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Transcript
The Expo function
2017-08-12
Copyright(c) by L.R.Linares 2003
1
Exponential functions
10

9
8
It appears often in the solution of
circuits in transient conditions.
i(t )  A e
7
6
5

4
3
2
0
0
t is the time constant. At t = t, the
function is only 37% of its original
amplitude. At t = 5t …
tau
1
5tau
5
2017-08-12
10
15
20
25
t /t
30
35
40
45
50
…the value is so small that
by engineering standards we
consider that i(t) is zero!
Copyright(c) by L.R.Linares 2003
2
…exponential function

The derivative of an exponential
function is proportional to itself.
d
A t /t
t /t
( Ae )   e
dt
t
2017-08-12
Copyright(c) by L.R.Linares 2003
3
Exercise (homework)


A current i(t) is known to be exponential.
Its value at t = 2.5 ms is 27 mA, its value
at t = 8 ms is
5 mA.
Determine A, and “tau” (the time constant)
in …
i(t )  A e

2017-08-12
t /t
How long before the current decays by
engineering standards?
Copyright(c) by L.R.Linares 2003
4
RC Circuits
ECE – UBC
2017-08-12
Copyright (c) by L.R.Linares,
2003.
5
Thevenin to RC
2017-08-12
Copyright (c) by L.R.Linares,
2003.
6
dvc
dv1
ic  C 
C
dt
dt
vc
2017-08-12
Copyright (c) by L.R.Linares,
2003.
vc
7
vc
vc
dvc
vs  vc  RC 
dt
dvc
RC 
 vc  vs
dt
2017-08-12
Copyright (c) by L.R.Linares,
2003.
8
2017-08-12
Copyright (c) by L.R.Linares,
2003.
9
2017-08-12
Copyright (c) by L.R.Linares,
2003.
10
Initial value of y
Final value of y
Time constant
2017-08-12
Copyright (c) by L.R.Linares,
2003.
11
0
0
0
2017-08-12
Copyright (c) by L.R.Linares,
2003.
12
0
0
0
2017-08-12
Copyright (c) by L.R.Linares,
2003.
13
a1
a0=1
Final voltage
in the capacitor
Initial voltage
in the capacitor
2017-08-12
tRC
Copyright (c) by L.R.Linares,
2003.
14
Observation:

The final voltage of the capacitor (if
we let it charge for a long time)…is
equal to the Thevenin Voltage!!!
VCF  VTH

So, the voltage in the C goes from
VCO up (or down) to VCF with the
time constant
t  RTH C
2017-08-12
Copyright (c) by L.R.Linares,
2003.
15
Solution of a RC case:

From the SS snapshot at t = “right
before switch(es) operates”…


Compute VCO.
With the switch(es) “operated”,
determine the Thevenin equivalent
“seen” by the capacitor: RTH and
VTH … then…
vC (t )  (VCO  VTH )  e t / RTH C  VTH
2017-08-12
Copyright (c) by L.R.Linares,
2003.
16
A capacitor was initially charged up to
3 volts. It is connected to a Thevenin
circuit with Vs = 12V and R = 100 ohms
at t = 0.
The difference in voltage across
the resistor R will (Toothpaste
rule) push current from left to
right, and the capacitor voltage
will increase…
When the voltage in the capacitor
matches that of the source, the
current will stop. So the final
Vcf = Vs.
2017-08-12
Copyright (c) by L.R.Linares,
2003.
17
A capacitor was initially charged up to 3
volts. It is connected to a Thevenin circuit
with Vs = 12V and R = 100 ohms at t = 0.
The difference in voltage across
the resistor R will (Toothpaste
rule) push current from left to
right, and the capacitor voltage
will increase…
When the voltage in the capacitor
matches that of the source, the
current will stop. So the final
Vcf = Vs.
2017-08-12
Copyright (c) by L.R.Linares,
2003.
18
t=0.01s
Initial value: VC0 = 3V
Final value: VCf = 12V
2017-08-12
Copyright (c) by L.R.Linares,
2003.
19
In Matlab…
2017-08-12
Copyright (c) by L.R.Linares,
2003.
20
Final value: VCf = 12V
12
10
Vc, volts
8
6
4
Initial value: VC0 = 3V
t=0.01s
2
0
2017-08-12
0
0.005
0.01
0.015
0.02
0.025
time, seconds
0.03
0.035
Copyright (c) by L.R.Linares,
2003.
0.04
0.045
0.05
21
Lab in Circuit Maker





2017-08-12
An RC circuit with R=4.7k and
C=1uF.
Enough time to charge (five times
tau = 5RC = 23.5ms)
Enough time to discharge (5RC =
23.5ms)
Period T = 2 * 5RC = 47ms
Frequency = 1/T = 1/47ms
= 21.2 Hz.
Copyright (c) by L.R.Linares,
2003.
22
R1
4.7k
0/5V
V1
1uF C1
21.2 Hz
2017-08-12
Copyright (c) by L.R.Linares,
2003.
23
Voltage in the Capacitor
A: r1_2
7.000 V
Final voltage (5V)
6.000 V
5.000 V
4.000 V
3.000 V
2.000 V
Initial voltage (0)
1.000 V
0.000 V
-1.000 V
-5.000ms
2017-08-12
5.000ms
15.00ms
25.00ms
Copyright (c) by L.R.Linares,
2003.
35.00ms
45.00ms
24
Voltage in the Capacitor
A: r1_2
7.000 V
Initial voltage (5V)
6.000 V
5.000 V
4.000 V
3.000 V
2.000 V
1.000 V
0.000 V
-1.000 V
-5.000ms
5.000ms
15.00ms
25.00ms
35.00ms
45.00ms
Final voltage (0)
2017-08-12
Copyright (c) by L.R.Linares,
2003.
25
Current in the Capacitor
A: c1[i]
1.250mA
0.750mA
0.250mA
-0.250mA
-0.750mA
-1.250mA
40.00ms
2017-08-12
50.00ms
60.00ms
70.00ms
80.00ms
Copyright (c) by L.R.Linares,
2003.
90.00ms
100.0ms
26
-25.00ms
0.000ms
25.00ms
50.00ms
75.00ms
100.0ms
A: c1[i]
Current in the capacitor
B: r1_2
125.0ms
150.0ms
1.250mA
-1.250mA
5.000 V
Voltage in the capacitor
0.000 V
5.000 V
C: v1_1
Voltage applied by the source
0.000 V
2017-08-12
Copyright (c) by L.R.Linares,
2003.
27