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Home / Technical Articles / Exciting and inrush currents in transformers that often make protection relays go crazy

Transformer exciting and inrush currents

If normal voltage is impressed across the primary terminals of a transformer with its secondary open-circuited, a small exciting current flows. This exciting current consists of two components, the loss component and the magnetizing component. The loss component is in phase with the impressed voltage, and its magnitude depends upon the no-load losses of the transformer.

Exciting and inrush currents in transformers that often make protection relays go crazy
Exciting and inrush currents in transformers that often make protection relays go crazy (photo credit: Andrew George Cowan via Flickr)

The magnetizing component lags the impressed voltage by 90 electrical degrees, and its magnitude depends upon the number of turns in the primary winding, the shape of the transformer saturation curve and the maximum flux density for which the transformer was designed.

A brief discussion of each of these components follows:

  1. Magnetizing Component of Exciting Current
  2. Loss Component of Exciting Current
  3. Total Exciting Current
  4. Typical Magnitudes of Exciting Current
  5. Inrush Current
  6. Determination of Current Inrush

1. Magnetizing Component of Exciting Current

If the secondary of the transformer is open, the transformer can be treated as an iron-core reactor. The differential equation for the circuit consisting of the supply and the transformer can be written as follows:

e = R×i + n1×dΦ/dt

where,

  • e = instantaneous value of supply voltage
  • i = instantaneous value of current
  • R = effective resistance of the winding
  • Φ = instantaneous flux threading primary winding
  • n1 = primary turns

Normally the resistance, R, and the exciting current, i are small. Consequently the R×i term in the above equation has little effect on the flux in the transformer and can, for the purpose of discussion, be neglected. Under these conditions Equation (1) can be rewritten:

e = n1× dΦ/dt (equation 1)

If the supply voltage is a sine wave voltage,
e = √2 E sin (ωt+λ) (equation 2)

where, E is a rms value of supply voltage, ω = 2πf

Substituting in Equation (1):
√2 E sin (ωt+λ) = n1× dΦ/dt

Solving the above differential equation,
Φ = (− √2 E cos (ωt+λ)) / ωn1 + Φ(equation 3)

In this solution, (− √2 E cos (ωt+λ)) / ωn1 is the normal steady-state flux in the transformer core. The second term, Φt represents a transient component of flux the magnitude of which depends upon the instant at which the transformer is energized, the normal maximum flux and the residual flux in the core at the time the transformer is energized.

Under steady-state conditions this component is equal to zero; the magnitude of Φt is discussed in Section 5.

From Equation (3) it can be seen that the normal steady-state flux is a sine wave and lags the sine wave supply voltage by 90 degrees. The supply voltage and the normal flux are plotted in Figure 1 as a function of time.

Impressed voltage and steady-state flux
Figure 1 – Impressed voltage and steady-state flux

If there were no appreciable saturation in the magnetic circuit in a transformer, the magnetizing current and the flux would vary in direct proportion, resulting in a sinusoidal magnetizing current wave in phase with the flux. However, the economic design of a power transformer requires that the transformer iron he worked at the curved part of the saturation curve, resulting in appreciable saturation.

Under this condition the magnetizing current is not a sine wave, and its shape depends upon the saturation characteristics (the B-H curve) of the transformer magnetic circuit. The shape of the current wave can be determined graphically as shown in Figure 2 below.

In Figure 2(b) are shown the impressed voltage and the flux wave lagging the voltage by 90 degrees. For any flux the corresponding value of current can be found from the B-H curve. Following this procedure the entire current wave can be plotted. The current found in this manner does not consist of magnetizing current alone but includes a loss component required to furnish the hysteresis loss of the core.

However, this component is quite small in comparison to the magnetizing component and has little effect on the maximum value of the total current.

Graphical method of determining magnetizing current
Figure 2 – Graphical method of determining magnetizing current

A study of Figure 2 shows that although the flux is a sine wave the current is a distorted wave. An analysis of this current wave shows that it contains odd-harmonic components of appreciable magnitude; the third harmonic component is included in Figure 2.

In a typical case the harmonics may be as follows: 45 percent third, 15 percent fifth, three percent seventh, and smaller percentages of higher frequency. The above components are expressed in percent of the equivalent sine wave value of the total exciting current. These percentages of harmonic currents will not change much with changes in transformer terminal voltage over the usual ranges in terminal voltage.

In Figure 3 are shown the variations in the harmonic content of the exciting current for a particular grade of silicon steel.

Harmonic content of exciting current for a particular grade of silicon steel
Figure 3 – Harmonic content of exciting current for a particular grade of silicon steel

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2. Loss Component of Exciting Current

The no-load losses of a transformer are the iron losses, a small dielectric loss, and the copper loss caused by the exciting current. Usually only the iron losses, i.e., hysteresis and eddy current losses, are important. These losses depend upon frequency, maximum flux density, and the characteristics of the magnetic circuit. In practice the iron losses are determined from laboratory tests on samples of transformer steel.

However, the formulas given below are useful in showing the qualitative effect of the various factors on loss.

  • Iron loss= Wh + We
  • Wh = Kh f Bxmax watts per lib
  • We = Ke f2t2 B2max watts per lib

Where,

  • Wh = hysteresis loss
  • We = eddy current loss
  • f = frequency
  • f = thickness of laminations
  • Bmax = maximum flux density

Kh, Ke and x  are factors that depend upon the quality of the steel used in the core. In the original derivation of the hysteresis loss formula by Dr. Steinmetz, x was 1.6. For modern steels x may have a value as high as 3.0. The iron loss in a 60-cycle power transformer of modern design is approximately one watt per pound.

Tile ratio of hysteresis loss to eddy current loss will be on the order of 3.0 with silicon steel and 3,6 with oriented steel. These figures should be used as a rough guide only, as they vary considerably with transformer design.

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3. Total Exciting Current

As discussed above, the total exciting current of a transformer includes a magnetizing and a loss component. The economic design of a transformer dictates working the iron at the curved part of the saturation curve at normal voltage; hence any increase in terminal voltage above normal will greatly increase the exciting current.

In Figure 4 the exciting current of a typical transformer is given as a function of the voltage applied to its terminals. The exciting current increases far more rapidly than the terminal voltage.

For example. 108-percent terminal voltage results in 200-percent exciting current.

Exciting current vs. terminal voltage.
Figure 4 – Exciting current vs. terminal voltage. The above curve applies for one particular design of transformer; the shape of the curve may vary considerably depending upon the grade of steel and the transformer design.

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4. Typical Magnitudes of Exciting Current

The actual magnitudes of exciting currents vary over fairly wide ranges depending upon transformer size, voltage class, etc. In Table 1 are given typical exciting currents for power transformers. The exciting currents vary directly with the voltage rating and inversely with the kVA rating.

Table 1 – Typical exciting current values for single-phase power transformers (in percent of full load current)

3-phase kVAVoltage Class (Full Insulation)
2.5 kV15 kV25 kV69 kV138 kV161 kV230 kV
5003.7%3.7%3.8%4.9%
1 0003.3%3.3%3.6%4.3%
2 5003.1%3.2%3.8%
5 0002.8%3.1%2.5%4.1%
10 0003.0%3.1%2.4%3.6%4.0% *
25 0002.2%2.4%3.1%3.9%3.5% *
50 0003.1%3.9%2.8% *

* Reduced insulation


The following values should be considered as very approximate for average standard designs and are predicated on prevailing performance characteristics. Test values will as a rule come below these values but a plus or minus variation must be expected depending upon purchaser’s requirements.

Should closer estimating data be required, the matter should be referred to the proper manufacturer’s design engineers.

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5. Inrush Current

When a transformer is first energized, a transient exciting current flows to bridge the gap between the conditions existing before the transformer is energized and the conditions dictated by steady-state requirements.

For any given transformer this transient current depends upon the magnitude of the supply voltage at the instant the transformer is energized, the residual flux in the core, and the impedance of the supply circuit.

Often the magnitude of this transient current exceeds full-load current and may reach 8 to 10 times full-load current. These high inrush currents are important principally because of their effect on the operation of relays used for differential protection of transformers.

In studying the phenomena that occur when a transformer is energized it is more satisfactory to determine the flux in the magnetic circuit first and then derive the current from the flux. This procedure is preferable because the flux does not depart much from a sine wave even though the current wave is usually distorted.


The total flux in a transformer core is equal to the nor-mal steady-state flux plus a transient component of flux, as shown in Equation (3). This relation can be used to determine the transient flux in the core of a transformer immediately a/2 after the transformer is energized.

As √2E/ωn1 represents the tons crest of the normal steady-state flux, Equation (3) can be rewritten as:

Φ = − Φm cos (ωt+λ) + Φt (Equation 4)

where Φm = √2E/ωn1

at t = 0 and Φ0 = − Φm cos λ + Φt0

where,

  • Φ0 = transformer residual flux
  • − Φm cos λ = steady-state flux at t = 0
  • Φt0 = initial transient flux
In the above equation the angle λ depends upon the instantaneous value of the supply voltage at the instant the transformer is energized. If the transformer is energized at zero voltage, λ is equal to zero, whereas if the trans-former is energized where the supply voltage is at a positive maximum value, λ is equal to 90 degrees.

Assume that a transformer having zero residual flux is energized when the supply voltage is at a positive maximum. For these conditions Φ0 and cos λ are both equal to zero so Φt0 also equal to zero. The transformer flux therefore starts out under normal conditions and there would be no transient.

However, if a transformer having zero residual is energized at zero supply voltage the following conditions exist:

λ = 0
− Φm cos λ = − Φm
Φ0 = 0
Φt0 = Φm

Substituting in Equation (4): Φ = − Φm cos (ωt) + Φm (Equation 5)

The flux wave represented by Equation (5) is plotted in Figure 5(a). The total flux wave consists of a sinusoidal flux wave plus a DC flux wave and reaches a crest equal to twice the normal maximum flux.

Transformer flux during transient condition: Primary Closed at Zero Voltage-Zero Residual Flux
Figure 5(a) – Transformer flux during transient condition: Primary Closed at Zero Voltage-Zero Residual Flux

In this figure the transient flux has been assumed to have no decrement; if loss is considered the transient flux decreases with time and the crest value of the total flux is less than shown.

In Figure 5(b) similar waves have been plotted for a transformer having 60 percent positive residual flux and energized at zero supply voltage. Sixty percent residual flux has been assumed for illustration only. Flux waves for any other initial conditions can be calculated in a similar manner using Equation (4).

Transformer flux during transient condition: Primary Closed at Zero Voltage - 60% Positive Residual Flux
Figure 5(b) – Transformer flux during transient condition: Primary Closed at Zero Voltage – 60% Positive Residual Flux

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6. Determination of Current Inrush

After the flux variation has been determined by the method described, the current wave can be obtained graphically as shown in Figure 6. In the case illustrated it was assumed that a transformer having zero residual flux was energized at zero supply voltage; the flux there-fore is equal to twice normal crest flux. For any flux the corresponding current can be obtained from the transformer B-H curve.

Although the maximum flux is only twice its normal value, the current reaches a value equal to many times the maximum value of the normal trans-former exciting current. This high value of current is reached because of the high degree of saturation of the transformer magnetic circuit.

In the above discussion loss has been neglected in order to simplify the problem. Loss is important in an actual transformer because it decreases the maximum inrush cur-rent and reduces the exciting current to normal after a period of time.

Graphical method of determining inrush current
Figure 6 – Graphical method of determining inrush current

The losses that are effective are the resistance loss of the supply circuit and the resistance and stray loosed in the transformer.

Figure 7 is an oscillogram of a typical exciting-current inrush for a single-phase transformer energized at the zero point on the supply voltage wave. The transient has a rapid decrement during the first few cycles and decays more slowly thereafter. The damping coefficient, R/L for this circuit is not constant because of the variation of the transformer inductance with saturation.

During the first few current peaks, the degree of saturation of the iron is high, making L low. The inductance of the transformer increases as the saturation decreases, and hence the damping factor becomes smaller as the current decays.

Current inrush for a particular transformer energized at zero voltage
Figure 7 – Current inrush for a particular transformer energized at zero voltage

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7. Estimating Inrush Currents

The calculation of the inrush current to a power transformer requires considerable detailed transformer design information not readily available to the application engineer. For this reason reference should be made to the manufacturer in those few cases where a reasonably accurate estimate is required.

An order of magnitude of inrush currents to single-phase, 60-cycle transformers can be obtained from the data in Table 2. The values given are based on the transformer being energized from the high-voltage side at the instant the supply voltage passes through zero. Energizing a core-form transformer from the low-voltage side may result in inrush currents approaching twice the values in the table.

Table 2 – Approximate inrush currents to 60-cycle power transformers energized from the high-voltage side

Transformer Rating kVACore FormShell Form
2 0005 – 8
10 0004 – 72.5 – 5
20 0002.0 – 4

The per unit inrush current to a shell-form transformer is approximately the same on the high- and low-voltage sides. The inrush currents in Table 2 are based on energizing a transformer from a zero-reactance source.

When it is desired to give some weight to source reactance, the inrush current may be estimated from the relation:

I = I0 / (1+I0X)

where

  • Io = Inrush current neglecting supply reactance in per unit of rated transformer current
  • X = Effective supply reactance in per unit on the transformer kva base.

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Source: T&D book by ABB

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Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry facilities. Professional in AutoCAD programming.

4 Comments


  1. Zenati
    Dec 10, 2020

    Thanks you for the good information


  2. Ikpotor Ewiteyetan Stanley
    Dec 09, 2020

    I need information about power systems protection.


  3. kalyan kr.Chatterjee
    Dec 08, 2020

    Discussion on inrush current&exciting current is very clear to understand &
    Mathematical deduction of differential equation is helpful for engineers at any stage.


  4. MD.Aminul Islam
    Dec 08, 2020

    Dhaka,Bangladesh.
    It’s an informative lesson.thaks for sharing.really I like it very much

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