I am the battery assassin

[Backinthesaddle]

Here goes, I have an 02 RSV and it assassinates batteries. This has happened three times in the past year. I come out to the bike and nothing. No lights, etc. I take the battery in and it will show fully charged, but the cold cranking amps way down. They give me a new battery in exchange and repeat the cycle (no pun intended). I am afraid to take the bike on any trips, because I don't know when it will fail next.

I have 44xxx miles on the 02. Should I just put a new stator and regulator on it? Any suggestions?

thanks, Paul

[MikeWa]

Let me make sure I understand this. The battery is fully charged but will not deliver the required cranking amps. Does it loose voltage quickly when testing CCA? Or does it continue to show good voltage? What kind of batteries are these?

 

Mike

[bongobobny]

Well, before you replace parts that may not need replacing, you need to do a little basic troubleshooting with a voltmeter. For starters, with the voltmeter on DC volts, measure across the battery with the bike running. At idle you should see around 12.8 to 13.2 volts. When you rev the engine up you should see the voltage rise, the higher the RPM's, the higher the voltage until you reach the "regulator operating voltage" which can be anything from around 13.6 to 14.8 volts, give or take. If you are seeing anything higher than 14.8 volts, chances are your regulator is defective and needs replacing!! Too high of a voltage will indeed cook a battery in a short time!

 

The next thing to do if your regulator voltage is good is to switch the voltmeter to AC voltage and do the same. Regardless of the RPM's you should not see any AC voltage other than possibly 1 volt or less. If you see a large AC voltage, the rectifier section of your rectifier/regulator is probably toasted. If your battery is seeing a large AC component that too will kill a battery. With AC applied to the battery it will be constantly charging and discharging. AC voltage is constantly changing polarity...

[Backinthesaddle]

Thank you both for the quick replies. The battery is an Interstate. Anyway just put the meter on and on enrichment high idle I was getting readings of 16-17. This then would point to the regulator correct?

[Backinthesaddle]

Ordered new voltage regulator

[Flyinfool]

That is definitely a bad regulator.

As long as you will be in there, do a test of the stator.

 

Disconnect the stator from the regulator.

 

With the meter set to Ohms, measure from each of the stator wires to a good ground, they should all read infinity.

 

Mark the 3 stator wires A, B, C.

With the bike running at idle and again at around 3000 RPM, Set the meter to the 200VAC scale, check voltage from A-B, B-C and A-C. At idle you should see around 15VAC, at 3000 RPM you should see 60 to 100VAC. What is important is that all 3 reading are very close to the same number.

[revinger]

Deja vu, at least you weren't 1000 miles from home like I was 2 weeks ago.

 

Ride it all day with a regulator like that and the battery gets so hot cant touch and whistles like a tea kettle.

 

Glad you are safe.

 

 

Sent from my iPhone using Tapatalk

[BlueSky]

Just tossing this out there. I checked the battery voltage in my truck one time and it checked 18 volts. After scratching my head a bit, I replaced the 9 volt battery in my multi-meter and the truck battery checked 12.6 volts.

[MikeWa]

Yes with voltage that high it sounds like the VR is cooking the battery.

[Backinthesaddle]

The 9v battery in the tester is new, but thanks for the heads up

[gibvel]

Just had the same thing happen and burnt up a battery. Changed the VR and everything is back to normal. If you ride at night much another sign is that you'll probably see your headlight change from dim to bright and or visa versa occasionally (without flipping the switch I mean.).

[Condor]

I've related this story before, and some of you are probably getting sick of me telling it, but I rode for 3 days on a DEKA AGM battery at 17vdc on a 1300 miles trip. When it died I stuck it on a charger and it took a full charge and it's still in the bike today. I'm gonna stick my neck out here and say that it's not the volts that will get you but the amps. 17vdc at .25 amps won't kill a battery. 17vdc at an uncontrolled amperage charge will BBQ one. Replaced the rectifier and everything is 'Hunky Dory'... :-)

[Flyinfool]

There is a direct mathematical relationship between amps and volts. Your 12V lead acid battery will never see a 17 volt .25 amp charge, it is mathematically impossible. As the volts go up the amps will also go up. If you do something to limit the amps that something will limit the amps by reducing the voltage.

 

There may have been other variables that got your battery to survive the couple of days with a runaway regulator. Most don't survive.

[bongobobny]

Try to think of it as trying to ram a size 18 quadruple E foot into a size 12 medium shoe! The shoe ain't gonna survive...

[Condor]

There is a direct mathematical relationship between amps and volts. Your 12V lead acid battery will never see a 17 volt .25 amp charge, it is mathematically impossible. As the volts go up the amps will also go up. If you do something to limit the amps that something will limit the amps by reducing the voltage.

 

There may have been other variables that got your battery to survive the couple of days with a runaway regulator. Most don't survive.

 

OK I 'd like to see the math that supports your statement. I've seen chargers or power supplies that will put out 12vdc@1amp, and others that will put out 12vdc@50amps. I have a battery charger that puts out 1a,10a,50a@12vdc. 17vdc pressure is not that much greater than 12vdc...

 

So my thinking is the pipe is only dripping?? :-)

[Flyinfool]

It is called Ohms Law. E ÷ R = I

E is voltage

R is resistance

I is current

 

Ohms Law is usually very close to the very first thing you learn about if you were to take a course in electronics because it governs almost everything in DC electronics and has a strong influence in AC electronics.

 

In this case since you are charging a battery which also has a voltage you need to use the difference in voltage between the battery and the charger as E in the equation. The resistance of the battery increases as the battery charges and the voltage difference gets smaller. That is why the amps go down as the battery charges until at full charge the amps are near zero and the battery voltage is at its max

 

 

Have you ever measured the actual voltage of your charger. It is not 12V it is much higher. Have you ever measured the actual charge current of your charger it will be less than the chargers rating. Chargers never tell you what voltage they put out they simply are stating that this charger will charge a 12V battery at up to X amps.

 

On your selectable current charger. Hook up a volt meter and an amp meter and connect it to a battery, as you change to the different switch positions you will see that as the volts go up the amps also go up.

 

If you do have a charger that is able to maintain say 10 amps thru the whole charge cycle, you will notice that the voltage is contaly climbing as the battery charges to be able to satisfy the requirements of Ohms Law.

[Condor]

Try to think of it as trying to ram a size 18 quadruple E foot into a size 12 medium shoe! The shoe ain't gonna survive...

I got all that resistence-pipe size,volts-pressure, and amps-water flow analogy down fairly good. The 18 foot will only go into the 12 shoe just so much in and then it won't fit any longer. Unless of course you try to jamb the rest of the foot in by increasing pressure and then the resistence..wire.. can't handle it can't handle it and burns up. But... if the control tap-rectifier will only let a small amount of water thru to the flowers the chance of the hose bursting does't exist.... IMHO

[Flyinfool]

I got all that resistence-pipe size,volts-pressure, and amps-water flow analogy down fairly good. The 18 foot will only go into the 12 shoe just so much in and then it won't fit any longer. Unless of course you try to jamb the rest of the foot in by increasing pressure and then the resistence..wire.. can't handle it can't handle it and burns up. But... if the control tap-rectifier will only let a small amount of water thru to the flowers the chance of the hose bursting does't exist.... IMHO

 

You are correct, if you use a regulator to reduce the flow of water to the flowers, the pressure in the hose will also be a lot lower than the pressure on the other side of regulator. Now if the regulator is broken and allowing more pressure thru you will have more flow and also have a greater risk of the hose bursting from that increased pressure.

 

To use real numbers

A fully charged battery is 12.6V, The output voltage on the charger is ~14.0V, that is a difference of 1.4V. You want the battery charging at 10A.

Plug that into the formula and you have

 

1.4V ÷ R = 10A

 

add some algebra (bet you never thought you would need that again after high school) and you get 1.4V ÷ 10A = .14 ohms

 

So now in that same scenario the regulator fails and the charge voltage goes to 17V?

We already calculated the resistance of the battery in this situation as .14Ω

The battery is at 12.6V so the difference between charger and battery is 4.4V

So plug that into our formula.

 

You get 4.4V ÷ .14Ω = 31.42875A

 

By having the regulator now outputting 17V that same battery is now being pumped 31.43 amps. It will be cooking.................

[Condor]

You are correct, if you use a regulator to reduce the flow of water to the flowers, the pressure in the hose will also be a lot lower than the pressure on the other side of regulator. Now if the regulator is broken and allowing more pressure thru you will have more flow and also have a greater risk of the hose bursting from that increased pressure.

 

To use real numbers

A fully charged battery is 12.6V, The output voltage on the charger is ~14.0V, that is a difference of 1.4V. You want the battery charging at 10A.

Plug that into the formula and you have

 

1.4V ÷ R = 10A

 

add some algebra (bet you never thought you would need that again after high school) and you get 1.4V ÷ 10A = .14 ohms

 

So now in that same scenario the regulator fails and the charge voltage goes to 17V?

We already calculated the resistance of the battery in this situation as .14Ω

The battery is at 12.6V so the difference between charger and battery is 4.4V

So plug that into our formula.

 

You get 4.4V ÷ .14Ω = 31.42875A

 

By having the regulator now outputting 17V that same battery is now being pumped 31.43 amps. It will be cooking.................

 

That's way above my paygrade... :-) However I have a question. Correct me if I'm wrong, but aren't the charge systems on the 1st and 2ndGens rated at 26amps?? OK I lied...one more. Why didn't I cook the battery on the 99RSV over three solid days of riding at 17vdc?? Rather, I depleted it down to an El Cheapo meter reading of 8.9vdc....

[Oldodge]

@Flyinfool , I just want to know where that ohm symbol is from. Is it hidden on the keyboard?????????? Inquiring (twisted) minds want to know.

[snyper316]

ΩΩΩΩΩΩΩ lol

 

Sent from my LG-K371 using Tapatalk

[snyper316]

ΩΩΩ¥alt234

 

Sent from my LG-K371 using Tapatalk

[Flyinfool]

That's way above my paygrade... :-) However I have a question. Correct me if I'm wrong, but aren't the charge systems on the 1st and 2ndGens rated at 26amps?? OK I lied...one more. Why didn't I cook the battery on the 99RSV over three solid days of riding at 17vdc?? Rather, I depleted it down to an El Cheapo meter reading of 8.9vdc....

 

With out doing some testing to see what all was going on in your specific case It would be purely guessing at what happened. There are a lot of possibilities that are not to far fetched.

 

I do not know what our electrical systems are rated for. Just that I wish it were more.

If a system is rated for 26A that means it can do 26A forever with no harm to the components, there is always the ability to output more than what it is rated for but that higher output is stressing the components and likely shortening there useful life span. It may only mean that you get 4 years out of the battery instead of 5, or a light bulb may burn out after only 8 years instead of 10. Even though it did not completely kill the battery right away it might have still done damage, same for all other other parts of the bikes electrical system.

 

It is also possible that the 17V was intermittent and was sometimes running off the battery and sometimes cooking things. Lots of realistic scenarios and even some unrealistic ones.

[Flyinfool]

:hijacked:

 

Flyinfool , I just want to know where that ohm symbol is from. Is it hidden on the keyboard?????????? Inquiring (twisted) minds want to know.

 

Using the Alt key along with various number combinations on the numeric keypad will find all kinds of neat symbols.

 

Like Alt 234 = Ω the ohm sysbol

Alt 1 = ☺ an open smiley face

Alt 2 + ☻ a filled smiley face

Alt246 = ÷ a divide symbol

Alt 248 = ° A degree symbol

 

There are hundreds of special characters that can be accessed this way.

 

OK back to your normally schedualed thread.

:hijacked:

[snyper316]

ω≈ç√∫ ̃μ≤≥÷