Lecture 7: 555 Timer
Energy storage,
Periodic Waveforms, and
One of the most useful electronic devices
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Examples of Physical Periodic Motion
•
•
•
•
•
Pendulum
Bouncing ball
Vibrating string (stringed instrument)
Circular motion (wheel)
Cantilever beam (tuning fork)
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Other Periodic Phenomena
•
•
•
•
•
•
•
Daily cycle of solar energy
Annual cycle of solar energy
Daily temperature cycle
Annual temperature cycle
Monthly bank balance cycle
Electronic clock pulse trains
Line voltage and current
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Daily Average Temperature
Albany-Troy-Schenectady
90
80
70
60
50
Series1
Series2
40
30
20
10
2850
2773
2696
2619
2542
2465
2388
2311
2234
2157
2080
2003
1926
1849
1772
1695
1618
1541
1464
1387
1310
1233
1156
1079
1002
925
848
771
694
617
540
463
386
309
232
155
1
-10
78
0
• Data (blue) covers 1995-2002
• Note the sinusoid (pink) fit to the data
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Using Matlab to Produce Audio
Signal from Daily Average Temps
Original data (normalized)
Sinusoid fit to data
0.8
0.5
0.6
0.4
0.2
0
0
-0.2
-0.4
-0.6
-0.8
-1
0
200
400
600
-0.5
0
200
400
600
• Data is normalized to mimic sound
• Data is filtered to find fundamental
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Matlab Window
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Periodic Pulse Train from a 555 Timer
• This circuit puts out a steady state train of
pulses whose timing is determined by the
values of R1, R2 and C1
• The formula has a small error as we will see
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Using Models
• Recall that you should use a model that you
understand and/or know how to properly
apply
• To use it properly
 Check for plausibility of predicted values (ballpark
test). Are the values in a reasonable range?
 Check the rate of changes in the values (checking
derivative or slope of plot).
 Are the most basic things satisfied?
• Conservation of energy, power, current, etc.
• Developing a qualitative understanding of
phenomena now will help later and with
simulations.
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Charging a Capacitor
10V
TCLOSE = 0
1
U1
R1
2
8V
V
V
1
V
1k
6V
U2
V1
TOPEN = 0
Voltage
C1
4V
2
10V
Capacitor
1uF
2V
0V
0
0s
1ms
V(U2:1)
V(R1:2)
2ms
3ms
4ms
5ms
6ms
7ms
8ms
9ms
10ms
V(V1:+)
• Capacitor C1 is charged up by current flowing
through R1
V1 V
10  V
Time
I
CAPACITOR
R1

• As the capacitor charges up, its voltage
increases and the current charging it
decreases, resulting in the charging rate
shown
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CAPACITOR
1k
9
Charging a Capacitor
10mA
10V
8mA
8V
6mA
Capacitor
and
Resistor
6V
Current
Capacitor
4mA
4V
2mA
2V
0A
Voltage
0V
0s
1ms
I(R1)
2ms
3ms
4ms
5ms
6ms
7ms
8ms
9ms
10ms
I(C1)
0s
1ms
V(U2:1)
V(R1:2)
2ms
3ms
4ms
5ms
6ms
8ms
9ms
10ms
Time
t
• Capacitor Current
I  Ioe
• Capacitor Voltage
V  Vo 1  e
• Where the time constant
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7ms
V(V1:+)
Time

t



  RC  R1 C1  1ms
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Charging a Capacitor
10V
8V
6V
Capacitor
Voltage
4V
2V
0V
0s
1ms
V(U2:1)
V(R1:2)
2ms
3ms
4ms
5ms
6ms
7ms
8ms
9ms
10ms
V(V1:+)
Time
• Note that the voltage rises to a little
above 6V in 1ms.
1
(1  e ) .632
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Charging a Capacitor
• There is a good description of capacitor
charging and its use in 555 timer circuits at
http://www.uoguelph.ca/~antoon/gadgets/555/555.html
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2 Minute Quiz
Name______Section_____Date_____
True or False?
• If C1 < C2, for a fixed charging current, it will
take longer to charge C1 than C2
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555 Timer
• At the beginning of the cycle, C1 is charged through
resistors R1 and R2. The charging time constant is
  ( R1  R2)C1
• The voltage reaches (2/3)Vcc in a time
  0.693( R1  R2)C1
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555 Timer
• When the voltage on the capacitor reaches
(2/3)Vcc, a switch is closed at pin 7 and
the capacitor is discharged to (1/3)Vcc, at
which time the switch is opened and the
cycle starts over
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555 Timer
• The capacitor voltage cycles back and
forth between (2/3)Vcc and (1/3)Vcc at
 1  0and
.693( R1  R2)C1
times
 2  0.693( R2)C1
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555 Timer
• The frequency is then given by
1
144
.
f 

0.693( R1  2  R2)C1 ( R1  2  R2)C1
Note the error in the figure
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Inside the 555
• Note the voltage divider inside the 555
made up of 3 equal 5k resistors
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7
DIS
8
VCC
R
4
555 Timer
6
2
5
THR
TR
CV
3
GND
Q
1
NE555
• These figures are from the lab write up
• Each pin has a name (function)
• Note the divider and other components inside
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Astable and Monostable Multivibrators
5V
8
V CC
4
R
LED
5
NE555
C
THR
TR
CV
GND
CV
Q
6
2
3
LED
NE555
1
C
0.01 uF
5
THR
TR
3
1
6
2
GND
Q
DIS
1K
1
Rb
7
0.01 uF
DIS
2
7
R
V CC
R
4
Ra
8
5V
• Astable puts out a continuous sequence
of pulses
• Monostable puts out one pulse each
time the switch is connected
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Astable and Monostable Multivibrators
• What are they good for?
 Astable: clock, timing signal
 Monostable: a clean pulse of the correct
height and duration for digital system
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Optical Transmitter Circuit
Astable is used to produce carrier pulses at a
frequency we cannot hear (well above 20kHz)
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Optical Receiver Circuit
• Receiver circuit for transmitter on previous slide
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2 Minute Quiz
Name______Section_____Date_____
True or False?
• If R1 < R2, for a fixed charging voltage, it will
take longer to charge a given capacitor C
through R1 than R2
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Clapper Circuit
• Signal is detected by microphone
• Clap is amplified by 741 op-amp
• Ugly clap pulse triggers monostable to
produce clean digital pulse
• Counter counts clean pulses and triggers
relay through the transistor
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555 Timer Applications
• 40 LED bicycle light with 20 LEDs
flashing alternately at 4.7Hz
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555 Timer Applications
• 555 timer is used to produce an oscillating
signal whose voltage output is increased by
the transformer to a dangerous level,
producing sparks.
• DO NOT DO THIS WITHOUT SUPERVISION
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Tank Circuit: A Classical Method Used to
Produce an Oscillating Signal
• A Tank Circuit is a combination of a
capacitor and an inductor
• Each are energy storage devices
1
1 2
2
WE  WC  CV
WM  WL  LI
2
2
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Tank Circuit: How Does It Work?
TOPEN = 0
TCLOSE = 0
1
U1
1
U2
2
2
V
V1
L1
10V
C1
10uH
1uF
0
• Charge capacitor to 10V. At this point, all of the
energy is in the capacitor.
• Disconnect voltage source and connect capacitor to
inductor.
• Charge flows as current through inductor until
capacitor voltage goes to zero. Current is then
maximum through the inductor and all of the energy
is in the inductor.
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Tank Circuit
TOPEN = 0
TCLOSE = 0
1
U1
1
U2
2
2
V
V1
L1
10V
C1
10uH
1uF
0
• The current in the inductor then
recharges the capacitor until the cycle
repeats.
• The energy oscillates between the
capacitor and the inductor.
• Both the voltage and the current are
sinusoidal.
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Tank Circuit Voltage and Current
4.0A
Current
0A
-4.0A
I(L1)
10V
Voltage
0V
SEL>>
-10V
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
V(C1:1)
Time
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100us
Tank Circuit
4.0A
Current
0A
-4.0A
I(L1)
10V
Voltage
0V
SEL>>
-10V
0s
10us
20us
30us
40us
50us
60us
70us
80us
90us
100us
V(C1:1)
• There is a slight decay due to finite wire
resistance.
1
• The frequency is given by f 
(period is about 10ms)
2 LC
Time
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Tank Circuit
•Tank circuits are the basis of most oscillators. If such a
combination is combined with an op-amp, an oscillator
that produces a pure tone will result.
•This combination can also be used to power an
electromagnet.
•Charge a capacitor
•Connect the capacitor to an electromagnet
(inductor). A sinusoidal magnetic field will result.
•The magnetic field can produce a magnetic force
on magnetic materials and conductors.
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2 Minute Quiz
Name______Section_____Date_____
True or False?
• When a capacitor C is connected to a battery
through a resistor R, the charging current will
be a maximum at the moment the connection
is made and decays after that.
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Tank Circuit Application
• In lab 10 we will be using the circuit from a
disposable camera.
• We can also use this type of camera as a
power source for an electromagnet.
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Disposable Camera Flash Capacitor
Connected to a Small Electromagnet
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Disposable Camera Flash Experiment/Project
• A piece of a paperclip is placed part way into the
electromagnet.
• The camera capacitor is charged and then triggered to
discharge through the electromagnet (coil).
• The large magnetic field of the coil attracts the paperclip to
move inside of the coil.
• The clip passes through the coil, coasting out the other side
at high speed when the current is gone.
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Coin Flipper and Can Crusher
• The can crusher device crushes a soda
can with a magnetic field.
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Can Crusher and Coin Flipper
• This is an animation a student made as
a graphics project a few years ago
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Can Crusher and Coin Flipper
• For both the can crusher and coin flipper, the coil fed
by the capacitor acts as the primary of a transformer.
• The can or the coin acts as the secondary.
• A large current in the primary coil produces an even
larger current in the can or coin (larger by the ratio of
the turns in the primary coil)
• The current in the coin or can is such that an
electromagnet of the opposite polarity is formed
(Lenz’ Law) producing two magnets in close proximity
with similar poles facing one another.
• The similar poles dramatically repel one another
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Magnetic Launchers
• Coilguns/Railguns
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Coilguns & Railguns
• Two types of launchers are being developed
for a variety of purposes.
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Where Will You See This Material Again?
• Electromagnetic Fields and Forces:
Fields and Waves I
• 555 Timers: Many courses including
Analog Electronics and Digital
Electronics
• Oscillators: Analog electronics
• Clocks, etc: Digital Electronics,
Computer Components and Operations,
and about half of the ECSE courses.
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Appendix
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Using Conservation Laws to Derive
Fundamental Equations
• Energy stored in capacitor plus inductor
Energy  WTOTAL
1 2 1
 LI  CV 2
2
2
• Total energy must be constant, thus
dWTOTAL
1
dI 1
dV
 0  L2 I
 C 2V
dt
2
dt 2
dt
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Using Conservation Laws
• Simplifying
dWTOTAL
dI L
dVC
0L
IL  C
VC
dt
dt
dt
• This expression will hold if
dI L
VL  L
dt
dVC
IC  C
dt
• Noting that
VC  VL
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IC  I L
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Using Conservation Laws
+
VC
+I
VL
• Note that for the tank circuit
 The same current I flows through both
components
 The convention is that the current enters
the higher voltage end of each component
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Using Conservation Laws
• Experimentally, it was also determined
that the current-voltage relationship for
a capacitor is
dVC
IC  C
dt
• Experimentally, it was also determined
that the current-voltage relationship for
an inductor is
dI L
VL  L
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dt
48
Using Conservation Laws
• Applying the I-V relationship for a
capacitor to the expressions we saw
before for charging a capacitor through
a resistor
t
t

dVC
IC  C
I  I o e  V  Vo 1  e  
dt
• We see that
IC  Ioe
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 t


dVC
 t 

C
 CVo  0   1 e 
dt
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Using Conservation Laws
• Simplifying
IC  I o e
t

dVC
C
 CVo
dt
   
1
e
t

• Which is satisfied if
  RC
Vo
Io 
R
• The latter is the relationship for a
resistor, so the results work.
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