Transcript No Slide Title

CIRCUITS 1
DEVELOP TOOLS FOR THE ANALYSIS AND DESIGN OF
BASIC LINEAR ELECTRIC CIRCUITS
A FEW WORDS ABOUT ANALYSIS
USING MATHEMATICAL MODELS
BASIC STRATEGY USED IN ANALYSIS
MATHEMATICAL ANALYSIS
DEVELOP A SET OF MATHEMATICAL
EQUATIONS THAT REPRESENT THE CIRCUIT
- A MATEMATICAL MODEL LEARN HOW TO SOLVE THE MODEL TO
DETERMINE HOW THE CIRCUIT WILL BEHAVE
IN A GIVEN SITUATION
THIS COURSE TEACHES THE BASIC TECHNIQUES
TO DEVELOP MATHEMATICAL MODELS FOR
ELECTRIC CIRCUITS
THE MODELS THAT WILL BE DEVELOPED HAVE
NICE MATHEMATICAL PROPERTIES.
IN PARTICULAR THEY WILL BE LINEAR WHICH
MEANS THAT THEY SATISFY THE PRINCIPLE OF
SUPERPOSITION
T ( 1u1   2u2 )   1T (u1 )   2T (u2 )
Principle of Superposition
y  Tu
Model
THE MATHEMATICS CLASSES - LINEAR ALGEBRA,
DIFFERENTIAL EQUATIONS- PROVIDE THE TOOLS
TO SOLVE THE MATHEMATICAL MODELS
FOR THE FIRST PART WE WILL BE EXPECTED
TO SOLVE SYSTEMS OF ALGEBRAIC EQUATIONS
12V1  9V2  4V3  8
 4V1  16V2  V3  0
 2V1  4V2  6V3  20
LATER THE MODELS WILL BE DIFFERENTIAL
EQUATIONS OF THE FORM
dy
y f
dt
d2y
dy
df

4

8
y

3
4f
2
dt
dt
dt
3
ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS
2 TERMINALS COMPONENT
a
b
NODE
characterized by the
current through it and
the voltage difference
betweeb terminals
NODE
L
R1
R2
vS

TYPICAL LINEAR
CIRCUIT
vO

+
-
C
LOW DISTORTION POWER AMPLIFIER
BASIC CONCEPTS
LEARNING GOALS
•System of Units: The SI standard system; prefixes
•Basic Quantities: Charge, current, voltage, power and energy
•Circuit Elements: Active and Passive
http://physics.nist.gov/cuu/index.html
Information at the f
modern science an
from the Physics L
Detailed contents
Values of the constants
and related infor
Searchable bibliography
on the constant
In-depth information on the SI
, the mode
metric system
Guidelinesfor the expression
SI DERIVED BASIC ELECTRICAL UNITS
ONE AMPERE OF CURRENT CARRIES ONE COULOMB OF CHARGE EVERY SECOND.
ACs
1 COULOMB  6.28 1018 (e)
(e) IS THE CHARGE OF ONE ELECTRON
VOLT IS A MEASURE OF ENERGY PER CHARGE.
TWO POINTS HAVE A VOLTAGE DIFFERENCE OF ONE VOLT IF ONE COULOMB OF CHARGE
GAINS ONE JOULE OF CHARGE WHEN IT IS MOVED FROM ONE POINT TO THE OTHER.
V
J
C
OHM IS A MEASURE OF THE RESISTANCE TO THE FLOW OF CHARGE.
THERE IS ONE OHM OF RESISTENCE IF IT IS REQUIRED ONE VOLT OF ELECTROMOTIVE FORCE
TO DRIVE THROUGH ONE AMPERE OF CURRENT

V
A
IT IS REQUIRED ONE WATT OF POWER TO DRIVE ONE AMPER OF CURRENT AGAINST AN
ELECTROMOTIVE DIFFERENCE OF ONE VOLTS
W V  A
CURRENT AND VOLTAGE RANGES
Strictly speaking current is a basic quantity and charge is derived. However,
physically the electric current is created by a movement of charged particles.
What is the meaning of a negative value for q(t)?




q(t )
PROBLEM SOLVING TIP
IF THE CHARGE IS GIVEN DETERMINE THE CURRENT BY
DIFFERENTIATION
IF THE CURRENT IS KNOWN DETERMINE THE CHARGE BY
INTEGRATION
A PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRIC
CURRENTS IS THAT OF WATER FLOW.
CHARGES ARE VISUALIZED AS WATER PARTICLES
EXAMPLE
EXAMPLE
 0 t 0
i (t )   2 t
e mA t  0
q(t )  4 103 sin( 120 t )[C ]




i (t )  4 103 120 cos(120 t ) [ A]
i (t )  0.480 cos(120 t ) [mA ]
FIND THE CHARGE THAT PASSES
DURING IN THE INTERVAL 0<t<1
1
q(t )
q  e
0
2 x
1
1
1
1
dx   e 2 x   e 2  ( e 0 )
2
2
2
0
1
q  (1  e 2 )
2
Units?
FIND THE CHARGE AS A FUNCTION OF TIME
q(t ) 
t
t
 i ( x)dx   e

2 x
dx

t  0  q( t )  0
t
1
t  0  q(t )   e 2 x dx  (1  e 2t )
2
0
And the units for the charge?...
DETERMINE THE
CURRENT
Charge(pC)
Here we are given the
charge flow as function
of time.
30
20
10
 10
 10  1012  10  1012 C
9
m


10

10
(C / s )
3
s
2  10  0
1 2 3 4 5 6
Time(ms)
Current(nA )
To determine current we
must take derivatives.
PAY ATTENTION TO
UNITS
40
30
20
10
 10
 20
1 2 3 4 5 6
Time(ms)
CONVENTION FOR CURRENTS
THE DOUBLE INDEX NOTATION
IT IS ABSOLUTELY NECESSARY TO INDICATE
THE DIRECTION OF MOVEMENT OF CHARGED
PARTICLES.
THE UNIVERSALLY ACCEPTED CONVENTION IN
ELECTRICAL ENGINEERING IS THAT CURRENT IS
FLOW OF POSITIVE CHARGES.
AND WE INDICATE THE DIRECTION OF FLOW
FOR POSITIVE CHARGES
-THE REFERENCE DIRECTIONA POSITIVE VALUE FOR
THE CURRENT INDICATES
FLOW IN THE DIRECTION
OF THE ARROW (THE
REFERENCE DIRECTION)
IF THE INITIAL AND TERMINAL NODE ARE
LABELED ONE CAN INDICATE THEM AS
SUBINDICES FOR THE CURRENT NAME
a
5A
b
I ab  5 A
a 3A b a  3A b
I ab  3 A
I ab  3 A
a  3A b a 3A b
A NEGATIVE VALUE FOR
THE CURRENT INDICATES
FLOW IN THE OPPOSITE
DIRECTION THAN THE
REFERENCE DIRECTION
I ba  3 A
POSITIVE CHARGES
FLOW LEFT-RIGHT
I ba  3 A
POSITIVE CHARGES
FLOW RIGHT-LEFT
I ab   I ba
I  2 A
a
2A
I
b
I cb  4 A
I ab 
c
3A
This example illustrates the various ways
in which the current notation can be used
CONVENTIONS FOR VOLTAGES
ONE DEFINITION FOR VOLT
TWO POINTS HAVE A VOLTAGE DIFFERENTIAL OF
ONE VOLT IF ONE COULOMB OF CHARGE GAINS
(OR LOSES) ONE JOULE OF ENERGY WHEN IT
MOVES FROM ONE POINT TO THE OTHER
b IF THE CHARGE GAINS
ENERGY MOVING FROM
a TO b THEN b HAS HIGHER
VOLTAGE THAN a.
IF IT LOSES ENERGY THEN
b HAS LOWER VOLTAGE
THAN a
 a
1C
DIMENSIONALLY VOLT IS A DERIVED UNIT
VOLT 
JOULE
N m

COULOMB A  s
VOLTAGE IS ALWAYS MEASURED IN A RELATIVE FORM AS THE VOLTAGE DIFFERENCE
BETWEEN TWO POINTS
IT IS ESSENTIAL THAT OUR NOTATION ALLOWS US TO DETERMINE WHICH POINT
HAS THE HIGHER VOLTAGE
THE + AND - SIGNS
DEFINE THE REFERENCE
POLARITY
V
IF THE NUMBER V IS POSITIVE POINT A HAS V
VOLTS MORE THAN POINT B.
IF THE NUMBER V IS NEGATIVE POINT A HAS
|V| LESS THAN POINT B
POINT A HAS 2V MORE
THAN POINT B
POINT A HAS 5V LESS
THAN POINT B
THE TWO-INDEX NOTATION FOR VOLTAGES
INSTEAD OF SHOWING THE REFERENCE POLARITY
WE AGREE THAT THE FIRST SUBINDEX DENOTES
THE POINT WITH POSITIVE REFERENCE POLARITY
WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROM
POINT B TO POINT A IN THE CIRCUIT?
V AB  2V
V
W
 W  VQ  240J
Q
THE CHARGES MOVE TO A POINT WITH HIGHER
VOLTAGE -THEY GAINED (OR ABSORBED) ENERGY
THE VOLTAGE
DIFFERENCE
IS 5V

5V
VAB  5V
V AB  VBA
VBA  5V

WHICH POINT HAS THE
HIGHER VOLTAGE?
V AB  5V
EXAMPLE
A CAMCODER BATTERY PLATE CLAIMS THAT
THE UNIT STORES 2700mAHr AT 7.2V.
WHAT IS THE TOTAL CHARGE AND ENERGY
STORED?
ENERGY AND POWER
2[C/s] PASS
THROUGH
THE ELEMENT
CHARGE
THE NOTATION 2700mAHr INDICATES THAT
THE UNIT CAN DELIVER 2700mA FOR ONE
FULL HOUR
s
C 
Q  2700 103    3600
1Hr
Hr
S
 9.72 103[C ]
TOTAL ENERGY STORED
THE CHARGES ARE MOVED THROUGH A 7.2V
VOLTAGE DIFFERENTIAL
J 
W  Q[C ] V    9.72 103  7.2[ J ]
C 
 6.998 10 [ J ]
4
EACH COULOMB OF CHARGE LOSES 3[J]
OR SUPPLIES 3[J] OF ENERGY TO THE
ELEMENT
THE ELEMENT RECEIVES ENERGY AT A
RATE OF 6[J/s]
THE ELECTRIC POWER RECEIVED BY THE
ELEMENT IS 6[W]
IN GENERAL
P  VI
t2
w (t 2 , t1 )   p( x )dx
t1
HOW DO WE RECOGNIZE IF AN ELEMENT
SUPPLIES OR RECEIVES POWER?
PASSIVE SIGN CONVENTION
POWER RECEIVED IS POSITIVE WHILE POWER
SUPPLIED IS CONSIDERED NEGATIVE
 Vab 
a
b
I ab
P  Vab I ab
IF VOLTAGE AND CURRENT
ARE BOTH POSITIVE THE
CHARGES MOVE FROM
HIGH TO LOW VOLTAGE
AND THE COMPONENT
RECEIVES ENERGY --IT IS
A PASSIVE ELEMENT
A CONSEQUENCE OF THIS CONVENTION IS THAT
THE REFERENCE DIRECTIONS FOR CURRENT AND
VOLTAGE ARE NOT INDEPENDENT -- IF WE
ASSUME PASSIVE ELEMENTS
GIVEN THE REFERENCE POLARITY
 Vab 
a
b
REFERENCE DIRECTION FOR CURRENT
THIS IS THE REFERENCE FOR POLARITY

a

b
I ab
IF THE REFERENCE DIRECTION FOR CURRENT
IS GIVEN
EXAMPLE
 Vab 
2A
a I
b
ab
Vab  10V
THE ELEMENT RECEIVES 20W OF POWER.
WHAT IS THE CURRENT?
SELECT REFERENCE DIRECTION BASED ON
PASSIVE SIGN CONVENTION
20[W ]  Vab I ab  (10V ) I ab
I ab  2[ A]
UNDERSTANDING PASSIVE SIGN CONVENTION
We must examine the voltage across the component
and the current through it
I
A
A’

V
S1
B
Current A - A'
positive
positive
positive negative
negative positive
negative negative
Voltage(V)
PS1  V AB I AB

S1
supplies
receives
receives
supplies
S2
PS 2  V A'B ' I A'B '
B’
S2
ON S1
ON S2
receives VAB  0, I AB  0 VA B  0, I A B  0
supplies
ON S2
supplies
VA'B '  0, I A'B '  0
receives
'
'
'
'
CHARGES RECEIVE ENERGY.
THIS BATTERY SUPPLIES ENERGY
CHARGES LOSE ENERGY.
THIS BATTERY RECEIVES THE ENERGY
WHAT WOULD HAPPEN IF THE CONNECTIONS ARE REVERSED
IN ONE OF THE BATTERIES?
DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWER
AND HOW MUCH
a
a
Vab  2V
I ab  4 A
2A
I ab  2 A
Vab  2V
P  8W
SUPPLIES POWER
b
P  4W
RECEIVES POWER
WHEN IN DOUBT LABEL THE TERMINALS
OF THE COMPONENT
1
1
2
2
V12  12V , I12  4 A
V12  4V , I12  2 A
b
I  8[ A]
VAB  4[V ]
 20[W ]  VAB  (5 A)




SELECT VOLTAGE REFERENCE POLARITY
BASED ON CURRENT REFERENCE DIRECTION
40[W ]  (5V )  I
WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION
V1  20[V ]
40[W ]  V1  (2 A)
 2A
I  5[ A]
SELECT HERE THE CURRENT REFERENCE DIRECTION
 50[W ]  (10[V ])  I
BASED ON VOLTAGE REFERENCE POLARITY
COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT
P1  (6V )( 2 A)
2 A  6V 



1
24V
+
-
3
2
2A
P1 = 12W
P2 = 36W
P3 = -48W
18V

P2  (18V )( 2 A)
P3  (24V )( 2 A)  (24V )( 2 A)
IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT
CIRCUIT ELEMENTS
PASSIVE ELEMENTS
VOLTAGE
DEPENDENT
SOURCES
UNITS FOR  , g , r ,  ?
INDEPENDENT SOURCES
CURRENT
DEPENDENT
SOURCES
EXERCISES WITH DEPENDENT SOURCES
FIND VO
VO  40[V ]
FIND I O
I O  50mA
DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES
40[V ]
P  (40[V ])( 2[ A])  80[W ]
TAKE VOLTAGE POLARITY REFERENCE
P  (10[V ])( 4  4[ A])  160[W ]
TAKE CURRENT REFERENCE DIRECTION
POWER ABSORBED OR SUPPLIED BY EACH
ELEMENT
USE POWER BALANCE TO COMPUTE Io
 12W
(12)( 9)
(6)( I O )
(10)( 3)
(4)(8)
P1  (12V )( 4 A)  48[W ]
P2  (24V )( 2 A)  48[W ]
P3  (28V )( 2 A)  56[W ]
PDS  (1I x )( 2 A)  (4V )( 2 A)  8[W ]
P36V  (36V )( 4 A)  144[W ]
NOTICE THE POWER BALANCE
POWER BALANCE
I O  1[ A]
(8  2)(11)