Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

[email protected] • ENGR-43_Lec-10-2_Transformers.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp 8 [3-4]Chp 8 [3-4]Magnetic Magnetic CouplingCoupling



Outline – Magnetic CouplingOutline – Magnetic Coupling

Mutual Inductance• Behavior of inductors sharing a common

magnetic field

Energy Analysis• Used to establish relationship between

mutual reluctance and self-inductance



Outline – Magnetic Coupling cont.Outline – Magnetic Coupling cont.

Ideal Transformer• Device modeling of components used to

change voltage and/or current levels

Safety Considerations• Important issues

for the safe operation of circuits with transformers



The Ideal TransformerThe Ideal Transformer Consider Now two Coils Wrapped Around a

Closed Magnetic (usually iron) Core.

The Iron Core Strongly confines the Magnetic Flux, , to the Interior of the Closed Ring• All Turns, N1 & N2, of Both Coils are Linked by the Core Flux

– Again, this is a NONconductive (no wires) connection

A Area



Ideal X-Former Physics/MathIdeal X-Former Physics/Math The Coils, N1 & N2, are

Flux-Linked: = N Then the Ratio of v1:v2

Next Apply Ampere’s Law (One of Maxwell’s Eqns)

2211 iNiNidlH encl

dt

Nd

dt

dv

dt

Nd

dt

dv

22

11

By Faraday’s Induction Law for Both Coils

2

1

2

1

2

1

N

N

dtd

dtd

N

N

v

v

Where• H Magnetic Field Strength

(Amp/m)



Ideal X-Former Physics cont.1Ideal X-Former Physics cont.1 Ampere’s Law H=0 in Ampere’s Law

Now Manipulate Ampere’s Law Eqn The Path for the Closed

Line Integral is a path Within the Iron Core

If the Magnetic Core is IDEAL, then

2211 iNiNdlH

0H

1

2

2

1

2211 0

N

N

i

i

iNiN

or

0

0

211

211

22111

1

ivN

Niv

soiNiNN

v



Ideal X-Former Physics cont.2Ideal X-Former Physics cont.2 The Ideal Xformer But v•i = POWER, and

by the previous Eqn the total power used by the Xformer is ZERO

Thus in Ideal Form, a Transformer is LOSSLESS• Thus the INput Power =

OUTput Power

But By Flux Linkage

0or

0

2211

22

12

1

211

iviv

iN

Nv

N

Niv

2121 NNvv So in the ideal Case

Ampere’s Law



Ideal X-Former Circuit SymbolIdeal X-Former Circuit Symbol From The Device Physics;

The Ideal Xformer Eqns The Circuit Symbol

Since an Xformer is Two Coupled Inductors, Need the DOT Convention to Track Polarities

2

1

2

1

2211 0

N

N

v

v

iNiN

The Main practical Application for This Device:• TRANSFORM one AC

Voltage-Level to Another

Iron Core



Transformer ApplicationTransformer Application When a Voltage is

Transformed, Give the INput & OUTput sides Special Names• INput, v1, Side

PRIMARY Circuit

• OUTput, v2, Side SECONDARY Circuit

The Voltage Xformer

Then the Circuit Symbol Usage as Applied to a Real Circuit

Pictorial Representation

Load

Src



Transformer Practical AppTransformer Practical App The Actual Ckt

Symbol As used On an Engineering Dwg

The Practical Symbol does NOT Use DOTS• NEMA has Established

Numbering Schemes That are Functionally Equivalent to the Dots

The Multiple “Taps” on the Primary Side Allow The Transformation of More Than One Voltage Level

See next Slide for a REAL Xformer Design

The Parallel (||) lines Between the Coils Signify the Magnetic Core



208Vac, 80 kVA Input

208:115 Vac StepDown Xformer




Sign (Dot) ConventionsSign (Dot) Conventions Have TWO Choices for

Polarity Definitions• Symmetrical

Thus The Form of the Governing Equations Will Depend on the Assigned:• DOT POSITION

• VOLTAGE POLARITY

• CURRENT DIRECTION

• INput/OUTput (I/O)



Phasor AnalysisPhasor Analysis In Practice, The Vast

Majority of Xformers are Used in AC Circuits

Recall The Symmetrical Ideal-Xformer Eqns

Illustration: InPut IMPEDANCE = V1/I1

These are LINEAR in i & v, so PHASOR Analysis Applies

Notice That This is an I/O Model; So the Eqns

1

2

2

1

2

1

2

1

2211

;

0

N

N

i

i

N

N

v

v

iNiN

1

2

2

1

1

2

2

1

2

1

2

1 ;

N

N

N

N

N

N

I

I

I

I

V

V



Phasor Analysis: Input ZPhasor Analysis: Input Z An I/O Xformer in

Phasor Domain Solve for V1

Now Apply Ohm’s Law to the Load, ZL

Now The Input Impedance Z1

2IZV2 L Now Sub for V2 and I2

From I/O Xformer Eqns

2

11

1

2122 N

N

N

NLL IZVIZV

1

2

2

11 IZV LN

N

LN

NZZ

I

V2

2

11

1

1

• For 10X stepDOWN (N1:N2 = 10:1) Z1 is 100X ZL



Phasor Analysis: Input Z cont.1Phasor Analysis: Input Z cont.1 An I/O Xformer Phasor

Domain Input Impedance

Thus ZL is Said to be REFLECTED to the Input Side (by [N1/N2]2)

For Future Reference

For a LOSSLESS Primary/Secondary Xformer the INput impedance is a Fcn ONLY of the LOAD Impedance, ZL

Ideal Xformer Phasor Eqns

LN

NZZ

I

V2

2

11

1

1

2*22

*

1

22

2

12

*111 S

N

N

N

NS

IVIVIV

ratio turns1

2 N

Nn

2121

212

1

SSn

nn

L

ZZ

IIV

V



Numerical ExampleNumerical Example Given the Ckt Below

Find all I’s and V’s Using

Note: n = N2/N1 = 1/4

Game Plan• reflect impedance into

the primary side and make the transformer “transparent to Source”

Find I1 by Ohm

21 nLZZ

LZ

16321 jZ

5.1333.25.1342.51

0120

4181632

0120

211 jj

S

ZZ

VI

stepDOWNXformer



Numerical Example cont.1Numerical Example cont.1 The Intermediate Ckt

About the Same Hassle-Factor; use ZI

Now Find V1 by Ohm or V-Divider

Next Determine SIGNS

LZ

16321 jZ

SVZZ

ZIZV

21

1111

Which is Easier?

07.1336.83

5.1333.257.2678.351V

1205.1342.51

16320120

5.1333.2)1632(

21

1

11

j

j

ZZ

Z

IZ

Z2

stepDOWNXformer



Numerical Example cont.2Numerical Example cont.2 The Original Ckt

Compare Current Case to I/O Model• Voltage-2 is OPPOSITE

(NEGATIVE at Dot)

• Current-2 is OPPOSITE (INTO Dot)

Then In This Case

Recall Now the I/O Model Eqns

nnN

N

nnN

N

121

2

2

1

122

1

2

1 1

III

I

VVV

V

n

n

12

12 ;

II

VV

IN OUT

stepDOWNXformer



Numerical Example cont.3Numerical Example cont.3 The Original Ckt The Output Voltage

Using the Signs as Determined by Dots and Polarities

93.166875.20

Quadrant 3rd720.433.20

88.1832.8125.0

13.0783.490.2525.0

2

112

V

VVV

j

j

n

5.16632.9

Quadrant 2nd176.2065.9

5439.0266.24

5.1333.244

2

11

2

I

II

I

j

jn

Then Output Current

stepDOWNXformer

Note: On Calculator aTan(–4.72/–20.33) = 13.07°• Recall RANGE of aTan = –90° to + 90°



IllustrationIllustration Given the Ckt Below Find I1 & Vo

Note: n = N2/N1 = 2/1

Game Plan• reflect impedance into

the primary side and make the transformer “transparent to Src”

Again using

2n

21 nLZZ

5.01

12

22221 j

jZ

1Z

5.23 jZ i

stepUPXformer



Illustration cont.1Illustration cont.1 Given the Ckt Below Find I1 & Vo

Thus I1 by Ohm

2n

n

II 1

2

5.0122

012

121 jj

VS

ZZ

VI

81.3907.3

81.3991.3

0V121 AI

Next Find Vo by I2 and Ohm’s Law• Define I2 Direction per I/O

Model (V2 is ok)

81.3907.32

2

so and 2

10

201

2

V

L

IV

IZVI

I

Z2

stepUPXformer



Xformer Thevenin EquivalentXformer Thevenin Equivalent Given I/O XFormer Ckt

Find the The Thevenin Equivalent at 2-2’

First Find the OPEN Ckt Voltage at 2-2’• Note: The Dots &

Polarities Follow the I/O Model

00

121

2

III

I

n 11 SVV

112

11SOC

S nn

VVVV

VV



Xformer Thevenin Equiv. cont.1Xformer Thevenin Equiv. cont.1 Now find ZTH at

Terminals 2-2’

“Back Reflect” Impedance into SECONDARY

Thus the Thevenin Equivalent at 2-2’

THZ

12ZZ nTH

The Xformer has been “made Transparent” to the Secondary Side• Next: Find

Thevenin Equiv at Terminals 1-1’



Thevenin Equiv. from PrimaryThevenin Equiv. from Primary Given I/O XFormer Ckt

Find the The Thevenin Equivalent at 1-1’

Then The Open Ckt Voltage Depends on VS2

As in Open Ckt

Thevenin impedance will be the Secondary impedance reflected into the PRIMARY Ckt

nSOC 2VV

0 and 0 21 II

22

n

ZZTH



Primary v. Secondary TheveninPrimary v. Secondary Thevenin

Thevenin From Primary

Equivalent circuit reflecting into primary

Thevenin From Secondary

Equivalent circuit reflecting into secondary

The Base Ckt



Exmpl: Draw Thevenin Equiv’sExmpl: Draw Thevenin Equiv’s2n

Equivalent circuit reflecting into SECONDARY

Equivalent circuit reflecting into PRIMARY



ExampleExample Given the Ckt Below

Find I1

Note The Dot Locations

Note: n = N2/N1 = 2/1

Game Plan Find Thevenin Looking

from PRIMARY Side• Draw the Ckt

22 j

036

5.0)(22 j

036

1I

25.2

060361 j

I

66.38119.13

86.362016.3

0421I

V

nS 6

2

0122

V

1

1’



Example: Safety ConsiderationsExample: Safety Considerations Two Houses Powered

By DIFFERENT XFormers

Utility Circuit Breaker X-Y OPENS: Powering DOWN House-B

The Well-Meaning Neighbor Runs Extension Cord House-A → House-B• This POWERS the

2nd-ary Side of the House-B Pole Xformer



Example: Safety Consid cont.1Example: Safety Consid cont.1

Transformers are BIdirectional Devices• They can step-UP or

step-DOWN Voltages

Thus the 120Vac/15A Extension Cord Produces 7200 Vac Across Terminals X-Z

The Service Engineer (SE) Now Goes to the Breaker to ReMake the Connection to House-B

The SE expects ZERO Volts at X-Z; If She/He Does NOT Check by DMM, then He/She Could Sustain a Potentially FATAL 7.2 kV Electric-Shock!



Exmple Exmple Power Transmission Power Transmission At the Generating

Facility (e.g. Diablo Canyon) Electricity is Generated at 15-25 kVac

But Xformers are used to Set-UP the Voltage-Level to 400-765 kVac

Why? → Line SIZE (and others)



Exmpl – Power Xmission cont.1Exmpl – Power Xmission cont.1 Case Study: Transmit

• 225 MW over

• 100 Miles of Wire

• 2 Conductors

• 95% Efficiency

• Cu Wire w/ Resistivity– ρ = 80 nΩ-m

Find the Wire Diameter, d, for:a) V = 15 kVac

b) V = 500 kVac



Exmpl – Power Xmission cont.2Exmpl – Power Xmission cont.2 By Solid-State Physics

(c.f. ENGR-45)

• Where for the Wire– ρ Resistivity (Ω-m)

– l Length (m)

– A X-Section Area (m2)

In This Case

Then the Power Loss

By Power Rln for Resistive Ckts

Solve for d

Al Rwire

4

1609351002dA

mmil

MWMW Pwire 25.11225%5

4

25.11

22

2

d

lAlPVR

MWPRV P

wirewire

wirewireloss

Vd

V

lPd

P

V

d

l

11920 :Values Subbing

144 2

2



Exmpl – Power Xmission cont.3Exmpl – Power Xmission cont.3 Finally Solve for

Transmission Cable Diameter

md

md

00384.0500000

11920

128.015000

11920

500

15

d15 = 5.03”

• Pretty BIG & HEAVY

d500 = 0.15”

• MUCH Better



Summary: Ideal TransformerSummary: Ideal Transformer Consider Now two Coils Wrapped Around a Closed

Magnetic (usually iron) Core.

The Iron Core Strongly confines the Magnetic Flux, , to the Interior of the Closed Ring• All Turns, N1 & N2, of Both Coils are Linked by the Core Flux

A Area



Ideal X-Former Circuit SymbolIdeal X-Former Circuit Symbol The Ideal Xformer Eqns

• Faraday’s Law

The Circuit Symbol

As Two Coupled Inductors, Xformers Use the DOT Convention to Track Polarities

2

1

2

1

N

N

v

v

The Main practical Application for This Device:• TRANSFORM one AC

Voltage-Level to Another

Iron Core

02211 iNiN

• Ampere’s Law



Transformer ApplicationTransformer Application When a Voltage is

Transfrormed, Give the INput & OUTput sides Special Names• INput, v1, Side

PRIMARY Circuit

• OUTput, v2, Side SECONDARY Circuit

The Voltage Xformer

Then the Circuit Symbol Usage as Applied to a Real Circuit

Pictorial Representation

Load



Sign (Dot) ConventionsSign (Dot) Conventions Have TWO Choices for

Polarity Definitions• Symmetrical

The Form of the Governing Equations

• INput/OUTput (I/O)

2

1

2

1

2211 0

N

N

v

v

iNiN

2

1

2

1

2211

N

N

v

v

iNiN



Phasor AnalysisPhasor Analysis In Practice, The Vast

Majority of Xformers are Used in AC Circuits

Ideal-Xformer Eqns Are LINEAR in i&v so PHASOR Analysis Applies

Illustration: InPut IMPEDANCE

Notice That This is an I/O Model; So the Eqns Yield

LN

NZZ

I

V2

2

11

1

1

(symm) 0

(I/O)

1

2211

2211

2

1

2

1

II

II

V

V

NN

NN

nN

N



WhiteBoard WorkWhiteBoard Work

Let’s Work This Nice Problem



208Vac, 80 kVA Input



Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Documents