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#1 |
Nomadic Tribesman
Join Date: Nov 2008
Location: Brampton, Canada
Moto: '09 ER-6n
Posts: 11,150
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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" ![]() 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.
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"Everything's better with pirates." - Lodge, "Dorkness Rising" http://www.morallyambiguous.net/ Last edited by Papa_Complex; 02-21-2011 at 10:51 PM.. |
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#2 |
~Italian Stallion~
Join Date: Mar 2008
Location: North West Georgia
Moto: 2000 Yamaha YZF-R6
Posts: 446
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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.
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#3 |
Nomadic Tribesman
Join Date: Nov 2008
Location: Brampton, Canada
Moto: '09 ER-6n
Posts: 11,150
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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.
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"Everything's better with pirates." - Lodge, "Dorkness Rising" http://www.morallyambiguous.net/ |
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#4 |
~Italian Stallion~
Join Date: Mar 2008
Location: North West Georgia
Moto: 2000 Yamaha YZF-R6
Posts: 446
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Okay so i just need to stick a resistor before each signal?
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#5 |
Nomadic Tribesman
Join Date: Nov 2008
Location: Brampton, Canada
Moto: '09 ER-6n
Posts: 11,150
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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.
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"Everything's better with pirates." - Lodge, "Dorkness Rising" http://www.morallyambiguous.net/ |
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#6 |
~Italian Stallion~
Join Date: Mar 2008
Location: North West Georgia
Moto: 2000 Yamaha YZF-R6
Posts: 446
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![]() ![]() http://www.amazon.com/PR-TURN-SIGNAL.../dp/B002KR51SM Is this something that your suggesting i put before each signal? |
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#7 | |
Semi-reformed Squid
Join Date: Nov 2008
Posts: 531
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![]() Quote:
![]() ![]() 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! ![]() |
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