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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
Review of charging and discharging in RC Circuits
• Lectures 2 and 3: Gate Delay Ideas
• Lecture 4: Basic properties of circuit elements
• Today: Finish up circuit elements
• How to find delays in circuits which have to charge
capacitors (which every circuit effectively has)
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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
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
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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
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
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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
IDEAL CURRENT SOURCE
“Complement” or “dual” of the voltage source: Current though
branch is independent of the voltage across the branch
+
note unassociated
v
i
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)
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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
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
If V is positive and I is
only positive
i
absorbing
power
releasing
power


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
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE & CURRENT
SOURCES (con’t)
+

absorbing power
i
v
_
i

releasing power
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.
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
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?
I (mA)
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
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most often approximate it to be zero.
EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
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.
A
or
Symbol
d
i
a
Constitutive relationship: Q = C (Va  Vb).
(Q is positive on plate a if Va > Vb)
dQ
dv
a
But i 
so i  C
dt
dt
where we use the associated reference directions.
+
Vab
b

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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
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
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
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
i
+
v

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
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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
INDUCTORS
Inductors are the dual of capacitors – they store energy in magnetic fields
that are proportional to current.
Capacitor
Inductor
dV
dt
1
E  CV 2
2
v L
iC
E
di
dt
1 2
LI
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 continous
I can jump
Do not short circuit a
charged capacitor
(produces  current)
Inductor
i is continous
V can jump
Do not open an inductor
with current flowing
(produces  voltage)
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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
SWITCHING PROPERTIES OF L, C
Rule: The voltage across a capacitor must be continuous and differentiable.
For an inductor the current must be continuous and differentiable.
Basis: The energy cannot “jump” because that would require infinite energy
flow (power) and of course neither the current or voltage can be infinite. For
a capacitor this demands that V be continuous (no jumps in V); for an
inductor it demands i be continuous (no jumps in i ).
Capacitor
v is continous
I can jump
Do not short circuit a
charged capacitor
(produces  current)
Inductor
i is continous
V can jump
Do not open an inductor
with current flowing
(produces  voltage)
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
The RC Circuit to Study
(All single-capacitor circuits reduce to this one)
Input node
R
Output node
C
ground
• R represents total resistance (wire plus whatever drives
the input)
• C represents the total capacitance from node to the
outside world (from devices, nearby wires, ground etc)13
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
RC RESPONSE EXAMPLE
Case 1 Capacitor uncharged: Apply + voltage step of 2V
5
R
Input node
Vin
C
Vin
-
0
0
Output node
Vout
+
ground
time
5
The capacitor voltage, Vout, is
initially zero and rises in response to
the step in input.
Its initial value, just after the step, is
still zero WHY?
Its final value, long after the step is
is 5V WHY?
Time period during
which the output
changes: transient
Vout
EECS 40
0
0
time
We need to find the solution
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during the transient.
EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
RC RESPONSE
Case 1 – Rising voltage. Capacitor uncharged: Apply + voltage step
V1
Vin
Input node
Output node
+
Vout
Vin
-
0
0
R
time
Vout
C
ground
• Vin “jumps” at t=0, but Vout cannot “jump” like Vin. Why not?
 Because an instantaneous change in a capacitor voltage would require
instantaneous increase in energy stored (1/2CV²), that is, infinite
power. (Mathematically, V must be differentiable: I=CdV/dt)
V does not “jump” at t=0 , i.e. V(t=0+) = V(t=0-)
Therefore the dc solution before the transient tells us the capacitor voltage
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at the beginning of the transient.
EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
RC RESPONSE
Case 1 – Capacitor uncharged: Apply voltage step
V1
Vin
Input node
+
Vout
Vin
-
0
0
time
R
Output node
Vout
C
ground
• Vout approaches its final value asymptotically (It never quite gets
to V1, but it gets arbitrarily close). Why?
• After the transient is over (nothing changing anymore) it means
d(V)/dt = 0 ; that is all currents must be zero. From Ohm’s law, the
voltage across R must be zero, i.e. Vin = Vout.
 That is, Vout  V1 as t  . (Asymptotic behavior)
Again the dc solution (after the transient) tells us (the asymptotic limit of) the
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capacitor voltage during the transient.
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
RC RESPONSE: Case 1 (cont.)
Vin
V1
Vout
?
R
Input node
Output node
Vout
+
C
Vin
0
ground
time
0
Exact form of Vout?
Equation for Vout: Do you remember
general form?
Vout = V1(1-e-t/t)
V1
0
0
Exponential!
time
Vout
V1
Vout
EECS 40
0
0
t
time
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
Review of simple exponentials.
Rising Exponential from Zero
Falling Exponential to Zero
Vout = V1(1-e-t/t)
Vout = V1e-t/t
at t = 0,
Vout = V1 , and
at t  ,
Vout  V1 also
at t  ,
at t = t,
Vout  0, also
Vout = 0.63 V1
at t = t,
Vout = 0.37 V1
Vout
8
Vout = 0 , and
8
at t = 0,
Vout
V1
V1
.63V1
.37V1
0
0
t
time
0
0
t
time
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
Further Review of simple exponentials.
Rising Exponential from Zero
Falling Exponential to Zero
Vout = V1(1-e-t/t)
Vout = V1e-t/t
We can add a constant (positive or negative)
Vout = V1e-t/t + V2
Vout = V1(1-e-t/t) + V2
Vout
Vout
V1 + V2
V1 + V2
.63V1+ V2
.37V1 + V2
V2
0
V2
0
t
time
0
0
t
time
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
Further Review of simple exponentials.
Rising Exponential
Falling Exponential
Vout = V1
(1-e-t/t)
+ V2
Vout = V1e-t/t + V2
Both equations can be written in one simple form:
Initial value (t=0) : Vout = A + B.
Vout = A + Be-t/t
Final value (t>>t): Vout = A
Thus: if B < 0, rising exponential; if B > 0, falling exponential
Vout
Vout
A
A+B
Here B > 0
Here B < 0
A+B
0
A
0
time
0
0
time
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
RC RESPONSE: Proof
R
Vin
Vout
iR
iR 
Vin  Vout
(Ohm' s law)
R
But iR  iC !
Thus,
C
iC
iC  C
Vin  Vout
dVout
C
R
dt
dVout
(capacitance law)
dt
or
dVout
1

( Vin  Vout )
dt
RC
RC is “time constant” t !
Large R or large C => slow rise and fall
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EECS 40
Fall 2001 Lecture 5
Copyright Regents of University of California
W. G. Oldham
RC RESPONSE: Proof
We know
dVout
1

( Vin  Vout )
dt
RC
Is Vout(t) = Vin + [Vout(t=0)-Vin]e-t/RC a solution? Substitute:
dVout
d
1
 t / RC
 Vin  [ Vout ( t  0)  Vin ]e

[ Vout ( t  0)  Vin ]e  t / RC
dt
dt
RC

1
[Vin - Vout (t  0)]e - t/RC
RC
1
1
(Vin - Vout (t)) 
[Vin - (Vin  [Vout (t  0) - Vin ]e -t/RC )]
RC
RC
1

[Vin - Vout (t  0)]e - t/RC
RC
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
Example
The simple RC circuit with a step
input has a universal exponential
solution of the form:
Input node
+
Vout = A + Be-t/RC
-
R
Output node
Vout
C
Vin
ground
Example: R = 1K, C = 1pF, Vin steps
from zero to 10V at t=0:
Vin
1) Initial value of Vout is 0
2) Final value of Vout is 10V
10
Vout
6.3V
3) Time constant is 10-9 sec
4) Vout reaches 0.63 X 10 in 10-9 sec
DRAW THE GRAPH
Now just write the equation:
0
0
1nsec
time
Vout = 10 - 10e-t/1nsec
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EECS 40
Fall 2001 Lecture 5
W. G. Oldham
Copyright Regents of University of California
Charging and discharging in RC Circuits
(The official EE40 Easy Method)
Method of solving for any node voltage in a single capacitor circuit.
1) Simplify the circuit so it looks like one resistor, a source, and a
capacitor (it will take another two weeks to learn all the tricks to do
this.) But then the circuit looks like this:
2) The time constant of the
Input node
transient is t = RC.
+
3) Solve the dc problem for the
Vin
capacitor voltage before the transient.
This is the starting value (initial value)
for the transient voltage.
R
Output node
Vout
C
ground
4) Solve the dc problem for the capacitor voltage after the transient is over.
This is the asymptotic value.
5) Sketch the Transient. It is 63% complete after one time constant.
6) Write the equation by inspection.
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