Expect current draw. Mk1 cab?
Posted
#992887
(In Topic #117748)
Settling In
Expect current draw. Mk1 cab?
My battery is going flat rather quickly. Using an amp meter measured draw of around 35->50 mAh with ignition off. Car has had clifford alarm/imobiliser fitted so disconnected 25pin wire from control unit and draw went down to 27 mAh.
This still seems high. Can anyone suggest if this is acceptable? If deemed too high were should I start looking for faults?
Battery is 44Ah, so if fully charged:
44000/27 = 1629 hours = 68 days
Correct?
Thanks,
Chris
This still seems high. Can anyone suggest if this is acceptable? If deemed too high were should I start looking for faults?
Battery is 44Ah, so if fully charged:
44000/27 = 1629 hours = 68 days
Correct?
Thanks,
Chris
Posted
Local Hero
Your figures look fine. Disconnect the clock and the radio too, if there's still a current draw there's an issue somewhere (make sure the courtest light isn't on).
Posted
Old Timer
Re: Expect current draw. Mk1 cab?
cocjh1 said
44000/27 = 1629 hours = 68 days
Correct?
Thanks,
Chris
Not exactly its 27ma/sec not 27ma/h =) so you'd have to times that by 60.
Your drawing 1.6A per hour, so thats roughly 38A per day. And the 45AH battery is the maximum current draw (or something like that) it'll be something like 200ish AH (capacity), which gives you probably about 4days charge.
*edit* Actually thats wrong let me think about it, its 27ma/sec anyway =)
Just use this calculator its easier http://lib.store.yahoo…rts/BatterydrainCalc.html
~Madferret
Mk1 1457cc 5door GX '83
Mk1 1457cc 5door GX '83
Posted
Local Hero
Power = current x voltage
Power = energy / time
in other words, the rate of using* energy
*converting it from one form, eg chemical, to another, eg an internal loss in the alarm will be heat
The 27mAh is a red herring, in fact I think the OP would have measured the current using a multimeter as 27mA.
The battery's figure of 44Ah is rate x time, ie the whole quantity of energy it stores. So it can supply 44A for 1 hour, or 1A for 44 hours. So it would supply 0.027A for 1629 hours, or 68 days.
The only inaccuracy is that once the battery is 50%-75% drained, its voltage will be low, the cells will chemically alter and it will start draining itself, in other words a normal car battery is good at supplying high current for short time, not low current for long times. Compare this with a 'deep cycling' leisure battery (which is designed to supply low current for long periods between charging) or a more modern battery design which can do a bit of both (eg AGM).
Power = energy / time
in other words, the rate of using* energy
*converting it from one form, eg chemical, to another, eg an internal loss in the alarm will be heat
The 27mAh is a red herring, in fact I think the OP would have measured the current using a multimeter as 27mA.
The battery's figure of 44Ah is rate x time, ie the whole quantity of energy it stores. So it can supply 44A for 1 hour, or 1A for 44 hours. So it would supply 0.027A for 1629 hours, or 68 days.
The only inaccuracy is that once the battery is 50%-75% drained, its voltage will be low, the cells will chemically alter and it will start draining itself, in other words a normal car battery is good at supplying high current for short time, not low current for long times. Compare this with a 'deep cycling' leisure battery (which is designed to supply low current for long periods between charging) or a more modern battery design which can do a bit of both (eg AGM).
Posted
Old Timer
Its more complicated than that, hence using the calculator, can't remember why it was a good 10 years ago when I did it @ university something to do with internal resistance and the load comparitivly compared to it.
Lookup Thevinens and Nortons theorys.
Lookup Thevinens and Nortons theorys.
~Madferret
Mk1 1457cc 5door GX '83
Mk1 1457cc 5door GX '83
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