 VRE Xn Re-arranging equation 2-31 gives the value of Xx (2:32) From figure 2-14(¢), it follows that the turns ratio is given by: (2:33) Short-circuit test If the secondary terminals are short-circuited, the ideal transformer in figure 2-14 can be replaced by a short circuit, so the equivalent circuit takes the form shown in figure 2-16(a). In a typical power transformer, the shunt elements R. and Xj. are at Teast 100 times larger than the series elements Re and x,. Consequently, the shunt elements can be neglected, and the circuit reduces to the form shown in figure 2-16(b). R. Ske. Xm (a) Tne RP + Vive (b) : Short-circuit test. The values of the elements R, and x, can be determined from measurements of the input voltage Vigq Current /\,- and power P,.as follows. The input power is entirely dissipated in the resistance R,, giving: (2:34) 8 Electrical Machines and Systems Course NotesRe-arranging equation 2-36 gives the value of x: 237) In practice, the open-circuit testis usually made on the low-voltage side of the transformer to minimise the value of V,,, and the short-circuit test is made on the high-voltage side to minimise the value of [,.. The resulting parameter values are then referred to the primary side of the transformer. 2.6 Transformer performance Consider a transformer with an impedance Z connected to the secondary. With the approximate equivalent circuit, this may be represented by the circuit diagram of figure 2-17. MN Figure 2-17: Transformer with a load. ‘The load impedance can be referred to the primary side of the ideal transformer element, giving the circuit shown in figure 2-18, Jeep, nk: 1 R Figure 2-18: Circuit with referred impedance. solved for the currents Ip and 1’, The referred secondary voltage is given by V2= Zil's, and the secondary terminal quantities are given by V2= nV'2, = V2 n. Voltage regulation and efficiency When a load is connected to the secondary of a transformer, there will be a voltage drop in the series elements R, and x,, so the secondary terminal voltage will change. The voltage regulation is defined as: Vani = (2-38) V; where Vaqris the magnitude of the no-load secondary terminal voltage, and Vaq is the corresponding full-load voltage. Power is lost as heat in the windings and core of the transformer, represented by the resistance elements R, and R-in the equivalent circuit. The efficiency is defined in the usual way as: Pp. oe (239) P. ) where P, is the power input to the primary and Phar is the power output from the secondary. The power lost as heat in the transformer is: Pog = Ping — Poe (2-40) so we have the following alternative forms of equation 2-39: P, ut + Poss Large transformers are very efficient. Even a 2 KVA transformer can have an efficiency of about 95%, Above 25 kVA, the efficiency usually exceeds 99%, It is very difficult to make an accurate measurement of efficiency by direct measurement of Pay and Py, since this would require an accuracy of measurement of the order of 0.01%, Instead, the normal practice is to determine the losses from measurements, and calculate the efficiency from one of the alternative expressions in equation 2-41. The losses can be calculated with high accuracy from the equivalent-circuit parameters determined from tests on the transformer. Maximum efficiency ‘The power loss in a transformer has two components: the core loss, given by Vi / Rand Transformers 9the PR loss, given by /°R,. The core loss will be constant if the primary voltage V; is constant, but the FR loss will vary with the secondary current. ‘When the current is low, the output power will be ow, but the core loss remains at the normal value, Consequently, the efficiency of the transformer will be low under these conditions. It may be shown that the efficiency is a maximum when the secondary current is such that the variable FR loss is equal to the fixed core loss. This result also applies to other devices where the losses have fixed and variable components. Transformers are usually designed to have maximum efficiency at the normal operating value of secondary current, which may be less than the maximum rated current Power relationships When calculating the transformer performance, the following power relationships can be useful. The complex power S is given by [2] S=P+jQ=vIr (2-42) where V is the voltage phasor, I* is the complex conjugate of the current, P is the real power, and Q is the reactive power. If gis the phase angle, then: P=Scos =Vicos$=Re(VI*) (2-43) Q=Ssing =Vising =Im(VI*) (2-44) |e? <0! If the voltage phasor V is chosen as the reference quantity, and defined to be purely real (V = V+j0), then the power relationships take a simple form: P=Vicos$=VRe(I) (2-46) S= (2-45) Q=Vising=-VIm) 2-47) Worked example 2-1 A2KVA, 50 Hz, power transformer has the following equivalent-circuit parameter values referred to the primary: R.= 0.682 Q, x,= 0.173 2, R, X= 657 2, Nz / Ny = 0.472, If the primary is connected to 230 V 50 Hz supply, and a load impedance (6.0 + j2.5) is connected to the secondary, determine: (a) the secondary current magnitude, 1080.9, (b) the secondary terminal voltage magnitude, (©) the primary current magnitude, (@ the voltage regulation, (©) the efficiency of the transformer. (a) Secondary current The load impedance referred to the primary is: Z, _6.0+ j25 nn (0.472)? 26.9 + j.22 The secondary current referred to the primary is: ye Z.+2, 230+ j0 ~ (0.682 + j0.173) + (26.9 + j11.2) 230. jo 27.64 jll4 = [7.12 - j2.94]=7.70 A =7.12- j294A The secondary current magnitude is: I, 7.70 _ n 0472 6.34 (b) Secondary voltage The secondary voltage magnitude is: Vp =fyZz =16.3|6.0+ 2.5] =16.3x6.5=106.0V (c) Primary current The no-load current is: Vi, Vi _ 230+ j0 | 230+ j0 RX» 1080” j657 =0.213— j0.350A The primary current is: =h+h = (0.213 = 0.350) + (7.12 = j2.94) 33 j3.29A 33— j3.29|=8.03 A h (d) Voltage regulation On no load, the secondary voltage is: V, = nV, = 0.472 x 230 = 108.6 V 10 Electrical Machines and Systems Course Notes‘The voltage regulation is therefore: ¢ Vani ~ Vo _ 108.6-106.0 1 108.6 2.33% (c) Efficiency ‘The output power is: Pay = BR, = (16:31)? 6.0 =1597 W ‘The power loss is: We Pay = + IPR, oss _ 230.0)? 1080 =894W (7.699)? «0.682 ‘The input power is therefore: P, + Py, = 1597 + 89.4 = 1686 W Alternatively, from equation 2-46 the input power B,, = V, Re(I,) = 230.0 x7.330 = 1686 W The efficiency is therefore: Pow P. 2.7 Current transformers ‘The use of transformers for measuring current has. been introduced in Electromechanics (1], where the danger of open-circuiting the secondary has been explained, This section introduces the important topic of measurement errors. Current transformer errors Ina well-designed current transformer, the core ‘lux density is low and the core is made from a high-quality magnetic material. Under normal ‘operating conditions, the core loss will be negligibly small, so the core loss resistance R-can bbe omitted from the equivalent circuit. A circuit ‘model for a current transformer connected to a load therefore takes the form shown in figure 2-19. Lor, NuNy Figure 2-19: Current transformer circuit model. Only the relationship between currents is of interest, so the primary impedance (R; + jx:) can be disregarded. It is now convenient to refer the primary quantities to the secondary side, giving the circuit model shown in figure 2-20. iy Figure 2-20: Modified circuit model. ‘The circuit acts as a current divider, where the current in the secondary branch is given by: IX nM L at Dt Xm (2-48) where I”; is the primary current referred to the secondary, and Z2 = R: + jr. Ina well-designed transformer, the secondary impedance Zais very small in comparison with the referred magnetising reactance X"q, $0 this term introduces very little error. Equation 2-48 shows that itis desirable to keep the load impedance Z, as small as possible if the error is to be minimised. In practice, current transformers are designed for a specified maximum secondary voltage at the rated secondary current. This defines a maximum, apparent power for the secondary load, or burden. ‘Typically, a small current transformer will have a rated secondary burden of 5 VA. With the usual secondary current rating of 5 A, this implies that the maximum secondary voltage is | V, and the maximum impedance magnitude is 0.2.9. Transformers 11