MAGNETICALLY COUPLED NETWORKS

LEARNING GOALS

Mutual Inductance
Behavior of inductors sharing a common magnetic field

The ideal transformer

Device modeling components used to change voltage and/or
current levels
BASIC CONCEPTS A REVIEW

Total magnetic
Magnetic field
flux linked by N-
turn coil
Amperes Law
(linear model)

Faradays
Induction Law

Assumes constant L
and linear models!

Ideal Inductor
MUTUAL INDUCTANCE
Overview of Induction Laws Induced links
on second
Magneti coil
c ( )
2

flux

Total fluxlinkage
N (webers)
If linkage is created by a current flowing
through the coils
Li (Amperes Law)

The voltage created at the terminals of

the components is
di
vL (Faradays Induction Law)
dt

One has the effect of mutual inductance

TWO-COIL SYSTEM (both currents contribute to flux)

Self-induced Mutual-induced

Linear model simplifying

notation
THE DOT CONVENTION
COUPLED COILS WITH DIFFERENT WINDING CONFIGURATION

Dots mark reference

polarity for voltages
induced by each flux
THE DOT CONVENTION REVIEW LEARNING EXAMPLE

Currents and voltages follow

passive sign convention

(v2 (t ))

i1 (t ) i2 ( t )

di di
Flux 2 induced v1 (t ) L1 1 M 2
voltage has + at dot dt dt
di di
v2 (t ) M 1 L2 2
di1 di dt dt
v1 (t ) L1 (t ) M 2 (t )
dt dt
di1 di
di di v1 L1 M 2
v2 ( t ) M 1 (t ) L2 2 ( t ) dt dt
dt dt
di di
v2 M 1 L2 2
For other cases change polarities or dt dt
current directions to convert to this
basic case
LEARNING EXAMPLE

Mesh 1
LEARNING EXAMPLE - CONTINUED

Mesh 2 Voltage Terms

PHASORS AND MUTUAL INDUCTANCE

di1 di
v1 (t ) L1 (t ) M 2 (t )
dt dt
di di
v2 (t ) M 1 (t ) L2 2 (t )
dt dt

Assuming complex V1 jL1 I1 jMI 2 Phasor model for mutually

exponential sources V2 jMI1 jL2 I 2 coupled linear inductors
LEARNING EXAMPLEThe coupled inductors can be connected in four different w
Find the model for each case

CASE I Currents into dots

V V1 V2
I I
V1 jL1 I jMI
V1 V2
V2 jMI jL2 I
V j ( L1 L2 2 M ) I j Leq I

CASE 2 Currents into dots V V1 V2

I I V1 j L1I j MI
V2 j MI j L2 I
V1 V2
V j ( L1 2 M L2 ) I
Leq
Leq 0 imposesa physicalconstrain
on the valueof M
CASE 3 I I1 I 2 I 2 I I1
Currents into dots
V jL1 I1 jMI 2
I1 I2 V jMI1 jL2 I 2
V V V jL1 I1 jM ( I I1 )
V jMI1 jL2 ( I I1 )
V j ( L1 M ) I1 jMI /( L2 M )
V j ( L2 M ) I1 jL2 I /( L1 M )

L1 L2 M 2 ( L1 L2 2 M )V j M ( L2 M ) L2 ( L1 M ) I
V j I
L1 L2 2 M

CASE 4
Currents into dots

I1 I2

V ( V ) I I1 I 2

V jL1 I1 jMI 2 L1 L2 M 2
V j I
V jMI1 jL2 I 2 L1 L2 2M
LEARNING EXAMPLEFIND THE VOLTAGEV
0

I2
I1 1. Coupled inductors. Define the
voltages and currents

V1 V2
2. Write loop equation
KVL: 2430 2 I1 V1 in terms of coupled
VS KVL: - V2 j 2 I 2 2 I 2 0 inductor voltages
MUTUALINDUCTANCECIRCUIT 3. Write equations fo
V1 j 4 I1 j 2( I 2 ) V0 2I 2 coupled inductors
V2 j 2 I1 j 6( I 2 )

VS ( 2 j 4) I1 j 2 I 2 / j2 4. Replace into loop equation

0 j 2 I1 (2 j 2 j 6) I 2 / 2 j 4 and do the algebra

j 2VS 4 (2 j 4) 2 I 2
j 2VS j 2VS
I2
8 j16 j 16 8 j
VS 2430
V0 2 I 2 5.373.42
4 2 j 4.4726.57
LEARNING EXAMPLEWrite the mesh equations

3. Write equations for coupled inductor

I1 I 2 I 2 I3
V1 jL1 ( I1 I 2 ) jM ( I 2 I 3 )
V2 jM ( I1 I 2 ) jL2 ( I 2 I 3 )

V1 V2 4. Replace into loop equations and
rearrange terms

1
V R1 jL1 I1
j C1

1
jL1 jM I 2 jMI 3
1. Define variables for coupled inductors j C
1
2. Write loop equations in terms of coupled
inductor voltages 1
0 jL1 jM I1
I1 I 2 j C 1
V R1 I1 V1
jC1 1
I 2 I1 jL2 jM R2 jM jL2 R3 I 2
V1 R2 I 2 V2 R3 ( I 2 I 3 ) 0 j C1
jC1
jM jL2 R3 I 3
I3
V2 R4 I 3 R3 ( I 3 I 2 ) 0 0 jMI1 jL2 jM R3 I 2
j C 2
1
jL2 R4 R3 I 3
j C 2
 VS
Zi
LEARNING EXAMPLE DETERMINE IMPEDANCE SEEN BY THE SOURCE
I1
Z S 3 j1()
Z L 1 j1()
Z S jL1 I1 ( jM ) I 2 VS /( Z L jL2 )
I1 I2
( jM ) I1 ( Z L jL2 ) I 2 0 / jM
V1 V2 ( Z S
jL1 )( Z L jL2 ) ( jM ) 2 I1
( Z L jL2 )VS
jL1 j 2() jL2 j 2()
jM j1() VS ( j M ) 2
Zi ( Z S jL1 )
I1 Z L jL2
1. Variables for coupled inductors
( j1) 2 1 1 j
2. Loop equations in terms of coupled Z i 3 j3 3 j3
1 j1 1 j 1 j
inductors voltages
Z S I1 V1 VS 1 j
Z i 3 j3 3.5 j 2.5()
V2 Z L I 2 0 2
Z i 4.3035.54()
3. Equations for coupled inductors
V1 jL1I1 jM ( I 2 )
V2 jMI1 jL2 ( I 2 )
4. Replace and do the algebra
 THE IDEAL TRANSFORMER

1 N1 2 N 2

Insures that no magnetic flux

d goes astray
v1 ( t ) N1 (t ) v N
dt 1 1 First ideal transformer
d
v2 (t ) N 2 (t ) v2 N 2 equation
dt
v1 (t )i1 (t ) v2 (t )i2 (t ) 0 Ideal transformer is lossless
i1 N
2 Second ideal transformer
i2 N1 equations

v1 N1 i1 N 2
;
Circuit Representations v2 N 2 i2 N 1
REFLECTING IMPEDANCES For future reference
*
N N
S1 V1 I1* V2 1 I 2 2 V2 I 2* S 2
N2 N1
N2
n turnsratio
N1

V1 N1
(both signsat dots) Phasor equations for ideal transformer
V2 N 2 V2
V1
I1 N 2 n
(CurrentI 2 leavingtransforme
r)
I 2 N1 I1 nI 2
V2 Z L I 2 (Ohm's Law) ZL
Z1
2 n2
N N N
V1 2 Z L I1 1 V1 1 Z L I1 S1 S2
N1 N2 N2
2
V1 N
Z1 1 Z L
I1 N2
Z1 impedance,Z L , reflected
intotheprimaryside
LEARNING EXAMPLE Determine all indicated voltages and currents
n 1 / 4 0.25
1200 1200
I1 2.33 13.5
50 j12 51.4213.5
V1 Z1 I1 (32 j16) 2.33 13.5
83.3613.07

Strategy: reflect impedance into the

primary side and make transformer
transparent to user. ZL
Z1
n2 CAREFUL WITH POLARITIES AND
CURRENT DIRECTIONS!
ZL
I1
I2 4 I1 (current into dot)
n
V2 nV1 0.25V1 ( is oppositeto dot)

Z1 32 j16
USING THEVENINS THEOREM TO SIMPLIFY CIRCUITS WITH IDEAL TRANSFORME

Replace this circuit with its Thevenin

Reflect impedance into ZTH
equivalent
secondary
ZTH n 2 Z1

I2 0
I1 0 V1 VS1
I1 nI 2 Equivalent circuit with transformer
V1 VS1 made transparent.
VOC nVS1
V2 nV1 One can also determine the Thevenin
equivalent at 1 - 1
To determine the Thevenin impedance...
 USING THEVENINS THEOREM: REFLECTING INTO THE PRIMARY

Z2
ZTH Find the Thevenin equivalent of Equivalent circuit reflecting
n2 this part into primary
In opencircuitI1 0 and I 2 0 VS 2
VOC
n
Thevenin impedance will be the the
secondary mpedance reflected into
the primary circuit

Equivalent circuit reflecting

into secondary
LEARNING EXAMPLE Draw the two equivalent circuits

n2

Equivalent circuit reflecti

into secondary

Equivalent circuit reflectin

into primary
LEARNING EXAMPLEFindVo
Thevenin equivalent of this part

To computeVo is betterto reflectintosecondary

But before doing that it is better to simplify the primary using Thevenins The

VOC Vd 4 90
Vd j4 24 90
Vd 240
4 j4 1 j
VOC 14.42 33.69(V )
j16 8 j8 j16
ZTH 2
4 j4 4 j4 ZTH 2 (4 || j 4) ZTH 4 j 2()
2 j6 1 j 8 j4 This equivalent circuit is now transferred t
ZTH
1 j 1 j 2 the secondary
LEARNING EXAMPLE (continued)
n2

Thevenin equivalent of primary side

Equivalent circuit reflectin

into secondary

Circuit with primary transferred to secondary

2 2 28.84 33.69
Vo 28.84 33.69
20 j 5 20.62 14.04
LEARNING EXAMPLE Find I1 , I 2 ,V1 ,V2 2V1 V2 2 I1 100 I1 50
V1 (1 j )V2 2 I 2 0
V2 2V1
n2
I1 2 I 2 I 2 2.50
V1 (1 j )(2V1 ) 50

50 50
V1 563.43
1 j 2 2.24 63.43
Nothing can be transferred. Use
transformer equations and circuit V2 2 563.43
analysis tools

Phasor equations for ideal transformer

V2
V1
n
I1 nI 2
V 100 V1 V2
@ Node1: 1 I1 0
2 2
V V V
@Node2: 2 1 2 I 2 0
2 j2
4 equations in 4 unknowns!