Ok, so last year I got a new battery. Winter was long and I am forgetting now but pretty sure it was on a tender. Anyway, bike had issues starting. Replaced battery again.
Now, if it sits for a 1-2 days, it wont start up. It just cant quite make it. Not sure what the issue is.
I have not ridden much this year, so I dont know if its an issue Ive had for a while and did not know since I was riding every day when I did not have a car.
The only electronics I have are the tender lead, a usb lead for my phone and denali 2.0 lights. Im not sure if there is a drain or something is broken, but need some direction. I am unsure how to troubleshoot it since the battery is brand new and now the second one.
If you have a meter, real simple, most meters have 3 lead connections, unless it is a cheap fixed lead meter. I will start with a cheap fixed lead meter. Several things I want to point out, if you have a cheap battery tender, plug it in and check the volts DC output without connecting it to the battery, if you have 13 VDC or higher great
, this is strictly so you don't need to reset your clock,
if you don't care then ignore this step, or if you have a smart tender that only powers up on correct polarity , again forget this step.
#1- With tender ( one that outputs DC without a battery connected ) hook the positive to the battery positive and the negative to frame ground Make sure you have it plugged in and your voltage is rising above what you measured at the battery before connecting your tender. Loosen off the negative battery post.
#2 -Next many meters do not have alligator, if you have a 2 lead meter select the highest DC amps, Do Not
hook across the battery, connect the positive meter lead to the lead clamp and the negative to the battery post, you can use some heavy elastics for hand free use, one on the post and one on the clamp / negative lead of the wiring. Your meter should show a reversed current which is you tender charge current, next disconnect the positive battery tender lead, your meter should show less than 0.013 amps and it should change rapidly to less than 0.005 amp , every so often it will be at around 0.013 amp for 1-3 seconds, this is normal, this is the clock circuit and part of the ECU, however most of the ECU is powered by a ECU relay which is keyed on power. If you have anything above these values, you have a drain, very easy to prove, as I said if you followed my advice on the Denali it shouldn't be a problem, however you can simply pull the included fuse of the Denali to prove if indeed you have a current drain.
Before I go further best to wait and see if you have a drain and what value it is, please post it.
For 3 lead connection meters
, you should have a common, a positive ?ohms / VDC/ VAC and a current positive AC amps and DC amps, you may also have 2 current ranges , pick the highest one. FYI when you are done make sure you put the positive current lead back in the volts / ohms socket, many fuses have been blown by forgetting, I have probably blown 4 in my lifetime, that is what happens after working 14 hours and are both physically and mentally exhausted and are not paying attention.
FYI anyone who has installed a series regulator
and not followed my install post, be aware that no matter what make of series regulator, you will have a continuous drain of 0.036 amp which is the headlight relay.After about 6 days you will be at less than 50% amp hour value of your AGM battery, at this level it may not start, at 9 days you will be at 20%, almost guaranteed not to start.
To calculate time:
our battery is 10 amp hour, what that means is 10 amp for 1 hour or theoretically 10 X 12.4 = 124 watts for 1 hour
Take discharge current , say 0.050 amp, 10 divided by 0.015 =200 hours theoretically divided by 24 hour days= 8.33 days, so 50% would be approximately 4 days, there are other factors to consider which is why I say theoretically , because after 2 or 3 days the battery voltage will have dropped and therefore the discharge current will be less except for devices such as USB power supply which will actually increase current as voltage drops.