Lock switch position

[t160hq]

Originally posted by HeartofDarkness
Hello -

I'm new to the forum. I just bought a BeV3100 receiver that has a Digital Lock aand an EEPROM lock already installed. But, the switches aren't labeled. How do I know which is the "locked" positions? Am I correct to assume that for the Digital Lock, the "lock" position is such that pin 1 of the 74AC244 buffer IC is shorted to ground? For the EEPROM lock, is the "lock" position such that pin 7 is connected to pin 8 through the 1K ohm resistor? BTW, there is no 1K ohm resistor installed on my EEPROM lock switch - pin 8 is connected directly to the switch. Should I disconnect it and put a 1K ohm resistor in series?

Thanks.

Sounds like the two IC version of the lock. If it is

Pin 1 shorted to ground is the unlocked position.

Pin 1 no connection is the locked position.

The locked position for the EEPROM is the
(still connected to the printed circuit board)
pin 8 of the IC (5 volt supply). Connected to
Pin 7 of the IC (the pin of the IC lifited from the
printed circuit board). Not the pin 7 connection
on the printed circuit board trace.

Unlocked would be pin 7 of the IC to the pin
7 connection on the printed circuit board.

The use of the resistor depends on where
you tapped the 5 volt supply. If you got the
5 volts from the power supply of the receiver
then you need the resistor. If you tapped
the 5 volts from the resistor going to the
EEPROM pin 8. Then you already have a
resistor hooked up.

If in doubt put one in anyways. It won't
affect the operation of the lock. It will save the
power supply in the receiver in case of
a short.

t160hq

[t160hq]

Yep all EEPROM's no matter the model are 5V.

TSOP's vary by model. 3.3V in the newer models.
Older models use 5V.

t160hq

Don't take my word for it though. It's a hobby
after all and I heartly encourage hands on
experience. Get out a multimeter and start
checking it out. :-)