Download Discussion #10 - Energy Storage

Schedule…
Date
Day
6 Oct
Mon
7 Oct
Tue
8 Oct
Wed
9 Oct
Thu
10 Oct
Fri
11 Oct
Sat
12 Oct
Sun
13 Oct
Mon
Class
No.
10
Title
Energy Storage
Chapters
HW
Due date
3.7, 4.1
Lab
Due date
NO LAB
NO LAB
11
Dynamic Circuits
Recitation
12
4.2 – 4.4
HW 4
Exam 1 Review
LAB 4
14 Oct
ECEN 301
Exam
EXAM 1
Tue
Discussion #10 – Energy Storage
1
Energy
Moro. 7: 48
48 Wherefore, my beloved brethren, pray unto the
Father with all the energy of heart, that ye may be
filled with this love, which he hath bestowed upon all
who are true followers of his Son, Jesus Christ; that ye
may become the sons of God; that when he shall
appear we shall be like him, for we shall see him as he
is; that we may have this hope; that we may be
purified even as he is pure. Amen.
ECEN 301
Discussion #10 – Energy Storage
2
Lecture 10 – Energy Storage
Maximum Power Transfer
Capacitors
Inductors
ECEN 301
Discussion #10 – Energy Storage
3
Maximum Power Transfer
Equivalent circuit representations are important
in power transfer analysis
RT
i
+
vT +
–
v
Load
–
Ideally all power from source is absorbed by the load
But some power will be absorbed by internal circuits
(represented by RT)
ECEN 301
Discussion #10 – Energy Storage
4
Maximum Power Transfer
Efficiently transferring power from source to load
means that internal resistances (RT) must be
minimized
i.e. for a given RL we want RT as small as possible!
RT
vT +
–
iL
RL
For a given RT there is a specific RL that will
maximize power transfer to the load
ECEN 301
Discussion #10 – Energy Storage
5
Maximum Power Transfer
Consider the power (PL) absorbed by the load
RT
vT +
–
ECEN 301
iL
RL
Discussion #10 – Energy Storage
6
Maximum Power Transfer
Consider the power (PL) absorbed by the load
PL  iL RL
2
RT
vT
+
–
BUT
iL
RL
vT
iL 
RL  RT
Therefore
2
vT
PL 
RL
2
RL  RT 
ECEN 301
Discussion #10 – Energy Storage
7
Maximum Power Transfer
The maximum power transfer can be found by
calculating dPL/dRL = 0
dPL vT RL  RT   2vT RL RL  RT 

0
4
dRL
RL  RT 
2
RT
2
2
OR
vT +
–
iL
RL
RL  RT 2  2 RL RL  RT   0
OR
NB: assumes RT is fixed
and RL is variable
ECEN 301
RL  RT
Discussion #10 – Energy Storage
8
Maximum Power Transfer
0.30
Normalized Power (PL/VT2)
0.25
When RL < RT:
PL is lowered since vL goes
down
0.20
0.15
When RL > RT:
0.10
PL is lowered since less
current (iL) flows
0.05
0.00
0.0
1.0
2.0
3.0
4.0
5.0
6.0
Normalized Resistance (RL/RT)
To transfer maximum power to the load, the source and load
resistors must be matched (i.e. RT = RL). Power is attenuated
as RL departs from RT
ECEN 301
Discussion #10 – Energy Storage
9
Maximum Power Transfer
Maximum power theorem: maximum power is
delivered by a source (represented by its Thévenin
equivalent circuit) is attained when the load (RL) is
equal to the Thévenin resistance RT
2
RT
2
vT +
–
iL
RL
vT
PL 
RL
2
RL  RT 
PLMax
vT

4 RL

RL i T
2
4
NB: Substitute (RT = RL) to get PLMax
ECEN 301
Discussion #10 – Energy Storage
10
Maximum Power Transfer
Efficiency of power transfer (η): the ratio of the power
delivered to the load (Pout), to the power supplied by
the source (Pin)
pout

pin
NB: Pout = PL
ECEN 301
Discussion #10 – Energy Storage
11
Maximum Power Transfer
Efficiency of power transfer (η): the ratio of the power
delivered to the load (Pout), to the power supplied by
the source (Pin)
RT
vT +
–
ECEN 301
iL
RL
Discussion #10 – Energy Storage
12
Maximum Power Transfer
Efficiency of power transfer (η): the ratio of the power
delivered to the load (Pout), to the power supplied by
the source (Pin)
Pin  iL vT
RT
vT
+
–
iL
RL
 vT 

 vT 
 RT  RL 
2
vT
PinMax 
2 RL
RL iT

2
2
NB: Substitute (RT = RL) to get PinMax
ECEN 301
Discussion #10 – Energy Storage
13
Maximum Power Transfer
Efficiency of power transfer (η): the ratio of the power
delivered to the load (Pout), to the power supplied by
the source (Pin)
 max
RT
vT +
–
PoutMax

PinMax
2
iL
RL
vT 2 RL

 2
4 RL vT
1

2
ECEN 301
Discussion #10 – Energy Storage
In other words, at
best, only 50%
efficiency can be
achieved
14
Maximum Power Transfer
Example1: Find the maximum power delivered to RL
vs = 18V, R1 = 3Ω, R2 = 6Ω, R3 = 2Ω
R1
vs
+
–
ECEN 301
R3
R2
RL
Discussion #10 – Energy Storage
15
Maximum Power Transfer
Example1: Find the maximum power delivered to RL
vs = 18V, R1 = 3Ω, R2 = 6Ω, R3 = 2Ω
1.
Find Thévenin equivalent circuit
RT
vT +
–
ECEN 301
RL
R2
vT 
vs RT  R3  R1 || R2
R1  R2
 4
 12V
Discussion #10 – Energy Storage
16
Maximum Power Transfer
Example1: Find the maximum power delivered to RL
vs = 18V, R1 = 3Ω, R2 = 6Ω, R3 = 2Ω
1.
2.
3.
RT
vT +
–
RL
Find Thévenin equivalent circuit
Set RL = RT
Calculate PLMax
PL max  0.5 Pin max
2
vT
 0 .5
2 RL
(12) 2
 0 .5
8
 9W
ECEN 301
Discussion #10 – Energy Storage
17
Dynamic Energy Storage Elements
Capacitor and Inductor
ECEN 301
Discussion #10 – Energy Storage
18
Storage Elements
To this point, circuits have consisted of either:
DC Sources (active element, provide energy)
Resistors (passive element, absorb energy)
Some passive (non-source) elements can store energy
Capacitors – store voltage
Inductors – store current
+
C
–
+
C
–
+C–
+
L
–
Capacitor symbols
ECEN 301
Discussion #10 – Energy Storage
+ L –
Inductor symbols
19
Ideal Capacitor
A capacitor (C) is composed of:
2 parallel conducting plates
• cross-sectional area A
Separated distance d by either:
+
• Air
• Dielectric (insulating) material
A
d
C
A
Dielectric (or air)
–
d
C – Capacitance
ε – relative permittivity
(dielectric constant)
ECEN 301
NB: the parallel plates do not
touch – thus under DC current a
capacitor acts like an open circuit
Discussion #10 – Energy Storage
20
Ideal Capacitor
Capacitance: a measure of the ability to store energy in
the form of separated charge (or an electric field)
q  Cv
The charge q+ on one plate
is equal to the charge q– on
the other plate
+ + + + ++
+ + ++ +
– –– –– –– ––
– – – –––
Farad (F): unit of capacitance.
1 farad = 1 coulomb/volt (C/V)
ECEN 301
Discussion #10 – Energy Storage
21
Ideal Capacitor
 No current can flow through a capacitor if the voltage across it
is constant (i.e. connected to a DC source)
 BUT when a DC source is applied to a capacitor, there is an
initial circuit current as the capacitor plates are charged
(‘charging’ the capacitor)
dq(t )
i (t ) 
dt
Charging the capacitor: a current
flows as electrons leave one plate to
accumulate on the other.
This occurs until the voltage across
the capacitor = source voltage
ECEN 301
vs
i
+
–
Discussion #10 – Energy Storage
++ + +
– – –– –
–
22
Ideal Capacitor
 Capacitor with an AC source: since AC sources periodically
reverse voltage directions, the capacitor plates will change
charge polarity accordingly
 With each polarity change, the capacitor goes through a ‘charging’
state, thus current continuously flows
 Just as with a DC source, no electrons cross between the plates
 Just as with a DC source, voltage across a capacitor cannot change
instantaneously
i
vs(t)
+
–
ECEN 301
i
++ + +
– – –– –
–
vs(t)
–
+
Discussion #10 – Energy Storage
–– – –
+ + +++
23
Ideal Capacitor
i – v characteristic for an ideal capacitor
Differentiate
both sides
q (t )  Cv(t )
dq(t ) dCv(t )

dt
dt
dv(t )
i (t )  C
dt
ECEN 301
OR
BUT
dq(t )
i (t ) 
dt
1 t
v(t )   iC ( )d
C 
Discussion #10 – Energy Storage
24
Ideal Capacitor
i – v characteristic for an ideal capacitor
1 t
v(t )   iC ( )d
C 
NB: Assumes we know the value of the
capacitor from time (τ = -∞) until time (τ = t)
Instead
1 t
v(t )   iC ( )d  v(t0 )
C t0
Initial condition
Initial time (often t0 = 0)
ECEN 301
Discussion #10 – Energy Storage
25
Ideal Capacitor
 Example2: calculate the current through the capacitor C = 0.1μF
with the voltage as shown: v(t) = 5(1-e-t/10-6)
5.0
voltage (V)
4.0
3.0
2.0
1.0
0.0
0.00
2.00
4.00
6.00
time (us)
ECEN 301
Discussion #10 – Energy Storage
26
Ideal Capacitor
 Example2: calculate the current through the capacitor C = 0.1μF
with the voltage as shown: v(t) = 5(1-e-t/10-6)
5.0
dv(t )
i (t )  C
dt
voltage (V)
4.0
3.0
2.0
1.0
0.0
0.00
2.00
4.00
time (us)
ECEN 301

5 t /106
 (10 ) 6 e
10
7
6.00
 0.5e
Discussion #10 – Energy Storage
t / 106

A
27
Ideal Capacitor
5.0
0.5
4.0
0.4
current (A)
voltage (V)
 Example2: calculate the current through the capacitor C = 0.1μF
with the voltage as shown: v(t) = 5(1-e-t/10-6)
3.0
2.0
1.0
0.0
0.00
0.3
0.2
0.1
2.00
4.00
6.00
0.0
0.00
time (us)
4.00
6.00
time (us)
NB: the capacitor’s current jumps ‘instantaneously’ to 0.5A.
The ability of a capacitor’s current to change instantaneously is
an important property of capacitors
ECEN 301
2.00
Discussion #10 – Energy Storage
i(t )  0.5e
t /106
A
28
Ideal Capacitor
Example3: find the voltage v(t) for a capacitor C =
0.5F with the current as shown and v(0) = 0
1.2
current (A)
1.0
0.8
0.6
0.4
0.2
0.0
0.0
1.0
2.0
3.0
time (s)
ECEN 301
Discussion #10 – Energy Storage
29
Ideal Capacitor
Example3: find the voltage v(t) for a capacitor C =
0.5F with the current as shown and v(0) = 0
1.2
current (A)
1.0
0.8
0.6
0.4
0.2
0.0
0.0
1.0
2.0
time (s)
3.0
Interval
t0
0  t 1
1 t  2
2t
current i(t)
0
t
1
0
1 t
v(t )   id   v(0)
C 0
ECEN 301
Discussion #10 – Energy Storage
30
Ideal Capacitor
Example3: find the voltage v(t) for a capacitor C =
0.5F with the current as shown and v(0) = 0
1.2
Interval
t0
current (A)
1.0
0.8
0.6
0  t 1
0.4
0.2
0.0
0.0
1.0
2.0
time (s)
3.0
1 t  2
2t
voltage v(t)
0
t
2 d  0
0
t
2 (1)d  v(1)
0
0  v(2)
1 t
v(t )   id   v(0)
C 0
ECEN 301
Discussion #10 – Energy Storage
31
Ideal Capacitor
Example3: find the voltage v(t) for a capacitor C =
0.5F with the current as shown and v(0) = 0
1.2
current (A)
1.0
0.8
0.6
0.4
0.2
0.0
0.0
1.0
2.0
time (s)
ECEN 301
3.0
Interval
t0
0  t 1
1 t  2
2t
Discussion #10 – Energy Storage
voltage v(t)
0
t2
2t  1
3
32
Ideal Capacitor
1.2
3.5
1.0
3.0
voltage (V)
current (A)
Example3: find the voltage v(t) for a capacitor C =
0.5F with the current as shown and v(0) = 0
0.8
0.6
0.4
0.2
2.5
2.0
1.5
1.0
0.5
0.0
0.0
0.0
1.0
2.0
3.0
0.0
1.0
time (s)
3.0
4.0
time (s)
NB: The final value of the capacitor voltage after
the current source has stopped charging the
capacitor depends on two things:
1. The initial capacitor voltage
2. The history of the capacitor current
ECEN 301
2.0
Interval
t0
0  t 1
1 t  2
2t
Discussion #10 – Energy Storage
voltage v(t)
0
t2
2t  1
3
33
Series Capacitors
Capacitor Series Rule: two or more circuit elements are
said to be in series if the current from one element
exclusively flows into the next element.
Capacitors in series add the same way resistors in parallel add
1
1
1 1
1

  
C EQ C1 C2 C3
CN
C1
C2
C3
C EQ 
1
1
1 1
1
  
C1 C2 C3
CN
Cn
∙∙∙
CN
∙∙∙
CEQ
ECEN 301
Discussion #10 – Energy Storage
34
Parallel Capacitors
Parallel Rule: two or more circuit elements are said to
be in parallel if the elements share the same terminals
Capacitors in parallel add the same way resistors in series add
N
C EQ   C n
n 1
C1
ECEN 301
C2
C3
Cn
CN
Discussion #10 – Energy Storage
CEQ
35
Parallel and Series Capacitors
Example4: determine the equivalent capacitance CEQ
C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF,
C6 = 1/3mF
C1
C4
C3
C2
ECEN 301
C6
C5
Discussion #10 – Energy Storage
36
Parallel and Series Capacitors
Example4: determine the equivalent capacitance CEQ
C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF,
C6 = 1/3mF
C1
C4
C3
C2
ECEN 301
C4C5C6
C EQ1 
C4C5  C4C6  C5C6
C6
C5
(1 / 3)(1 / 3)(1 / 3)
(1 / 3)(1 / 3)  (1 / 3)(1 / 3)  (1 / 3)(1 / 3)
1 / 27

1/ 3
1
 mF
9

Discussion #10 – Energy Storage
37
Parallel and Series Capacitors
Example4: determine the equivalent capacitance CEQ
C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF,
C6 = 1/3mF
C EQ 2  C3  C EQ1
C1
C3
CEQ1
C2
1
C EQ1  mF
9
ECEN 301
Discussion #10 – Energy Storage
1
 (1)   
9
10
 mF
9
38
Parallel and Series Capacitors
Example4: determine the equivalent capacitance CEQ
C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF,
C6 = 1/3mF
C1
C EQ 
CEQ2
C2
C EQ 2
ECEN 301
10
 mF
9
C1C2C EQ 2
C1C2  C1C EQ 2  C2C EQ 2
(2)( 2)(10 / 9)
(2)( 2)  (2)(10 / 9)  (2)(10 / 9)
40 / 9

4  20 / 9  20 / 9
10
 mF
19

Discussion #10 – Energy Storage
39
Parallel and Series Capacitors
Example4: determine the equivalent capacitance CEQ
C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF,
C6 = 1/3mF
CEQ
ECEN 301
C EQ
10
 mF
19
Discussion #10 – Energy Storage
40
Energy Storage in Capacitors
Capacitor energy WC(t): can be found by taking the
integral of power
Instantaneous power
PC = iv
t
WC (t )   PC ( )d

t
  vC ( )iC ( )d

t
  vC ( )C

 C
v (t )
vdv
v (  )
1
 Cv 2
2
ECEN 301
v (t )
dvC ( )
d
d
Since the capacitor is uncharged at t = -∞,
v(- ∞) = 0, thus:
1
WC (t )  Cv(t ) 2 J
2
v (  )
Discussion #10 – Energy Storage
41
Ideal Inductor
Inductors are made by winding a coil of wire around a
core
The core can be an insulator or ferromagnetic material
+
N A
2
L
l  0.45d
N turns
l
–
L – Inductance
μ – relative permeability
• Cross-sectional area A
• Diameter d
ECEN 301
Discussion #10 – Energy Storage
42
Ideal Inductor
Inductance: a measure of the ability of a device
to store energy in the form of a magnetic field
Ideally the resistance through an inductor is zero
(i.e. no voltage drop), thus an inductor acts like
a short circuit in the presence of a DC source.
i
BUT there is an initial voltage across the
inductor as the current builds up (much like
‘charging’ with capacitors)
Henry (H): unit of inductance.
1 henry = 1 volt-second/ampere (V-s/A)
ECEN 301
Discussion #10 – Energy Storage
43
Ideal Inductor
 Inductor with an AC source: since AC sources periodically
reverse current directions, the current flow through the inductor
also changes
 With each current direction change, the current through the inductor
must ‘build up’, thus there is a continual voltage drop across the inductor
 Just as with a DC source, current across an inductor cannot change
instantaneously
i
i
vs(t)
+
–
ECEN 301
vs(t)
–
+
Discussion #10 – Energy Storage
44
Ideal Inductor
I – v characteristic for an ideal inductor
di (t )
v(t )  L
dt
1 t
i (t )   vL ( )d
L 
NB: note the duality between inductors and capacitors
ECEN 301
Discussion #10 – Energy Storage
45
Ideal Inductor
i – v characteristic for an ideal inductor
1 t
i (t )   vL ( )d
L 
NB: Assumes we know the value of the
inductor from time (τ = -∞) until time (τ = t)
Instead
1 t
i (t )   vL ( )d  i (t0 )
L t0
Initial condition
Initial time (often t0 = 0)
ECEN 301
Discussion #10 – Energy Storage
46
Ideal Inductor
Example5: find the voltage across an inductor L =
0.1H when the current is: i(t) = 20 t e-2t
current (A)
4.0
2.0
0.0
0.0
0.5
1.0
1.5
time (s)
ECEN 301
Discussion #10 – Energy Storage
47
Ideal Inductor
Example5: find the voltage across an inductor L =
0.1H when the current is: i(t) = 20 t e-2t
current (A)
4.0
2.0
0.0
0.0
0.5
1.0
1.5
di (t )
v(t )  L
dt
d
 0.1 (20te 2t )
dt
 2(2te 2t  e  2t )
 2e  2t (1  2t )V
time (s)
ECEN 301
Discussion #10 – Energy Storage
48
Ideal Inductor
Example5: find the voltage across an inductor L =
0.1H when the current is: i(t) = 20 t e-2t
1.5
voltage (V)
current (A)
4.0
2.0
0.0
0.0
0.0
0.5
1.0
1.5
0.5
1.5
time (s)
time (s)
NB: the inductor’s voltage jumps ‘instantaneously’ to
2V. The ability of an inductor’s voltage to change
instantaneously is an important property of inductors
ECEN 301
1.0
-0.5
Discussion #10 – Energy Storage
v(t )  2e 2t (1  2t )V
49
Series Inductors
Series Rule: two or more circuit elements are said to
be in series if the current from one element
exclusively flows into the next element.
Inductors in series add the same way resistors in series add
N
LEQ   Ln
n 1
L1
L2
L3
Ln
∙∙∙
LN
∙∙∙
LEQ
ECEN 301
Discussion #10 – Energy Storage
50
Parallel Inductors
Parallel Rule: two or more circuit elements are said to
be in parallel if the elements share the same terminals
Inductors in parallel add the same way resistors in parallel add
1
1 1 1
1
   
LEQ L1 L2 L3
LN
L1
ECEN 301
L2
L3
LEQ 
Ln
1
1 1 1
1
  
L1 L2 L3
LN
LN
Discussion #10 – Energy Storage
LEQ
51
Energy Storage in Capacitors
Capacitor energy WL(t): can be found by taking the
integral of power
Instantaneous power
PL = iv
t
WL (t )   PL ( )d

t
  vL ( )iL ( )d

t
 L

 L
i (t )
diL ( )
iL ( )d
d
idi
i (  )
1
 Li 2
2
ECEN 301
i (t )
Since the inductor is uncharged at t = -∞,
i(- ∞) = 0, thus:
1
WL (t )  Li(t ) 2 J
2
i (  )
Discussion #10 – Energy Storage
52
Energy Storage in Capacitors
 Example6: calculate the power and energy stored in a 0.1-H
inductor when:
 i = 20 t e-2t A (i = 0 for t < 0)
 V = 2 e-2t (1- 2t)V (for t >= 0)
ECEN 301
Discussion #10 – Energy Storage
53
Energy Storage in Capacitors
 Example6: calculate the power and energy stored in a 0.1-H
inductor when:
 i = 20 t e-2t A (i = 0 for t < 0)
 V = 2 e-2t (1- 2t)V (for t >= 0)
PL (t )  vL (t )iL (t )


 2e 2t (1  2t ) 20te2t

 40te4t (1  2t )W
ECEN 301
Discussion #10 – Energy Storage
54
Energy Storage in Capacitors
 Example6: calculate the power and energy stored in a 0.1-H
inductor when:
 i = 20 t e-2t A (i = 0 for t < 0)
 V = 2 e-2t (1- 2t)V (for t >= 0)
PL (t )  vL (t )iL (t )


 2e 2t (1  2t ) 20te2t
 4t
 40te (1  2t )W
ECEN 301

1
WL (t )  LiL (t ) 2
2
1
 (0.1) 20te 2t
2
 20t 2e  4t J
Discussion #10 – Energy Storage


2
55
Ideal Capacitors and Inductors
Inductors
Passive sign
convention
Capacitors
+C–
+ L –
i
i
Voltage
di (t )
v(t )  L
dt
1 t
v(t )   iC ( )d  v(t0 )
C t0
Current
1 t
i (t )   vL ( )d  i (t0 )
L t0
Power
di (t )
PL (t )  Li(t )
dt
dv(t )
i (t )  C
dt
dv(t )
PC (t )  Cv(t )
dt
ECEN 301
Discussion #10 – Energy Storage
56
Ideal Capacitors and Inductors
Inductors
Energy
Capacitors
1
1
2
WL (t )  Li(t ) WC (t )  Cv(t ) 2
2
2
An instantaneous change
is not permitted in:
Current
Voltage
Will permit an
instantaneous change in:
Voltage
Current
With DC source element
acts as a:
Short Circuit
Open Circuit
ECEN 301
Discussion #10 – Energy Storage
57