Theory of
Operation—MicroLab
|
Instruction
When
the
TXD
output
of
U2040
is
low,
a "0"
state,
the
J-K
flip-flop
pair act
as
a
divide-by-four counter.
U7090B
will
output
a
2400
Hz
square
wave.
The
output
of
U7090A
will
be
1200
Hz.
The
data
that
entered the
modulator
as
a
series
of
“1's” and “0's” has been transformed
into
square-wave
signals alternating between 1200
Hz
and
2400
Hz.
The capacitor
and
resistor
at
the
output
of
U7090A round
off
the corners the square waves,
making the signal more
acceptable
to
a
tape
recorder.
Resistor
R8092
also
protects
U7O90A
from
accidental
shorting
at
P8098.
The
Demodulator.
The
Cassette
Demodulator
does the
reverse
of
the
modulator. The Demodulator
receives the
1200
Hz
and
2400
Hz
tones
and converts them
to
0's”
and
tg
As
the
signal
comes
in
from
the
recorder,
C8097
removes
any dc
component that
may exist. The two diodes
connected to the
incoming
line
limit
the
signal
swing
to
a
positive voltage
less than
+5
volts.
U2090A
and
its
surrounding
circuitry
act
as
an amplifier
with
a
gain
of 10.
Capacitor
C2097
is
a
high
frequency
roll-
off
capacitor
that
prevents the
amplifier from breaking into
oscillation.
The
amplifier’s
output centers around
+2.5
volts,
which
is
about
a
volt
too high
for
the
input to U2050B.
C2084 and
R2079 solve
this
problem.
C2084
removes
the
dc
component
from
the output
of
U2090A.
Then R2079
works
in
combination with
a
20K-ohm
resistor
within U2050B to
create
a voltage divider. The voltage divider lowers
the
center
voltage to
about
+1.2
volts.
U2050B
cleans
up
the
incoming
signal
to make
a
sharp-cornered square
wave.
The next
stage
of
the
Demodulator
is
made up
of
U1050B,
U2050D,
and
U1050C.
These devices act as
an edge-
triggered one-shot
multivibrator at both
the
rising
and
falling
edges
of
the
input from
U2050B
that
generates
a
pulse. The output
frequency
of
U1050C
will
be
twice
that
of
the
input
frequency.
This
is
how
it
works:
When
the
input
on
pin
4
of
U1050B
goes
high,
pin
6
goes
high.
At
the same
time,
the
input
signal
is
fed to
pin
10
of U2050D.
Notice that
the output
of
U1050B
is
fed
through
an
RC
network and
inverted
by
U2050D.
These
components
tend
to
delay
the
signal from
U1050B
pin
6
slightly,
so
that
it
arrives at
pin 9
of
U2050D
a
little
after
the
signal
at
pin
10.
Because
of
this
delay, every
transition
of
the
input to
pin
4
of
U1050B
causes
a
low-going
output
pulse
at
pin
8
of
U1050C. The
output pulse
has
a
duration
of
about
5
microseconds.
When
a
1200
Hz
signal
is
received,
a
pulse
is
produced
about
every
416
microseconds.
When
a
2400
Hz
signal
is
received,
the
pulse
is
produced
about
every
208
microseconds.
The low-going
pulse
output at
pin
8
of
U1050C
does
two
things:
the
pulse
is
turned
into
data
by
U1040; and, the
pulse
is
used
by
U1020
and
U1030
to
create
the
receive
clock
used
by
U2040.
Let's
talk
about the
receive
clock
first.
Remember
that
the
pulse
frequency
is
twice the frequency
coming from
the
tape
recorder. Also
remember
that
the
time between pulses
will
vary
with
the
incoming
signal
frequency.
Now
notice
that
the
pulse
is
fed to the
reset
input
of
U1020,
which
is
a
4-bit binary counter. The
clock
input
for
U1020
is
from
the 2400
baud
clock,
and
operates
at
a
frequency
of
38.4
KHz.
The
period
of
the
clock
input
is
26 microseconds.
As you
can
see,
all
the
load
inputs to U1020
are
tied
to
ground,
and
so
the
counter
counts
from
"0000" to"1111",
then
starts
at "0000"
again.
The
A,
B,
C,
and
D
outputs
of
the
counter
are fed to U1030, which
is a
demultiplexer.
When
the output
of
U1020
equals "1010"
(a
decimal
10),
the
Y2
output
of
U1030 goes
low.
When
the
output
of
U1020
equals “1011”
(
a
decimal
11),
the
Y3
output
of
U1030 goes
low. So,
the time
from
a
“0000” output
to
either
a Y2
or
a Y3
can be
calculated
by
multiplying
the
input
clock
period
by
10 or 11.
From
the time
U1020
is
reset
until
the
time
it
reaches
a
Y2
output
is
260
microseconds.
The
time
it
takes
to
reach
a Y3 output
is
286
microseconds.
Now,
let's
put
all
these
numbers together
and
see what
happens.
Let's say
we
get
a
low-going pulse out
of
U1050C
at
“time
0”.
This pulse
resets
U1020,
and the counter
begins
counting.
Now,
if
the
signal
we're
getting
from
the tape
recorder
is
at
2400
Hz
(a
“1”),
another
pulse
will
come
along
in
208
microseconds. So at "time
O”
plus 208
microseconds,
the
PULSE
line
goes
low
at
pin 12
of
U1050D.
Pin 13
is high
because
our
counter hasn't
reached
a
"1010"
state,
and
the
Y2
output
of
U1030
is
high.
At
“time
0”
plus
213
microseconds,
the
PULSE
line
goes low again,
and the output
of
U1050D
goes
high.
The
high clocks
the
RXC
input to
U2040.
At
the
same
time,
the
negative-going edge
of
PULSE
resets
U1020, and the
counter
starts
counting
again.
As long
as
a
2400
Hz
signal
is
being received,
the
PULSE
line
will
always beat the
counter
to U1050
and
create
an
RXC
clock.
An
remember,
the 2400
Hz
signal
from
the
recorder
is
doubled
by
U1050,
so the
frequency fed
to
the
RXC
input
ot
U2040
is
4800
Hz
—
exactly the
clock
frequency
required
by
U2040
to
communicate
at
300
baud.
What happens
if
the
incoming signal
from
the tape
recorder
is
1200
Hz?
Much
the
same
thing.
Recall
that
at
1200
Hz,
the
PULSE
line
will go low
once every
416
microseconds. This
rate
is
not
fast enough
to provide
4800
Hz
clock
input to
U2040. But,
the counter
takes
Summary of Contents for 067-0892-00
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