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EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
Lecture 4
•
Last time
–
–
–
–
•
Gate delay: why it happens (Capacitance! Resistance!)
Voltage rise and fall through logic circuits is gradual
t (delay between input and output at 50% of final value)
found graphically
Used clock signal to prevent false output (due to signals
delayed different amounts)
Now we will
–
–
Get to calculating delays for circuits (RC analysis)
Review fundamental knowledge of electric circuits
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
REVIEW OF ELECTRICAL QUANTITIES AND BASIC
CIRCUIT ELEMENTS
Free Charge
Most matter is macroscopically electrically neutral most of the time.
Exceptions: clouds in thunderstorm, people on carpets in dry weather,
plates of a charged capacitor, etc.
Microscopically, of course, matter is full of charges. Consider solids:
• Solids in which all charges are bound to atoms are called insulators.
•
Solids in which outer-most atomic electrons are free to move around are
called metals.
– Metals typically have ~ 1 “free electron” per atom (~ 5 X1022/cm3)
– Charge on a free electron is ”e” or “q”, where |e| = 1.6 x 10-19 C
•
Semiconductors are insulators in which electrons are not tightly
bound and thus can be easily “promoted” to a free state (by
heat or even by “doping” with a foreign atom).
Al or Cu – good
metallic conductor
– great for wires
Quartz – good
insulator – great
for dielectric
Si or GaAs – classic
semiconductors
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
CHARGE (cont.)
Charge flow  Current
Charge storage  Energy
Definition of current i (or I)
i (in Amperes)  flow of 1 coulomb per second
i (A) 
dq  C 
 
dt  S 
where q is the charge in coulomb and t is the
time in sec
Note: Current has sign
Examples:
(a)
105 positive unit charges of value e  1.6 1019C
flow to right (call this  direction) every nanosecond
5 1.6 1019
10
i
 1.6 105 A  16A (left to right)
109
(b)
1010 electrons flow to right in a wire every microsecond
10
16
i  -10 1.610
 1.6103  1.6mA (left to right)
106
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
CURRENT & CURRENT DENSITY
Define Current Density by Example:
Semiconductor with free
electrons
Wire attached to
ends
2 cm
C1
1 cm
C2
10 cm
X
Suppose we force a + 1A current from C1 to C2. Then
•Current of 1A flows in the semiconductor (in +X direction)
•Current Density = Current / (Cross-Sectional area current goes thru)
•Current density of above = 1 A / (2 cm • 1 cm) = 0.5 A/cm2
•Electron flow (in negative X direction) consists of
 1C / sec
18 electrons

6
.
25

10
 1.6  10 19 C / electron
sec
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
CURRENT & CURRENT DENSITY (cont.)
• A wire carrying 1 amp has a 1 mm2  10 2 cm2 cross-section,
then the current density is
1 100 A
102
cm2
Note: Typical dimensions of integrated circuit components are in the
range of 1m(104cm) . What is the current density in a component
with [1m]² area carrying 5A?
Answer
5106A  500 A/cm2
(1104cm)2
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
POSSIBLE CONCEPTUAL ISSUES
1 How does charge move through the wire?
Remember a wire has a huge number of free carriers moving
very fast but randomly (because of thermal energy)
Drift concept: Now add even a modest electric field
Carriers “feel” an electric field along the wire and tend to drift
with it (+ sign charge) or against it ( charge carrier). This
drift is still small compared to the random motion.
2 Sign of the charge carriers: It is often negative (for metals); in silicon, it
can be either negative or positive…we don’t care for circuit
analysis.
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
POSSIBLE CONCEPTUAL ISSUES (con’t)
3. Electric field is in the direction positive charge carriers move. Thus
if we have an electrical field in the Z direction, positive charges (ions,
positrons, whatever) will experience a force in the positive Z direction.
Negatively charged particles will experience the force in the –Z
direction.
Thus the free carriers we will be concerned with (electrons with
negative charge and “holes” with positive charge) will move against
and with the electric field respectively.
4. Field or Current can have positive or negative sign.
Examples: I = 0.2 mA
same as

going left to right
I = 0.2 mA

going right to left
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
POSSIBLE CONCEPTUAL ISSUES (con’t)
5. When we have unknown quantities such as current or voltage, we of
course do not know the sign. A question like “Find the current in the
wire” is always accompanied by a definition of the direction:
I
I
(or)
In this example if the current turned out to be 1mA, but flowing to the
left we would merely say I = -1mA.
6. To solve circuits, you may need to specify reference directions for
currents. But there is no need to guess the reference direction so that the
answer comes out positive….Your guess won’t affect what the charge
carriers are doing! Of course you will find that your intuition and
experience will often guide you to define a current direction so that the
answer comes out positive.
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
THINKING ABOUT VOLTAGE
a

Vab
b

“Vab” means the
potential at a
minus the potential
at b.
Generalized circuit element with two terminals (wires) a and b, with a
potential difference v ab across the element
Potential is always referenced to some point (Vab in the example; Va is
meaningless without an understood reference point)
If a conducting path exists between A and B, and there is a potential
difference, charges will drift due to the electric field  current flows
but potential difference may be present whether or
not a conducting path is present…
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
SIGN CONVENTIONS
Suppose you have an unlabelled battery and you measure its voltage with
a digital voltmeter. It will tell you magnitude and sign of the voltage.
a
1.401
DVM

b
b

DVM
a

Now you
make a
change.
What would
this circuit
measure?
With this circuit, you are measuring
Va  V (or V ). DVM indicates 1.4,
b
ab
so Va  Vb by 1.4 V. Which is the
positive battery terminal?
Answer: terminal b ( Vb  Va )
Answer: +1.401V
Note that we have used “ground”
symbol ( ) for reference node on
DVM. Often it is labeled “C” or
“common.”
EECS 40 Fall 2002 Lecture 4
SIGN CONVENTIONS (cont.)
W. G. Oldham and S. Ross
Lets put a bunch of batteries, say 1.5V and 9V in series to see what we already
know about sign conventions:
How is VAD related to VAB, VBC etc ?
Example 1
A
1.5V
+
B
1.5V
What is VAD ?
1.5V
-
V1
+
VAB = -1.5, VBC = -1.5,
and VAC = -3 so clearly
VAC =VAB+VBC etc.
D
9V
Answer: Clearly A is above D by 9V – 3V or 6V. That is
consistent with VAB+VBC + VCD= -1.5 –1.5 + 9 = 6V
How are single-subscript voltages
related to double-subscript voltages?
Example 2
A
+
C
+
B
1.5V +
C
+
D
9V
+
+
VX
Answer: Clearly
V1 = 1.5 = - VAB = VBA
VX = -6 = VDA
EECS 40 Fall 2002 Lecture 4
SIGN CONVENTIONS (cont.)
W. G. Oldham and S. Ross
Just as in the physical world, we do not know a priori the sign (or magnitude) of
voltages or currents; we don’t know them in the theoretical world until solving for
them, so we just define voltages and currents and accept that half of the time they
will be negative (quite analogously to probing with a DC meter).
ix
Example 1
Solve for
+
B
a
t
t
+
1.5
1.5K
Vx

Example 2
B
a
t 1.5
t
+
ix
Solve
for ix
+
1.5K
ix
Remembering Ohm’s law,
ix 
Here
1.5V
1.5  10
3
 1mA
i x  1mA
Vx
Note that in this case, the
current flows counterclockwise,
i.e., opposite to arrow defining i x

Here i x  1mA
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
KEEPING THE VOLTAGE SIGNS STRAIGHT
Labeling Conventions
• Indicate + and  terminals clearly; or label terminals with letters
• The + sign corresponds to the first subscript; the  sign to the second
subscript. Therefore, Vab = - Vba
Note: The labeling convention has nothing to do with whether or not v > 0
or v < 0

+
2V
Using sign conventions:
a
c
+
+
vcd
1V


b
+
v bd

Vab  1 ; Vca  2, thus
Vcb  2  1  1V
d
Obviously Vcd +Vdb = Vcb too. Then if Vbd = 5V, what is Vcd?
Answer: Vcd -5 = -1
so Vcd = 4V
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
POWER IN ELECTRIC CIRCUITS
Power: Transfer of energy per unit time (Joules per second = Watts)
Concept: in falling through a positive potential drop V, a positive charge q
gains energy
• potential energy change = qV for each charge q
• Rate is given by # charges/sec
Power = P = V (dq/dt) = VI
P=VI
Volt  Amps = Volts  Coulombs/sec = Joules/sec = Watts
Circuit elements can absorb or release power (i.e., from or to the rest of the
circuit); power can be a function of time.
How to keep the signs straight for absorbing and releasing power?
+ Power  absorbed into element
 Power  delivered from element
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
“ASSOCIATED REFERENCE DIRECTIONS”
It is often convenient to define the current through a circuit element as positive
when entering the terminal associated with the + reference for voltage
This box represents
the rest of the circuit
i
+
V
-
Here, I and V are
“associated”
Circuit element
For positive current and positive voltage, positive charge “falls down” a
potential “drop” in moving through the circuit element: it absorbs power
• P = VI > 0 corresponds to the element absorbing power if the
definitions of I and V are associated.
How can a circuit element absorb power?
By converting electrical energy into heat (resistors in toasters); light (light
bulbs); acoustic energy (speakers); by storing energy (charging a
battery).
Negative power  releasing power to rest of circuit
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
“ASSOCIATED REFERENCE DIRECTIONS” (cont.)
We know i = +1 mA
i
+
vB
+
1.5

vR
1.5K

A) For Resistor: P = i vR
(because we are using
“associated reference directions”)
Hence, for the resistor, P = i vR
= +1.5mW (absorbed)
B) For battery, current i is
opposite to associated
So, P  (i)vB  1.5 mW
(delivered out of battery)
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
“ASSOCIATED REFERENCE DIRECTIONS” (cont.)
iB
iR
Again, iR = 1mA
+
vB
therefore iB = -1mA
1.5
R

A) Resistor: P = iB vB = +1.5mW
B) Battery: iB and vB are
associated, therefore P= iB vB .
Thus
P  1.5  (  1 10  3 )  1.5 mW
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
EXAMPLES OF CALCULATING POWER
Find the power absorbed by each element
 +
a

2V c
+
3 mA
3V

+

0.5 mA
b
2.5 mA
+
1V
1V




3 mA
Element  : flip current direction:
Element  :
3V
 2V(3 mA)  6 mW

Element  :  1V(0.5 mA)  0.5 mW
Element  :
 1V(2.5 mA)  2.5 mW
 3V( 3 mA)  9 mW
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
BASIC CIRCUIT ELEMENTS
• Voltage Source
• Current Source
• Resistor
• Capacitor
• Inductor
(like ideal battery)
(always supplies some constant
given current)
(Ohm’s law)
(capacitor law – based on energy
storage in electric field of a
dielectric)
(inductor law – based on energy
storage in magnetic field produced
by a current)
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
DEFINITION OF IDEAL VOLTAGE SOURCE
Symbol

V


Note: Reference direction for voltage
source is unassociated (by convention)
Examples:
1) V = 3V
2) v = v(t) = 160 cos 377t
Special cases:
upper case V  constant voltage … called “DC”
lower case v  general voltage, may vary with time
Current through voltage source can take on any value
(positive or negative) but not infinite
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
IDEAL CURRENT SOURCE
“Complement” or “dual” of the voltage source: Current though branch
is independent of the voltage across the branch
+
i
note unassociated
v
direction

Actual current source examples – hard to find except in electronics
(transistors, etc.), as we will see
upper-case I  DC (constant) value
lower-case implies current could be time-varying i(t)
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE &
CURRENT SOURCES
Describe a two-terminal circuit element by plotting current vs. voltage

V
Ideal voltage source


Assume unassociated
signs
i
absorbing
power
releasing
power
If V is positive and I is
only positive


releasing power
V
absorbing power
(charging)
But this is arbitrary; i
might be negative so we
extend into 2nd quadrant
But this is still arbitrary, V
could be negative; all four
quadrants are possible
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE &
CURRENT SOURCES (con’t)
Ideal current
source
absorbing power
releasing power
i
+
i
v
 

If i is positive then we are
confined to quadrants 4 and 1:
Remember the voltage across the
current source can be any finite value
(not just zero)
And do not forget i can be positive or negative. Thus we can be in
any quadrant.
V
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
RESISTOR
i
+
v
R

If we use associated current and
voltage (i.e., i is defined as into +
terminal), then
v = iR (Ohm’s
law)
Question: What is the I-V characteristic for a 1K resistor? Draw on axis
I (mA)
below.
Answer: V = 0  I = 0
3
2
1
1
2
3
Slope = 1/R
V = 1V  I = 1 mA
V = 2V  I = 2 mA
V
etc
Note that all wires and circuit connections have resistance, though we will most
often approximate it to be zero. But we can (and do) deliberately construct circuit
elements with some desired resistance, even very large values such as 10M.
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
CAPACITOR
Any two conductors a and b separated by an insulator with a difference in
voltage Vab will have an equal and opposite charge on their surfaces whose
value is given by Q = CVab, where C is the capacitance of the structure, and
the + charge is on the more positive electrode.
We learned about the parallel-plate
capacitor in physics. If the area of the
plate is A, the separation d, and the
dielectric constant of the insulator is
, the capacitance equals C = A /d.
Symbol
A
or
d
i
a
Constitutive relationship: Q = C (Va  Vb).
(Q is positive on plate a if Va > Vb)
dQa
But i 
so
dt
dv
dt
where we use the associated reference directions.
iC
+
Vab
b

EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
ENERGY STORED IN A CAPACITOR
You might think the energy (in Joules) is QV, which has the dimension of
joules. But during charging the average voltage was only half the final
value of V.
Thus, energy is 1 QV 
2
1
CV 2
2
.
EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
ENERGY STORED IN A CAPACITOR (cont.)
More rigorous derivation: During charging, the power flow is
v  i into the capacitor, where i is into + terminal. We
integrate the power from t = 0 (v = 0) to t = end (v = V). The
integrated power is the energy
t  end
end dq
vV
E   v  i dt   v dt   v dq
0
0 dt
v0
but dq = C dv. (We are using small q instead of Q to remind us
that it is time varying . Most texts use Q.)
vV
1
E   Cv dv  CV 2
2
v0
i
+
v

EECS 40 Fall 2002 Lecture 4
W. G. Oldham and S. Ross
INDUCTORS
Inductors are the dual of capacitors – they store energy in magnetic
fields that are proportional to current.
Capacitor
dV
iC
dt
1
E  CV2
2
Inductor
di
vL
dt
1
E  LI 2
2
Switching properties: Just as capacitors demand v be continuous
(no jumps in V), inductors demand i be continuous (no jumps in i ).
Reason? In both cases the continuity follows from non-infinite, i.e.,
finite, power flow.
Capacitor
v is continuous
i can jump
Do not short circuit a
charged capacitor
(produces  current)
Inductor
i is continuous
V can jump
Do not open an inductor
with current flowing
(produces  voltage)