8.1.1.5. Alarm Signal Sending Circuit
This section explains about the operation of the circuit that sends AC alarm/DC alarm.
1) Dropper Circuit
The main converter output voltage is dropped to 27V by the constant voltage circuit consisting of Q521 and zener diode D521
to feed the power supply.
Refer to Fig. 5 (C).
2) AC Alarm Signal Sending Circuit
When input AC voltage is high, a current flows to the diode of PC2 in the AC detection circuit (Refer to AC Input ~ PFC Output
(P.17)) to turn ON the transistor of PC2. Therefore, output voltage from the dropper circuit is fed to the AC alarm sending circuit.
The transistor Q532 (on Sub board) turns ON, and then AC alarm sending output becomes LOW.
On the other hand, when input AC voltage is low, PC2 turns OFF, transistor Q532 (on Sub board) turns OFF, and then AC alarm
sending output becomes OPEN.
Refer to Fig. 5 (D).
3) DC Alarm Signal Sending Circuit
When 15VPT, 30V and 41V output voltage normally, transistors Q541 and Q544 (on Sub board) in output voltage detection
circuit turn ON, then the voltage fed from dropper circuit turns ON Q544 and PC1 (or Q506*1), transistor Q540 in DC alarm
sending circuit becomes ON, then DC alarm sending output becomes LOW.
When any of 15VPT, 30V and 41V output becomes abnormal, transistors Q541 and Q544 (on Sub board) turn OFF, Q540
becomes OFF, and then DC alarm sending output becomes OPEN.
(*1) AC input operation: PC1, Battery operation: Q506
Refer to Fig. 5 (A), (B).
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