EE212 Passive AC Circuits

Lecture Notes 3

Transformers

1EE 212 2010-2011

Magnetic Circuit

• What is the relationship between magnetic flux and magnetomotive force, Fm?

• What is the relationship between electric

current, i, and magnetomotive force?

Magnetic field strength, H = Fm / l

2EE 212 2010-2011

Magnetic Circuit

• What is Faraday’s Law? – The voltage induced in an electric circuit is

proportional to the rate of change of the magnetic flux linking the circuit

3EE 212 2010-2011

Magnetic Circuit

Consider a coil around a magnetic core. If a current i flows through the coil, a magnetic flux is generated in the core.

N = number of turns in the coilRm = constant known as reluctance(depends on the magnetic path of the flux)

a

b

i

Lvab = L didt

= N i in webers (Wb)mR

1

Direction of flux by

Right-Hand Rule

Fingers curled around coil – direction of current

Thumb – direction of flux

Rm = in ampere-turns/WbA.

l

l = length of magnetic path A = cross-section aream = permeability

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B-H Curve

Flux Density, B = in teslas, TA

Magnetic field strength, H = in AT/m

liN.

B = m H

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Coupled Circuits

Circuits that affect each other by mutual magnetic fields

a

b

i1

f1

L1vab

c

d

vcd

f2

L2 i2

The flux f2 generated by current i2 in Coil 2

induces a voltage in Coil 1, and vice-versa.

vab = L1dtdi1 ± M

dtdi2

vcd = L2dtdi2 ± M

dtdi1

± depending on whether the fluxes add or oppose each other

L1, L2: self inductance

M: mutual inductance ratio of induced voltage in one circuitto the rate of change of current in another circuit

6EE 212 2010-2011

Coupled Circuits in Phasors

If input signals are sinusoidal waveforms, coupled circuits can be in phasor representation

vab = L1dtdi1 ± M

dtdi2

vcd = L2dtdi2 ± M

dtdi1

± depending on flux directions

a

b

Vab

I1

L1

c

d

Vcd

I2

L2

M

7EE 212 2010-2011

Equivalent Circuit with Dependent Sources

8EE 212 2010-2011

Dot Convention

Dots are placed at one end of each coil, so that currents entering the dots produce fluxes that add each other.

a

b

i1

f1

L1

c

d

f2

L2 i2

a

b

I1

L1

c

dI2

L2

(currents entering the dots produce upward fluxes)

The dots provide information on how the coils are wound with respect to each other.

A current i entering a dotted terminal in one coil

induces a voltage M with a positive polarity

at the dotted terminal of the other coil.dtdi

Vab = (jwL1) I1 + (jwM) I2 for the currents as shown

(+) if both currents enter the dotted terminals (or the undotted terminals).

(-) if one current enters a dotted terminal and the other current enters an undotted terminal.

9EE 212 2010-2011

Coefficient of Coupling, k

k = 0 ≤ k ≤ 1

k depends on the magnetic properties of the flux path.

When k = 0, no coupling

k = 0.01 to 0.1, loosely coupled

k > 0.5, close coupled, e.g. air core

k ≈ 1.0, e.g. power transformerall the flux generated by one coil is linked

to the other coil (i.e. no leakage flux)

k = 1.0 ideal transformer

21LL

M

10EE 212 2010-2011

E1

N2

E2

N1

If+

-

V1

+

-

V1 = -E1

If f

E1

E2

Transformer

Faraday’s Law

fif

v1N1

+

-e2N2

+

-

e1

primarywinding

secondarywinding

dt

de2 = N2

When a voltage V1 is applied to the primary winding, an emf e2 is induced in the secondary winding. The induced emf lags the inducing flux by 900.

11EE 212 2010-2011

Transformer Application in Power System

Instrument Transformers:CT (current transformer)PT or VT (voltage transformer)

- V/I step up/down- Y/D conversion- circuit (dc) isolation- Z matching (for max power transfer, min. reflection from load)

12EE 212 2010-2011

No leakage flux

dt

d and e2 = N2

2

1

ee

= 2

1

NN

= a (turns ratio)

1

2

ii

= 2

1

ee

= 2

1

NN

= a

Ideal Transformer

No Losses No voltage drops in the windings: V1 = - e1

Instantaneous powers in primary and secondary are equal (i.e. all the energy from the primary is transferred to the secondary winding).

e1 i1 = e2 i2 Therefore,

Coupling Coefficient, k = 1 , i.e. the same flux f

goes through both windings

e1 = N1 dt

d

13

=>

Turns ratio (a or n) is also known as the transformation ratio.

EE 212 2010-2011

Transformer Loading

• a secondary current I2 is drawn by the load

• I2 generates a flux that opposes the mutual flux f

(Lenz’s Law: effect opposes the cause)• reduction in f would reduce induced emf e1

• since source voltage V1 is constant, and V1 = - e1, f must remain constant

• the primary winding must draw an additional current I’1 from the source to neutralize the demagnetizing effect from the secondary

e1

N2

e2

N1

I1

V1

+

-

ZLV2

I2

Primary current I1 = I’1 + If14EE 212 2010-2011

Ideal Transformer

• No leakage flux, i.e., k = 1

• Self-inductance, L1 = L2 = ∞ i.e., magnetizing current = 0

• Coil losses are negligible• With polarities, dots and currents as shown:

V2 = V1 /a

I2 = a I1

Z1 = a2 Z2 where Z2 is the load impedance

15

a

b

Vab

I1

L1

c

d

Vcd

I2

L2

M

● ●+

-

+

-EE 212 2010-2011

Actual Transformer

fi1

v1

N1

+

-e2

N2

+

-e1 ZL

R1 R2

v2

- resistance in primary and secondary windings- leakage reactance in pr. and sec. windings- voltage drops in both windings (leakage impedance)

- losses- copper loss primary: I1

2·R1 secondary: I22·R2

- iron loss (core loss) Core loss depends on voltage and frequency

eddy current loss hysteresis loss 16EE 212 2010-2011

Eddy Current Loss:

emf induced in core generates eddy currents which circulate in the core material, generating heat.

laminations (silica sheets between core layers) – to reduce eddy current, and minimize loss

Iron (Core) Losses

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Hysteresis Loss:

The direction of the magnetic flux in the core changes every cycle. Power is consumed to move around the magnetic dipoles in the core material, and energy is dissipated as heat.

Hyst. loss (vol. of core) x (area of hyst. loop)

Iron (Core) Losses

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Transformer Construction

• Coil Winding • Core Assembly • Core-Coil Assembly • Tank-up • Accessories Mounting and Finishing

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Core Assembly

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Core-Coil Assembly• Core vertical sides – limbs, top horizontal side – yoke • Yoke is removed to insert the coils into the limbs • LV coil is first placed on the insulated core limbs • Insulating blocks are placed at the top and bottom of the

LV coil • Cylinder made out of corrugated paper is placed over

the LV Coil • HV coil is placed over the cylinder • The top yoke is fixed in position • LV and HV windings are connected as required

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Tank-up

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Accessories Mounting

• Connections of LV and HV coil ends to the terminal bushings are made

• Transformer tap changer and protection accessories (e.g. Buchholz relay, Conservator, Breather, temperature indicator, etc.) are installed

• Tank is closed

Functions of Transformer Oil • Cooling• muffle noise• displace moisture (avoid insulation degradation)

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Transformer Cooling

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Typical Power Transformers

Pole-mounted Single-phase Transformer

Three-phase Transformer

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Transformer Rating

• Rated kVA• Rated Voltage primary and secondary: (transformers are normally operated

close to their rated voltages)

• Rated Current (FL Current) is the maximum continuous current the transformer can withstandFor single phase transformer:Rated primary current = Rated VA / Rated Pr Voltage

When rated current flows through a transformer, it said to be fully loaded.

The actual current through a transformer varies depending on the load connected at different times of the day.

26EE 212 2010-2011

Equivalent Circuit• Represent inductively coupled circuits by a conductively connected circuit• Equivalence in terms of loop equations• Assume a load impedance is connected to the secondary.

V1

+

-L22

+

-

R1R2

V2

M

N2N1

L11I1 I2

inductively coupled circuits

KVL at Loop 1:

KVL at Loop 2:

-V1 + R1I1 + jwL11I1 - jwM I2 = 0

R2I2 + V2+ jwL22I2 - jwM I1 = 0

The loop equations are:turns ratio,

2

1

N

N = a

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Equivalent Circuit (continued)

conductively connected circuit

Consider following substitutions:

The loop equations remain the same:

V1

- V2

R1 + jwL11 - jwM

- jwM R2 + jwL22

=I1

I2

V1

+

-

+

-aV2

aMI1

R1L1 a2R2

a2L2

M a·M L2 a2·L2 R2 a2·R2

V2 a·V2 I2 a

I2

Let L11 = L1 + aM

and L22 = L2 + M/a

KVL at Loop 1:

KVL at Loop 2:

-V1 + R1I1 + jwL1I1 + jwaM (I1-I2/a) = 0

jwaM (I2/a - I1) + jw a2L2.I2/a + jw a2R2.I2/a + aV1 = 0

28EE 212 2010-2011

Transformer Equivalent Circuit

equivalent circuit referred to the primary side

a2ZLV1

+

-

I1

R1X1 a2R2

a2X2

Ie

IfIc

Rc Xf

R1, R2= primary, secondary winding resistanceX1, X2= primary, secondary leakage reactanceI1, I2= primary, secondary current

ZL = load impedance

a2I

Ie = excitation currentIf = magnetizing currentXf = magnetizing reactance

Rc = core loss resistance (equivalent resistance contributing to core loss)Ic = core loss equivalent current 29EE 212 2010-2011

Example: Transformer Equivalent Circuit

A 50-kVA, 2400/240-V, 60-Hz distribution transformer has a leakage impedance of (0.72 + j0.92) W in the high voltage (HV) side and (0.0070 + j0.0090) W in the low voltage (LV) winding. At rated voltage and frequency, the admittance of the shunt branch of the equivalent circuit is (0.324 – j2.24)x10-2 S [siemens] when viewed from the LV side. Draw the circuit:

a) viewed from the LV sideb) viewed from the HV side

Turns ratio, a = 2400/240 = 10

0.72/a2 W j0.92/a2 W 0.0070 W j0.0090 W

j44.6429 W308.642 W

a)

0.72 W j0.92 W 0.0070a2

Wj0.0090a2 W

308.642a2 W j44.6429 a2 W

b)

30EE 212 2010-2011

Transformer: Approximate Equivalent Circuit

Re Xe

Rc Xf

Re Xe

Xe

Re = R1 + a2R2

Xe = X1 + a2X2

where, R2 and X2 are referred to the primary side

Re and Xe obtained from Short Circuit TestRc and Xf obtained from Open Circuit Test

For transformers operating close to full load

For transformers in a large power network analysis31EE 212 2010-2011

Short Circuit Test

LVHV

Vsupply

+

-

W A

V

Connect meters on HV side as shownShort circuit LV sideEnergize HV side with a variable voltage source and increase voltage gradually to get rated current reading on the AmmeterTake the V, I and P readings from meters

Re Xe

RcXf

I

V

+

-

Equivalent circuit referred to HV side

P = copper loss = I2 Re Re = 2I

P

Ze = Xe = I

V 2e

2e RZ

R1 ≈ a2 R2 ≈ 2

Re X1 ≈ a2 X2 ≈ 2

Xe

32EE 212 2010-2011

Open Circuit Test

Connect meters on LV side as shownOpen circuit HV sideEnergize LV side with rated voltageTake the V, I and P readings from meters

Equivalent circuit referred to LV side

Vsupply

W A

V

+

-LVH

V

Rc Xf

I

V

+

-

IfIe

Ic

P = core loss = Rc = c

2

R

V

P

V2

Ie = I , Ic = cR

V

If = Xf = 2c

2e II

φI

V

33EE 212 2010-2011

Transformer Efficiency

Efficiency, h = Power Output / Power Input

Power Output = |VL| |IL| cos = |IL|2 RL

Power Input = Power Output + Cu losses + Core losses

Cu Loss – varies with load current

Core Loss – depends on voltage (usually a constant for practical purposes)

Transformer efficiency is maximum when Cu loss = core loss

34EE 212 2010-2011

Transformer Efficiency (continued)

Power transformers: - usually operate at rated capacity, - designed to have max. h at full load

Distribution transformers: carry a widely varying load - designed to have max. h at less than full load - always energized despite load levels – designed to have low core loss

35EE 212 2010-2011

Example: Transformer

Find the equivalent circuit of a 50-kVA, 2400/240-V transformer. The following readings were obtained from short circuit and open circuit tests:

If rated voltage is available at the load terminals, calculate the transformer efficiency at (a) full load with 0.8 p.f. lagging

(b) 60% load with 0.8 p.f. lagging. What is the voltage at the HV terminals of the transformer?

Test V-meter (V) A-meter (A) W-meter (W)

Short Circuit 48 20.8 617 Open Circuit 240 5.41 186

If rated voltage is applied at the primary terminals, calculate the transformer efficiency at (c) full load with 0.8 p.f. lagging

36EE 212 2010-2011

The transformer is used to step down the voltage at the load end of a feeder whose impedance is 0.3 + j1.6 ohm. The voltage at the sending end of the feeder is 2400 V. Find the voltage at the load terminals when the connected load is

(a) Zero

(b)60% load at 0.8 p.f. lagging

(c) Full load at 0.8 p.f. laging

(d)Full load at 0.5 p.f. leading

What will be the current on the low voltage side if a short circuit occurs at the load point?

Transformer Example 2

37EE 212 2010-2011

When a constant rated voltage is applied to the primary,

At no load- no current, and therefore no voltage drop in transformer- secondary voltage, Vsec = rated voltage

As load (resistive or inductive) increases- voltage drop in transformer (Vdrop) increases

- secondary voltage decreases (reverse is the case with capacitive load increase)

Voltage Regulation

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To maintain rated voltage at the secondary,

At no load, primary voltage required, Vpr = rated voltage

At a certain load, required Vpr = rated voltage + Vdrop in transformer

Voltage regulation is the change in primary voltage required to keep the secondary voltage constant from no load to full load, expressed as a percentage of rated primary voltage.

Load p.f. has a big effect on voltage regulation.

Voltage Regulation (continued)

39EE 212 2010-2011

A 50-kVA, 2400/240-V, 60-Hz transformer has a leakage impedance of (1.42+j1.82) ohms on the HV side. The transformer is operating at full load in all 3 cases below.

1. Calculate the voltage regulation ata) 0.8 p.f. lagging b) unity p.f. c) 0.5 p.f. leading

40EE 212 2010-2011

Example

- Use of transformer tap changers: on-load/off-load

- Injecting reactive power (Vars): series/shunt compensation

Voltage Control

The allowable voltage variation at the customer load point is usually ± 5% of the rated (nominal) value. Control measures are taken to maintain the voltage within the limits.

41EE 212 2010-2011

Autotransformers

• Used where electrical isolation between primary side and secondary side is not required.

• Usually relatively low power transformers• Can be readily made to be variable.• Can think of it as two separate windings

connected in series.• Usually it is a single winding with a tap point.• Can be used as either step-up or step-down

transformer

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Autotransformers (continued)

• Schematic diagrams:

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From J.D. Irwin, “Basic Engineering Circuit Analysis”, 3rd ed. Macmillan

Autotransformers (continued)

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From Jackson et al. “Introduction to Electric Circuits” 8th ed., Oxford

Autotransformers (continued)

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Assume N1 = 200and N2 = 100

Let Vsource = 120 V

Then what is V2?

Assume I2 is 15 A. Then Sload = ?, and Ssource = ?

Autotransformers (continued)

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What is the node equation at the tap?

Node equation: I1 + IZY = I2 therefore, IZY =

From transformer action:N1I1 = N2IZY

therefore,

That is, only part of the load current is due to the magnetomotive force that the current in the primary coil exerts on the secondary coil. The remaining portion is direct conduction of the source current to the load.