555 audio oscillator
PARTS AND MATERIALS
-
Two 6 volt batteries
-
One capacitor, 0.1 �F, non-polarized
(Radio Shack catalog # 272-135)
-
One 555 timer IC (Radio Shack catalog #
276-1723)
-
Two light-emitting diodes (Radio Shack
catalog # 276-026 or equivalent)
-
One 1 MΩ resistor
-
One 100 kΩ resistor
-
Two 510 Ω resistors
-
Audio detector with headphones
-
Oscilloscope (recommended, but not
necessary)
A oscilloscope would be useful in analyzing
the waveforms produced by this circuit, but it is not
essential. An audio detector is a very useful piece of test
equipment for this experiment, especially if you don't have
an oscilloscope.
CROSS-REFERENCES
Lessons In Electric Circuits, Volume
4, chapter 10: "Multivibrators"
LEARNING OBJECTIVES
SCHEMATIC DIAGRAM
ILLUSTRATION
INSTRUCTIONS
The "555" integrated circuit is a
general-purpose timer useful for a variety of functions. In
this experiment, we explore its use as an astable
multivibrator, or oscillator. Connected to a capacitor and
two resistors as shown, it will oscillate freely, driving
the LEDs on and off with a square-wave output voltage.
This circuit works on the principle of
alternately charging and discharging a capacitor. The 555
begins to discharge the capacitor by grounding the Disch
terminal when the voltage detected by the Thresh
terminal exceeds 2/3 the power supply voltage (Vcc).
It stops discharging the capacitor when the voltage detected
by the Trig terminal falls below 1/3 the power
supply voltage. Thus, when both Thresh and Trig
terminals are connected to the capacitor's positive
terminal, the capacitor voltage will cycle between 1/3 and
2/3 power supply voltage in a "sawtooth" pattern.
During the charging cycle, the capacitor
receives charging current through the series combination of
the 1 MΩ and 100 kΩ resistors. As soon as the Disch
terminal on the 555 timer goes to ground potential (a
transistor inside the 555 connected between that terminal
and ground turns on), the capacitor's discharging current
only has to go through the 100 kΩ resistor. The result is an
RC time constant that is much longer for charging than for
discharging, resulting in a charging time greatly exceeding
the discharging time.
The 555's Out terminal produces a
square-wave voltage signal that is "high" (nearly Vcc)
when the capacitor is charging, and "low" (nearly 0 volts)
when the capacitor is discharging. This alternating high/low
voltage signal drives the two LEDs in opposite modes: when
one is on, the other will be off. Because the capacitor's
charging and discharging times are unequal, the "high" and
"low" times of the output's square-wave waveform will be
unequal as well. This can be seen in the relative brightness
of the two LEDs: one will be much brighter than the other,
because it is on for a longer period of time during each
cycle.
The equality or inequality between "high"
and "low" times of a square wave is expressed as that wave's
duty cycle. A square wave with a 50% duty cycle is
perfectly symmetrical: its "high" time is precisely equal to
its "low" time. A square wave that is "high" 10% of the time
and "low" 90% of the time is said to have a 10% duty cycle.
In this circuit, the output waveform has a "high" time
exceeding the "low" time, resulting in a duty cycle greater
than 50%.
Use the audio detector (or an oscilloscope)
to investigate the different voltage waveforms produced by
this circuit. Try different resistor values and/or capacitor
values to see what effects they have on output frequency or
charge/discharge times. |