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Epson PX-8 - 3.7 The CP;M Built-In Commands

Epson PX-8
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REV.-A
2.2.5.2
LCD
Drive
Source
Regulator
Two
voltage
supplies are required
to
drive
the
LCD display;
+5V
is required
for
the
logic
circuit, and
20V
is required
for
the
X-V
drivers. A
total
potential
dif-
ference
of
20V
is obtained
by
subtracting
the
-1
5V
output
voltage
of
this
regulator
from
the
+5V
of
the
logic
circuit
power
supply. Fig.
2-18
shows
the
regulator circuit.
Circuit
Operation
The
oscillator
circuit
generates a
clock
of
ap-
proximately
35
kHz
when
the
POWER
switch
is
turned on. This
clock
is fed
to
pin 7
of
the
IC
140
through
R55, C27, and R54. The inverted
output
at
pin 6 is
input
to
the
base
of
the
tran-
sistor
029
through
R20,
switching
it
on and
off.
The
emitter
of
029
is
connected
to
the
+5V
lo-
gic
circuit
power
supply
and
the
collector
is
connected
to
ground
through
inductance L3.
As
the
transistor
is
switched
on and
off
by
the
clock
signal, a voltage, as
shown
in Fig.
2-20
,
which
is
the
counter
electromotive
force across
L3, appears
at
the
collector
of
029.
While
the
collector
voltage
swings
negative, a
current
flows
in
through
diode
06,
generating a nega-
tive
voltage
at
the
negative side
of
capacitor
C 17.
This
output
is used as
the
LCD drive
source voltage.
It
is also fed
to
the
constant
voltage
circuit
which
connects
the
LCD
drive
voltage
to
the
+5V
logic
circuit
line
through
the
resistor
R147
and
the
zener
diode
Z020.
The
zener
diode
has a
breakdown
voltage
of
20V.
Thus,
when
the
output
voltage
rises
to
-15V
or
LCD
regulator
output
voltage
35
kHz clock
output
Return voltage
Oscillator
circuit
Constant-voltage
R
27
R90
68K
2
converter
GND-----+----------
Fig.
2-18
LCD
Drive
Source
Regulator
Circuit
above,
the
zener
diode
breaks
down,
raising
the
base
of
the
transistor
034
to
the
high level
of
+5V. This
puts
the
transistor
in
conduction
and
its
collector
is driven negative,
disabling
the
clock
signal
to
029.
This
stops
switching
of
029
and
thereby
lowers
the
output
voltage.
This
state
is
maintained until
the
zener
breakdown
comes
to
an
end.
At
that
time,
029
switches
again.
The
circuit
repeats
this
operation
to
produce a stable
voltage
of
-15V.
The signal generated
at
the
collector
of
029
is fed
to
the
diode
014
through
the
capacitor
C35.
The negative
component
of
the
signal is removed
by
a
current
supply
from
ground
through
diode
014,
while
the
positive
voltage
is fed back
to
the
auxiliary
battery
through
014.
2-24

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