Date: Wed, 26 Oct 2005 01:45:02 +0100 From: Pooh Bear <rabbitsfriendsandrelationsnospamail.com> Subject: Re: expected current draw when car is off
Mike Deskevich wrote: > I have an '87 C900 that seems to be killing batteries. After being > parked for 2 weeks while I was out of town, the battery was dead > (didn't even light up the dash board a little when I turned the key). > I wasn't too surprised, beacuse it had been getting hard to start over > the last month or so and the battery was 5 years old. So I replaced > the battery. Since my truck sits outside in the cold weather, I bought > a new battery for the truck and took the 2 year old battery out of the > truck and put it in the Saab. Everything was fine for another month or > so, and then I let the Saab sit for a week. It was very hard to start, > but it finally turned over. Then the Saab sits for another week and > now the battery is dead again. I realize that I put an old battery in > the Saab as a replacement, but I do know the battery was good. It > never gave me any problems in the truck. Well...... Batteries sometimes do just die like that. > Before I buy another battery for the Saab I want to make sure there is > no short or something wierd going on. The first thing I want to look > as is how much current the car is drawing when the key is in the off > position. The only thing that I can imagine drawing power is the > radio to help it remember presets, and the ECU to help it remember > things that it need to remember. Everything else should be > disconnected. Does anyone know how much current should be drawn from > the battery when the key is off? I don't have a FSM or anything for > the Saab (and I know the FSM for my Toyota doesn't give this kind of > info). > > I don't want to put a new battery in to have it get killed so soon, > nor do I want to take it to $80/hr saab mechanics to look for a short. > So I want to do as much diagnosis as I can by myself first. For a tiny fraction of $80 you can buy a basic multimeter ( on Ebay for example ) that will measure the current for you. Graham