Delay Circuit, Reset Circuit; Shift - Xerox OLYMPIA ELECTRONIC COMPACT Training Program

3.06 Delay circuit, Reset circuit
(Circuit Diagram F)
Since the voltage UL +12V and then UC +SV are not soon available when
switching on the machine, a switching-on delay circuit is effected to
UC +SV which activates the motors and the hammer solenoid.
The CPU (IC 4) has H-level at its pin 33 and the NOR gate IC 7 at
pin 13 when the machine is switched on. During increasing of the
UL to +12V, the Zener diode ZD 1 gives 4.SV (UL-7.SV) to the base
of transistor Tr 3 . And its collector comes to L-level when the
UL reaches 10V or more. The L-level goes to the pin 12 of NOR gate
IC 7 • The L-level is led to the base of Tr 31 from the output of
IC 7 and the Tr 31 is opened. In the meantime Tr 6 for the driver
circuit still keeps opening.
Then the capacitor C 2 is charged, the reset input pin 4 of CPU
comes to H-level and the CPU initiates its programme. After the
CPU has initiated the programme, the pin 33 comes to L-level and the
pin 13 of IC 7 also comes to L-level. The output of IC 7 comes to
H-level. Then the transistor Tr 31 and Tr 6 open and the control
voltage +SV is switched through to the driver circuit.
When the machine is switched off or when the UL drops below 9V,
Tr 3 opens and pin 12 of IC 7 comes to H-level. The output of IC 7
comes to L-level. Then the transistors Tr 31 and Tr 6 open and the
control voltage +SV for the driver circuit is switched off.
The switched-through transistor (Tr 3) gives H-level to the base
of Tr 4, which makes the pin 4 of CPU L-level via the reset line.
Then the reset is effected by means of the capacitor C 2.
3.07 Shift
L
(Circuit Diagram F)
The emitter of transistor Tr 1 has H-level from the CPU (IC 4) via
the pin 9 of IC 5.
When pressing the shift lock key, +SV goes to the pin 1 of IC 7 via
the cone 1 pin 9 and the cone 2 pin 1, and the flip flop is set.
Since the output of flip flop comes to H-level and one of the TRA 1
is inverted, L-level goes to the LED via the cone 1 pin 10 and the
LED lights on. In the meantime H-level goes to the keyboard diode
D
G
via the collector of Tr 1 and cone 2 pin 3, and goes to the
keyboard buffer IC 1 via the cone 1 pin4. Thus the key scanning
from CPU detects the shift lock key being depressed.
When one of the shift keys is depressed, +SV goes to the pin 6 of
IC 7 via the cone 2 pin 1, and the flip flop is reset. Since +SV
goes through the diode D 2 and inverts one of the TRA 1, L-level
goes to the LED via the cone 1 pin 10 and the LED lights on. And
the key scanning from CPU detects the shift key being depressed.
While the shift key is depressed, the current keeps flowing via D 2
to TRA 1 and the LED keeps lighting on. However, the output of IC 7
keeps L-level.
When the shift key is released, the input of the IC 7 comes to L-
level and the output is keeping its L-level. Since the base of
Tr 1 comes to H-level and it opens, the LED has no voltage and
lights off. In the meantime L-level goes to the keyboard diode
D
0
via the cone 2 pin 3, and goes to the keyboard buffer IC 1 via
9~