When a transformer is disconnected from its power source, the current is interrupted and the magnetic field or mmf driving flux through the core is reduced to zero. As we have seen in Section 2.4, the core retains a residual induction, called the remanence, when the hysteresis path is on the positive descending (negative ascending) branch of a normal loop. A residual induc- tion also can occur in other cases, but it will not have as high a magnitude. To drive the core to the zero-magnetization state, the peak induction should be lowered gradually while cycling the field. Because the intrinsic normal hysteresis loops for oriented Si-Fe have fairly flat tops or bottoms, the rema- nence is close to the peak induction; however, the presence of gaps in the core reduces this somewhat. When the unit is reenergized by a voltage source, the flux change must match the voltage change according to Faraday’s law:
V= −Nd dtΦ (2.22) Heff H1 Heff H1
Intrinsic loop Jointed loop
B
H
Figure 2.8
For a sinusoidal voltage source, the flux is also sinusoidal:
Φ Φ= psin(ω ϕt+ ) (2.23)
where Φp is the peak flux. Substituting this into Equation 2.22, we get
V= −NωΦpcos(ω ϕt+ ) (2.24)
Hence, the peak voltage, using ω = 2πf, where f is the frequency in hertz, is
Vp= 2π Φ fN p (2.25)
Thus to follow the voltage change, the flux must change by ±Φp over a cycle. If, in a worst case scenario, the voltage source is turned on when the voltage is at a value that requires a negative Φp value and the remanent induction has a positive value of Br so that the associated core flux is BrAc, where Ac is the core area, then the flux will increase to 2Φp + BrAc when the voltage reaches a value corresponding to +Φp. If we assume the flux change occurs in the core, as it normally does, so that Φp = BpAc, then the core induction will increase to 2Bp + Br. Since Bp is usually ∼10–20% below saturation in typical power trans- formers, the core will be driven strongly into saturation, which requires a very high exciting current. This exciting current is called the inrush current and can be many times the normal load current in a transformer.
As saturation is approached, the flux will no longer remain confined to the core, but will spill into the air or oil space inside the coil that supplies the excit- ing current. Thus, beyond saturation, the entire area inside the exciting coil, including the core, must be considered the flux-carrying area with an incre- mental relative permeability of 1. Let Br be the residual induction in the core, which, without loss of generality, we can assume to be positive. We will assume that this is the remanence. Thus, the residual flux is Φr = BrAc. Let ΔΦ be the flux change required to bring the voltage from its turn-on point up to its maximum value, in the same sense as the residual flux. ΔΦ could be positive, negative, or zero, depending on the turn-on point. We assume it is positive here.
Part of the increase in ΔΦ will simply bring the induction up to the satura- tion level, expending little exciting power or current. This part is approxi- mately given by (Bs − Br)Ac, where Bs is the saturation induction. Beyond this point, further increases in ΔΦ occur with the expenditure of high exciting current since the incremental relative permeability is 1. Beyond saturation, the core and air or oil have the same permeability, so the incremental flux density will be the same throughout the interior of the coil, ignoring end effects. Letting the interior coil area up to the mean radius Rm be A(A = πRm2), the incremental flux density is
B B B A
A
The incremental magnetic field inside the coil, ignoring end effects, is Hinc =
NI/h, where NI are the exciting amp-turns and h is the coil height. We ignore the exciting amp-turns required to reach saturation, as these are compara- tively small. Because the permeability for this flux change is μ0, we have Binc = μ0Hinc = μ0NI/h, which implies, using Equation 2.26:
NI h
A B B A
= −
(
−)
µ0
∆Φ s r c (2.27)
Assuming, in the worst case, that the voltage is turned on at a correspond- ing flux density that is at the most negative point in the cycle, we have, using Equation 2.25: ∆Φmax=2∆Φ = = 2 2 2 p p p c V fN B A π (2.28)
In the third equality in Equation 2.28, we compute this flux change math- ematically as if it all occurred in the core (as it would under normal conditions) even though we know this is not physically true. Thus, substi- tuting the maximum flux change ΔΦmax for ΔΦ for a worst case situation, Equation 2.27 becomes NI hA A B B B max=µ c
(
p+ r− s)
0 2 (2.29)As a numerical example, consider transformers with stacked cores and step- lapped joints where Br ≈ 0.9 Bp. Taking h = 2 m, Ac/A = 0.5, Bp = 1.7 T, Br = 0.9,
Bp = 1.53 T, and Bs = 2 T, we obtain NImax = 2.33 × 106 A • T. For N = 500, Imax = 4660 A. This is very high for an exciting current.
Although the voltage is constrained to be sinusoidal, the exciting current will be distorted mainly due to saturation effects. Even below saturation, there is some distortion due to nonlinearity in the B-H curve. Figure 2.9a illustrates the situation on inrush. The sinusoidal voltage is proportional to the incremental induction, which is displaced by the remanent induction. It requires high H values near its peak and comparatively small to zero H values near its trough. This is reflected in the exciting current, which is pro- portional to H. This current appears as a series of positive pulses separated by broad regions of near zero value as shown in Figure 2.9a. The positive current pulses appear undistorted because they are associated with the satu- rated portion of the B-H curve, where the relative permeability is close to 1. The high exciting inrush current will decrease with time due to resistive effects, as shown in Figure 2.9b.