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02-21-2011, 10:05 PM | #1 |
Semi-reformed Squid
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Nah, not bizzaro - pretty typical diode application, similar to how the AC current from the stator is 'rectified' to DC current by only allowing the 'positive' voltage through (but with only 2 legs instead of 3).
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02-21-2011, 10:10 PM | #2 |
Nomadic Tribesman
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I say 'bizarro' because, if you follow the path that current would take (well actually opposite to the path, since flow is from negative to positive opposite to the old theory, that all of the component symbols were designed to follow), current feeds from the display straight back into the opposite signal through that bridge. Dumb, if you're designing for an LED system, but they weren't designing for LEDs.
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02-21-2011, 10:28 PM | #3 |
Semi-reformed Squid
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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. |
02-21-2011, 10:42 PM | #4 |
Nomadic Tribesman
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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" 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.
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02-21-2011, 10:57 PM | #5 |
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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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02-21-2011, 10:59 PM | #6 |
Nomadic Tribesman
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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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02-21-2011, 11:09 PM | #7 |
~Italian Stallion~
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Okay so i just need to stick a resistor before each signal?
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02-21-2011, 11:11 PM | #8 |
Nomadic Tribesman
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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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02-21-2011, 11:14 PM | #9 |
~Italian Stallion~
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http://www.amazon.com/PR-TURN-SIGNAL.../dp/B002KR51SM Is this something that your suggesting i put before each signal? |
02-21-2011, 11:23 PM | #10 | |
Semi-reformed Squid
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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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