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Bus’ battery doesn’t survive light use

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  • 18-01-2022 11:31pm
    #1
    Registered Users Posts: 3,818 ✭✭✭


    My brother has a bus for his job as a private tour guide. It’s being practically idle at my da’s for years since COVID. The bus’ keeps needing new batteries or at least a jump. My da starts it and lets it run for up to three hours every week but a battery has gone flat even after this.

    it’s the tachograph killing the battery apparently but still…. Is this normal? Is it the cold weather? Seems odd to me but I’m no expert.

    he can’t switch off the tachograph apparently, it’s illegal or something…although if the battery is flat it’s off anyway.



Comments

  • Registered Users Posts: 2,882 ✭✭✭cute geoge


    Cut the power from battery when bus is stoped wirh an isolator switch .The cheapest type are on a peoguet partner from mid noughties ,it is just a simple clamp on the leed for the positive on the battery just clip it off when not in use and clamp it back on when you want to start



  • Registered Users Posts: 3,818 ✭✭✭Did you smash it




  • Posts: 468 [Deleted User]


    There must be massive parasitic draw or alternator does not work.



  • Registered Users Posts: 7,247 ✭✭✭MrMusician18


    I was thinking the same. If I was to bet though, there is something that was fitted to the vehicle like the tachograph, alarm, tracker or other device that was wired direct to the battery and is a constant draw.



  • Registered Users Posts: 14,906 ✭✭✭✭CJhaughey


    If you want to ensure a decent charge maybe consider a maintenance charger that you leave on the battery all the time its not running.

    These are common on motorcycles and will keep the battery topped up so its always ready to go.

    It does sound as though there is some kind of parasitic draw maybe the tacho maybe something else, but the only way to find that is to start probing the fusebox and see which fuse the power is coming through.



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  • Registered Users Posts: 3,471 ✭✭✭John.G


    Get hold of or buy a cheap mutimeter for €30 or so like my one and check charging voltage with the engine idling with no main electrical loads on, the check it with dipped beams on, should get 14/14.5V, if not alternator regulator not working properly. If voltage OK you can then check parasitic loss by removing the battery negative strap, move the M.meter lead to the "10A max", turn the selector switch to "10" (6 oclock position) connect the red lead to the (removed) battery strap and the black lead to the battery negative terminal, let the reading settle down for 5 minutes or so then read it off, if very low select the next lower scale 0 to 200ma and read off, if reading less than 200ma remove the (m.meter) red lead and put it back in its original position and repeat the tests.


    Post edited by John.G on


  • Registered Users Posts: 3,818 ✭✭✭Did you smash it


    Thanks to those that have replied. I’ll pass onto my brother.



  • Posts: 468 [Deleted User]


    Cheapest multimeter is 2.5eur. Multimeter series parasitic draw test is so old school and impossible to do by modern technology. Just 10A is not enough for inrush current. Set multimeter to millivolt range and measure the across the fuses. I attached 4 different type fuse charts.




  • Registered Users Posts: 3,471 ✭✭✭John.G


    A €2.5 multimeter must come in a lucky bag! or else I was fleeced in paying €30 for mine 10 years ago. My M.meter has no problem in dealing with any inrush current if set to 10A and red lead in the 10A terminal, it certainly has never damaged it and I have tested a dozen cars or more, after 5 mins or so I change the scale to 200ma and read it off.

    I am a bit skeptical of the accuracy of the measurements of how many? fused circuits to summate them to the actual parasitic loss using the volt drop method.



  • Posts: 468 [Deleted User]


    2.5eur is these yellow/orange multimeters what they selling here about 10-15eur. You can test only older cars by that 10A range series. Modern car actuate many things soon as you complete the circuit. That chart is good enough to show to you the current draw. Simple ohm law I=V/R



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