Voltage loss battery to charge post?
08-09-2014, 02:56 PM
#1
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Voltage loss battery to charge post?
Does anyone know the typical voltage loss when measured directly from the battery and then from the charge post?
Measured at the battery, mine was 12.8 volts and at the post 12.58 volts.
I am guessing that is not too bad for the length and gauge of wire involved.
Thanks
Measured at the battery, mine was 12.8 volts and at the post 12.58 volts.
I am guessing that is not too bad for the length and gauge of wire involved.
Thanks
08-09-2014, 03:39 PM
#2
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The conditions under which you measure are critical - there should be ~no voltage loss with the car off. With ignition on engine off you will see a positive loss (jump post will be lower) with engine on at cold idle you will likely see a negative loss (battery will be lower). How much equipment is on/active and the battery state of charge will directly affect both these cases - so there can be no simple answer.
The connections: Alternator to jump post, jump post to starter, starter to battery and the associated ground connections: battery to ground point and engine to ground point all factor into the measurements.
Each of these segments has a resistance - which will vary slightly with temperature and may also vary mechanically (ground strap). Even the battery itself also has an internal resistance and this varies depending on its health and state of charge. The voltage dropped across each of these resistances is dependant on current (V = I x R). You are always looking at several or more of these segments combined in any measurement you take.
Alan
The connections: Alternator to jump post, jump post to starter, starter to battery and the associated ground connections: battery to ground point and engine to ground point all factor into the measurements.
Each of these segments has a resistance - which will vary slightly with temperature and may also vary mechanically (ground strap). Even the battery itself also has an internal resistance and this varies depending on its health and state of charge. The voltage dropped across each of these resistances is dependant on current (V = I x R). You are always looking at several or more of these segments combined in any measurement you take.
Alan
08-09-2014, 03:48 PM
#3
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I do have a very simple remote door unlock/alarm hot wired to the battery, would that could explain the small voltage drop?
Wouldn't the analog clock draw some voltage as well the hot wired portion of the stereo?
Thanks
Wouldn't the analog clock draw some voltage as well the hot wired portion of the stereo?
Thanks
08-09-2014, 04:06 PM
#4
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So you seem to be implying this was engine off key out... in that case your results make no sense.
V = I x R So: R = V/I
Normal parasitic leakage is <50mA = 0.05A, you are showing a total 0.22v drop => 4.4 ohms
(This is a huge huge value in this context)
Now if we are trying to supply say a modest 15A over that same segment 15 x 4.4 = 66v! => it's not even close to possible to do with a 12v system. So something does not compute here...
Do the measurements again (use the same ground point for both - get a longer lead) or evaluate what parasitic current you actually have? and/or get the battery tested
This is most likely just a measurement technique issue
Alan
V = I x R So: R = V/I
Normal parasitic leakage is <50mA = 0.05A, you are showing a total 0.22v drop => 4.4 ohms
(This is a huge huge value in this context)
Now if we are trying to supply say a modest 15A over that same segment 15 x 4.4 = 66v! => it's not even close to possible to do with a 12v system. So something does not compute here...
Do the measurements again (use the same ground point for both - get a longer lead) or evaluate what parasitic current you actually have? and/or get the battery tested
This is most likely just a measurement technique issue
Alan