QuestionThere is a seperate trigger for the 2nd step. This was going to be used in the single wire scheme as mentioned before. The 1st step I was going to do just as you mentioned, but in this case cutthe lead to drivers side motor control beyond the OEM lock module. Then connect the 2 ends to terminal 87A and common, with 12V at 87 and 85, switched using the 1st step signal.
|---| single wire scheme
|---|-|-12V from here with lock
(1st)--| | and 2nd step unlock
|---|---------------------| |
|---------------| | |
| | |
DRIVERS LOCK MOTOR ------|- cut -|-|----------------|
| |OEM DOOR MODULE |
|----------------------------------|----------------|
To reiterate the wiring spec calls for only ONE resistor (1.5K) only for the UNLOCK, so since this is a single wire system this is essentially shorting, usless the intention is not to use relay for LOCK and connect normally using the module lock signal line, however since you have probably done this on countless GM's it is likly that the wiring spec provided is missleading or inaccurate.
"[2] Lock requires a negative trigger, and unlock requires a negative trigger through a 1,500 ohm resistor"
So your recomendation is that I really need two resistors. One value was provided in the spec the other I'll have to figure out... Corrrect?
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Followup To
Question -
Thanks, now to add upon this design do you see any potential problems if I adapt a 2 step unlock process with this design. In otherwards drivers side unlock using a relay beyond the OEM lock module and applying the schematic you provided for the 2nd step unlock and lock.
PS. It is interesting that the wiring spec (2001 Malibu) never recommended any resistor for the lock trigger. Does the lock trigger expect to see a short? Shorts in my book are almost always bad.
I suppose I'll have to introduce a pot to figure out what this unknown, unmentioned resistance is.
~Steve
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Followup To
Question -
My wiring diagram for installing a keyless entry system on a 2001 Chev Malibu notes as follows:
[2] Lock requires a negative trigger, and unlock requires a negative trigger through a 1,500 ohm resistor. Must use relays.
If I am understanding this correctly, lock/unlock are both (-), but from the addition of the resistor, the unlock is less (-) by the addition of the potential on the resistor. Having the resistor simply connected in series on the unlock signal does not seem to warrant a relay, especially since the signal goes through the OEM lock/unlock relay triggers anyway.
What are the aformentioned relays in this case used for, and what would the wiring schematic look like.
Answer -
This is Chevys single wire negative system
and needs the relays since most alarms will not
provide enough ground output without relays
to do this correctly.
My friends at designtech has made up a nice display
page at:
http://www.designtech-intl.com/Content/techsupport/documents/section04/note211.p
Go have a look at it.
ALARMTEK
Answer -
Check the alarm/keyless setting for your alarm. It may already address this issue by providing a 2nd pulse
out put for drivers door priority locks. Not really
certain of how to do a single wire/drivers door
priority system.
Normally with a 2 wire lock, you would cut the unlock
and put in a negative relay to differciate from fisrt
pulse to the second. I would be interested as to if
you solve this so I can share it with others.
I would not catogorize it as "shorting" it out. More
like designation the proper resistance to
pulse the lock to respond.
Let's see what you come up with.
Thank you for the interesting question.
ALARMTEK
AnswerI am not certain this needs 2 seperate resistors.
I would say trial and error is the way to go since
I have not tried this method. We don't attempt drivers
door priority with single wire systems here.
Let me know what you find. It's an interesting
twist on this set up, and I'd like to pass
you sokution on to the next guy if
it works.
ALARMTEK
