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Most of the ideas in this chapter can be
found on the pages of this website, but just in case you
want to go over the capabilities of the '508A, we have
brought them together.
Quite often when you are programming,
the first thing you will run out of is output lines.
Many projects need lots of drive lines and if you need
more than about 8, you should go to another
micro-controller.
Don't expect an 8-pin chip to perform
the impossible.
The designers of the '508A have done an
amazing job providing 5 output lines (and one input
line) in an 8-pin chip, but even so, many projects run
out of drive-lines.
On other pages of this course we have
shown how to expand the drive lines with a binary
counter or shift register. This can increase the lines
to more than 10 however if you want to add just one or
two more devices than the chip is directly capable of
handling (5), there are clever ways to connect them to
the chip.
One of the simplest combinations is a
LED and Piezo on a single drive-line as shown in fig: 1.
These can be combined because the requirements of a LED
are different to a piezo. A LED requires a constant HIGH
for it to illuminate while a piezo requires a
HIGH-LOW-HIGH waveform at approx 3kHz to produce a tone.
If the mark-space ratio of this waveform
is kept short as shown in fig: 2, the LED will only
illuminate very dimly. A short mark-space ratio means
the "mark" is very small compared to the "space". A very
short on-time (mark) and a long off-time (space) will
not affect the tone from the piezo but will deliver very
little energy to the LED and this is exactly what we
want.
On the other hand, each time the LED is
activated, only a very small click will be heard, and
this will hardly be noticeable. In this way the two
devices can be combined on the same line.
In this book, we show how to connect two
and up to five or more push buttons on a single input
line and generally you will not have any problems adding
buttons to a project. But if you want to add a secret
reset button (or a "cheat" button, for example), it can
be added across an existing LED as shown in fig: 3.
The resistor between the switch and
micro acts as a safety resistor to prevent the output of
the chip being damaged if the switch is pressed when the
LED is activated, and it also acts as a dropper resistor
for the LED.
These two items will work in combination
because the impedance of the LED is very high when no
voltage is across it and when the micro turns the line
into an input line, it sees the LED as a high impedance.
In other words it is not detected so that when the
switch is pressed the micro only sees the switch as a
LOW.
Up to 5 different programs can be burnt
into a single '508A and the required program can be
accessed by soldering a resistor between one of the
outputs and the "input-only" line - GP3, as shown in
fig: 4.
During turn-on, a special program will
put a HIGH on each output in turn and the output
containing the resistor will determine the program.
Combining 5 programs in one chip will
reduce inventory costs as the required program can be
selected by fitting the resistor in the appropriate
place on the board.
If a device is connected between two
output lines as shown in fig: 5, a program can be
written so that the device sees a voltage reversal. When
one output is HIGH the other is LOW and this is then
reversed.
The drive lines have a maximum output
current of 25mA and this is enough to drive a number of
different devices. If a red LED is connected in one
direction and a green LED in the opposite direction,
they can be turned on and off individually, as shown in
fig: 6.
If the two LEDs are placed near each
other or combined in the one LED (called a tri-coloured
LED), they will produce a number of colours including
orange, depending on the mark-space waveform delivered
to each LED.
A single LED containing red and green
chips is available in 2 or 3 lead versions. The wiring
for a 3-leaded tri-colour LED is shown in fig: 7. The
tri-leaded version is shown in fig: 6.
Tri-coloured LEDs are fairly expensive
but if the project can cover the expense, they can be
the basis of "running message" displays and simple TV
screens.
If you connect a piezo to two
out-of-phase lines as shown in fig: 8b, the sound
produced will be slightly louder than the arrangement in
fig: 8a.
When we talk about a piezo we really
mean a PIEZO DIAPHRAGM. A piezo diaphragm is a passive
device and is very similar to a capacitor as far as the
circuit is concerned. Ceramic substrate on a metal
diaphragm causes the metal to "dish" and bend to produce
a high pitched sound. The size of the voltage (the
amplitude) determines the intensity of the sound and the
frequency of the waveform determines the tone.
The voltage across the piezo from one
drive line is about 5v whereas the voltage seen by the
piezo from two reversing lines is about 10v.
Unfortunately this doesn't produce twice the sound
output but the sound is slightly louder. If you want a
louder output you should use a better-quality
high-output diaphragm (such as from a Christmas card).
The loudest output is a piezo siren and
this is an active device containing a transistor
oscillator and choke. These units operate from 5v to 15v
and produce a very loud output while consuming only
about 10mA to 15mA.
Each output line of a '508A can only
deliver about 25mA. This current is determined by the
size of the transistor delivering the current. The
transistor inside the chip is only very tiny and if a
higher current
is drawn, it may be damaged.
When a resistance of 200 ohms is
connected from output to ground, a current of 25mA flows
(when the output is HIGH). If the resistance is reduced,
a higher current flows. This means a resistance of 200
ohms or higher is required to make sure the current flow
is less than 25mA.
But if a LED is placed on the output,
how is the resistance worked out?
A LED drops a voltage across it
according to its colour. This is called the
CHARACTERISTIC voltage or the CHARACTERISTIC VOLTAGE
DROP. This voltage is constant, no matter how bright the
LED is illuminated.
For a red LED the characteristic voltage
is 1.7v.
For an orange LED the characteristic
voltage is 1.9v.
For a green LED the characteristic
voltage is 2.1v.
LEDs cannot be connected directly to the
output of a drive-line without a voltage-dropping
resistor.
The reason is very technical but
basically a red LED does not turn on AT ALL until
exactly 1.7v is placed across it and if the voltage
tries to rise above 1.7v, the LED will glow brighter,
allow a very high current to flow and will be damaged.
The only way to prevent damaging the LED
is to provide it with a very accurate supply voltage or
simply connect a resistor in series. If the value of the
resistance is worked out, an accurate current can be
delivered to the LED and everything will be ok. The LED
will last 100 years!
Suppose you want to deliver 25mA to a
LED.
If we take a red LED, the value of
resistance can be worked out by Ohms law. The voltage
across the resistor is: 5v - 1.7v = 3.3v
I = V/R
0.025 = 3.3/R
R = 132 ohms Use 130R resistor.
If a number of LEDs are required to be
driven from a single output, you will need to connect
them in parallel or series. There are limitations,
however, as you will see.
Firstly it is ok to connect two LEDs in
series, provided you work out the value of the dropper
resistor.
If a single red LED is connected to an
output and supplied with 25mA via a 130 ohm resistor,
when two LEDs are connected in series, the second LED
will drop 1.7v and the supply voltage will be 5v - 3.4v
= 1.6v. In this case the two LEDs will receive a current
of 12mA from the 130 ohm resistor and they may be a lot
duller than expected.
To supply them with 25mA, the dropper
resistor must be:
R = V/I
= 1.6/0.025
= 64 ohms
Use 68R resistor as shown in fig: 10.
If three LEDs are connected in series,
the total characteristic voltage drop will be 1.7v +
1.7v + 1.7v = 5.1v This is higher than the maximum
voltage on the output line and in theory, the LEDs will
not illuminate AT ALL, no matter what dropper resistor
is used!
This means only two LEDs can be
connected in series to an output line.
LEDs can be connected in parallel AND
series as shown in fig: 11. Four LEDs is the maximum
that can be driven from a single output line and this
will deliver about 12mA to each LED.
You will notice a separate dropper
resistor is required for each column of LEDs because
LEDs cannot be operated in parallel due to the 1.7v
characteristic voltage required across each for perfect
operation.
For example: one LED may have a
characteristic of 1.75v and another may have 1.65v
characteristic. The 1.65v LED will rob the other of
voltage and prevent it operating. More on this in our
Notebook series.
If more than 4 LEDs are required to be
driven, a buffer transistor will be required as shown in
fig: 12. This transistor will allow the LEDs to be
driven from a 12v supply (or higher) and the number of
LEDs can be increased to 6 per column for 12v.
If the transistor can handle 100mA, four
columns can be made, allowing 24 LEDs to be illuminated.
In this way segments of a large pattern
can be illuminated and by referring to some of the
projects we have included, the lines of the '508A can be
expanded to twelve or more and an impressive display can
be created.
A globe is a device that requires a very
high start-up current. This is something you may not be
aware of. The start-up or warm-up current for a globe is
about 6 times its operating current and although this
current is drawn for only a very short period of time,
it is one of the reasons why a globe does not work in
some circuits. The high start-up current prevented one
of our flip-flop circuits working. The resistance of the
leads from the project to the battery was sufficient to
prevent the circuit starting-up. That's why it is
important to remember everything we discuss.
To drive a globe from an output of the
microcontroller, a buffer transistor is needed.
Any type of transistor will be suitable
providing its current handling ability is about 600mA
for each 100mA of operating current for the globe. A
suitable drive circuit is shown in fig: 13.
The globe can be dimmed by delivering a
variable mark-space waveform. Fig:2 shows the type of
waveform with the on-time represented by the "mark"
portion of the waveform.
Most relays require more than 25mA for
operation and need a 12v supply.
For this a buffer transistor is
required. Fig: 14 shows how a relay is connected to an
output of the '508A. The diode across the relay prevents
voltage from the relay getting into any of the supply
lines and affecting the operation of the microprocessor.
When the relay is turned off (de-energised)
the collapsing magnetic field of the coil generates a
very high voltage and this can be passed to the supply
rail if it is not "snubbed." The diode absorbs (snubs)
this voltage.
When designing an alarm project, the
wiring between the sensors and the control panel must
not be able to be cut otherwise the alarm system is
useless!
Making the wiring FULLY SECURE is very
complex but a simple way to prevent the wires being cut
or joined together (without detection) involves a
line-sensing feature called LINE REVERSAL.
Most alarms consist of pin switches or
reed switches with the windows and doors physically
keeping the switch closed or having a magnet to keep the
contacts of a reed-switch closed. If the door or window
is opened, the circuit becomes open and the alarm is
activated.
This is called a CLOSED LOOP detection
circuit.
It's very easy to see that if the two
wires going to the alarm panel are joined together near
the alarm panel, the alarm will not activate when a
window or door is opened.
To overcome this problem a diode is
placed in the line near one of the sensors as shown in
fig: 15.
The alarm is required to send out a HIGH
on one line and detect the high on the other line. The
alarm then sends out a HIGH on the other line and since
the diode is reverse biased for this condition, the
alarm must sense a low on the first line.
This "line testing" is done many times
per second and if the line is shorted, the program will
detect the interference. The only limitation to this
system is the micro will not detect a diode fitted
across the two lines near the alarm panel. The thief has
to know of this limitation and the diode has to be
fitted around the correct way to defeat the system.
How many thieves carry a diode with them
and know how to fit it?
In theory you can connect a motor to two
drive-lines and get forward and reverse operation.
But unfortunately a motor requires a
very high start-up current and the drive lines of a
'508A are not capable of delivering this current. The
solution is to use transistor buffers in the bridge
arrangement
of fig: 16.
Lines A and B are two outputs from a
'508A. There is an important point to remember when
programming the output lines to prevent a short-circuit
occurring with the bridge.
You will notice that if both A and B are
HIGH at the same time, transistors L, M and X, Y will be
turned on at the same time and a short-circuit will
occur on the power rail.
To prevent this from happening lines A
and B must be LOW at the beginning of the program.
By taking line A HIGH, transistors M and
X are turned on and this delivers voltage to the motor
to turn it in the clockwise direction.
To control the RPM of the motor, line A
can be given a variable mark-space ratio.
To reverse the motor, line A must be
taken low and after a short delay, line B can be taken
HIGH. This will deliver voltage to the motor via
transistors Y and L and cause the motor to revolve in
the opposite direction. Reverse RPM can be adjusted with
a variable mark-space pulse.
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