| Author | Message | ||
     Steve899 |
Where is it? I can't find it. anyone know | ||
| Jules |
There isn't one (that's why you can't find it) the functionality is built into the IC | ||
| Steve899 |
oh! so I guess I should bolt the engine back in now, I got this bike all apart trying to find it, thanks Jules, I knew I should have went to Bad Web first. | ||
| Froggy |
Right here:  | ||
| Steve899 |
can you slow the blink rate | ||
| Avc8130 |
Are you using aftermarket signals? Throw some resistors in parallel. ac | ||
| Steve899 |
ft turn signals are on the fairing( aftermarket), no mirrors ( one left mirror bar end). parallel where do Install that? | ||
| Avc8130 |
You need to install resistors in parallel with the signals. This will involve soldering. ac | ||
| Steve899 |
Trying to increase or decrease the resistance, also resistor on each turn signal? | ||
| Avc8130 |
Increase. There is NO way to add resistors to decrease resistance. Yes, you will need one for left and one for right. ac | ||
| Jdugger |
> There is NO way to add resistors to decrease resistance Parallel vs. series? | ||
| Avc8130 |
Oh yeah, that whole R= sum of 1/R. Stupid electronics. It doesn't really exist. Just keep the smoke in an everything is fine. ac (Message edited by avc8130 on May 16, 2010) | ||
| Tibman260 |
AC, you are both right and wrong. In circuit that you can add resistors in parallel to the existing resistance then you can decrease the resistance. In this case though since the circuit is internal to the IC then you can only add res in series thus increasing impedance. As for adding resistance to decrease flash rate, that will not work. In fact if you add enough you will get the reverse effect and the flash rate will increase due to the cluster seeing a higher load on the circuit. The cluster is specifically calibrated to LEDs and any deviation from the expected ranges of current it wants to see will cause the cluster to flash the turn signal at the higher rate to signify there's an issue. | ||
| Ratbuell |
Actually, the cluster is calibrated to the BUELL led's, which have resistance. Otherwise, how would they have a fast-flash-light-out warning feature? LED's have zero resistance by their nature. The circuitry in the OEM signals has resistance built in specifically for this purpose. Is it as much as a 10w incandescent? I doubt it, but it could be (witness the '10 1125 bikes' change to incandescent rear signals). Actually, now that I think about it more...it probably is equal to a 10w bulb. Someone with a stock signal (either front or rear, or ideally both) that's not currently installed, and a voltmeter - set it on ohms for us and measure the resistance across the connector wires. I have a pair of mirrors out there somewhere, I'll measure those if I can find 'em... Since Steve is using aftermarket signals in front, he's probably not running enough resistance in THOSE signals, and the cluster thinks that means a bulb is out. Add a resistor to each side and the rate should go back to normal. | ||
| Tibman260 |
ratbuell, the cluster is looking at current draw, not resistance. You are correct though that as I said the cluster is calibrated to a specific range that is centered around the stock signals current draw. The draw is less than a 10W bulb. The 08 and 09 clusters are made to handle LEDs on the front and rear. The 10 cluster is made for 10 watt bulbs on the front and rear hence the LED load module on the 10 bikes to make the front LEDs look like bulbs. | ||
| Avc8130 |
Correct, nothing is looking for resistance, it is looking for current...however 12v is there so V=IR prevails. V~12 R=? This will give current. Let's use the 10W bulbs. There are 2 on each circuit. P=VI 20W=12(I) I=1.67A V=IR 12=1.67(R) R=7.2Ohm The stock signals (LED) do have a resistance. Measure it and add resistors to match above calculations and I would imagine you will be successful. ac | ||
| Tibman260 |
AC. Yes V=IR does prevail but it has nothing to do with the flash freq. It does affect the brightness of the LEDs though. So again you can not decrease the flash rate by adding resistance. You are either going to have the stock flash rate if your current draw is within the range of what the cluster is looking for or an increased rate if you are outside the expected range. | ||
| Avc8130 |
That is my point. You just have to match the current draw. You cannot force current. You place 12V on the circuit and the current is a result of the resistance. ac | ||
| Tibman260 |
Right that is what I have been saying as well. I guess then I am confused, I took what you said earlier as a blanket statement that if you increase the current draw you will decrease the flash rate. That statement is false. If you meant that in the event his turn signals weren't puling enough current and he needed to add resistors in parallel to bring the total current draw of the circuit within the range the cluster is looking for, then yes that is correct. All I'm saying is if your turn signals are flashing at their correct rate then you can't add resistance to decrease the flash rate. | ||
| Avc8130 |
We agree now  ac | ||
| Dktechguy112 |
can you use ohms law, v=ir for diodes? i thought they are nonohmic. if you look at a v vs I curve for a diode, it is not linear at all. And for ohms law to apply, the v vs i curve must be linear, (ie same slope) in which the slope is the resistance. | ||
| Sportster_mann |
I fitted 2010 bulb indicators to my 08 R as I got fed up with the rears failing - they work fine without any modifications. So I have LED fronts and bulb rears. | ||
| Avc8130 |
I am not surprised. Buell probably did the simple thing and chose internal resistors to make the LED signals "act" like bulbs. ac | ||
| 1324 |
can you use ohms law, v=ir for diodes? i thought they are nonohmic. if you look at a v vs I curve for a diode, it is not linear at all. And for ohms law to apply, the v vs i curve must be linear, (ie same slope) in which the slope is the resistance. You are correct. A diode is nothing more than a switch (one way). The rest of the circuit is where the resistance comes from. This includes the circuitry inside the turn signal and the rest of the harness that the signal is plugged into (negligible on the bike for the sake of this discussion). The total resistance of the circuit is what allows the flasher to operate. No different than a bulb with the exception that the bulb is the primary source of resistance. | ||
| Renzo750 |
Today I got a new issue I have 1100 miles on my CR and stock signals - my left one is "normal" - my right one started flashing VERY fast today...bugs me more than it should...anybody else seen this? I see a trip to the dealer in my future for warranty work. | ||
| Steve899 |
cool thread, added resistors in parallel and they blink at a good rate, Thanks everyone. Renzo750 , I know the rear turn signal go out , Bought my 2008 1125R in Oct 09 with 4 miles on it and both rear turn signal were out , dealer switched them with another unit before I took the bike home. | ||
| Steve899 |
Tibman260, I increased the current draw and the flash rate slowed. I put a 3.8watt bulb in Parallel with each turn signal. My ft mirror turn signals are off the bike, I have 09R6 14 led turn signals mounted on the fairing, stock t signals on the rear. Without the 3.8 watt bulb on each turn signal they flash real fast ( really too fast), but with the 3.8w bulb on each Turn signal it flashes perfect. so did increased amp flow to slow the flash rate, Yes! Thats my story and Iam sticking to it... | ||
| Tibman260 |
Steve, read the discussion, you likely put the proper load on the turn signal circuit. There is only one story and that is how electricity works. | ||
| Ratbuell |
Renzo - my guess is you have a rear signal out (you can see the fronts from the drivers seat so it'd be obvious). | ||
| Renzo750 |
Duh! I never thought to look...i just stood up while riding and saw the front was flashing...I'll check tonight, but I bet that is it! Thank You :-) |
