AIRBAG light
. Can this be edited/scanned/autotapped out so that it does not return, or is the resistor to fool the system into thinking the airbag is still in place the only solution?? Any really good GM PCM/BCM electrical techs out there? In other words, can DTCB0670 be edited out/eliminated in any way/shape or form, or will it keep lighting up dash lights forever? Trending Topics
I think the resistance of the driver's side is 6 ohms.
http://www.ls7power.com/diagrams/air...int.system.pdf
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I think the resistance of the driver's side is 6 ohms.
http://www.ls7power.com/diagrams/air...int.system.pdf
If the circuit has less than .7 volts the resistance is infinite. If more....usually the resistance is close to nothing. Hard to say the exact amount.
I would just put the LED back in and try to trick the airbag module into thinking the airbag is still there. If you read through the document above it has the conditions the computer will check for.
) and fail me instantly.

Here is an LED: http://www.radioshack.com/product/in...ductId=2102852
Here is the resistor: http://www.radioshack.com/product/in...entPage=search
You can get them at many electronics stores. Make sure the LED is white when lit not just when off.

With the yellow airbag connector... It won't work unless both sides are plugged in. I would take the other side of the connector, cut it off, put the resistor on that, and then plug it back in.
The airbag system will run the test cycle when you throw the key. It wants more than 3.7 ohms and less than 6. That can be done with two 10 ohm resistors in parellel. That would be 5 ohms. Unless, you can just find a 5 ohm resistor and use that. If you can, use 1/2 watt resistors instead of the 1/4 watt. I think the 1/4 watt will not be enough.
In the event that the airbag module decides to active the deployment loop:
volts times current equals power
(23 volts) times (4.6 amp) = 105.6 watts
So the resistors will instantly burn up and open the circuit again and you will have to replace them.
There would be .4 watts applied in the circuit during the test cycle. That is more than 1/4 watt but the test cycle is only for 1/3 of a second. The strain is smaller because it lasts for such a short time. Plus the .4 watts is going to be across both resistors so it is half as much.
Always over design.
Last edited by 2002_Z28_Six_Speed; Jan 10, 2006 at 10:07 PM.
As I stated in reply #16 I was able to make a resistance that the computer would accept putting it in place of the airbag and leaving out the steering column completely.
The value was somewhere around 1 to 1.5 ohms. I remember having to dial it in. Wasn't too bad. Just a trip to Radio Shack.
It fooled the computer and the light never came back on again until I had the time to replace my broken clockwork spring. Unfortunaley, I have since sold the device to a friend and there is no market for them so I never made another one.
Wonder if dailydriver ever got his fixed.
It has gone through 2 inspections (even with our ***** here) without a question being asked. I guess when they plug into the OBD2 port it DOES NOT connect/interface with the BCM like it does with the ECM.
I may still try the solution you came up with, but it may not work since my AIRBAG bulb is broken out of the IP, with no "jumper" or new bulb in it's place.
One question; do you think with your solution that the system would be fooled to the point of keeping the passenger airbag functional/deployable, EVEN with NO airbag on the driver's side???
As I stated in reply #16 I was able to make a resistance that the computer would accept putting it in place of the airbag and leaving out the steering column completely.
The value was somewhere around 1 to 1.5 ohms. I remember having to dial it in. Wasn't too bad. Just a trip to Radio Shack.
It fooled the computer and the light never came back on again until I had the time to replace my broken clockwork spring. Unfortunaley, I have since sold the device to a friend and there is no market for them so I never made another one.
Wonder if dailydriver ever got his fixed.
i went out and bought two 10ohm 10watt resistors.
well i also have 5 1/4watt 10 ohm resistors...
i guess i could just slap em all together in parallel.
please clarify.
thanks






