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CAPACITANCE AND INDUCTANCE

Introduces two passive, energy storing devices: Capacitors and Inductors CAPACITORS Store energy in their electric field (electrostatic energy) Model as circuit element INDUCTORS Store energy in their magnetic field Model as circuit element CAPACITOR AND INDUCTOR COMBINATIONS Series/parallel combinations of elements

CAPACITORS

First of the energy storage devices to be discussed

Typical Capacitors Basic parallel-plates capacitor CIRCUIT REPRESENTATION NOTICE USE OF PASSIVE SIGN CONVENTION

C

d A

 Dielectric constant of material in gap PLATE SIZE FOR EQUIVALENT AIR-GAP CAPACITOR 55

F

 8 .

85 1 .

 10  12 016  10  4

A

A

 6 .

3141  10 8

m

2 Normal values of capacitance are small.

Microfarads is common.

For integrated circuits nano or pico farads are not unusual

Basic capacitance law

Q

f

(

V C

)

Linear capacitors obey Coulomb’s law

Q

CV C

C is called the CAPACITANCE of the device and has units of

charge voltage

One Farad(F)is the capacitance of a device that can store one Coulomb of charge at one Volt.

Farad

Coulomb Volt

Linear capacitor circuit representation EXAMPLE Voltage across a capacitor of 2 micro Farads holding 10mC of charge

V C

 1

C Q

 1 2 * 10  6 10 * 10  3  5000

V Capacitance in Farads, charge in Coulombs result in voltage in Volts Capacitors can be dangerous!!!

Capacitors only store and release ELECTROSTATIC energy. They do not “create” The capacitor is a passive element and follows the passive sign convention Linear capacitor circuit representation

i

(

t

) 

C dv

(

t

)

dt

Q C

CV C

Capacitance Law If the voltage varies the charge varies and there is a displacement current One can also express the voltage across in terms of the current … Or one can express the current through in terms of the voltage across

V C

(

t

)  1

C Q

 1

C

t

  

i

C

(

x

)

dx

Integral form of Capacitance law

i

C

dQ

C dV

C

dt dt

Differential form of Capacitance law The mathematical implication of the integral form is ...

V C

(

t

 ) 

V C

(

t

 ); 

t

Voltage across a capacitor MUST be continuous Implications of differential form??

V C

Const

i C

 0 DC or steady state behavior A capacitor in steady state acts as an OPEN CIRCUIT

CAPACITOR AS CIRCUIT ELEMENT

i C

v C

i C v C

(

t

) (

t

) 

C dv c

(

t

)

i R v R

 1

v R Ri R

LEARNING EXAMPLE

C

 5 

F

DETERMINE THE CURRENT

i

(

t

) 

C dv

(

t

)

dt

 60

mA i

 5  10  6 [

F

]  24 6  10  3  

V s i

(

t

)     0 20

mA elsewhere

CAPACITOR AS ENERGY STORAGE DEVICE

i C

Instantaneous power

p C

(

t

) 

v C

(

t

)

i C

(

t

)

W

i

C

(

t

) 

C dv

c v C

(

t

)

dt

v C

(

t

)  1

C

t

 

i C

(

x

)

dx

q C

(

t

)

C

p

C

(

t

) 

Cv

C

(

t

)

dv

c

dt

p C

(

t

) 

C d dt

1 2

v C

2 (

t

)

p C

(

t

)  1

C q C

(

t

)

dq C dt

Energy is the integral of power

w C

(

t

2 ,

t

1 ) 

t

1

t

 2

p C

(

x

)

dx p C

(

t

)  1

C d dt

(

t

) 1 2

q c

2 (

t

) If t1 is minus infinity we talk about “energy stored at time t2.” If both limits are infinity then we talk about the “total energy stored.”

w C

(

t

2 ,

t

1 )  1 2

Cv C

2 (

t

2 )  1 2

Cv C

2 (

t

1 )

w C

(

t

2 ,

t

1 )  1

C q

2

C

(

t

2 )  1

C q

2

C

(

t

1 )

EXAMPLE

C

 5 

F

Energy stored in 0 - 6 msec

w C

( 0 , 6 )  1 2

Cv C

2 ( 6 )  1 2

Cv C

2 ( 0 )

w C

( 0 , 6 )  1 2 5 * 10  6 [

F

] * ( 6 ) 2 [

V

2 ] Charge stored at 3msec

q C

( 3 ) 

Cv C

( 3 )

q C

( 3 )  5 * 10  6 [

F

] * 12 [

V

]  60 

C

C

 4 

F

.

FIND THE VOLTAGE

v

( 0 )  0

v

(

t

) 

v

( 0 )  1

C

0 

t i

(

x

)

dx

;

t

 0

v

(

t

) 

v

( 2 )  1

C

2 

t i

(

x

)

dx

;

t

 2 0 

t

 2

v

(

t

)   2

t

 8  10  3 [

V

] 2 

t

 4

ms

Flux lines may extend beyond inductor creating stray inductance effects

INDUCTORS NOTICE USE OF PASSIVE SIGN CONVENTION

Circuit representation for an inductor A TIME VARYING FLUX CREATES A COUNTER EMF AND CAUSES A VOLTAGE TO APPEAR AT THE TERMINALS OF THE DEVICE

A TIME VARYING MAGNETIC FLUX INDUCES A VOLTAGE 

v

L

d

 Induction law

dt

FOR A LINEAR INDUCTOR THE FLUX IS PROPORTIONAL TO THE CURRENT

Li

L

v L

L di L dt

DIFFERENTIAL FORM OF INDUCTION LAW THE PROPORTIONALITY CONSTANT, L, IS CALLED THE INDUCTANCE OF THE COMPONENT INDUCTANCE IS MEASURED IN UNITS OF henry (H). DIMENSIONALLY HENRY  Volt Amp sec INDUCTORS STORE ELECTROMAGNETIC ENERGY.

THEY MAY SUPPLY STORED ENERGY BACK TO THE CIRCUIT BUT THEY CANNOT CREATE ENERGY.

THEY MUST ABIDE BY THE PASSIVE SIGN CONVENTION Follow passive sign convention

v

L

L di

L

dt

Differential form of induction law

i L

(

t

)  1

L t

  

v L

(

x

)

dx

Integral form of induction law

i L

(

t

) 

i L

(

t

0 )  1

L

0

t t

v L

(

x

)

dx

; A direct consequence of integral form

i L

(

t

 ) 

i L

(

t

 ); 

t t

t

0 Current MUST be continuous A direct consequence of differential form

i

L

Const

.

v

L

 0

Power and Energy stored DC (steady state) behavior

p L

(

t

) 

v L

(

t

)

i L

(

t

)

w L

(

t

2 ,

t

1 ) 

t

1

t

 2

d dt

W 1 2

Li L

2 (

x

) 

dx

p L

(

t

) 

L di L dt

(

t

)

i L

(

t

) 

d dt

J 1 2

Li L

2 (

t

) Current in Amps, Inductance in Henrys yield energy in Joules

w

(

t

2 ,

t

1 )  1 2 2

Li L

(

t

2 )  1 2 2

Li L

(

t

1 )

w L

(

t

)  1 2

Li L

2 (

t

) Energy stored on the interval Can be positive or negative

EXAMPLE FIND THE TOTAL ENERGY STORED IN THE CIRCUIT In steady state inductors act as short circuits and capacitors act as open circuits

W C

1 2

CV C

2

W L

1 2

LI L

2 @

A A

V A

9

V A

6

9

0

V

A

 81 [

V

5 ]

I L

1

3

A

I L

2

I L

1

 

1.2

A V C

1

I L

1

V C

1

16.2

V V C

2

6 6

3

V A

10.8

V I L

2

V A

9

1.8

A

EXAMPLE L=10mH. FIND THE VOLTAGE

v

(

t

) 

L di dt

(

t

) ENERGY STORED BETWEEN 2 AND 4 ms

m

 20  10  3 2  10  3

s A

 10  

A s

 

m

  10  

A s

  THE DERIVATIVE OF A STRAIGHT LINE IS ITS SLOPE

di dt

    10 ( 10 (

A A

0 / /

s

)

s

) 0 2  

t t

 

elsewhere

2

ms

4

ms di dt L

(

t

 )  10 10 (

A

 10  3 /

s H

)   

v

(

t

)  100  10  3

V

 100

mV w

( 4 , 2 )  1 2

Li L

2 ( 4 )  1 2

Li L

2 ( 2 )

w

( 4 , 2 )  0  0 .

5 * 10 * 10  3 ( 20 * 10  3 ) 2 J THE VALUE IS NEGATIVE BECAUSE THE INDUCTOR IS SUPPLYING ENERGY PREVIOUSLY STORED

CAPACITOR SPECIFICATIONS CAPACITANC E RANGE

p F

IN STANDARD VALUES 

C

 50

mF

STANDARD CAPACITOR RATINGS 6 .

3

V

 500

V

STANDARD TOLERANCE  5 %,  10 %,  20 % INDUCTOR SPECIFICATIONS INDUCTANCE RANGES  1

nH

IN STANDARD VALUES 

L

 100

mH

STANDARD INDUCTOR RATINGS 

mA

  1

A

STANDARD TOLERANCE  5 %,  10 %

C

v

i L i

v

IDEAL AND PRACTICAL ELEMENTS 

i

(

t

)

i

(

t

)

i

(

t

)

i

(

t

)  

v

(

t

)  IDEAL ELEMENTS 

v

(

t

) 

i

(

t

) 

C dv

(

t

)

dt v

(

t

) 

L di dt

(

t

)

v

(

t

)

v

(

t

) 

i

( CAPACITOR/INDUCTOR MODELS INCLUDING LEAKAGE RESISTANCE

t

) 

v

(

t

)

R leak

C dv

(

t

)

dt

MODEL FOR “LEAKY” CAPACITOR 

v

(

t

) 

R leak i

(

t

) 

L di dt

(

t

) MODEL FOR “LEAKY” INDUCTORS

SERIES CAPACITORS NOTICE SIMILARITY WITH RESITORS IN PARALLEL

C s

C C

1

C

2 1 

C

2 Series Combination of two capacitors 6 

F

3 

F C S

2 

F

i

(

t

) PARALLEL CAPACITORS

i k

(

t

) 

C k dv

(

t

)

dt C P

F

SERIES INDUCTORS

v k

(

t

) 

L k di dt

(

t

)

L eq

 7

H v

(

t

) 

L S di dt

(

t

)

PARALLEL INDUCTORS

i

(

t

)

i

(

t

0 ) 

j N

  1

i j

(

t

0 ) 4

mH i

(

t

0 )  3

A

 6

A

 2

A

  1

A

INDUCTORS COMBINE LIKE RESISTORS CAPACITORS COMBINE LIKE CONDUCTANCES 2

mH

LEARNING EXAMPLE FLIP CHIP MOUNTING IC WITH WIREBONDS TO THE OUTSIDE GOAL: REDUCE INDUCTANCE IN THE WIRING AND REDUCE THE “GROUND BOUNCE” EFFECT A SIMPLE MODEL CAN BE USED TO DESCRIBE GROUND BOUNCE

MODELING THE GROUND BOUNCE EFFECT

L ball

 0 .

1

nH V GB

(

t

) 

L ball di G dt

(

t

)

m

 40 40  10  3  10  9

A s

IF ALL GATES IN A CHIP ARE CONNECTED TO A SINGLE GROUND THE CURRENT CAN BE QUITE HIGH AND THE BOUNCE MAY BECOME UNACCEPTABLE USE SEVERAL GROUND CONNECTIONS (BALLS) AND ALLOCATE A FRACTION OF THE GATES TO EACH BALL