Both Blinkers blink when i turn on a turn signal - Page 4

Kerry_129 · 02-21-2011, 11:28 PM

Gotcha - yep, even simple circuitry can become really confusing shit. I remember the electron flow being opposite conventional circuit flow throwing me for a loop back in school - one reason I stuck with nuts & bolts!

As you pointed out, the circuit was designed for incandescent bulbs which draw more current & thus cause more voltage drop - so the indicator diodes 'see' less voltage normally. I'm thinking that the higher voltage present on the indicator circuit due to lower drop across the LED's is a bit above the 'breakdown voltage' (point at which the diodes quit blocking) - so there's the 'leakage' to the opposite signals.

Trick is, figuring out how much additional resistance is required in the circuit to drop that voltage below the threshhold.

Full Throttle · 02-21-2011, 11:32 PM

The fronts don't do it only the backs. and they are all LED.

Here is a video of what i'm talking about. Excuse the loud music lol. I forgot it was playing when i was recording.

http://s156.photobucket.com/albums/t...afilter=videos

Papa_Complex · 02-21-2011, 11:42 PM

It's actually almost the opposite situation, in theory. An incandescent lamp provides low voltage drop, unless it's heated up. It takes a fair bit of current to heat one up enough, that it lights. As it heats up, it has a higher resistance to current flow. That's why incandescent lights usually blow when you first turn them on; low resistance equals high current flow, so they blow like a fuse.

LEDs are a different animal. Once they reach a certain threshold current, they try to force a fixed voltage drop across them. For a basic red electronics LED, that voltage is roughly 1.5 volts. They effectively become a variable resistance in the circuit.

"Breakdown voltage" is more of a thing for Zener diodes, which are designed to let current through once they hit a certain voltage. If you hit breakdown on a regular diode, or a LED, then it's usually followed by "burnt chip smell"

Leakage current is generally pretty damned low, though it can obviously be significant in some applications.

I actually went to school for this crap, though I never use it these days

*EDIT* Actually you're right about the 'breakdown voltage' thing. I forgot that they also use the term when talking about both forward AND reverse current.

Full Throttle · 02-21-2011, 11:57 PM

Well with all that said i understand very little lol. What i think i was able to grasp was i need to maybe add a diode to the circuit somewhere.

Papa_Complex · 02-21-2011, 11:59 PM

Quote:

Originally Posted by Full Throttle

Well with all that said i understand very little lol. What i think i was able to grasp was i need to maybe add a diode to the circuit somewhere.

A diode or a low value, high wattage resistor in the path back from the display, will likely do the job for you. Barring that, just toss it inline with the signal diodes.

Full Throttle · 02-22-2011, 12:09 AM

Quote:

Originally Posted by Papa_Complex

A diode or a low value, high wattage resistor in the path back from the display, will likely do the job for you. Barring that, just toss it inline with the signal diodes.

Okay so i just need to stick a resistor before each signal?

Papa_Complex · 02-22-2011, 12:11 AM

Quote:

Originally Posted by Full Throttle

Okay so i just need to stick a resistor before each signal?

That would be my guess. You're just trying to kill off a little current, that's leaking back through the circuit. That should be all it takes.

Full Throttle · 02-22-2011, 12:14 AM

http://www.amazon.com/PR-TURN-SIGNAL.../dp/B002KR51SM

Is this something that your suggesting i put before each signal?

Papa_Complex · 02-22-2011, 12:17 AM

Not much detail there, but it looks like what I would suggest.

Kerry_129 · 02-22-2011, 12:23 AM

Quote:

Originally Posted by Papa_Complex

It's actually almost the opposite situation, in theory. An incandescent lamp provides low voltage drop, unless it's heated up. It takes a fair bit of current to heat one up enough, that it lights. As it heats up, it has a higher resistance to current flow. That's why incandescent lights usually blow when you first turn them on; low resistance equals high current flow, so they blow like a fuse.

Ahh - that makes sense, thanks for the illumination...

As to only the rears only doing it - I bet the fronts have a bit higher value inline resistors which are dropping the 'leakage' below their threshhold on that branch. I bet upping the inline resistor to the rear leds (easily accessible, since it's a homemade deal, likely) just a bit will square it away. Shouldn't need to be high-wattage since they draw so little current - the little $2 a pack RadioShack 1/2-watt jobbies should work, I think. Just have to ID the resistance of the existing ones (http://www.csgnetwork.com/resistcolcalc.html - or just disconnect & measure), and bump it up a bit (might take an assortment & a some trial & error). Or you could say 'fuck all this lectrical shit' and just let 'em flash!

02-21-2011, 11:32 PM	#32
Full Throttle ~Italian Stallion~ Join Date: Mar 2008 Location: North West Georgia Moto: 2000 Yamaha YZF-R6 Posts: 446	The fronts don't do it only the backs. and they are all LED. Here is a video of what i'm talking about. Excuse the loud music lol. I forgot it was playing when i was recording. http://s156.photobucket.com/albums/t...afilter=videos __________________ Last edited by Full Throttle; 02-21-2011 at 11:40 PM..

02-21-2011, 11:42 PM	#33
Papa_Complex Nomadic Tribesman Join Date: Nov 2008 Location: Brampton, Canada Moto: '09 ER-6n Posts: 11,150	It's actually almost the opposite situation, in theory. An incandescent lamp provides low voltage drop, unless it's heated up. It takes a fair bit of current to heat one up enough, that it lights. As it heats up, it has a higher resistance to current flow. That's why incandescent lights usually blow when you first turn them on; low resistance equals high current flow, so they blow like a fuse. LEDs are a different animal. Once they reach a certain threshold current, they try to force a fixed voltage drop across them. For a basic red electronics LED, that voltage is roughly 1.5 volts. They effectively become a variable resistance in the circuit. "Breakdown voltage" is more of a thing for Zener diodes, which are designed to let current through once they hit a certain voltage. If you hit breakdown on a regular diode, or a LED, then it's usually followed by "burnt chip smell" Leakage current is generally pretty damned low, though it can obviously be significant in some applications. I actually went to school for this crap, though I never use it these days EDIT Actually you're right about the 'breakdown voltage' thing. I forgot that they also use the term when talking about both forward AND reverse current. __________________ "Everything's better with pirates." - Lodge, "Dorkness Rising" http://www.morallyambiguous.net/ Last edited by Papa_Complex; 02-21-2011 at 11:51 PM..

02-21-2011, 11:57 PM	#34
Full Throttle ~Italian Stallion~ Join Date: Mar 2008 Location: North West Georgia Moto: 2000 Yamaha YZF-R6 Posts: 446	Well with all that said i understand very little lol. What i think i was able to grasp was i need to maybe add a diode to the circuit somewhere. __________________

02-22-2011, 12:14 AM	#38
Full Throttle ~Italian Stallion~ Join Date: Mar 2008 Location: North West Georgia Moto: 2000 Yamaha YZF-R6 Posts: 446	http://www.amazon.com/PR-TURN-SIGNAL.../dp/B002KR51SM Is this something that your suggesting i put before each signal? __________________

02-22-2011, 12:17 AM	#39
Papa_Complex Nomadic Tribesman Join Date: Nov 2008 Location: Brampton, Canada Moto: '09 ER-6n Posts: 11,150	Not much detail there, but it looks like what I would suggest. __________________ "Everything's better with pirates." - Lodge, "Dorkness Rising" http://www.morallyambiguous.net/

02-21-2011, 11:28 PM	#31
Kerry_129 Semi-reformed Squid Join Date: Nov 2008 Posts: 531	Gotcha - yep, even simple circuitry can become really confusing shit. I remember the electron flow being opposite conventional circuit flow throwing me for a loop back in school - one reason I stuck with nuts & bolts! As you pointed out, the circuit was designed for incandescent bulbs which draw more current & thus cause more voltage drop - so the indicator diodes 'see' less voltage normally. I'm thinking that the higher voltage present on the indicator circuit due to lower drop across the LED's is a bit above the 'breakdown voltage' (point at which the diodes quit blocking) - so there's the 'leakage' to the opposite signals. Trick is, figuring out how much additional resistance is required in the circuit to drop that voltage below the threshhold.