📄 traffic.htm
字号:
<html><head>
<meta http-equiv="content-type" content="text/html; charset=ISO-8859-1">
<title>555 Timer</title></head><body bgcolor="#ffffff">
<blockquote>
<a name="tone.gif"></a>
<center><h3>555 Tone Generator (8 ohm speaker)</h3></center>
<br>This is a basic 555 squarewave oscillator used to produce a 1 Khz
tone from an 8 ohm speaker. In the circuit on the left, the speaker is
isolated from the oscillator by the NPN medium power transistor which also
provides more current than can be obtained directly from the 555
(limit = 200 mA). A small capacitor is used at the transistor base to slow
the switching times which reduces the inductive voltage produced by the
speaker. Frequency is about 1.44/(R1 + 2*R2)C where R1 (1K) is much smaller
than R2 (6.2K) to produce a near squarewave. Lower frequencies can be
obtained by increasing the 6.2K value, higher frequencies will probably
require a smaller capacitor as R1 cannot be reduced much below 1K. Lower
volume levels can be obtained by adding a small resistor in series with
the speaker (10-100 ohms). In the circuit on the right, the speaker
is directly driven from the 555 timer output. The series capacitor (100 uF)
increases the output by supplying an AC current to the speaker and driving
it in both directions rather than just a pulsating DC current which would
be the case without the capacitor. The 51 ohm resistor limits the current
to less than 200 mA to prevent overloading the timer output at 9 volts.
At 4.5 volts, a smaller resistor can be used.
<br><img src="TRAFFIC_files/tone.gif">
<center><a href="http://ourworld.compuserve.com/homepages/Bill_Bowden/homepage.htm#menu">Menu</a></center>
<hr size="1">
<a name="5volts.gif"></a>
<center><h3>Generating -5 Volts From a 9 Volt Battery</h3></center>
<br>A 555 timer can be used to generate a squarewave to produce
a negative voltage relative to the negative battery terminal. When the timer
output at pin 3 goes positive, the series 22 uF capacitor charges through
the diode (D1) to about 8 volts. When the output switches to ground, the
22 uF cap discharges through the second diode (D2) and charges the 100 uF
capacitor to a negative voltage. The negative voltage can rise over
several cycles to about -7 volts but is limited by the 5.1 volt zener
diode which serves as a regulator. Circuit draws about 6 milliamps
from the battery without the zener diode connected and about 18 milliamps
connected. Output current available for the load is about 12 milliamps.
An additional 5.1 volt zener and 330 ohm resistor could be used to
regulate the +9 down to +5 at 12 mA if a symmetrical +/- 5 volt supply
is needed. The battery drain would then be around 30 mA.
<br><img src="TRAFFIC_files/5volts.gif">
<center><a href="http://ourworld.compuserve.com/homepages/Bill_Bowden/homepage.htm#menu">Menu</a></center>
<hr size="1">
<a name="outline.gif"></a>
<center><h3>Transistor / Diode / IC (DIP) Outlines</h3></center>
<br>
<img src="TRAFFIC_files/outline.gif">
<center><a href="http://ourworld.compuserve.com/homepages/Bill_Bowden/homepage.htm#menu">Menu</a></center>
<hr size="1">
<a name="15flash.gif"></a>
<center><h3>1.5 Volt LED Flashers</h3></center>
<br>The LED flasher circuits below operate on a single 1.5 volt battery.
The circuit on the upper right uses the popular LM3909 LED flasher IC and
requires only a timing capacitor and LED.
<p>
The top left circuit, designed by Andre De-Guerin illustrates using a
100uF capacitor to double the battery voltage to obtain 3 volts for the LED.
Two sections of a 74HC04 hex inverter are used as a squarewave oscillator
that establishes the flash rate while a third section is used as a buffer
that charges the capacitor in series with a 470 ohm resistor while the
buffer output is at +1.5 volts. When the buffer output switches to ground
(zero volts) the charged capacitor is placed in series with the LED and
the battery which supplies enough voltage to illuminate the LED. The LED
current is approximately 3 mA, so a high brightness LED is recommended.
</p><p>
In the other two circuits, the same voltage doubling principle is used
with the addition of a transistor to allow the capacitor to discharge
faster and supply a greater current (about 40 mA peak). A larger capacitor
(1000uF) in series with a 33 ohm resistor would increase the flash duration
to about 50mS. The discrete 3 transistor circuit at the lower right would
need a resistor (about 5K) in series with the 1uF capacitor to widen the pulse
width.
</p><p><img src="TRAFFIC_files/15flash.gif">
</p><center><a href="http://ourworld.compuserve.com/homepages/Bill_Bowden/homepage.htm#menu">Menu</a></center>
<hr size="1">
<a name="lineled.gif"></a>
<center><h3>AC Line powered LEDs</h3></center>
<br>
The circuit below illustrates powering a LED (or two) from the
120 volt AC line using a capacitor to drop the voltage and a small
resistor to limit the inrush current. Since the capacitor must
pass current in both directions, a small diode is connected
in parallel with the LED to provide a path for the negative
half cycle and also to limit the reverse voltage across the
LED. A second LED with the polarity reversed may be subsituted
for the diode, or a tri-color LED could be used which would appear
orange with alternating current. The circuit is fairly efficient
and draws only about a half watt from the line. The resistor value
(1K / half watt) was chosen to limit the worst case inrush current to about
150 mA which will drop to less than 30 mA in a millisecond as the capacitor
charges. This appears to be a safe value, I have switched the circuit
on and off many times without damage to the LED. The 0.47 uF capacitor
has a reactance of 5600 ohms at 60 cycles so the LED current is about
20 mA half wave, or 10 mA average. A larger capacitor will increase the
current and a smaller one will reduce it. The capacitor must be a
non-polarized type with a voltage rating of 200 volts or more.
<p>
The lower circuit is an example of obtaining a low regulated voltage
from the AC line. The zener diode serves as a regulator and also provides
a path for the negative half cycle current when it conducts in the forward
direction. In this example the output voltage is about 5 volts and will
provide over 30 milliamps with about 300 millivolts of ripple. Use caution
when operating any circuits connected directly to the AC line.
</p><p><img src="TRAFFIC_files/lineled.gif">
</p><center><a href="http://ourworld.compuserve.com/homepages/Bill_Bowden/homepage.htm#menu">Menu</a></center>
<hr size="1">
<a name="ledlamp.gif"></a>
<center><h3>Line Powered White LEDs</h3></center>
<br>
The LED circuit below is an example of using 25 white LEDs in series
connected to the 120VAC line. It can be modified for more or less
LEDs by adjusting the resistor value. The exact resistance will depend
on the particular LEDs used. But working out the resistor value is a bit
complicated since current will not continously flow through
the resistor.
<p>
In operation, the output of the bridge rectifier will be about
120 DC RMS or 170 volts peak. If we use 25 white LEDs with a
forward voltage of 3 volts each, the total LED voltage will be
75 volts. The peak resistor voltage will be 170- 75 or 95 volts
but the resistor voltage will not be continous since the input
must rise above 75 before any current flows. This (dead time)
represents about 26 degrees of the 90 degree half wave rectified
cycle, (asin) 75/170 = (asin) .44 = 26 degrees. This means the
resistor will conduct during 90-26 = 64 degrees, or about
71 percent of the time.
</p><p>
Next we can work out the peak LED current to
determine the resistor value. If the LED current is 20mA RMS,
the peak current will be 20*1.414 or 28mA. But since the
duty cycle is only 71 percent, we need to adjust this figure
up to 28/0.71 = 39mA. So, the resistor value should be
95/.039 = 2436 ohms (2.4K) and the power rating will be
.02^2 *2400= .96 watts. A two watt size is recommended.
</p><p>
Now this circuit can also be built using 2 diodes
and resistor as shown in the lower drawing. The
second diode in parallel with the LEDs is used to avoid
a reverse voltage on the LEDs in case the other diode
leaks a little bit. It may not be necessary but I thought
it was a good idea.
</p><p>
Working out the resistor value is similar to the other
example and comes out to about half the value of the
full wave version, or about 1.2K at 1 watt in this case.
But the peak LED current will be twice as much or about 78mA.
This is probably not too much, but you may want to look
up the maximum current for short duty cycles for the LEDs
used and insure 79mA doesn't exceed the spec.
</p><p>
</p><p><img src="TRAFFIC_files/ledlamp.gif">
</p><p><img src="TRAFFIC_files/ledlamp2.gif">
</p><center><a href="http://ourworld.compuserve.com/homepages/Bill_Bowden/homepage.htm#menu">Menu</a></center>
<hr size="1">
<a name="traffic.gif"></a>
<center><h3>LED Traffic Lights</h3></center>
<br>
<p>
The LED traffic Light circuit controls 6 LEDs (red, yellow and green)
for both north/south directions and east/west directions. The timing
sequence is generated using a CMOS 4017 decade counter and a 555 timer.
Counter outputs 1 through 4 are wire ORed using 4 diodes so that the
(Red - North/South) and (Green - East/West) LEDs will be on during the
first four counts. The fifth count (pin 10) illuminates
(Yellow - East/West) and (Red - North/South). Counts 6 through 9 are
also wire ORed using diodes to control (Red - East/West) and
(Green - North/South). Count 10 (pin 11) controls (Red - East/West) and
(Yellow - North/South). The time period for the red and green lamps will
be 4 times longer than for the yellow and the complete cycle time
can be adjusted with the 47K resistor. The eight 1N914 diodes could be
subsituted with a dual 4 input OR gate (CD4072).
</p><p><img src="TRAFFIC_files/traffic.gif">
</p><center><a href="http://ourworld.compuserve.com/homepages/Bill_Bowden/homepage.htm#menu">Menu</a></center>
<hr size="1">
<a name="sunset.gif"></a>
<center><h3>Photo Electric Street Light</h3></center>
<br>
<p>
This is basically a Schmitt Trigger circuit which receives input from
a cadmium sulfide photo cell and controls a relay that can be used to
switch off and on a street lamp at dawn and dusk. I have built the
circuit with a 120 ohm/12 volt relay and monitored performance using a
lamp dimmer, but did not connect the relay to an outside light.
</p><p>
The photo cell should be shielded from the lamp to prevent feedback
and is usually mounted above the light on top of a reflector and
pointed upward at the sky so the lamp light does not strike the
photo cell and switch off the lamp.
</p><p>
The photo cell is wired in series with a potentiometer so the voltage
at the junction (and base of transistor) can be adjusted to about half
the supply, at the desired ambient light level. The two PNP transistors
are connected with a common emitter resistor for positive feedback so as
one transistor turns on, the other will turn off, and visa versa.
Under dark conditions, the photo cell resistance will be higher than
the potentiometer producing a voltage at Q1 that is higher than the
base voltage at Q2 which causes Q2 to conduct and activate the relay.
</p><p>
The switching points are about 8 volts and 4 volts using the resistor
values shown but could be brought closer together by using a lower
value for the 7.5K resistor. 3.3K would move the levels to about
3.5 and 5.5 for a range of 2 volts instead of 4 so the relay turns
on and off closer to the same ambient light level. The potentiometer
would need to be readjusted so that the voltage is around 4.5 at the
desired ambient condition.
</p><p><img src="TRAFFIC_files/sunset.gif">
</p><center><a href="http://ourworld.compuserve.com/homepages/Bill_Bowden/homepage.htm#menu">Menu</a></center>
<hr size="1">
</blockquote>
</body></html>⌨️ 快捷键说明
复制代码
Ctrl + C
搜索代码
Ctrl + F
全屏模式
F11
切换主题
Ctrl + Shift + D
显示快捷键
?
增大字号
Ctrl + =
减小字号
Ctrl + -