Slide 1 EE42/43/100 Summer 2013 Prof.

Chang-Hasnain
EE 42/43/100
Introduction to Digital
Electronics

Lecture 3
6/28/13
Instructors:
Prof. Connie Chang-Hasnain
Dr. Wenbin Hsu

Slide 2 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Lecture 3
Outline
Equivalent Circuits
Resistors in Series Voltage Divider
Conductances in Parallel Current Divider
Delta-Y Transformation
Wheatstone Bridge
Superposition
Thvenin equivalent circuits
Norton equivalent circuits

Slide 3 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Equivalent Circuit Concept
Two circuits are equivalent between a pair of nodes if
the i-v characteristics are identical.
The circuits can be a network of voltage sources,
current sources, and resistors can be replaced by an
equivalent circuit without affecting the operation of
the rest of the circuit.
+
v
A
_
network A
of
sources
and
resistors
i
A

+
v
B
_
network B
of
sources
and
resistors
i
B
i
A
(v
A
) = i
B
(v
B
)
Slide 4 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Consider a circuit with multiple resistors connected in series.
Find their equivalent resistance.
KCL tells us that the same
current (I ) flows through
every resistor
What does KVL tell us?
Equivalent resistance of resistors in series is the sum
R
2
R
1
I
R
3
Resistors in Series

=
1
+
2
+
3

V
SS


+
+

V
1
+

V
2
+

V
3
Slide 5 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
SS
4 3 2 1
2
2
V
R R R R
R
V
+ + +
=
Correct, if nothing else
is connected to nodes



SS
4 3 2 1
2
2
V
R R R R
R
V
+ + +

When can the Voltage Divider Formula be Used?
+

V
2
R
2
R
1
V
SS

I
R
3
R
4

+
R
2
R
1
V
SS

I
R
3
R
4

+
R
5
+

V
2
Slide 6 EE42/43/100 Summer 2013 Prof. Chang-Hasnain


What is V
2
?
SS
eq 2 1
2
2
V
R R R
R
V
+ +
=
When can the Voltage Divider Formula be Used?
R
2
R
1
V
SS

I
R
3
R
4

+
R
5
+

V
2
Slide 7 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
KVL tells us that the
same voltage is dropped
across each resistor
V
x
=I
1
R
1
=I
2
R
2
What does KCL tell us?
R
2
R
1
I
SS

I
2
I
1
x
Resistors in Parallel
Consider a circuit with two resistors connected in parallel.
Find their equivalent resistance.
Slide 8 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
What single resistance R
eq
is equivalent to three resistors in parallel?
+

V
I
V
+

I
R
3
R
2
R
1

R
eq

eq

Parallel Resistors and Current Divider
Equivalent conductance of resistors in parallel is the sum
|
|
.
|

\
|
+ |
.
|

\
|
+ |
.
|

\
|
=
3 2 1
R
1
R
1
R
1
I
V
(

+ +
= =
3 2 1
3
3
3
1/R 1/R 1/R
1/R
I
R
V
I
Generalized Current Divider Formula
Slide 9 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
To measure the voltage drop across an element in a
real circuit, insert a voltmeter (digital multimeter in
voltage mode) in parallel with the element.
Voltmeters are characterized by their voltmeter input
resistance (R
in
). Ideally, this should be very high
(typical value 10 MO)
Ideal
Voltmeter
R
in
Measuring Voltage
Slide 10 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
(

+
=
2 1
2
SS 2
R R
R
V V
(

+
=
'
1 in 2
in 2
SS 2
R R || R
R || R
V V
Example: V 1 V K 900 R , K 100 R , V 10 V
2 1 2 SS
= = = =
V
SS

R
1
R
2
2
10 , ?
in
R M V
'
= =
Effect of Voltmeter
undisturbed circuit
circuit with voltmeter inserted
_
+
+


V
2
V
SS

R
1
R
2 R
in
_
+
+


V
2

Compare to R
2
Slide 11 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
To measure the current flowing through an element in a
real circuit, insert an ammeter (digital multimeter in
current mode) in series with the element.
Ammeters are characterized by their ammeter input
resistance (R
in
). Ideally, this should be very low (typical
value 1O).
Ideal
Ammeter
R
in
Measuring Current
Slide 12 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
R
in
V
1
I
meas
R
1
R
2
ammeter
circuit with ammeter inserted
_
+
V
1
I
R
1
R
2
undisturbed circuit
Example: V
1
= 1 V, R
1
= R
2
= 500 O, R
in
= 1O
2 1
1
R R
V
I
+
=
in 2 1
1
meas
R R R
V
I
+ +
=
1
1 , ?
500 500
meas
V
I mA I = = =
O+ O
Effect of Ammeter
Measurement error due to non-zero input resistance:
_
+
Compare to
R
2
+ R
2
Slide 13 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Simplify a circuit before applying KCL and/or KVL:

+
7 V
Using Equivalent Resistances
R
1
= R
2
= 3 kO
R
3
= 6 kO
R
4
= R
5
= 5 kO
R
6
= 10 kO
I
R
1
R
2
R
4
R
5
R
3
R
6
Example: Find I
Slide 14 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Y-Delta Conversion
These two resistive circuits are equivalent for
voltages and currents external to the Y and A
circuits. Internally, the voltages and currents
are different.
R
1
R
2
R
3
c
b
a
R
1
R
2
R
3
c
b
R
1
R
2
R
3
c
b
a
R
c
a b
c
R
b
R
a
R
c
a b
c
R
b
R
a
a b
c
R
b
R
a
R
1
=

R
b
R
c

R
a
+ R
b
+ R
c
R
2
=

R
a
R
c

R
a
+ R
b
+ R
c
R
3
=

R
a
R
b

R
a
+ R
b
+ R
c
Slide 15 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
In order for the Delta interconnection to be equivalent
to the Wye interconnection, the resistance between
corresponding terminal pairs must be the same
R
ab
= = R
1
+ R
2
R
c
(R
a
+ R
b
)
R
a
+ R
b
+ R
c
R
bc
= = R
2
+ R
3
R
a
(R
b
+ R
c
)
R
a
+ R
b
+ R
c
R
ca
= = R
1
+ R
3
R
b
(R
a
+ R
c
)
R
a
+ R
b
+ R
c
R
c
a b
c
R
b
R
a
R
1
R
2
R
3
c
b
a
Slide 16 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
A-Y and Y-A Conversion Formulas
R
1
R
2
R
3
c
b
a
R
1
R
2
R
3
c
b
R
1
R
2
R
3
c
b
a
R
1
=

R
b
R
c

R
a
+ R
b
+ R
c
R
2
=

R
a
R
c

R
a
+ R
b
+ R
c
R
3
=

R
a
R
b

R
a
+ R
b
+ R
c
Delta-to-Wye conversion Wye-to-Delta conversion
R
a
=

R
1
R
2
+ R
2
R
3
+ R
3
R
1

R
1
R
b
=

R
1
R
2
+ R
2
R
3
+ R
3
R
1

R
2
R
c
=

R
1
R
2
+ R
2
R
3
+ R
3
R
1

R
3
R
c
a b
c
R
b
R
a
R
c
a b
c
R
b
R
a
a b
c
R
b
R
a
Slide 17 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Circuit Simplification Example
Find the equivalent resistance R
ab
:
2O

a
b
18O

6O

12O

4O

9O


2O

a
b
4O

9O

Slide 18 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
The Wheatstone Bridge
Circuit used to precisely measure resistances in
the range from 1 O to 1 MO, with 0.1% accuracy
R
1
and R
2
are resistors with known values
R
3
is a variable resistor (typically 1 to 11,000O)
R
x
is the resistor whose value is to be measured
+
V

R
1
R
2
R
3
R
x
current detector
battery
variable resistor
Slide 19 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Finding the value of R
x
Adjust R
3
until there is no current in the detector
Then,
+
V

R
1
R
2
R
3
R
x
R
x
= R
3
R
2
R
1
Derivation:

i
1
i
2
i
x
i
3
Typically, R
2
/ R
1
can be varied
from 0.001 to 1000 in decimal steps
Slide 20 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Finding the value of R
x
Adjust R
3
until there is no current in the detector
Then,
+
V

R
1
R
2
R
3
R
x
R
x
= R
3
R
2
R
1
Derivation:
i
1
= i
3
and i
2
= i
x
i
3
R
3
= i
x
R
x
and i
1
R
1
= i
2
R
2
i
1
R
3
= i
2
R
x




KCL =>
KVL =>
R
3
R
1
R
x
R
2
=
i
1
i
2
i
x
i
3
Typically, R
2
/ R
1
can be varied
from 0.001 to 1000 in decimal steps
Slide 21 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Some circuits must be analyzed (not amenable to simple inspection)
-
+
R
2
R
1
V
I
R
4
R
3
R
5
Special cases:
R
3
= 0 OR R
3
=
R
1


+
R
4

R
5

R
2

V
R
3

Identifying Series and Parallel Combinations
Slide 22 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Superposition
A linear circuit is one constructed only of linear
elements (linear resistors, and linear capacitors and
inductors, linear dependent sources) and
independent sources. Linear
means I-V charcteristic of elements/sources are
straight lines when plotted
Principle of Superposition:
In any linear circuit containing multiple
independent sources, the current or voltage at
any point in the network may be calculated as
the algebraic sum of the individual
contributions of each source acting alone.
Slide 23 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Voltage sources in series can be replaced by an
equivalent voltage source:




Current sources in parallel can be replaced by
an equivalent current source:
Source Combinations
i
1
i
2

i
1
+i
2

+

+
v
1
v
2


+
v
1
+v
2
Slide 24 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Superposition
Procedure:
1. Determine contribution due to one
independent source
Set all other sources to 0.
Replace independent voltage source by
short circuit.
Replace independent current source by
open circuit.
2. Repeat for each independent source
3. Sum individual contributions to obtain
desired voltage or current

Slide 25 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Open Circuit and Short Circuit
Open circuit i=0 ; Cut off the branch
Short circuit v=0 ; replace the element by wire
Turn off an independent voltage source means
V=0
Replace by wire
Short circuit
Turn off an independent current source means
i=0
Cut off the branch
open circuit

Slide 26 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Superposition Example
Find V
o

+
24 V

2 O

4 O

4 A

4 V

+
+
V
o

Slide 27 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Thvenin Equivalent Circuit
Any* linear 2-terminal (1-port) network of indep. voltage
sources, indep. current sources, and linear resistors can
be replaced by an equivalent circuit consisting of an
independent voltage source in series with a resistor
without affecting the operation of the rest of the circuit.
network
of
sources
and
resistors


+
V
Th
R
Th
R
L
i
L
+
v
L

a
b
R
L
i
L
+
v
L

a
b
Thvenin equivalent circuit
load resistor
Slide 28 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
I-V Characteristic of Thvenin Equivalent
The I-V characteristic for the series combination of
elements is obtained by adding their voltage drops:

+
V
Th
R
Th
a
b
i
i
+
v
ab

v
ab
=V
Th
- iR
I-V characteristic
of resistor: v = iR

I-V characteristic of voltage source: v =V
Th
For a given current i, the voltage drop
v
ab
is equal to the sum of the voltages
dropped across the source (V
Th
)
and across the resistor (iR
Th
)
v
Slide 29 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Norton equivalent circuit
Norton Equivalent Circuit
Any* linear 2-terminal (1-port) network of indep. voltage
sources, indep. current sources, and linear resistors can
be replaced by an equivalent circuit consisting of an
independent current source in parallel with a resistor
without affecting the operation of the rest of the circuit.
network
of
sources
and
resistors

R
L
i
L
+
v
L

a
b
a
R
L
i
L
+
v
L

i
N
b
R
N
Slide 30 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
I-V Characteristic of Norton Equivalent
The I-V characteristic for the parallel combination of
elements is obtained by adding their currents:
i
i =I
N
-Gv
I-V characteristic
of resistor: i=Gv

I-V
characteristic
of current
source: i =-I
N
For a given voltage v
ab
, the current i is
equal to the sum of the currents in
each of the two branches:
v
i

+
v
ab

i
N
b
R
N
a
Slide 31 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Finding I
N
and R
N
=R
Th
I
N
i
sc
= V
Th
/R
Th
Analogous to calculation of Thevenin Eq. Ckt:
1) Find o.c voltage and s.c. current
2) Or, find s.c. current and Norton (Thev) resistance
Slide 32 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Finding I
N
and R
N
We can derive the Norton equivalent circuit from
a Thvenin equivalent circuit simply by making a
source transformation:
R
L
R
N
i
L
i
N
+
v
L

a
b

+
R
L
i
L
+
v
L

v
Th
R
Th
sc
Th
Th
N
sc
oc
Th N
; i
R
v
i
i
v
R R = = = =
a
b
Slide 33 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
Thvenin Equivalent Example
Find the Thevenin equivalent with respect to the terminals a,b:
Slide 34 EE42/43/100 Summer 2013 Prof. Chang-Hasnain
R
Th
Calculation Example #1
Set all independent sources to 0: