Circuits Lectures Mod7

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


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


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


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

+

+


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"


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.


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.


8/81

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


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

+ +

+ +


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


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


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.


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.


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 i i10 resistor watts ( ) ( . ) .= = =10 10 0 173 0 30041 22 2


Total power dissipated = .7 54417 5 029451846 0 3004 3 557037031

16 431

+ . + . + .

= . watts

Power Balance!


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.


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


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

+ +

+


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


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:


. .= = =5 5 8 4 42 22 W

Net generated power = 183 42 2 6 6 218 5 . . .+ = W


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!


19/81

Circuit Analysis Example 3: Node analysis (Kirchoffs Current Law)


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


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


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


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.


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.


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

+


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


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


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


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


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


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

+ +

=


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


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 /


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


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


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


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


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


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


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


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)


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)


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


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


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)


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)

+


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)


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.


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.


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


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.


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)


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)


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


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.


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.


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

+


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 )


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


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.


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


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


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


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


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


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


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


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


67/81


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


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


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


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


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)


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)


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)


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#[ ]


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!


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


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

*( )


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


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

*


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

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Circuits Lectures Mod7

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