circuits lectures mod7
TRANSCRIPT
-
8/11/2019 Circuits Lectures Mod7
1/81
Basic Concept 1: Analogy between the gravitational field
and the static electric field
m1 m
2 q
1 q
2
r
F = G m1m2
r2 r = g12 m2 r
r
F = q1 q2
4 0 r2 r = E12 q2 r
Newton's Law of gravitation Coulomb's Law (the force is always attractive) (attractive or repulsive force)
Both are inverse-square, central-force fields.
m q
Potential Potentialenergy = mgh h h energy = qEh
m q
h V
Both are conservative fields: The net change in absolute potentialaround any closed path = 0.
r
r
r
r
r
Fgrav = mr
gr
Felectric = qr
E
Absolutepotential = gh
Absolutepotential = Eh
Ground level (reference height = 0) Ground potential (reference voltage = 0)
r
Fgrav
r
Felectric
-
8/11/2019 Circuits Lectures Mod7
2/81
This leads to the fundamentalKirchoff's Voltage Law:
The sum of the rises and falls of electric potential (voltage)
around any closed loop in a circuit = 0.
Basic Concept 2: Current and conservation of charge
Define current i = q/t amperes = dq/dt in the infinitesimal limit
Circuit loopscomprised ofsources,resistors, etc.located on the"branches"
Surface S
q1
q2
q3
r
v1
r
v2
r
v3
q = q1 + q2 + q3coulombs move
through surface S
every t seconds
-
8/11/2019 Circuits Lectures Mod7
3/81
How about current flow leaving a closed surface?
S
At time t0: At time t0+ t:
Q(t0 ) = q4 + q5 + q6 + q7 Q(t0 + t) = q1 + q2 + q3 + q4 + q5
iinward = Q(t0 + t) Q(t0 )
t =
q1 + q2 + q3 q6 q7t
For static electric fields, iinward = , i.e., there is no net gain of charge inS over t seconds. Equivalently, ioutward = iinward = 0 .
At the green "node" (i.e., the
junction of circuit "branches"),
this yields the fundamental
Kirchoff's Current Law
which can be written here as
ileft + idown + iright + iup = 0
The sum of the electric currents leavingany circuit node = 0.
q1
q2
q3
q4
q5
q6
q7
q1
q2
q3
q4
q5
q6
q7
iup
idown
irightileft
S
S
-
8/11/2019 Circuits Lectures Mod7
4/81
Basic Concept 3: Resistance and Ohm's Law
Cylindrical bar of conducting material Let a potential of V volts be
applied between the flat parallel
faces of the bar.
Then, the internal electric field is
S1 S2rE = V/l in the direction shown,
and the force on an individual
A electron having the charge e
coulombs isrFe = e
rE .
V +
Let = characteristic time between randomizing collisions of an electron and the
nuclei of the atoms within the bar. Note that is a statistical average.
From Newton's Law,rFe = me
rae , so we have ae = Fe/me = eV/mel
This yields an average electron "drift velocity" ue = ae = e V /me l in the
rightward direction toward S2, the end of the bar having the higher voltage.
This electron drift velocity is analogous to the "terminal velocity" of raindrops falling
to the ground under the constant downward gravitational force. The raindrops reach a
limiting speed despite falling from great heights.
In the time period t = l/ue , all of the electrons in the cylinder volume lA areswept rightward through surface S2. This represents a total charge movement of
q = n electrons /m3( ) lA m3( ) e coulombs / electron( ) = en lA coulomb
r
Fe
+
l
rE
+
+
-
8/11/2019 Circuits Lectures Mod7
5/81
Therefore, the rightward current through S2 is given by
irightward = qt
= e nlAl/ue
= e nAue
= e nAe V
me l
= e2 nA
me l
V
We can now define the leftwardcurrent through S2 as
ileftward = irightward = +
e2 nA
me l
V
Solving for V, we obtain Ohm's Law:
V = ileftward me l
e2nA
= ileftwardR
V +
V = i R +
ileftward
R
i
V = i R where
R = m en e
2
.l
A
A
l
"Resistor"
-
8/11/2019 Circuits Lectures Mod7
6/81
Ohm's Law uses the following sign convention for i assumed entering a resistor:
V +
This will also be used in EECS 221 for other passive circuit elements such as
capacitors and inductors.
Basic Concept 4: Energy and power
Earlier, we hadq
We define 1 volt of absolute potential as follows:
Moving 1 coulomb of charge through a potential rise of
1 volt yields 1 joule of potential energy.
This is the action of a voltage source, i.e., an ideal battery, symbolized by
i
!
Felectric
= q!
E
Potential energy = qEh (joules)
Absolute potential = Eh (volts)
Ground potential (reference voltage = 0)
h
V+
+++
+N coulombs
For a total of N coulombs elevated
by a battery by V volts, there is
sourced NV joules.
-
8/11/2019 Circuits Lectures Mod7
7/81
For i coulombs/sec (amperes) elevated V volts, there is sourced iV joules/sec
(watts).
More generally, with this direction of current flow, a time-varying source generates:
p(t) = i(t) V(t) watts
What about the converse situation?
This dissipates iV watts of power as stored energy and heat within the voltage
source. We see that a voltage source such as a battery can either generate power or
absorb power, depending upon the direction of current flow. More generally:
V
+
+++
+icoulombs/sec
(amperes)
V+
++ +
+icoulombs/sec
(amperes)
V(t)
+
++ +
+
i(t)
p(t) = i(t) V(t) wattsare absorbed (dissipated) i
a circuit device where current enters the high-potentia
end of the device at time t. Interestingly, by our sign
convention, this holds for anyresistor. Thus,
resistors can only dissipate power. For a resistor
p(t) = i2(t)R since V(t) =i(t)R .
Here, i amperes entersthe +
terminal of the voltage source and
dropsin potential by V volts upon
leaving.
Thus, a voltage source emitting i
amperes from its + terminal
generates iV watts of power.
-
8/11/2019 Circuits Lectures Mod7
8/81
Circuit Analysis Example 1: Loop analysis (Kirchoffs Voltage Law)
Systematic step-by-step analysis procedure:
1. Assume directions of currents i1 and i2 . These directions are arbitrary.
There is no need to guess the correct directions of current flow before
beginning the analysis !! If the actual current flow is opposite to that chosen,
your answer for the current at the end of the analysis will simply be negative.
12 v+
+5 v9 6
10 3
12 v+
+5 v9 6
10 3
i1 i2
-
8/11/2019 Circuits Lectures Mod7
9/81
2. Obtain the current in the common branch. From Kirchoffs Current Law
implemented at the top of the 10 resistor, this is simply i1 i2 in the
downward direction.
3. Assign voltages and their associated polarities to the resistors according to
Ohms Law and the sign convention.
12 v+
+5 v9 6
10 3
i1 i2
i1i2
12+
+
59 i1 6 i1
10(i1 i2) 3 i2
i1 i2
i1i2
+ +
+ +
-
8/11/2019 Circuits Lectures Mod7
10/81
4. Walk around each loop and set the sum of the voltage rises and falls to
zero. The direction of your walk in each loop is arbitrary you simply
must return to your starting point after completely walking around the loop.
Left loop: Starting from the negative side of the 12 volt battery, we have
+ =12 9 6 10 01 1 1 2 ( )i i i i
This simplifies to
+ = 25 10 121 2i i
Call this Equation 1.
12+
+59 i1 6 i1
10(i1 i2) 3 i2
i1 i2
i1i2
+ +
+ +
Left loop walking path
-
8/11/2019 Circuits Lectures Mod7
11/81
Right loop: Starting from the bottom of the 10resistor, we have
+ + =10 5 3 01 2 2( )i i i
This simplifies to
10 13 51 2i i =
Call this Equation 2.
12+
+59 i1 6 i1
10(i1 i2) 3 i2
i1 i2
i1i2
+ +
+ +
Right loop walking path
-
8/11/2019 Circuits Lectures Mod7
12/81
5. Solve the system of loop equations simultaneously to obtain the desired
currents.
For convenience, we write both equations together:
+ =
=
25 10 12
10 13 5
1 2
1 2
i i
i i
In matrix form, we can write this as
=
25 10
10 13
12
5
1
2
i
i
A systematic approach to solve this system is to apply Cramers Rule:
i1
12 10
5 13
25 10
10 13
206
2250 915
det
det
.=
= amps = amps
i2
25 12
10 5
225
245
2251 08
det
.=
= amps = amps
i i1 2 0 915 1 08 0 173 . . . = = amps
With knowledge of these currents, the voltage across each resistor can be
inserted directly into the circuit diagram for Step 3.
-
8/11/2019 Circuits Lectures Mod7
13/81
6. If desired, we can obtain the power sourced or dissipated by each circuit
component and test for balance between the sourced and dissipated power.
Power associated with the batteries
From Step 5, we found that i1 is positive. Since i1 was assumed inStep 1 to flow outward from the + side of the 12-volt battery, this means
that this battery is generatingpower.
P i12-volt battery(generated)
watts . .= = =12 12 0 915 10 9861
From Step 5, we found that i2 is also positive. Since i2 was assumed in
Step 1 to flow outward from the + side of the 5-volt battery, this meansthat this battery is also generatingpower.
P i5-volt battery(generated)
watts . .= = =5 5 1 08 5 42
Total power generated = 10 986 5 4 16 431. . .+ = watts
Power dissipated by the resistors
Using the currents calculated in Step 5, we obtain
P i9 resistor watts . .= = =9 9 0 915 7 5441712 2
P i6 resistor watts . .= = =6 6 0 915 5 02945184612 2
P i i10 resistor watts ( ) ( . ) .= = =10 10 0 173 0 30041 22 2
P i3 resistor watts . .= = =3 3 1 08 3 55703703122 2
Total power dissipated = .7 54417 5 029451846 0 3004 3 557037031
16 431
+ . + . + .
= . watts
Power Balance!
-
8/11/2019 Circuits Lectures Mod7
14/81
Circuit Analysis Example 2: Loop analysis (Kirchoffs Voltage Law)
Only change from Example 1: Replace the 10 resistor in the common branch with a
9 A current source.
Systematic step-by-step analysis procedure:
1. Assume directions of currents i1 and i2 .
12 V+
+5 V9 6
9 A 3
12 V+
+5 V9 6
9 A 3
i1 i2
-
8/11/2019 Circuits Lectures Mod7
15/81
2. Note that i2 is no longer an independent unknown. From Kirchoffs Current
Law implemented at the top of the 9 A current source, i2 = i1+ 9 in the
rightward direction.
3. Assign voltages and their associated polarities to the resistors according to
Ohms Law and the sign convention.
We see that every resistor voltage in the circuit is a function only of i1.
Thus, we require only a single loop equationto solve the problem!
Note that the voltage across the current source is unknown at this point.
Thats OK! We dont need it to solve the problem. As an option, we can
obtain it later once we derive i1.
12 V+
+5 V9 6
9 A 3
i1 i1+9
12+
+59 i1 6 i1
3 (i1+ 9)
i1 i1+9
+ +
+
-
8/11/2019 Circuits Lectures Mod7
16/81
4. Define a superloopthat consists of the outer boundary of the combination
of the left and right loops. Walk around this superloop and set the sum of
the voltage rises and falls to zero. As in Example 1, the direction of your walk
is arbitrary you simply must return to your starting point after completely
walking around the superloop.
Starting from the negative side of the 12 volt battery, we have
+ + + =12 9 6 5 3 9 01 1 1 ( )i i i
This simplifies to
18 10 0 51 1 2 1i i i i .= = =, +9 = 8.4A A
With knowledge of i1, the voltage across each circuit element can be found
from the circuit diagram for Step 3. For example, the voltage across the
current source (in the direction of its current flow) is given by two
equivalent expressions:
V i
i i
currentsource
V
.
= +
= +
= +
3 5
12 9 6
20 3
2
1 1
(thru 3 resistor and 5V battery)
(thru 12V battery and 9 and 6 resistors)
12+
+59 i1 6 i1
3 (i1+ 9)
i1 i1+9
+ +
+
Superloop walking path
-
8/11/2019 Circuits Lectures Mod7
17/81
5. If desired, we can obtain the power sourced or dissipated by each circuit
component and test for balance between the sourced and dissipated power.
Power associated with the two batteries and the current source
From Step 4, we found that i1 is negative. Since i1 was assumed in
Step 1 to flow outwardfrom the + side of the 12-volt battery, this means
that i1 is really flowing intothe + side of the battery, which is
therefore dissipatingpower:
P i12-volt battery(dissipated)
. .= = =12 12 0 5 6 61 W
From Step 4, we found that the voltage across the current source in the
direction of its current flow is positive. This means that the current source
is generatingpower:
P V9-amp current source(generated)
currentsource
.= = =9 9 20 3 183 W
From Step 4, we found that i2 is positive. Since i2 was assumed inStep 1 to flow outward from the + side of the 5-volt battery, this means
that this battery is generatingpower:
P i5-volt battery(generated)
. .= = =5 5 8 4 42 22 W
Net generated power = 183 42 2 6 6 218 5 . . .+ = W
-
8/11/2019 Circuits Lectures Mod7
18/81
Power dissipated by the resistors
Using the current calculated in Step 4, we obtain
P i9 resistor ( . ) .= = =9 9 0 5 2 712 2
W
P i6 resistor ( . ) .= = =6 6 0 5 1 85112 2 W
P i3 resistor ( ) . .= + = =3 9 3 8 4 213 92512 2 W
Total power dissipated = + +. . .
.
2 7 1 851 213 925
218 5= W
Power Balance!
-
8/11/2019 Circuits Lectures Mod7
19/81
Circuit Analysis Example 3: Node analysis (Kirchoffs Current Law)
Systematic step-by-step analysis procedure:
1. Select one node as ground (zero potential, i.e., V = 0). Any node is OK, but
EEs usually select the node that has the maximum number of branches
connected to it. The ground node is often designated by the symbol
12 V+
4 7 5
3 A3 +
9 V
12 V+
4 7 5
3 A3 +
9 V
-
8/11/2019 Circuits Lectures Mod7
20/81
2. Locate and label all other nodes where the potential is unknown.
3. Write Kirchoffs Current Law (KCL) at Node #1:
i i i
V V V V
left down right1 1 1 + + =
+ + =
0
127
03 5
01 1 1 2
Clearing fractions, we obtain Equation 1:
71 21 1801 2V V =
12 V+
4 7 5
3 A3 +
9 V
V1 V2
12 V+
4 7 5
3 A3
+
9 V
V1 V2
ileft1 iright1
idown1
-
8/11/2019 Circuits Lectures Mod7
21/81
Now, write KCL at Node #2:
i i i
V V V
left down right2 2 2
( )
+ + =
+ +
=
0
53
9
402 1 2
Clearing fractions, we obtain Equation 2:
+ =4 9 151 2V V
4. Solve the system of node equations simultaneously to obtain the desired
voltages.
For convenience, we write both equations together:
71 21 1804 9 15
1 2
1 2
V V
V V =
+ =
12 V+
4 7 5
3 A3 +
9 V
V1 V2
ileft2 iright2
idown2
-
8/11/2019 Circuits Lectures Mod7
22/81
In matrix form, we can write this as
71 21
4 9
180
15
1
2
=
V
V
Applying Cramers Rule, we obtain:
V1
180 21
15 9
71 21
4 9
1935
5553 486
det
det
.=
= V V=
V2
71 180
4 15
555
1785
5553 216
det
.=
= V V=
With knowledge of these node voltages, the currents and powers
associated with each circuit element can be determined.
-
8/11/2019 Circuits Lectures Mod7
23/81
Circuit Analysis Example 4: Node analysis (Kirchoffs Current Law)
Only change from Example 3: Replace the 5 resistor in the top middle branch with a
6-volt battery, polarized as shown.
Systematic step-by-step analysis procedure:
1. Select one node as ground (zero potential, i.e., V = 0).
12 V+
4 7 6 V
3 A3 +
9 V
+
12 V+
4 7 6 V
3 A3 +
9 V
+
-
8/11/2019 Circuits Lectures Mod7
24/81
2. Locate and label all other nodes where the potential is unknown.
3. Note that V2 is no longer an independent unknown. From Kirchoffs Voltage
Law implemented at the 6-volt battery, V2 = V1+ 6.
12 V+
4 7 6 V
3 A3 +
9 V
+V1 V2
12 V+
4 7 6 V
3 A3 +
9 V
+V1 V1+6
-
8/11/2019 Circuits Lectures Mod7
25/81
4. Define a supernode that consists of the combination of Nodes 1 and 2.
Then, write KCL at this supernode.
i i i i
V V V
left down down right1 2
( ) ( )
+ + + =
+
+ +
+
=
0
12
7
0
33
6 9
401 1 1
This simplifies to:
61 81 1 3281 1 2 1V V V V .= = =, +6 = 7.328V V
With knowledge of V1 and V2 ,the currents and powers associated with
each circuit element can be determined.
12 V+
4 7 6 V
3 A3 +
9 V
+V1 V1+6
iright
idown2
ileft
idown1
-
8/11/2019 Circuits Lectures Mod7
26/81
Network Simplification
Series resistors and voltage sources
Assuming a clockwise current direction, we can implement Kirchoffs
Voltage Law:
+ + + =V V V R R R1 2 3 1 2 3 i i i 0
Grouping terms,
( ) ( )V V V R R R1 2 3
equivalent
1 2 3
equivalentV R
+ + = + +1 2444 3444 1 2444 3444
i
V1
+
R1+ + V2 V3 R2
R3
V1
+
+ + V2 V3
i
+
+ iR1
iR3
KVL walking path
+ iR2
-
8/11/2019 Circuits Lectures Mod7
27/81
We thus obtain an equivalent circuit having the same loop current as the
original:
Series voltage sources sum algebraically !
Series resistances sum algebraically !
Parallel resistors and current sources
To analyze this circuit, we apply Kirchoffs Current Law. We select the
lower node as ground and assign the upper node the potential V.
V1+V2+V3
+
i
R1 + R2 + R3
i1 R1i2 i3 R2 R3
-
8/11/2019 Circuits Lectures Mod7
28/81
V
+ + + =i i i1 2 3 R R R1 2 3 i i i 0
Using Ohms Law to express the resistor currents, we obtain
+
+
+
=i i i1 2 3V
R
V
R
V
R
0 0 00
1 2 3
Grouping terms,
( )i i iR R R
1 2 3
equivalent1 2 3
equivalent
i1/R
+ + = + +
1 244 344
1 2444 3444
V1 1 1
i1 R1i2 i3 R2 R3
iR1 iR2 iR3
-
8/11/2019 Circuits Lectures Mod7
29/81
We thus obtain an equivalent circuit having the same node voltage as the
original:
Parallel current sources sum algebraically !
The reciprocals of parallel resistances sum algebraicallyto yield the reciprocal of the equivalent resistance!
The second statement can be put more simply if we define the
conductance, G, of a resistor, R, as G= 1/R. By this definition, we have:
Conductances of parallel resistors sum algebraically !
For only tworesistors in parallel, we have
R
R R
R R
R Requivalent
1 2
1 2
1 2
=
+
=+
1
1 1
This product over sum rule should be memorized!
i1+ i2+ i3
V
1
1 1 1
R R R
R
1 2 3
equivalent
+ +
=
-
8/11/2019 Circuits Lectures Mod7
30/81
Voltage Divider
Series resistors and voltage source
Since series resistors sum algebraically, we have
ij
n
n
= + + + + + =
=
V
R R R R
V
RNN
1 2
1
L L
Then the voltage across thejth resistor is given by
V ij j
j
n
n
R
N
R V
R
R= =
=1This is the voltage-dividerformula, which should be memorized.
V+
R2 Rj
RN
R1
i
+
VRj
-
8/11/2019 Circuits Lectures Mod7
31/81
Current Divider
Parallel resistors and current source
. . . . . .
Since parallel conductances sum algebraically, we have
Vj
nn
=+ + + + +
=
=
i
G G G G
i
GN
N1 2
1
L L
where . Then the current through thejth resistor is given by
i V
Vj
j
j
j
n
n
RN
RG i
G
G= = =
=
1
This is the current-dividerformula, which should be memorized.
i R2 Rj RNR1
iRj
V
G Rj j= 1 /
-
8/11/2019 Circuits Lectures Mod7
32/81
Thevenin!s Theorem
The terminal behavior of a circuit containing an arbitraryconnection
of resistors, voltage sources, and current sources can be completelycharacterizedby a single voltage source VTh in series with a single
resistor RTh. Consider:
From the perspective of Circuit B, this is equivalent to:
if and are properly defined.
VTh+
b
a
Circuit A
Arbitrary arrangement ofresistors and sources
connected between nodesa and b
VTh
+
RTh
b
a
Circuit B
Connected to Circuit Aat nodes a and b
Circuit B
Connected to Circuit Aat nodes a and b
VTh RTh
Thevenin Equivalent of Circuit A
looking into
looking into
-
8/11/2019 Circuits Lectures Mod7
33/81
Use the method of external current injectionto characterize the terminal properties of
the Thevenin equivalent circuit:
Example: Find the Thevenin equivalent circuit looking into Node Pair abof Circuit A
Strategy: Connect current source I at Node Pair ab. Solve for the induced voltage
Vab as a function of I (kept in symbolic form). Then, compare term-by-term
with the terminal properties equationof the Thevenin equivalent circuit to obtain
VTh and RTh in one step.
RThI +
b
Vab
+
I
60 V
+
8 !
32 !
5 A
b
a
+
3 A
Terminal propertiesequation:
VTh
+
Vab
c a
Circuit A
Vab= VTh+ RThI
-
8/11/2019 Circuits Lectures Mod7
34/81
In this example, we choose to apply node equations to solve for Vab. We designate
Node bas the ground reference. Then Vab= Va. We see that there are two
unknown node voltages: Va and Vc. Now, we write KCL at Nodes a and c:
At Node a:
Va! V
c
8
"#$
%&' ! 5 ! I = 0
At Node c:
!3 +Vc! 6032
"#$
%&' +
Vc! V
a
8
"#$
%&' = 0
This system of equations can be written in matrix form as:
1 !1
!4 5
"
#$
%
&'
Va
Vc
"
#$
%
&' =
40 + 8I
156
"
#$
%
&'
60 V+
8 !
32 !
5 A
b
3 A I
Va= VabVc
ac
Va!V
c = 40 + 8I
!4Va+ 5V
c = 156
+
Vab
-
8/11/2019 Circuits Lectures Mod7
35/81
Applying Cramer!s Rule, we solve for Va:
Va =
det40+8 I !1
156 5
det1 !1
! 4 5
= 200 + 40 I + 1565! 4
Va = 356 + 40 I
This method of external current injectionisgeneraland can be applied to circuits
containing controlled voltage sourcesand controlled current sourceswhere other
methods to obtain the Thevenin equivalent circuit do not work.
Compare these results term-by-term
with the terminal properties equation
of a Thevenin circuit, and we can
immediately identify VTh and RTh.
b
a
VTh= 356 V
RTh= 40 !
+Thevenin
Equivalent Circuit
looking into
Node Pair ab
Of Circuit A
Thevenin Equivalent of Circuit A
Vab = VTh + RThI
-
8/11/2019 Circuits Lectures Mod7
36/81
Inductors and Capacitors: Energy Storage Elements
In addition to resistance to the flow of electric current, which is purely an
energy-loss phenomenon, circuit devices can also store magnetic andelectric field energy in a manner analogous to how a moving mass or a
spring stores mechanical energy.
Inductor (stores magnetic field energy)
v t di tdt
LLL volts( ) ( )= f t du t
dt( ) ( )= M newtons
This voltage-current relation for the inductor is useful in Kirchoffs Voltage
Law analyses. Equivalently, we can time-integrate this relation to obtain an
alternative expression that is useful in Kirchoffs Current Law analyses:
i t v t dt I v t dt t t t
t
t
L L LL L
for( ) ( ) ( )= = +
1 10 0
0
Here, I0 is the initial current flowing through the inductor at the starting time
t0 .
M f(t)
x(t) u t dx t dt( ) ( )/ =
vL(t)
+
iL(t)
L
Mechanical analog: A mass in motion
henrys
-
8/11/2019 Circuits Lectures Mod7
37/81
We see that the inductor current at any particular observation time
depends upon its initial current and the time history of the voltage across
the inductor. The inductor effectively integrates its terminal voltage
to develop its current; negative terminal voltage reduces the current flow.
The instantaneous power associated with the inductor is given by
P t v t i t di t
dti t
d
dti tL L L
LL LL L watts( ) ( ) ( )
( )( ) ( )= =
=
1
2
2
Integrating with respect to time to obtain the energy stored in the inductor
yields
W t i t L LL joules( ) ( )=1
2
2
-
8/11/2019 Circuits Lectures Mod7
38/81
Capacitor (stores electric field energy)
v t q t i t dt
V i t dt t t
t
t
t
C C C
C
1
C
1
C
volts
=1
C for
( ) ( ) ( )
( )
= =
+
0 0
0
Here, V0 is the initial terminal voltage of the capacitor at the starting time t0 .
We see that the capacitor voltage at any particular observation time
depends upon its initial voltage and the time history of the current flowingthrough the capacitor. The capacitor effectively integrates inflowing current
to store charge q; outflowing current reduces the stored charge.
Equivalently, we can time-differentiate the voltage-current relation for
the capacitor to obtain:
This alternative relation is useful in Kirchoffs Current Law analyses
involving the capacitor.
k
f(t)
x(t)
u t dx t dt ( ) ( ) / =
vC(t)
+
iC(t)
C
i t dv t
dtC CC amperes( )
( )
=
f t x t u t dtt
( ) ( ) ( )= =
k k newtons
farads
Mechanical analog: A stretched spring
-
8/11/2019 Circuits Lectures Mod7
39/81
The instantaneous power associated with the capacitor is given by
P t v t i t v t dv t
dt
d
dtv tC C C C
CCC C watts( ) ( ) ( ) ( )
( ) ( )= =
=
1
2
2
Integrating with respect to time to obtain the energy stored in the capacitor
yields
W t v t C CC joules( ) ( )=1
2
2
Analogy between Electrical and Mechanical Elements and Variables
Electrical System Mechanical System
Voltage, v (V) Force,f (N)
Current, i (A) Velocity, u (m/s)
Inductance, L (H) Mass, M (kg)
Capacitance, C (F) Reciprocal spring constant, 1/k (m/N)
Resistance, R (ohms) Dashpot damping, B (N-s/m)
-
8/11/2019 Circuits Lectures Mod7
40/81
Parallel Combination of Capacitors
From Kirchoffs Current Law, we have
i
C C C
C C C
C C C
equivalentC
= + +
= + +
= + +( )
i i i
dV
dt
dV
dt
dV
dt
dV
dt
1 2 3
1 2 3
1 2 31 244 344
Capacitances in parallel add!
i(t)
iC1(t)
C1
iC3(t)
C3
iC2(t)
C2
V(t)
-
8/11/2019 Circuits Lectures Mod7
41/81
Series Combination of Capacitors
From Kirchoffs Voltage Law, we have
V
C C C
C C C
C C C
1 2 3
1 2 3
equivalent1/C
( ) ( ) ( )
( )
= + +
= + +
= + +
V V V
i t dt i t dt i t dt
i t dt
t t t
t
1 2 3
1 1 1
1 1 1
1 2444 3444
The reciprocals of series capacitances sum algebraicallyto yield the reciprocal of the equivalent capacitance!
V(t)+
+
i(t)
C3
VC1(t)
+
VC2(t)
+
VC3(t)
C1 C2
-
8/11/2019 Circuits Lectures Mod7
42/81
Series Combination of Inductors
From Kirchoffs Voltage Law, we have
V
L L L
L L L
L L L
equivalentL
= + +
= + +
= + +( )
V V V
di
dt
di
dt
di
dt
di
dt
1 2 3
1 2 3
1 2 31 244 344
Inductances in series add!
V(t)+
+
i(t)
VL1(t)
+
VL2(t)
+
VL3(t)
L2
L3
L1
-
8/11/2019 Circuits Lectures Mod7
43/81
Parallel Combination of Inductors
From Kirchoffs Current Law, we have
i
L L L
L L L
L L L
1 2 3
1 2 3
equivalent1/L
( ) ( ) ( )
( )
= + +
= + +
= + +
i i i
V t dt V t dt V t dt
V t dt
t t t
t
1 2 3
1 1 1
1 1 1
1 2444 3444
The reciprocals of parallel inductances sum algebraically
to yield the reciprocal of the equivalent inductance!
i(t)
iL1(t)
L1
iL2(t)
L2
iL3(t)
L3
V(t)
-
8/11/2019 Circuits Lectures Mod7
44/81
Solution of Circuits Containing
Capacitors and Inductors
Circuits containing capacitors and/or inductors are described by differential
equations. For example, consider the following series RC circuit:
i t i t C R( ) ( )+ = 0
Cv (t)
RC C Sdv t
dt
v t( )
( )+
= 0
This yields
RC v (t)C C Sdv t
dtv t
( ) ( )+ =
which is a first-order ordinary differential equation with constant coefficients.
The source voltage v (t)S is assumed to be known. This permits solution
with any one of a variety of techniques that you learned in EA.
R
vC(t)
+
iC(t)
CvS(t)Apply KCL at thenode joining R and C.
iR(t)
+
-
8/11/2019 Circuits Lectures Mod7
45/81
In another example, consider the following series RLC circuit:
v t v t v t C L R Sv (t)( ) ( ) ( )+ + =
1
CL R vSi t dt
di t
dti t t
t
( )( )
( ) ( ) + + =
After time-differentiation, this yields
L RC
vSd i t
dt
di t
dti t
d t
dt
2
2
1( )
( ) ( )
( )+ + =
which is a second-order ordinary differential equation with constant
coefficients. EA methods again permit solution, assuming a known
excitation v (t)S .
R
vC(t)
+
CvS(t)
Apply KVL aroundthe loop.
vR(t) vL(t)+ +
L
+ i(t)
-
8/11/2019 Circuits Lectures Mod7
46/81
The Sinusoidal Steady State
For EEs, an important special case arises when the source exciting a
circuit is a continuous sinusoid, i.e., vS( ) cost V t= . Sinusoidal voltageand current excitations appear in all manner of EE technology ranging from
the ubiquitous 60-Hz alternating current used to power up our homes and
offices to the microwave signals used for our cellphone and satellite
communications.
How can we solve for the currents and voltages in an electric circuit
under sinusoidal excitation conditions? Well, we might try to directly solve
the circuit differential equation using EA-type methods. For example,
consider the series RC circuit. We can write the circuit differential equation
for the sinusoidally forced response as:
RC C Cdv t
dtv t V t
( ) ( ) cos+ =
We will assume that that the contribution to v tC( )generated by the
switching-on of the sinusoidal source has decayed to zero by the time that
we make our observation. In fact, this state is always achieved in passive
linear circuits if we simply wait long enough.
EEs would say that the circuits turn-on transient has decayed to zero,
and only the sinusoidal steady-stateresponse remains. Math professors
would say that we are throwing away the homogenous solution of the
differential equation and retaining only the particular solution that results
from a sinusoidal excitation.
-
8/11/2019 Circuits Lectures Mod7
47/81
The sinusoidal steady-state response of a linear circuit is simply a
sinusoidal time waveform having the same !frequencyas the source but,
in general, a different amplitudeAand different phase !:
vC(t) =Acos(!t+")
Aand !are initially unknown.
To determineAand !, we could substitute the above assumed solution
for vC(t) into the circuit differential equation for the sinusoidally forced
response. However, this is very tedious and non-intuitive, requiring the
use of lots of trig identities. Luckily, EE!s have developed a much bettermethod:
Phasor analysis!
which we will study next.
-
8/11/2019 Circuits Lectures Mod7
48/81
Introduction to Phasors and Impedance
Phasors provide an easy geometric interpretation of sinusoidal signals as
vectors in the complex plane. This eliminatesthe need to solve differential
equations when analyzing the sinusoidal steady-state behavior of circuits.
Eulers Identity (pronounced Oilers Identity)
This forms the basis of phasor notation. It defines a complex exponential
Aejas a vector in the complex plane which can be represented by real
and imaginary components:
We see that Eulers identity yields polar-to-rectangular and rectangular-to-
polar transformations for numbers in the complex plane:
A A jA, cos , sin [ ] [ ]
original polar- coordinaterepresentation of a complex number
equivalent rectangular-coordinate representation
123 1 2444 3444
Real
Imaginary
A
Acos
jAsin
Aej
Ae A jAj cos sin= +
Rea
Imaginary
tan-1(D/C)
C
jD
C + jD
C jD C D ej D C+ = +
tan ( / )2 2
1
C D2 2+
C D D C C jD2 2 1+[ ] [ ], tan ( / ) ,
equivalent polar-coordinate representation
original rectangular-coordinate representation of a complex number
1 24444 344441 24 34
-
8/11/2019 Circuits Lectures Mod7
49/81
For convenience, we will express the polar-coordinate representation of
a complex number as
Ae Aj
Note thatA, which represents the magnitude of the complex number, isalways a positive value. Now, using Eulers identity, familiar numbers can
be re-cast into a more general form:
1 1 00 = = ej
= = =
1 1 180
180 180
e e
j j
j ej = = 90 1 90
= = j e j 90 1 90
Relation of Phasors to Sinusoidal Signals
From Eulers identity, we see that a sinusoidal signal v t( )having a general
amplitudeA, angular frequency , and phase can be written as:
v t A t Ae Ae ej t j j t( ) cos( ) ( )= + = [ ] = [ ]+ Real Real
EEs denote the phasorassociated with v t( )as V( )j :
v t A t j Ae Aj( ) cos( ) ( )= + = = V
Note that V( )j contains only the amplitude and phase information of v t( ).
The ej t
term is factored outfor convenience.
-
8/11/2019 Circuits Lectures Mod7
50/81
The phasor representation of v(t)=Acos(!t+ ")is visualized as follows:
Note that a positive angle "rotates the phasor by "radians counterclockwise;
a negative "rotates the phasor clockwise. When "= 0, the cosine phasor lies
along the +Real axis. In fact, we might call the +Real axis the cosine axis.
The phasor representation of v(t)=Asin(!t+ ")is visualized as follows:
When "= 0, the sine phasor lies along the Imaginary axis, because
sin(!t) = cos(!t #/2). In fact, we might call the Imaginary axis the sine axis.
Imaginary (sine axis)
+Real (cosine axis)
A"
A!
A!"
/2 +
+Real (cosine axis)
-
8/11/2019 Circuits Lectures Mod7
51/81
Example: Addition of Sinusoids
The utility of phasors is first illustrated by considering the addition of two
60-Hz ac voltage sources in series:
v(t) = 3cos(377t) 4sin(377t)
Using the concept of the cosine and sine axes of the phasor diagram,
we can immediately draw the following arrows and add vectorially to
obtain the resultant, V:
Now, we translate Vback into the time domain to obtain v(t):
v(t) = Real[5
ej
53.1
ej
377t
]
= Real[5 ej(377t+ 53.1)]
= 5cos(377t+ 53.1)
+Real (cosine axis)
+Imaginary (sine axis)
-
8/11/2019 Circuits Lectures Mod7
52/81
Step 3: Translate Vinto the time domain to obtain v t( ):
v t e e t j j t( ) cos( . ). = [ ] = +Real 5 5 377 53 153 1 377
Impedance
Recall that for a resistor, we developed the concept of Ohms Law, which
provides a real number, R, as the ratio of the voltage across the resistor to
the current flowing through it. Now, using phasors, we can extend the
concept of resistance to obtain the voltage / current ratio for resistors,
capacitors, and inductors subjected to a sinusoidal steady-state excitation.
The extension of resistance to the sinusoidal case is called impedance.
The Ideal Resistor
Let the resistor voltage be v t A t R ( ) cos( )= . We identify this as the phasorvoltage VR= A 0. By Ohms Law:
( ) ( ) cos( ) i t v t A t AR R RR R R= = =
I 0
We can now define the resistor impedance ZR as
Z
R
RRR
R
= =
=
V
I
A
A
0
0
Thus, for a resistor, the sinusoidal steady-state impedance is independent
of frequency and is simply the dc resistance:
Z RR =
Impedance
-
8/11/2019 Circuits Lectures Mod7
53/81
The Ideal Capacitor
Let the capacitor voltage be v t A t C( ) cos( )= . We identify this as thephasor voltage VC= A 0 . By the defining relation for a capacitor:
i t dv t
dt
d
dtA t
A t A t A
CC
C
C C
C C C 90
( )( )
cos( )
sin( ) cos( )
= = [ ]
= = + } =
90 I
We can now define the capacitor impedance ZCas
ZC 90
1
C 90
1
C
1
C
CC
C
= =
=
= =
V
I
A
A
j
0
90
Thus, for a capacitor:
Z1
C
1
CC = =
j 90
We see that the capacitor impedance is a pure imaginary number having
a phase of 90 at all frequencies. The capacitor impedance becomes
infinite (an open circuit) at zero frequency (dc). The capacitor impedanceapproaches zero (a short circuit) at infinite frequencies.
In plain language, capacitors tend to block the passage of low-frequency
signals, but facilitate the passage of high-frequency signals.
-
8/11/2019 Circuits Lectures Mod7
54/81
The Ideal Inductor
Let the inductor current be i t A t L( ) cos( )= . We identify this as the phasorcurrent IL = A 0. By the defining relation for an inductor:
v t di t
dt
d
dtA t
A t A t A
LL
L
L L
L L L 90
( )( )
cos( )
sin( ) cos( )
= = [ ]
= = + } =
90 V
We can now define the inductor impedance ZL as
ZL 90
0L LL
L
L
= =
= =V
I
A
Aj90
Thus, for a inductor:
Z L LL = = j 90
We see that the inductor impedance is a pure imaginary number having
a phase of +90 at all frequencies. The inductor impedance becomes zero
(a short circuit) at zero frequency (dc). The inductor impedance approaches
infinity (an open circuit) at infinite frequencies.
In plain language, inductors tend to block the passage of high-frequency
signals, but facilitate the passage of low-frequency signals. This is opposite
the behavior exhibited by the capacitor.
-
8/11/2019 Circuits Lectures Mod7
55/81
Circuit Analysis Using Phasors and Impedance
Example: The Series RC Circuit
We can now use the concepts of phasors and impedance to analyze thesinusoidal steady-state behavior of circuits containing capacitors and / or
inductors. Consider the following example that we briefly touched on before
in the context of setting up a circuit differential equation: the series RC
circuit. Assuming a sinusoidal voltage source, this circuit is drawn as:
R
vC(t)
+
iC(t)
CAcos(t)
Transform to the phasordomain, assigningimpedances to all circuitcomponents.
iR(t)
+
R
VC
+
IC
1 /j CA0
Apply KCL at the nodejoining R and C:
IC+ IR= 0
IR
+
-
8/11/2019 Circuits Lectures Mod7
56/81
I IC R+ = 0
V VC C
CR1
00
j
A
+
=
Solving this algebraicequation for VC , we obtain the following ratio:
VCRC
= +A
j
0
1
Note that, whenever we have such a ratio of complex numbers, it is always
convenient to express the numerator and denominator in polar form.
Here, the numerator is already in polar form. Lets apply Eulers identity to
express the denominator in polar form:
1+ jRC
Hence,
1 1 2 1+ + j RC = RC RC( ) tan ( )
Real
Imaginary
RC
1
1 2+ ( )RC
tan1(RC / 1 )
-
8/11/2019 Circuits Lectures Mod7
57/81
Now, we can write VC as
VCRC RC / 1
RCRC)
( ) tan ( )
( )tan (
=
+
=+
A
A
0
1
1
2 1
2
1
The transition back to the time domain to obtain the final answer for the
capacitor voltage can now be made by inspection:
v t A
tCRC
RC)( )( )
cos tan (=+
[ ]1 2
1
If the capacitor current is also needed, we apply the definition of the
capacitor impedance, I VC C CZ= / :
I V
V VCC
C C
C
C C =
= = ( ) 1
1 90
j
j
and substitute our phasor solution for VC from the top of this page:
IC CRC
RC)
C
RCRC)
( )
tan (
( )
tan (
= ( ) +
=+
[ ]
1 901
190
2
1
2
1
A
A
-
8/11/2019 Circuits Lectures Mod7
58/81
The transition back to the time domain to obtain the final answer for the
capacitor current can now be made by inspection:
IC CRC
RC)( )
tan (= + [ ] A1 9021
i t A
tCC
RCRC)( )
( )cos tan (=
++ [ ]
190
2
1
Two notes of interest:
1) The magnitude of a capacitors sinusoidal current is simply the
magnitude of its sinusoidal voltage multiplied by C . EEs say that a
capacitors admittance is C . We see that this admittance is zero
at dc ( = 0 ) and approaches infinity at very high frequencies.
2) The phase angle of a capacitors sinusoidal current is always 90 plusthe phase angle of its sinusoidal voltage, regardless of the frequency, .
EEs say that a capacitors sinusoidal current always leads its
sinusoidal voltage by 90; or equivalently that a capacitors sinusoidal
voltage always lags its sinusoidal current by 90.
-
8/11/2019 Circuits Lectures Mod7
59/81
Circuit Simplification Via Combining Impedances
Series and parallel impedances combine just like series and parallel
resistors. Namely, series impedances add, and the reciprocals of parallelimpedances add to yield the reciprocal of the composite impedance.
Here is an example of combining the impedances of a resistor and a
capacitor connected in series to quickly obtain a phasor expression for the
generator current I:
I
( ) tan (
=
+
=
=+
[ ]
A
j
j A
j
A
0
1
190
2
1
RC
C
1 + RC
C
RCRC)
This is the same expression as that obtained earlier for IC using the node
equation method.
R
I
1 /j CA0+
Zj
equiv RC
= +1
-
8/11/2019 Circuits Lectures Mod7
60/81
Lets sketch the impedance magnitude behavior of this series R C
combination versus frequency:
R
Near dc (= 0), the combined impedance is dominated by the capacitor,
whose impedance goes to infinity (i.e., an open circuit) at exactly = 0.
As , the combined impedance approaches that of the resistor alonebecause the capacitors impedance asymptotically approaches zero
(i.e., a short circuit).
R
1 /j CZj
equiv RC
= +1
Zequiv
0
0
~ /1 C
R
-
8/11/2019 Circuits Lectures Mod7
61/81
Lets sketch the impedance magnitude behavior of a series L C
combination versus frequency:
Near dc (= 0), the combined impedance is dominated by the capacitor
whose impedance goes to infinity (i.e., an open circuit) at exactly = 0.
As , the combined impedance approaches that of the inductor alonebecause the capacitors impedance approaches zero (i.e., a short circuit).
At the resonant frequency = 1 / LC , the combined impedance is zero,
i.e., a short circuit.
j L
1 /j C
Z jj
j
equiv LC
1 LC
C
= +
=
1
2
Zequiv
0
0
1 / LC
~ /1 C
~ L
-
8/11/2019 Circuits Lectures Mod7
62/81
Finally, lets sketch the impedance magnitude behavior of a
parallel L Ccombination versus frequency:
Near dc (= 0), the combined impedance is dominated by the inductor
whose impedance goes to zero (i.e., a short circuit) at exactly = 0.
As , the combined impedance is dominated by the capacitor whoseimpedance approaches zero asymptotically.
At the resonant frequency = 1 / LC , the combined impedance is infinite,
i.e., an open circuit.
j L 1 /j C
11
2
/Z jj
Z j
equiv
equiv
CL
L
1 LC
= +
=
Zequiv
0
0
1 / LC
~ /1 C~ L
-
8/11/2019 Circuits Lectures Mod7
63/81
Transfer Functions and Filtering
Example: The Series RC Circuit
An important application of circuit analysis using phasors and impedanceis the derivation of the output / input relation of what EEs call two-port
networks. Consider the series RC circuit drawn in a slightly different way:
Here, the magnitude of the sinusoidal voltage exciting the RC circuit is
identified as Vin rather thanA, and the voltage across the capacitor is
identified as v tout ( ) rather than v tC( ).
R
vout(t)
+
CVin cos(t)+
vout(t)
+
CVin cos(t)+
Two-port network
-
8/11/2019 Circuits Lectures Mod7
64/81
We can use the results of our previous analysis of this circuit for VC
(Slides 55 57). Or, more simply, we can treat this circuit as a voltage
dividerwherein the voltage division is accomplished via ratios ofimpedances, rather than ratios of resistors:
We call this ratio H(!), the transfer functionof the two-port network.
The magnitude of H(!), given by
is simply the ratio of the amplitude of the output sine wave relative to the
input sine wave. The phase of H(!), given by
!H(") = !Vout
(") #!Vin
(") = # tan#1("RC)
tells us by how many radians (or degrees) the phase of the output sine
wave differs from the phase of the input sine wave.
Vout
Vin
=(1 /j!C)
R + (1 /j!C) =
1
1+j!RC
=
1
1+(!RC)2"#tan
#1(!RC)
H(!)
=
Vout
Vin
(!)
=
11+(!RC)
2
-
8/11/2019 Circuits Lectures Mod7
65/81
We can learn much about the behavior of this circuit by graphing H
and Hversus frequency. First, remembering that = 2 f , we have
H f
f
f f f B( )
( )
( ) ( ) ( / )= = + = +
V
V
out
in RC
1
1 2
1
12 2
= = = H f f f f f B( ) ( ) ( ) tan ( ) tan ( / )V Vout in RC1 12
where the characteristic frequency B = 1 2/ RCis defined as the circuitsbandwidth. Illustrative graphs of H f( ) and H f( ) follow:
f (Hz)
H f( )
1
0
1 2
B = 1 2/ RC0
H f( )
f (Hz)
B = 1 2/ RC0
45
90
passband stopband
-
8/11/2019 Circuits Lectures Mod7
66/81
With this circuit, input sinusoidal voltages having frequencies below
BHz appear at the output with relatively little attenuation, whereas input
sinusoidal voltages having frequencies aboveBare suppressed. In the limit
as the input signal frequency rises well aboveB, almost no output signal is
obtained.
This is the action of what EEs call a low-pass filter. The frequency
range belowBis called the passband, and the frequency range aboveB
is called the stopband.
Low-pass filters have important engineering applications in separating
desired signals from undesired (interfering) signals and noise. This action is
called windowing. For example, to optimally window a voice signal in a
telephone system, we would select R and C to yield B 3 kHz, since most
energy in the human voice is concentrated in sinusoidal components have
frequencies between 0 (dc) and 3 kHz. Here are the approximate
bandwidths for three other signals of interest that can be effectively
decomposed into low-pass collections of sinusoidal components:
audio from the Chicago Symphony Orchestra 20 kHz
color TV video signals 5 MHz
digital bit stream in your computer 10 GHz
-
8/11/2019 Circuits Lectures Mod7
67/81
-
8/11/2019 Circuits Lectures Mod7
68/81
R-L High-pass Filter
Vin
R
j!L Vout
Vout
Vin
=j!L
R + j!L
Vout
Vin
0
1
1
2
0!
R/L
PassbandStopband
3-db cutoff
-
8/11/2019 Circuits Lectures Mod7
69/81
R-L-C Band-pass (Peak) Filter
Vin
R
!L Vout1
j!C
Vout
Vin
=
ZL||C
R + ZL||C
Vout
Vin
0
1
1
2
0!
1
LC
3-db bandwidth
Passband
Stopband Stopband
-
8/11/2019 Circuits Lectures Mod7
70/81
R-L-C Band-stop (Notch) Filter
Vin
!L
Vout1
j!CR
Vout
Vin
=R
R + ZL||C
Vout
Vin
0
1
1
2
0!
1
LC
3-db bandwidth
Passband
Stopband
Passband
-
8/11/2019 Circuits Lectures Mod7
71/81
Typical Spectra of Radio Stations
0 db
10 db
20 db
30 db
40 db
50 db
Localradio
station
Distantradio
station #1
Distantradio
station #2
frequen
signalstrength
noise floor
-
8/11/2019 Circuits Lectures Mod7
72/81
Tuning In the Local Station
0 db
10 db
20 db
30 db
40 db
50 db
Localradio
station
Distantradio
station #1
Distantradio
station #2
frequen
signalstrength
noise floor
H(f)
-
8/11/2019 Circuits Lectures Mod7
73/81
Tuning In Distant Radio Station #1
0 db
10 db
20 db
30 db
40 db
50 db
Localradio
station
Distantradio
station #1
Distantradio
station #2
frequen
signalstrength
noise floor
H(f)
-
8/11/2019 Circuits Lectures Mod7
74/81
Tuning In Distant Radio Station #2
0 db
10 db
20 db
30 db
40 db
50 db
Localradio
station
Distantradio
station #1
Distantradio
station #2
frequen
signalstrength
noise floor
H(f)
-
8/11/2019 Circuits Lectures Mod7
75/81
Time-Average Power in theSinusoidal Steady State
Consider an arbitrary impedance,Z, in a
circuit that is operating in the sinusoidal
steady state:
Instantaneous power absorbed byZ:
vmax
cos!t
imaxcos(!t+")
Z
p(t) = vmaxcos!t"imaxcos(!t+#)
=vmaximax
2cos(2!t+#)+cos#[ ]
-
8/11/2019 Circuits Lectures Mod7
76/81
Time-average power absorbed byZ:
is often called the power factor.
This ranges from 1for a resistor, where the
voltage and current are in phase, to 0for a
capacitor or inductor, where the voltage and
current are 90 out of phase.
Hence, in the sinusoidal steady state, acapacitor or inductor has 0 time-average
power dissipation.
paverage =1T
p(t)dtT! } where Tis one sinusoidal period at "
=vmaximax
2Tcos(2"t+#)+cos#[ ]dt
T
!
=
vmaximax
2T0
+
Tcos#[ ]
=
vmaximax
2cos#
cos!
-
8/11/2019 Circuits Lectures Mod7
77/81
Another interpretation of time-average
power: Dot product of the voltage and
current phasors, and .!v !i
!v
!i!
real
imaginary
paverage =vmaximax
2cos! =
1
2!v"!i
-
8/11/2019 Circuits Lectures Mod7
78/81
We can expand this expression to derive the
universally used formula for the time-average
power dissipated by an impedance in the
sinusoidal steady state:
paverage
=1
2!v!!i =
1
2!vreal
!ireal
+!vimag
!iimag( )
=1
2Real !vreal + j!vimag( ) !ireal" j!iimag( )#$ %&
=
12Real !v !i
*( )
-
8/11/2019 Circuits Lectures Mod7
79/81
Maximum Power Transfer Theorem
Given the Thevenin equivalent circuit of asinusoidal steady-state power source with
fixed andZThand a variable load,Zload.
Then, the maximum time-average power
that can be transferred from the power
source to the load occurs whenZloadis set
equal to the complex conjugate ofZTh
, i.e.:
Zload
= ZTh
*
!vTh
ZTh
!vTh
Zload
(fixed)
(fixed)
(variable)
Powersource
-
8/11/2019 Circuits Lectures Mod7
80/81
Impedance Matching forMaximum Power Transfer
Often, all three quantities ( ,ZTh, andZload) are
fixed. Then, a lossless two-port matching
network can be inserted betweenZloadand the
power source as follows:
The matching network transforms the load
impedance such that the power sourcesenses at its terminals. Now, the
maximum possible time-average power is
transferred from the power source to the load.
!vTh
ZTh
!vTh
Zload
(fixed)
(fixed)
(fixed)
Powersource
Losslesstwo-port
matching
network
ZTh
*
ZTh
*
-
8/11/2019 Circuits Lectures Mod7
81/81
Example of impedance transformation: Using
two capacitors and one inductor, transform a
very low 1!antenna impedance to match the100!Thevenin impedance of a transmitter
operating at a frequency of 10 MHz:
Let C= 0.00159Fand L = 0.159 H. At 10 MHz,
ZC = j10! and ZL = j10!. This yields:
Zload= 1!
Lossless two-portmatching network
C C
LZtransformed
j10 j10
j10