Transcript W A T K I N S - J O H N S O N C O M P A N Y Semiconductor

Engineering 43
Capacitors &
Inductors
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitance & Inductance
 Introduce Two Energy STORING Devices
• Capacitors
• Inductors
 Outline
• Capacitors
– Store energy in their ELECTRIC field (electroStatic energy)
– Model as circuit element
• Inductors
– Store energy in their MAGNETIC field (electroMagnetic energy)
– Model as circuit element
• Capacitor And Inductor Combinations
– Series/Parallel Combinations Of Elements
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
The Capacitor
 First of the EnergyStorage Devices
 Basic Physical Model
 Circuit Representation
• Note use of the PASSIVE
SIGN Convention
 Details of Physical
Operation Described in
PHYS4B & ENGR45
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitance Defined
 Consider the Basic
Physical Model
 Where
• A The Horizontal
Plate-Area, m2
• d  The Vertical Plate
Separation Distance, m
• 0  “Permittivity” of Free
Space; i.e., a vacuum
– A Physical CONSTANT
– Value = 8.85x10-12
Farad/m
 The Capacitance, C,
of the Parallel-Plate
0A
Structure
C 
w/o Dielectric
d
Engineering-43: Engineering Circuit Analysis
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 Then What are the
UNITS of Capacitance, C
 Typical Cap Values →
“micro” or “nano”
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Circuit Operation
 Recall the Circuit
Representation
 LINEAR Caps Follow the
Capacitance Law; in DC
Q  CV c
 Where
 The Basic CircuitCapacitance Equation
q t   Cv c t 
Engineering-43: Engineering Circuit Analysis
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• Q  The CHARGE
STORED in the Cap,
Coulombs
• C  Capacitance, Farad
• Vc  DC-Voltage Across
the Capacitor
 Discern the Base Units
for Capacitance
Farad 
Coulomb
Volt
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
“Feel” for Capacitance
 Pick a Cap, Say 12 µF
 Recall Capacitor Law
Q  CV c
 Solving for Vc
Vc 
Q
C

15x10
12x10
6
3
Coul
Coul/Volt
V c  1250 V! !!
 Now Assume That The
Cap is Charged to hold
15 mC
• Find V c
Engineering-43: Engineering Circuit Analysis
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 Caps can RETAIN
Charge for a Long Time
after Removing the
Charging Voltage
• Caps can Be
DANGEROUS!
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Forms of the Capacitor Law
 The time-Invariant
Cap Law
 i  y dy
t
Q  CV c
 C
vC t 
dz
vC    

 If vC at − = 0, then the
traditional statement of
 Leads to DIFFERENTIAL
the Integral Law
Cap Law
i t  
dq t 
C
dv C t 
dt
dt
v C t  
1
C
 i  y dy
t

 The Differential Suggests  If at t0, vC = vC(t0) (a
SEPARATING Variables
KNOWN value), then the
Integral Law becomes
i t dt  Cdv C t 
 Leads to The
INTEGRAL form of the
Capacitance Law
Engineering-43: Engineering Circuit Analysis
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v C t  
1
C
1
 i  y dy  C  i  y dy
t0

v C t   v C t 0  
t
t0
1
C
 i  y dy
t
t0
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Integral Law
 Express the VOLTAGE
Across the Cap Using
the INTEGRAL Law
v C t  
q t 
C

1
C

t

iC  y dy
 If i(t) has NO Gaps in
its i(t) curve then
lim v C t   t   lim
t  0
t  0
1
C

t  t

i  y dy
 Even if i(y) has
VERTICAL Jumps:
lim v C t   t   v C t 
t  0
Engineering-43: Engineering Circuit Analysis
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 Thus a Major
Mathematical Implication
of the Integral law
 The Voltage Across a
Capacitor MUST be
Continuous
 An Alternative View
• The Differential Reln
i C t   C
dv C t 
dt
 If vC is NOT Continous
then dvC/dt → , and So
iC → . This is NOT
PHYSICALLYBrucepossible
Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Differential Law
 Express the CURRENT
“Thru” the Cap Using the
Differential Law
i C t  
dq t 
C
dv C t 
dt
dt
 If vC = Constant Then
iC  0
 This is the DC
Steady-State Behavior
for a Capacitor
Engineering-43: Engineering Circuit Analysis
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 Thus a Major
Mathematical Implication
of the Differential Law
 A Cap with CONSTANT
Voltage Across it
Behaves as an OPEN
Circuit
 Cap Current
• Charges do NOT flow
THRU a Cap
– Charge ENTER or EXITS
The Cap in Response to
Voltage CHANGES
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Current
 Charges do NOT flow THRU a Cap
 Charge ENTER or EXITS The Capacitor
in Response to the Voltage Across it
• That is, the Voltage-Change DISPLACES
the Charge Stored in The Cap
– This displaced Charge is, to the
REST OF THE CKT, Indistinguishable
from conduction (Resistor) Current
 Thus The Capacitor Current is Called
the “Displacement” Current
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Summary
 The Circuit Symbol
iC

 From Calculus, Recall an
Integral Property
 Note The Passive
Sign Convention
vC

 Compare Ohm’s Law
and Capactitance Law
Cap
Ohm
dv c
iC ( t )  C
vC (t ) 
1
C
(t )
dt
iR 
t
 iC ( x ) dx
R
vR
v R  Ri R

Engineering-43: Engineering Circuit Analysis
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1
t0
t

t0
t
 f  x dx


 f  x dx   f  x dx
 Now Recall the Long Form
of the Integral Relation
vC (t ) 
1
C
t0
i
C
( x ) dx 
1
C

t
i
C
( x ) dx
t0
 The DEFINITE Integral is
just a no.; call it vC(t0) so
vC (t )  vC (t 0 ) 
1
C
t
i
C
( x ) dx
t0
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Summary cont
 Consider Finally the Differential Reln
i C t   C
dv C t 
dt
 Some Implications
• For small Displacement Current dvC/dt is
small; i.e, vC changes only a little
• Obtaining Large iC requires HUGE Voltage
Gradients if C is small
 Conclusion: A Cap RESISTS
CHANGES in VOLTAGE ACROSS It
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
iC Defined by
Differential
C  5F
i ( t )  0 elsewhere
i (t )  C
dv
(t )
dt
 The fact that the Cap
Current is defined
through a
DIFFERENTIAL has
important
implications...
 Consider the Example
at Left
 60 mA
i  5  10
6
[F ]
24
6  10
3
V 
 20 mA
 s 
 Using the 1st Derivative
(slopes) to find i(t)
• Shows vC(t)
– C = 5 µF
• Find iC(t)
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Energy Storage
iC
 UNlike an I-src or
V-src a Cap Does
NOT Produce Energy
 For a Cap
 A Cap is a PASSIVE
Device that Can
STORE Energy
 Recall from Chp.1 The
Relation for POWER
 Recall also
dv C
iC ( t )  C
(t )
dt
 Subbing into Pwr Reln
p  vi

vC
p C t   v C t   iC t 
p C ( t )  Cv C ( t )

dv C
dt
 By the Derivative CHAIN
 Then the
d
d
RULE
INSTANTANEOUS Power

p t   v t   i t 
Engineering-43: Engineering Circuit Analysis
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d 1 2
 dt
p C (t )  C
 vC (t ) 
dt  2

Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
dv C

dv C
dt
Capacitor Energy Storage cont
 Again From Chp.1
Recall that Energy
(or Work) is the
time integral of Power
• Mathematically
p
C
( x ) dx
t1
 Comment on the Bounds
• If the Lower Bound is −
we talk about “energy
stored at time t2”
• If the Bounds are −  to
+ then we talk about the
“total energy stored”
Engineering-43: Engineering Circuit Analysis
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1
w C ( t1 , t 2 ) 
2
Cv ( t 2 ) 
2
C
1
2
Cv C ( t1 )
2
 Recall also
t2
w C ( t1 , t 2 ) 
 Integrating the “Chain
Rule” Relation
vC (t ) 
1
C
t
i
C
( x ) dx 
q C (t )
C

 Subbing into Pwr Reln
p C (t ) 
v C t 
1
C
q C (t )
dq C
(t )
dt
iC t 
 Again by Chain Rule d
1 d 1 2

p C (t ) 
q
(
t
)


c
C dt  2

Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
dt

d
dq C

dq C
dt
Capacitor Energy Storage cont.2
 Then Energy in
Terms of Capacitor
Stored-Charge
w C ( t1 , t 2 ) 
1
C
q (t 2 ) 
2
C
1
C
 Short Example
VC(t)
C = 5 µF
2
C
q ( t1 )
 The Total Energy Stored
during t = 0-6 ms
w C ( 0 ,6 ) 
w C ( 0 ,6 ) 
1
1
Cv ( 6 ) 
2
C
2
5 * 10
1
2
6
2
Cv C ( 0 )
2
2
[ F ] * ( 24 ) [V ]
2
 wC Units?
FV 
2
Coul
 V  Coul  V  J
2
V
 Charge Stored at 3 mS
q C ( 3 )  Cv C ( 3 )
q C ( 3 )  5 * 10
Engineering-43: Engineering Circuit Analysis
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6
[ F ] * 12 [V ]  60  C
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Some Questions About Example
 For t > 8 mS, What is
the Total Stored
CHARGED?
vC(t)
C = 5 µF
q C ( t  8 mS )  0
 For t > 8 mS, What is
the Total Stored
ENERGY?
w C ( t  8 mS )  0
w C (  , t 2 ) 
w C (  , t 2 ) 
1
2
Cv C t  9  
1
C
2
q C t 2  11  
2
1
 5F  0
DIScharging
Current
2
2
1
5F
Engineering-43: Engineering Circuit Analysis
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CHARGING
Current
0
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Summary: Q, V, I, P, W
 Consider A Cap
Driven by A
SINUSOIDAL V-Src
 Charge stored at a Given
Time q C ( t )  Cv C ( t )
q C 1 120  8 . 33 mS   2 * 10
6
[ C ] * sin(  )[ V ]  0
 Current “thru” the Cap

v (t )
i(t)
C  2 F
iC  C
dv C
(t )
dt

iC 1 120  8 . 33 mS   2 * 10
v ( t )  130 sin (120  t )
iC 1 120  8 . 33 mS    98
6
* 130 * 120  cos(  )
mA
 Energy stored at a given
time w ( t )  1 Cv 2 ( t )
 Find All typical Quantities
• Note
– 120 = 60∙(2) → 60 Hz
C
2
C
 1  1
6
2
2  
wC 
  2 * 10 [ F ] * 130 sin  
 240  2
2
w C  16.9 mJ
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitor Summary cont.
 Consider A Cap
Driven by A
SINUSOIDAL V-Src

v (t )
i(t)
C  2 F

v ( t )  130 sin (120  t )
 Electric power absorbed
by Cap at a given time
p C ( t )  v C ( t ) iC ( t )
Engineering-43: Engineering Circuit Analysis
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 At 135° = (3/4)
iC 1 160   2 * 10
6
* 130 * 120  cos( 0 . 75  )
iC 1 160    69 . 3 mA
v C 1 160   130 sin( 0 . 75  )  91 . 9V
p C 1 160   91 . 9V    69 . 3 mA    6371 mW
 The Cap is SUPPLYING
Power at At 135° = (3/4)
= 6.25 mS
• That is, The Cap is
RELEASING (previously)
STORED Energy at Rate
of 6.371 J/s
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
WhiteBoard Work
 Let’s Work this Problem
• The VOLTAGE across a
0.1-F capacitor is given
by the waveform in the
Figure Below. Find the
WaveForm Eqn for the
Capacitor CURRENT
v c(t)(V)
-12
0
1
2
3
4
5
t(s)
-12
Figure P5.14
ANOTHER PROB
0.5 μF
iC t   2 A  cos 50000 t 
+ vC(t) -
Engineering-43: Engineering Circuit Analysis
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See ENGR-43_Lec-061_Capacitors_WhtBd.ppt
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
The Inductor
 Second of the Energy-Storage Devices
 Basic Physical Model:
Ckt Symbol
 Details of Physical
Operation Described in
PHYS 4B or ENGR45
Engineering-43: Engineering Circuit Analysis
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 Note the Use of the
PASSIVE Sign
Convention
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Physical Inductor
 Inductors are Typically Fabricated by Winding Around
a Magnetic (e.g., Iron) Core a LOW Resistance Wire
• Applying to the Terminals a TIME VARYING Current Results
in a “Back EMF” voltage at the connection terminals
 Some Real
Inductors
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Inductance Defined
 From Physics, recall
that a time varying
magnetic flux, ,
Induces a voltage
Thru the Induction
Law
d
vL 
dt
 For a Linear Inductor The
Flux Is Proportional To
The Current Thru it
di L
  Li L  v L  L
dt
Engineering-43: Engineering Circuit Analysis
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 Where the Constant of
Proportionality, L, is called
the INDUCTANCE
 L is Measured in
Units of “Henrys”, H
• 1H = 1 V•s/Amp
 Inductors STORE
electromagnetic energy
 They May Supply Stored
Energy Back To The
Circuit, But They
CANNOT CREATE
Energy
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Inductance Sign Convention
 Inductors Cannot
Create Energy; They
are PASSIVE Devices
 All Passive Devices
Must Obey the
Passive Sign
Convention
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Inductor Circuit Operation
 Recall the Circuit
Representation
 Separating the Variables
and Integrating Yields
the INTEGRAL form
iL (t ) 
 Previously Defined the
Differential Form of the
Induction Law
di L
vL  L
dt
Engineering-43: Engineering Circuit Analysis
25
t
1
L
v
L
( x ) dx

 In a development Similar
to that used with caps,
Integrate − to t0 for an
Alternative integral Law
iL (t )  iL (t 0 ) 
1
t
v

L
L
( x ) dx ; t  t 0
t0
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Inductor Model Implications
 From the Differential
Law Observe That if iL
is not Continuous,
diL/dt → , and vL
must also → 
 This is NOT physically
Possible
 Thus iL must be
continuous
vL   
di L

dt
Engineering-43: Engineering Circuit Analysis
26
 Consider Now the
Alternative Integral law
iL (t )  iL (t 0 ) 
1
t
v

L
L
( x ) dx ; t  t 0
t0
 If iL is constant, say iL(t0),
then The Integral MUST
be ZERO, and hence vL
MUST be ZERO
• This is DC Steady-State
Inductor Behavior
– vL = 0 at DC
– i.e; the Inductor looks like
a SHORT CIRCUIT
to DC Potentials
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Inductor: Power and Energy
 From the Definition of
Instantaneous Power
p L (t )  v L (t )  iL (t )
 Subbing for the Voltage
by the Differential Law
p L (t )  L
di L
dt
(t )iL (t )
 Again By the Chain Rule
for Math Differentiation
d
dt

d

di
d 1 2 
p L t  
 Li L ( t ) 
dt  2

di dt
Engineering-43: Engineering Circuit Analysis
27
 Time Integrate Power to
Find the Energy (Work)
t2
w L ( t1 , t 2 ) 

t1
d 1 2

 Li L ( x )  dx
dx  2

 Units Analysis
• J = H x A2
 Energy Stored on Time
Interval
w ( t1 , t 2 ) 
1
2
Li L ( t 2 ) 
2
1
2
2
Li L ( t1 )
• Energy Stored on an Interval
Can be POSITIVE
or NEGATIVE
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Inductor: P & W cont.
 In the Interval Energy
Eqn Let at time t1
1 2
LiL (t1 )  0
2
 Then To Arrive At The
Stored Energy at a
later given time, t
1 2
w(t )  LiL (t )
2
 Thus Observe That the Stored Energy is
ALWAYS Positive In ABSOLUTE Terms as iL is
SQUARED
• ABSOLUTE-POSITIVE-ONLY Energy-Storage is
Characteristic of a PASSIVE ELEMENT
Engineering-43: Engineering Circuit Analysis
28
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Example
 Given The iL Current
WaveForm, Find vL for
L = 10 mH
 The Derivative of a Line
is its SLOPE, m
 Then the Slopes
 10 ( A / s ) 0  t  2 ms
di

   10 ( A / s ) 2  t  4 ms
dt

0 elsewhere

20  10
3
A
m 

10
3
 s 
2  10 s
A
m   10
A
 s 
 And the vL Voltage

( t )  10 ( A / s ) 
3
  v L ( t )  100  10 V  100 mV
dt
3
L  10  10 H 
di
 The Differential Reln
di L
vL  L
dt
Engineering-43: Engineering Circuit Analysis
29
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Example  Power & Energy
 The Energy Stored
between 2 & 4 mS
 The Value Is Negative
Because The Inductor
SUPPLIES Energy
PREVIOUSLY STORED
with a Magnitude of 2 μJ
 The Energy Eqn
w ( 2,4 ) 
1
Li ( 4 ) 
2
2
L
1
2
2
Li L ( 2 )
 Running the No.s

w ( 2 , 4 )  0   0 . 5 * 10 * 10
w ( 2 , 4 )   2 .0
( 20 * 10
3
)
2

µJ
Engineering-43: Engineering Circuit Analysis
30
3
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Numerical Example
 Given The Voltage Wave
Form Across L , Find iL if
• L = 0.1 H
• i(0) = 2A
 The PieceWise Function
t
v (t )  2 
 v ( x ) dx
 2t; 0  t  2
0
L  0 . 1 H  i ( t )  2  20 t ; 0  t  2 s
v ( t )  0 ; t  2  i ( t )  i ( 2 s ); t  2 s
v (V )
 A Line Followed by A
Constant; Plotting
2
2
t (s)
42
i( A)
 The Integral Reln
i (t )  i (0 ) 
1
t
v ( x ) dx

L
0
Engineering-43: Engineering Circuit Analysis
31
2
2
t (s)
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Numerical Example - Energy
 The Current
Characteristic
42
 The Initial Stored Energy
w ( 0 )  0 . 5 * 0 . 1[ H ]( 2 A )  0 . 2 J
2
 The “Total Stored Energy”
i( A)
w (  )  0 . 5 * 0 . 1[ H ] * ( 42 A )  88 . 2 J
2
 Energy Stored between 0-2
2
2
t (s)
w ( t1 , t 2 ) 
2
Li ( t 2 ) 
2
L
1
2
2
Li L ( 2 ) 
2
1
2
2
Li L ( 0 )
2
w ( 0 , 2 )  88 J
2
L
Li ( t1 )
• Energy Stored on Interval
Can be POS or NEG
Engineering-43: Engineering Circuit Analysis
32
1
w ( 0 , 2 )  0 . 5 * 0 . 1 * ( 42 )  0 . 5 * 0 . 1 * ( 2 )
 The Energy Eqn
1
w ( 2 ,0 ) 
 → Consistent with Previous
Calculation
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
2
Example
w L (t )
 Find The Voltage Across
And The Energy Stored
(As Function Of Time)
v (t )
 For The Energy Stored
Engineering-43: Engineering Circuit Analysis
33
 Notice That the
ABSOLUTE Energy
Stored At Any Given
Time Is Non Negative
(by sin2)
• i.e., The Inductor Is A
PASSIVE Element
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
L = 5 mH; Find the Voltage
m 
20 mA
m 
1ms
v  100 mV
10  20
v (t )  L
dt
2 1
v   50 mV
v  0V
3
Engineering-43: Engineering Circuit Analysis
34
(t )
( A / s)
m 0
v  5  10
di
m 
0  10
43
( A / s)
v   50 mV
( H )  20 ( A / s ); 0  t  1ms  100 mV
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Example  Total Energy Stored
 The Ckt Below is in
the DC-SteadyState
 Shorting-Caps;
Opening-Inds
 Find the Total
Energy Stored by
the 2-Caps & 2-Inds
 KCL at node-A
 Recall that at DC
 Solving for VA
• Cap → Short-Ckt
• Ind → Open-Ckt
Engineering-43: Engineering Circuit Analysis
35
I
out
 0  3 A 
VA 
81
VA
36
V  16 . 2 V
5
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt

VA  6
6
Example  Total Energy Stored
 Continue Analysis of
Shorted/Opened ckt
I L1  I L 2  3 A  1 .8 A  3 A
 I L1   1 .2 A
 VC2 by Ohm
V C 2  I L 2  6    1 . 8 A 6  
 Using Ohm and
VA = 16.2V
I L2 
16 . 2 V
3  6 
V  1 .8 A
 By KCL at Node-A
Engineering-43: Engineering Circuit Analysis
36
 V C 2  10 . 8 V
 VC1 by Ohm & KVL
V C 1  9 V  I L 1  6  
V C 1  9 V    1.2A    6  
V C 1  9 V  7 .2 V
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Example  Total Energy Stored
 Have all I’s & V’s:
−1.2 A
1.8 A
10.8 V
 Then the E- Storage
Calculations
16.2 V
 Next using the
E-Storage Eqns
wC
wL

1
2

1
2
2
CV C
w tot 
k
 13 . 46 mJ
2
LI L
Engineering-43: Engineering Circuit Analysis
37
w
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Caps & Inds  Ideal vs. Real
 A Real CAP has
Parasitic
Resistances &
Inductance:
 A Real IND has
Parasitic
Resistances &
Capacitance:
Generally
SMALL
Effect
Engineering-43: Engineering Circuit Analysis
38
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Ideal vs Real
 Practical Elements “Leak”
Thru Unwanted Resistance
 Ideal C & L

i (t )
i (t )
i (t )
i (t )



v (t )
v (t )


v (t )
v (t )


i (t )  C
dv
dt
(t )
v (t )  L
di
(t )
dt
Engineering-43: Engineering Circuit Analysis
39
i (t ) 
v (t )
R leak
C
dv
dt
(t )
v ( t )  R leak i ( t )  L
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
di
dt
(t )
Capacitors in Series
 By KVL for 1-LOOP ckt
 If the vi(t0) = 0, Then
Discern the Equivalent
Series Capacitance, CS
 CAPS in SERIES
Combine as Resistors
in PARALLEL
Engineering-43: Engineering Circuit Analysis
40
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Example
 Find
 Spot Caps in Series
• Equivalent C
• Initial Voltage
1
CS
1 F

1
2

1
3

1
6

3 2 1
6
 CS  1
 Or Can Reduce
Two at a Time
 Use KVL for Initial Voltage
v t 0   2V  4V  1V   3V
2 F
 This is the Algebraic Sum
of the Initial Voltages
• Polarity is Set by the
Reference Directions
noted in the Diagram
Engineering-43: Engineering Circuit Analysis
41
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Numerical Example
 Two charged
Capacitors Are
Connected As Shown.
 Find the Unknown
Capacitance

8V
+
-
12 V
30  F


4V
C

 Recognize SINGLE Loop
Ckt → Same Current in
Both Caps
• Thus Both Caps Accumulate
the SAME Charge
Q  ( 30  F )( 8V )  240  C
 And Find VC by KVL
• VC = 12V-8V = 4V
 Finally Find C by
Charge Eqn
C 
QC
VC
Engineering-43: Engineering Circuit Analysis
42

240 μC
 60 μC
4V
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Capacitors in Parallel
 By KCL for 1-NodePair ckt  Thus The Equivalent
Parallel Capacitance
 CAPS in Parallel
Combine as Resistors
in SERIES
Engineering-43: Engineering Circuit Analysis
43
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Complex Example → Find Ceq
6 F
3 F
C eq 
C eq 
2 F
Engineering-43: Engineering Circuit Analysis
44
4 F
3
2
4 F
F
12  F
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
3 F
Inductors in Series
 By KVL For 1-LOOP ckt
 Thus
 Use The Inductance Law
v k ( t )  Lk
di
(t )
dt
v (t )  LS
di
(t )
dt
 INDUCTORS in Series
add as Resistors
in SERIES
Engineering-43: Engineering Circuit Analysis
45
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Inductors in Parallel
 By KCL for 1-NodePair ckt  And
N
i (t0 ) 
 Thus
 i j (t0 )
j 1
 INDUCTORS in Parallel
combine as Resistors
in PARALLEL
Engineering-43: Engineering Circuit Analysis
46
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Example – Find: Leq, i0
4 mH
2 mH
i (t0 )  3 A  6 A  2 A  1 A
 Series↔Parallel Summary
• INDUCTORS Combine as do RESISTORS
• CAPACITORS Combine as do CONDUCTORS
Engineering-43: Engineering Circuit Analysis
47
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Inductor Ladder Network
 Find Leq for Li = 4 mH
 Place Nodes In Chosen
Locations
a
 Connect Between Nodes
6 mH
a 4 mH
d
4 mH
2 mH
c
L eq
c
d
2 mH
2 mH
b
2 mH
 When in Doubt, ReDraw
• Select Nodes
Engineering-43: Engineering Circuit Analysis
48
b
L eq  ( 6 mH || 4 mH )  2 mH  4 . 4 mH
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Find Leq for Li = 6mH
a
 ReDraw The Ckt for
Enhanced Clarity
a
6 || 6 || 6
2 mH
b
L eq
b
6 mH
c
 Nodes Can have
Complex Shapes
• The Electrical Diagram
Does NOT have to Follow
the Physical Layout
Engineering-43: Engineering Circuit Analysis
49
6 mH
c
6 mH
 It’s Simple Now
L eq  6  ( 6  2 ) || 6   6 
L eq 
66
48
14
mH
7
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
6
24
7
mH
C&L Summary
Engineering-43: Engineering Circuit Analysis
50
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
WhiteBoard Work
 Let’s Work This Problem
L  50  H
i t   0
t0
i t   2 te
t0
4t
 Find: v(t), tmax for imax, tmin for vmin
Engineering-43: Engineering Circuit Analysis
51
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
% Bruce Mayer, PE * 14Mar10
% ENGR43 * Inductor_Lec_WhtBd_Prob_1003.m
%
t = linspace(0, 1);
%
iLn = 2*t;
iexp = exp(-4*t);
plot(t, iLn, t, iexp), grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
%
i = 2*t.*exp(-4*t);
plot(t, iLn, t, iexp, t, i),grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
plot(t, i), grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
vL_uV =100*exp(-4*t).*(1-4*t);
plot(t, vL_uV)
i_mA = i*1000;
plot(t, vL_uV, t, i_mA), grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
%
% find mins with fminbnd
[t_vLmin vLmin] = fminbnd(@(t) 100*exp(-4*t).*(1-4*t), 0,1)
% must "turn over" current to create a minimum
i_mA_TO = -i*1000;
plot(t, i_mA_TO), grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
[t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1);
t_iLmin
iLmin = -iLmin_TO
plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid
Engineering-43: Engineering Circuit Analysis
52
By
MATLAB
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
% Bruce Mayer, PE * 14Mar10
% ENGR43 * Inductor_Lec_WhtBd_Prob_1003.m
%
t = linspace(0, 1);
%
iLn = 2*t;
iexp = exp(-4*t);
plot(t, iLn, t, iexp), grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
%
i = 2*t.*exp(-4*t);
plot(t, iLn, t, iexp, t, i),grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
plot(t, i), grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
vL_uV =100*exp(-4*t).*(1-4*t);
plot(t, vL_uV)
i_mA = i*1000;
plot(t, vL_uV, t, i_mA), grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
%
% find mins with fminbnd
[t_vLmin vLmin] = fminbnd(@(t) 100*exp(-4*t).*(1-4*t), 0,1)
% must "turn over" current to create a minimum
i_mA_TO = -i*1000;
plot(t, i_mA_TO), grid
disp('Showing Plot - Hit ANY KEY to Continue')
pause
[t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1);
t_iLmin
iLmin = -iLmin_TO
plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid
By
MATLAB
Engineering-43: Engineering Circuit Analysis
53
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
200
150
100
50
0
-50
0
0.1
0.2
0.3
Engineering-43: Engineering Circuit Analysis
54
0.4
0.5
0.6
0.7
0.8
0.9
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
1
Irw in P ro b 5 .2 6 : I(t) & v(t)
200
i(t) (m A )
175
150
C u rrent (m A )
125
100
75
50
25
0
-0 .1
0 .0
0 .1
0 .2
file = E n g r4 4 _ p ro b _ 5 -2 6 _ F a ll0 3 ..xls
Engineering-43: Engineering Circuit Analysis
55
0 .3
0 .4
0 .5
0 .6
0 .7
0 .8
0 .9
tim e (s )
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
1 .0
Irw in P ro b 5 .2 6 : I(t) & v(t)
200
120
i(t) (m A )
100
150
80
125
60
100
40
75
20
50
0
25
-2 0
0
-4 0
-0 .1
0 .0
0 .1
0 .2
file = E n g r4 4 _ p ro b _ 5 -2 6 _ F a ll0 3 .xls
Engineering-43: Engineering Circuit Analysis
56
0 .3
0 .4
0 .5
0 .6
0 .7
0 .8
0 .9
tim e (s )
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
1 .0
E lectrical P otential (µ V )
C u rrent (m A )
v(t) (µ V )
175
L = 10 mH; Find the Voltage
v   100 mV
v (t )  L
di
dt
v  10  10
3
[H ]
20  10
2  10
3
3
Engineering-43: Engineering Circuit Analysis
57
 A
 s 
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
(t )
Engineering 43
Appendix
Complex Cap
Example
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
58
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Numerical Example
 Given iC, Find vC
 The Piecewise Fcn for iC
C= 4µF
vC(0) = 0
>
v ( t )   2 t  8  10
v (t )  v (0) 
1
C
3
2  t  4 ms
[V ]
 Integrating & Graphing
t
 i ( x ) dx ; t  0
0
Linear
0t2
v (t )  v (2) 
1
C
t
 i ( x ) dx ;
t 2
2
Engineering-43: Engineering Circuit Analysis
59
Parabolic
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Power Example
 From Before the vC
 For The Previous
Conditions,
Find The POWER
Characteristic
• C = 4 µF
• iC by Piecewise curve
 Using the Pwr Reln
i ( t )  8  10
3
t
p ( t )  8 t , 0  t  2 ms
3
2  t  4 ms
p ( t )  0 , elsewhere
Engineering-43: Engineering Circuit Analysis
60
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Power Example cont
 Finally the Power
Characteristic
 Absorbing or
Supplying Power?
 During the
CHARGING Period
of 0-2 mS, the Cap
ABSORBS Power
 During DIScharge
the Cap
SUPPLIES power
• But only until the
stored charge is fully
depleted
Engineering-43: Engineering Circuit Analysis
61
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Energy Example
 For The Previous
Conditions, Find The
ENERGY Characteristic
 Now The Work
(or Energy) is the
Time Integral of Power
• C = 4 µF
• pC by Piecewise curve
 For 0  t  2 mS
p (t )  8t
Engineering-43: Engineering Circuit Analysis
62
3
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Energy Example cont
 For 2 < t  4 mS
2t
4
8   t  64 n  t  128 p
2
 Taking The Time Integral
and adding w(2 mS)
 Then the Energy
Characteristic
Engineering-43: Engineering Circuit Analysis
63
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
SAMPLE PROBLEM
If the current is known ...
Current through capacitor
iC

vC
C
 e  0 .5 t ; t  0
iC ( t )  
[ mA ]
 0; t  0

C  2F
Voltage at a given time t
vC (t ) 
1
C
t
i
v C ( 0 )  0[V ]
( x ) dx
C

Voltage at a given time t when voltage at time to<t is also known v C ( t )  v C ( t 0 ) 
vC ( 2 )  vC (0 ) 
1
C
2
e
0
 0 .5 x
dx 
2
1
2 * 10
6
Voltage as a function of time v C ( t ) 
C
q C ( 2 )  2 * 0 . 6321
t
v C (t )  0; t  0
 iC ( x ) dx

Energy stored in capacitor at a given time w ( t ) 
1
2
64
i
C
( x ) dx
t0
wT 
C
vC (t )  vC (0 ) 
1
C
1
2
2
Cv C ( t ) J
Cv C (  )
2
t
e
 0 .5 x
dx
0
 0 .5 t
10 (1  e
); t  0
vC (t )  
V
0; t  0

6
Electric power supplied to capacitor p C ( t )  v C ( t ) iC ( t ) W
“Total”Engineering-43:
energy stored
in the
capacitor
Engineering
Circuit
Analysis
C
1
1
 1  0 .5 x 
1
6

e

 0 .5
 2 * 10  6 0 . 5 1  e   0 . 6321 * 10 V

0
Charge at a given time q C ( t )  Cv C ( t )
1
t
1
1
PE 6
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[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
2
J
SAMPLE PROBLEM
Given current and capacitance
Compute voltage as a function of time
At minus infinity everything is zero. Since
current is zero for t<0 we have
0  t  5 m sec 
iC ( t ) 
VC (0)  0  VC (t ) 
5
10
3 * 10
4 * 10
15  A
3 t
6
t 3
5 ms
V C ( 5 ms ) 
q C ( t )  CV C ( t )
q ( 5 ms )  4 * 10
6
[F ]*
75 * 10
8
[ mV ]  V C ( t ) 
3
t  3 * 10
3
[ A / s]t
s
3
t [V ]; 0  t  5 * 10
2
3
V C ( 5 ms ) 
t (m sec)
5  t  10 ms  iC ( t )   10 [  A ]
75
A
3
[s]
8
0
In particular
Charge stored at 5ms
6
10
10
[V ]  3 * 10
 xdx
V C ( t )  0; t  0
75 * 10
8
3

3 * 10 * ( 5 * 10
3
)
2
[V ] 
75
8
t
1
4 * 10
6
[ mV ]
8
 (  10 * 10
5 *10
6
)[ A / s ]dx
3
3
[V ]
8
q ( 5 ms )  ( 75 / 2 ) [ nC ]
VC (t ) 
Total energy stored
E 
1
2
75 * 10
8
3

10
4
t  5 * 10  [V ] ; 5 * 10
3
3
 t  10 * 10
3
[s]
2
CV C
Total means at infinity. Hence
2
Before looking into a formal way to describe the current
we will look at additional questions that can be answered.
 25 * 10  3 

6
Bruce Mayer, PE
 Circuit
E T  0 . 5Engineering-43:
* 4 * 10  Engineering
[ J ] AnalysisNow, for a formal way to represent piecewise
functions....

65
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
8


Formal description of a piecewise analytical signal
t0
0;


3 2
0  t  5 ms
t ;

8

V c ( t )   75 10
t  5 ; 5  t  10 [ ms ]

 8
4

25
t  10 [ ms ]
;


8

Engineering-43: Engineering Circuit Analysis
66
[mV ]
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
Find Ceq for Ci = 4 µF
8 F
8 F
4 F
4 8
32
F
12
C eq
8
3
Engineering-43: Engineering Circuit Analysis
67
8
32
8 F
3
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06-1_Capacitors.ppt
8 F