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EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
CIRCUITS, NODES, AND BRANCHES
Last time:
• Looked at circuit elements, used to model the insides of logic gates
• Saw how the resistor and capacitor in the gate circuit model make the
change in output voltage gradual
• Looked at the mathematical equations that describe the output voltage
Today:
• Derive the mathematical equation for the change in gate output voltage
• Practice an easy method used to graph and write the equation for the
changing output voltage
• Look at how R and C affect the rate of change in output voltage
• Look at how R and C affect how quickly we can put new signals through
• Review ideas we will need to understand Kirchoff’s laws
1
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
RC RESPONSE
R
Internal Model
Vin
V out
of Logic Gate
C
Behavior of Vout after change in Vin
Vout
Vout
Vin
Vout(t=0)
Vin
Vout(t=0)
0
0
time
0
0
time
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
RC RESPONSE
We said that Vout(t) has the general exponential form
Vout(t) = A + Be-t/t
Initial value Vout(t=0) = A + B
Final value Vout(t→∞) = A
But we know Vout(t→∞) = Vin so rewrite
Vout(t) = Vin + [Vout(t=0)-Vin]e-t/t
We will now prove that time constant t is
t = RC
Remember, Vout(t) has gone 63% of the way
from initial to final value after 1 time constant.
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
RC RESPONSE: Proof of Vout equation
R
V in
V out
iR
iR =
iC
Vin  Vout
(Ohm' s law)
R
iC = C
C
dVout
(capacitance law)
dt
But iR = iC !
Vin  Vout
dVout
Thus,
=C
R
dt
or
dVout
1
=
( Vin  Vout )
dt
RC
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
RC RESPONSE: Proof
We know
Is
dVout
1
=
( Vin  Vout )
dt
RC
Vout(t) = Vin + [Vout(t=0)-Vin]e-t/RC a solution? Substitute:
dVout d
1
= Vin  [ Vout ( t = 0)  Vin ]e  t / RC = 
[ 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
This is the solution since it has the correct initial condition!
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
Example: Draw graph and write equation
Assume Vin has been zero for a
long time, and steps up to 10 V
at t=0.
Input node
+
R
Vout
C
Vin
-
Assume R = 1KΩ, C = 1pF .
Output node
ground
Ingredients for graph:
Vin
1) Initial value of Vout is 0
2) Final value of Vout is Vin=10V
10
Vout
6.3V
3) Time constant RC is 10-9 sec
4) Vout reaches 0.63 X 10 in 10-9 sec
Ingredients for equation:
Vout(t) = Vin + [Vout(t=0)-Vin]e-t/RC =>
0
0
1nsec
time
Vout(t) = 10 - 10e-t/1nsec
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
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 is t = RC. Input node
+
3) Find the capacitor voltage before
Vin
the input voltage changes. This is the
initial value for the output, Vout(t=0).
R
Output node
Vout
C
ground
4) Find the asymptotic value Vout(t→∞), in this case, equals Vin after step
5) Sketch the transient period. After one time constant, the graph has
passed 63% of the way from initial to final value.
6) Write the equation by inspection.
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
PULSE DISTORTION
Vin
What if I want to step up the input,
0
0
Vin
wait for the output to respond,
time
Vout
0
0
Vin
then bring the input back down for a
different response?
time
Vout
0
0
time
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
PULSE DISTORTION
R
Input node
The pulse width (PW)
must be greater than
RC to avoid severe
pulse distortion.
Output node
O
+
C
Vin
-
ground
PW = 0.1RC
PW = RC
6
5
4
3
2
1
0
PW = 10RC
6
5
4
3
2
1
0
Vout
Vout
Vout
6
5
4
3
2
1
0
0
1
2
Time
3
4
5
0
1
2
Time
3
4
5
0
5
10
Time
15
20
25
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
EXAMPLE
Suppose a voltage pulse of width
5 ms and height 4 V is applied to the
input of the circuit at the right.
Sketch the output voltage.
Vin
R
R = 2.5 KΩ
C = 1 nF
Vout
C
First, the output voltage will increase to approach the
4 V input, following the exponential form. When the input
goes back down, the output voltage will decrease back
to zero, again following exponential form.
How far will it increase? Time constant = RC = 2.5 ms
The output increases for 5ms or 2 time constants.
It reaches 1-e-2 or 86% of the final value.
0.86 x 4 V = 3.44 V is the peak value.
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
EXAMPLE
4
3.5
3
2.5
2
1.5
1
0.5
00
2
4
6
8
10
Just for fun, the equation for the output is:
Vout(t) =
4-4e-t/2ms for 0 ≤ t ≤ 5 ms
{
3.44e-(t-5ms)/2.5ms for t > 5 ms
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
APPLICATIONS
• Now we can find “propagation delay” tp; the time
between the input reaching 50% of its final value and the
output to reaching 50% of final value.
• Today’s case, using “perfect input” (50% reached at t=0):
0.5 = e-tp
tp = - ln 0.5 = 0.69
It takes 0.69 time constants, or 0.69 RC.
• We can find the time it takes for the output to reach other
desired levels. For example, we can find the time
required for the output to go from 0 V to the minimum
voltage level for logic 1.
• Knowing these delays helps us design clocked circuits.
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
Adding Voltages in Series
In electrical engineering we generally add voltage drops.
Example:
1. Go around the path that comprises Vx. Start at + terminal,
end at – terminal.
2. Look at the first sign you encounter at each element.
If +, add that voltage. If -, subtract.
3. The final sum is Vx. + 5 – 8 – 6 + 8 = -1 And that’s OK!
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
JUST PICK A POLARITY
• In the last slide, we “guessed Vx wrong”; the
bottom end was actually higher potential than
the top. WE DON’T CARE!
• If I need to find a voltage or current, I just give it
a name and write down a polarity/direction.
Whatever I feel like at the time. Then I solve for
the unknown. If the voltage/current “doesn’t
really go that way”, the answer will be negative.
SO WHAT.
• Just remember how to flip things if you need to:
Ix
Same
-Ix
-
Vx
+
+
-Vx
-
Thing!
EECS 40
Fall 2002 Lecture 6
W. G. Oldham and S. Ross
@#%*ING ASSOCIATED
REFERENCE CURRENTS
We only need to worry about associated reference
currents in 3 situations (here we need to have
associated reference current):
I
1.
2.
3.
Using Ohm’s law in a resistor
Using I = C dV/dt in a capacitor
Calculating power P=VI
+
V
Otherwise, forget about it! Work with currents
and voltage in whatever direction you want!
-