inductance notes

7/30/2019 Inductance Notes

1/12

EE201.3

Denard Lynch Page 1 of 12 Revised: October 2, 2006

INDUCTANCE

Resistance opposes current;Capacitance opposes changes in voltage;

Inductance opposes changes in current.

Recall Faradays Law:dt

dqE

M

!"= for the voltage

induce in a loop of wire.(Which resulted in: Em = lvelX B)

If we have N turns, its like multiplying the flux

linkages by N, so:

dt

Nd

dt

dNE

M

)( !=

!= , and since BA=!

dt

BAdNE

M

)(= ; and A is constant, so

!

)(

,define;)(

,constant,,;

)(

,;)(

;

22

dt

IdLE

l

ANL

dt

Id

l

ANE

lNdt

l

NId

NAE

l

NIHHlNI

dt

HdNAE

HBdt

dBNAE

M

M

M

M

M

=

==

=

===

==

Where the letter L is used to represent self inductance, usually referred to as just inductance,and the unit is the Henry (H). (E.g. A rate of change of current of 1 ampere per second in a 1 Henry

inductor will produce a voltage of 1 Volt across the inductor.)

Now looking at the expression for Inductance,

)valid!(always(2)thatso;thatrecalland;(1)22

!==!=N

LA

l

l

ANL

For an air-core coil, the reluctance outside the coil issmall compared to the reluctance inside the coil (Ri

>> Ro). Thus, for an air-core inductor where the lengthis >> than the diameter (i.e. l/d > 10), the inductance

can be estimated using:

4%.typicallyiserrorthe,10where;2

0

d

l

l

ANL

(Note: forl/d< 10, you can apply Nagoakas correction

factor)

V

V

RO

RI


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EE201.3


Now go back for a moment to:

L =N

2

"; and # = NI =$", (similar to V = IR) and substituting" =

NI

$

we are left with L =N$

I(also, always valid!).

Example 1:

An air-core coil find the inductance, L.

Example 2:

A Laminated sheet steel core (S.F. = .93) with dimensions as shown. Calculate the inductance, L.(Note: Assume we are operating in the linear region of the B-H curve).

1cm X 1cm

10cm

S.F. = .93

100t

12mm

15cm120t


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EE201.3


Example 3:

A laminated Sheet Steel C core inductor(S.F. = .96) with an air gap cut to prevent

saturation.

1cm X 1cm

1mm

6cm

6cm

100t


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EE201.3


Back to the expression for voltage across an inductor:

dt

diLV = .

For nice straight line (constant slope) changes in current, the calculation of voltage across aninductor is straight forward.

Example 4:

Determine the voltage across a 50mH inductor for the current shown in the following graph. Graphthe voltage on a comparable time-scale.

i(A)

T (msec)0 2 4 6 8 10

4

2

0

-2


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EE201.3


V

T (msec)0 2 4 6 8 10


6/12

EE201.3


Up til now weve considered inductors excited by directcurrent (D.C.) in situations where the circuit conditions

have reached steady state. Now lets look at thetransient response of inductors to step changes in current

(still using a D.C. source).

(Note: the derivations and expressions below apply to

a Thevenin Equivalentcircuit specifically as shown at

right. To apply most of these formulas, the circuit

should first be reduced to its Thevenin Equivalent.)

Using Kirchhoffs Voltage Law:

! !

!

" "

#=

$%

&'(

)#

==$%

&'(

)+=

+

#=

$%

&'(

)#

=

$%

&'(

)#

$

%

&'

(

)#

$%

&'(

)#=

#=

+=

===

I

0 0

LR

1

-E/Rb1,a,ln1

bau

dusidesbothintegratenow,

1,-bysideeachmultiplyand,

,dt

bysidesbothmultiplyand,

side...rightthefromL

Ranouttake,

..arranging.-re,

,Vand,Vthatrecalling;

t

dropsrises

dtL

Rdi

R

Ei

baua

dtL

R

R

Ei

di

dtL

R

iR

E

di

iR

Ei

R

E

L

R

dt

di

L

iRE

dt

di

dt

diLiRE

dt

diLiRVV

R

EL

@ t=0

i

+ -

+

-


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EE201.3


[]

( )

!!1I

constant,timethe,as

R

Ldefinenote,1I

sides...bothtoR

E-addand,

,

R

E-bysidesbothmultiply,

get...we,thatrecalland,

e''ofexponentanassideeachusenow,ln

this...reverse,)ln(lnlnrecall,lnln

,ln

)ln(

ln

0

0

!!

"

#

$$

%

&'=

!!

"

#

$$

%

&'='=

'='

=

'

'

==

'=

!!!!

"

#

$$$$

%

&

'

'

'=!"

#$%

&'=!

"

#$%

&''!

"

#$%

&'

'=(

)

*+,

-!"

#$%

&'

'

''

'

'

'!!!!

"

#

$$$$

%

&

'

'

.

.

t

tL

Rt

L

R

tL

R

tL

R

xt

L

R

R

E

R

EI

t

I

eR

E

eR

Ee

R

E

R

E

eR

E

R

EI

e

R

E

R

EI

xeee

tL

R

R

E

R

EI

bab

at

L

R

R

E

R

EI

tL

R

R

Ei


8/12

EE201.3


A similar derivation can be used to determine VL as a function of time:

VL = Ldi

dt, and using the expression for current derived above...

VL = L

dE

R1" e

"t

#

$

%

&&

'

(

))

$

%

&&

'

(

))

dt, taking out the constants...

VL =EL

R

d 1" e

"t

#

$

%

&&

'

(

))

$

%

&&

'

(

))

dt, and d(1) = 0, d e

"t

#

$

%

&&

'

(

))=

"1

#

e

"t

# , andL

RTh= #,

VL = E# 0""1

#

e

"t

#

$

%

&&

'

(

))

$

%

&&

'

(

)),

VL = Ee

"t

# !!

It can also be shown thatfor circuits under D.C. excitation, the decaying transientcan be describedby:

iL = ISteady"State e

"t

#

$

%

&&

'

(

)), where # again = L/RTh,

and

vL = "I0RTh e

"t

#

$

%

&&

'

(

)), where I0 is the steady state current,

and RTh is the effective resistance "seen" by the inductor.

Note: In such exponentially rising/decaying circuits, the varying quantity (i orv) is generallyassumed to have reached steady state after 5 time constants have elapsed (i.e. 99.3% of ultimate

value).


9/12

EE201.3


Consider the following example:Consider the circuit at right.

a) find vL, i, i1, and i2 at t=0.

b) Find all the above at t=5= .418s.

c) Find iL and vL as a function of timebefore and after the switch is opened.

E=10V L=1H

@ t=0

@ t>510

100

i

i2

i1


10/12

EE201.3


Energy stored in an Inductor:

Consider now the energy stored in an inductor. This can be thought of as the energy or work

required to establish a magnetic field, and since it is not lost in heating (in an ideal inductor at least),it is available to be returned to the system as the magnetic field collapses.

W(work,= energy) = (Power)(Time) = Pdt= VLILdtt1

t2

"0

#

" ,

substituting the expressions previously derived for V and I,

W = E e" t

#$

%&

'

()

E

R1" e

" t#

$

%&

'

()dt

0

*

+ , taking the constants out of the integration,

W =E

2

Re"t

#$

%&

'

()1" e

" t#

$

%&

'

()dt

0

*

+,

-.

/

01=

E2

Re"t

# "e"2t

#$

%&

'

()

0

*

+,

-.

/

01, and integrating the two terms separately,

W =E

2

Re"t

#$

%&

'

()dt" e

"2t#

$

%&

'

()dt

0

*

+0

*

+,

-.

/

01, recalling that

eax=

1

a" e

ax,a=

#1

$

,and#2

$

,

W =E

2

R"#e

"t# " "

#

2e

"2t#

$

%&

'

()

*

+,

-

./0

0

=E

2

R"#e

"t#+

#

2e

"2t#

$

%&

'

()

*

+,

-

./0

0, evaluating the integral,

W =E

2

R"#e

"$# " "

#

2e

"2$#

%

&'

(

)*

%

&

'(

)

*" "#e"0

# " "#

2e

" 2( )0#

%

&

''

(

)

**

%

&

''

(

)

**

+

,

-

-

.

/

0

0

W =E

2

R"#e

"$# " "

#

2e

"2$#

%

&'

(

)*

%

&'

(

)*" "#e

"0# " "

#

2e

" 2( )0#

%

&''

(

)**

%

&''

(

)**

+

,

--

.

/

00

recalling that:a"#

=

1

a#=

1

#= 0

, and (anything)0

= 1,

W =E

2

R"# 0( ) " "

#

20( )

$

%&

'

()

$

%&

'

()" "# 1( ) " "

#

21( )

$

%&

'

()

$

%&

'

()

*

+,

-

./ =

E2

R0 + # "

#

2

*

+,-

./=

E2

R

#

2, and # =

L

R,

W =E

2

2R

L

R=

L

2

E2

R2

, and sinceE

R= I

, then2

2LI

W =


11/12

EE201.3


Consider as an example the inductor and the circuit from the previous example. At steady state,the current in the Inductor was 0.8196A. This means that the energy stored in the inductor would be:

W =I2L

R=

.8196A2( ) iH( )2

= 0.336watt" sec(joules,J)

This energy is discharged as the current decays (i.e. the magnetic field collapses) and in this case

was dissipated as heat in the resistances in the circuit. This concept of energy temporarily stored inthe magnetic field of an inductor is also important in understanding how the Conservation of

Energy law is true in circuits with reactive components like capacitors and inductors. Energysupplied by the source would not equal the amount dissipated in other elements of the circuit without

considering the energy temporarily stored in these devices.

Inductors in Series or Parallel:

The inductance of a combination of inductors in series or parallel can be determined in a manner

very similar to that of resistors.

For inductors in series: LT = L1 + L2 + L3 + ...

For inductors in Parallel: ...LLLL

T

+++=

321

1111

Example:Find the total equivalent inductance of the circuit show below:


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EE201.3


Mutual Inductance:

The opposition to changes in current we have considered previously have been due to a devices

reaction to magnetic lines of flux produced by a current flowing in its own windings and theresulting induced voltage. This characteristic or parameter was called self-inductance, commonly

referred to a just inductance.

In a similar fashion, a coil of wire can react to the field established as a result of a current flowing inanother coil adjacent to, or at least in the vicinity of, the first (and vice-versa). This reaction isreferred to as mutual inductance M, and is measured in the same units as self inductance, the

Henry(H).

Mutual inductance is defined as:

M21 =N2"21

I1

where M21 is the mutual inductance of coil 2 with respect to coil 1; N2 is the number of turns in coil

2;21 is the flux produce by coil 1 that links coil 2; and I1 is the current in coil 1. Similarly, M12 is

the mutual inductance of coi1 with respect to coil 2:

M12 =N1"12

I2 The voltage induced in coil 2 as a result of flux created by coil 1:

V2 = "N1d#21

dt= "M21

dI1

dt,

and similarly for the voltage induced in coil 1 as a result of coil 2 is given by:

V1 = "N2d#

12dt

= "M12dI

2dt

,.

It can be shown that M12 = M21, and can be called simply M. Using this simplification, the inducedvoltages are:

V1 = "MdI2

dt, and V2 = "M

dI1

dt

If the coefficient of coupling between the two coils is 100% (i.e. 12 = 21), we can also write a

simplified expression for the mutual inductance, M:

M=N2"

I1

=

N1"

I2

While mutual inductance can be a nuisance in some circumstances (e.g. stray interference), it isexploited in devices called transformers, where the mutual coupling between two coils is, by design,

virtually 100%. This is usually accomplished by winding both coils on one continuous core so thatallthe flux produced by one coil links the other. By manipulating the above expressions, it can also

be shown that under these circumstances, the ratio of the induced or output voltage to the inputvoltage is the same as the turns ratio; a very useful result for electrical engineers.

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