how do i disable battery warning

I got a new battery for the motorcycle, unfortunatly its the wrong kind, in the fact that you cant screw the red dipper thing into it. without this in the battery it triggers the battery warning light. Its annoying the **** out of me. i need to know how to disable it so on the power on self test, it byspasses the battery.

 

here's a link that shows one way of bypassing the battery sensor
http://home.netcom.com/~rvtucker/battwarn.htm

 

The link Nick sent you is essentially correct, but I believe the logic to be just a bit off. Earlier Integrated Circuits (IC’s) did use a 5 VDC reference voltage and that voltage would have been more or less correct for the period. Not wanting to find out if it is, and blow up my computer (I kinda like it working), I have chosen to assume the path of least possible destruction and assume that a 3 VDC reference voltage was used (since the sensor is only immersed in one cell, it should by all accounts only read 2 VDC, 6 cells with an output of 12 VDC = approximately 2 volts per cell, well below the 5 VDC mentioned). This would lend credence to the newer types of logic circuits that use the 3 VDC reference, they also can read a 2 VDC signal. Their "cutoff" level is determined by the manufacturer but the general rule of thumb is half the reference voltage is all that is required to get that logic "1". The sensor allows a small amount of voltage to return to the computer, letting it know that it is there. Once the electrolyte drops below the sensor's range, the voltage is gone and flags the computer. To rid yourself of this pesky annoyance (Lord knows Yuasa batteries aren't cheap) one must return approximately 2 VDC back to the computer. This can be done by routing a 12 VDC return through a resistor back to the computer. I would do this non-destructively as you may wish to put it all back together correctly someday in the future (I hate getting hacked up wiring harnesses!!!) when you have the cash to spring for the correct battery. Let’s see here, 12 VDC down to 2 VDC would take a little figuring using Ohms Law. Let's assume that that little sensor is going to draw, say, a hundredth of an amp (which is really healthy and probably overkill for a circuit that only needs a thousandth of an amp to function) and we want 2 VDC. 2 divided by 0.001 equals 2000 so you would need a 2000 (or 2K, see it is close to what the link cites so they weren’t far off) ohm resistor to get your 2 VDC. I would suggest a carbon wound at about the 1/2 watt (more wattage is better than too little, so don’t be afraid of a 1 watt unit) range should be adequate. Another means would be to wire a 10,000 or 5,000 (sometimes referred to as 10 or 5 K) potentiometer or rheostat into the line and tweak it 'till the little light goes out, pull it out of the circuit and measure it to find your needed resistance. Good luck!

 

WOW!! I'm glad we have people on here who understand electrics. Not sure I understood all that, but good job Robert.

 

Many thanks Rich, but I owe my education to our "Uncle" and you, the US taxpayer. Hope you approve of your investment. It is supposed to serve the public good. Cheers!

 

damn, here i was hoping to be able to pull a plug that connected to the board to solve my problems. Ill have to give this a shot next weekend.

 

It's not terribly difficult Ziggy, you can build the little sucker in about 30 minutes. Using bullet connectors and a ring connector you can make a drop in adaptor that will not require any mods to the harness. Heck, if I wasn't so disorganized right now, I'd whip one out for you (moved in June and am still trying to get into the boxes to unpack!). Here is a picture to look at. Good luck!

 

Stupid picture won't attach. I'm working on it. More soon.

 

Sorry folks, the picture was too many pixels. I posted them in my album named XJ Tools. Check them out there. Feel free to add homemade toys as you find them. Thanks!

 

I have chosen to assume the path of least possible destruction and assume that a 3 VDC reference voltage was used (since the sensor is only immersed in one cell, it should by all accounts only read 2 VDC,


You are so right Robert. Here is Mike Oberle's site on the subject.

http://pw1.netcom.com/~rvtucker/battwarn.htm

 

Thanks Max, if you refer to an earlier post on this thread, Mike's site was mentioned by Nick. And I did mention that while Mike was headed in the right direction, the author of that article drew several conclusions that are in error. IC's do not sense current, they sense voltage. The chips used by Yamaha in the XJ series do not use a 5 VDC reference. And, well, you get the picture... I'm glad somebody took a stab at it, I simply pointed out the electronics of it all. Hope the details were useful. Did you get to see my pictures in the Gallery? They were to big to post here so I took the other route of posting them there. I hope we answered your question Ziggy. Give us the results of your efforts when you tackle this thing. Peace.

 

Robert, just curious. But i dont thinkt he voltage int he cell (2.1volts like you said) makes the least bit of diffrence, as the sensor does not have a negative and positive activated plate, and i beleive is only using the electrolyte like a mercury switch, in effect... not that i dont agree with your calculations, just nit picking the details out of curiosity for a better understanding of these bikes.

on second thought, i remember the sensor only having one lead wire... so i am unsure... however the sensor has no active material and therefor i dont see how it could produce voltage just from being in the electrolyte.

 

Fair question. If you were to ground your negative lead on your voltmeter to the negative post of the battery and dip the positive lead into the cell (I don't recommend this unless you don't really care much about the lead and have some baking soda handy to neutralize the acid afterward, come to think of it, wrap a paper clip around your lead and use it instead), you would find that the voltage is measured with respect to how many cells are between the cell you are testing and the negative post of the battery. Only one line going in. Remember the electrolyte acts as a conductor and is a return path internally. If you measure from the cell closest to the negative post, you should read a little more than 2 volts. Come to think of it, you may have just unknowingly pointed out a flaw in my logic of how much voltage is being felt by the sensor. The sensor is inserted into the third or middle cell isn't it? It has been a while since I reviewed this. That would mean the sensor is reading a little bit over 6 volts with respect to ground which would tend to favor Mike's specs of a 5 VDC signal. I failed to mention that the potential was 2 VDC per cell. My bust. However, I hope I answered your excellent question.

 

Another observation folks. Mike's specks for a 4.4Kohm resistor would be a sound resistance to use. Seeing the effort to logically test his theories and the subsequent result is proof positive of that. I should offer an apology to him for putting my foot in my mouth. Given the change in voltage that is present at the sensor, my math should have been 6VDC divided by 1 one thousandth of an ampere to equal 6,000 ohms. To further explain the phenomenon of electrolyte Brian, consider the lemon cell. You take a lemon, press a strip of copper into it. On the other side of the lemon, you press in a strip of zinc (or a zinc galvanized nail). Wrap copper wires around theses strips and you should easily see a few volts. What is supporting the current flow inside of the lemon? The acids inside that lemon act as an electrolyte and react in a similar manner as a battery in a car or bike. There is a fantastic book on understanding electricity for Joe Commonman at Radio Shack called "Getting Started In Electronics" by Forrest M. Mimms III. Their catalog number is 62-5004. They stopped carrying it in the stores a few years ago due to lack of demand but it still can be ordered. It's about $15 and worth every penny. The knowledge I gained from reading this book helped me land a $1,000DUS raise if that isn't enough incentive. Hope that helps. All the best, Robert. P.S. Mike, I'm very sorry if I offended, I failed to ponder the question long enough to catch my glaring error. I also stand corrected on the transistor being a voltage sensing device and not a current sensor. It is capable of both (I had always been told that it was a voltage switch).

 

good point on the lemon, i remember doing that so i see now that you are infact right. Irtonicly it had occured to me that you are 3 cells from the negative post, so it woulod be 6.3 volts, but didnt mention it as i figured it wasnt sensing the voltage.

good posts

 

i just placed a piece of black tape over the red flashing light on my seca and called it good.

 

lol i was tempted to paint black over the red lense. I first wanted too see if it would be difficult to do it the correct way.

 

Well is it? Did you see the picture I drew? Details please! I want to know if you where successful.

 

Have you tried forcing the sensor in anyway?
Most that ask about the mod have lost or broken sensor.
Since you still have your old one, just force it in.
Works that way here with a non correct battery from Walmart.
Heck, if you wanted to make it perfect, just tap the battery hole with new threads.
Seems much easier to keep the sensor in working condition than bypassing it. (that is the correct way).

 

DO NOT UNDER ANY CIRCUMSTANCES MODIFY A BATTERY CASE, and for heavens sake DONT force anythign in there htat doesnt fit anyways.

well... on second thought dow aht you want, BUT hey, it snot MY bike, OR my leg, OR my acid... so do waht you will