Epson PX-8 Technical Manual page 48

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