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LED Replacements for 1157's?

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  #21  
Old 05-15-2011, 03:51 PM
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Originally Posted by fenderzig
I just went out and snapped a few pics for you. The bike is sitting in my garage and door is wide open so the garage is pretty bright right now. This is without a flash on the camera. I tried to capture the LEDs and represent them well. They are brighter than the stockers.

First pic is the running lights:



Second pic is after applying the brake:

Wow those do look bright....thanks for the pic, I'll grab a set soon.
 
  #22  
Old 05-16-2011, 01:21 AM
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Originally Posted by fenderzig
I'm running these.

http://www.amazon.com/gp/product/B000TK3TPO

direct fit and no load equalizer needed.



Those aren't listed to fit Dyna's newer than 2006, but I see you have them on a 2009. I know the rears use the 1156 on my '07, will they fit?
 
  #23  
Old 05-16-2011, 06:40 AM
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Originally Posted by HemiOrange
Those aren't listed to fit Dyna's newer than 2006, but I see you have them on a 2009. I know the rears use the 1156 on my '07, will they fit?
I don't know for sure that they will fit an '07. Someone else on the forum recommended them to me and even though my '09 isn't listed I bought them and they work great.
 
  #24  
Old 05-18-2011, 04:17 PM
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I received them today. I had to trim them slightly to fit under the factory red lens, but they work well. They do go into the "malfunction mode", as though a lamp is burned out, but they still work. I have yet to install resistors on the brake/signal circuit. They're plenty bright!

The smoked lenses might arrive this afternoon. I figure I'll have to keep the red lenses in case of some crazy law of which I'm currently unaware. I'll put the red lenses back on when inspection time rolls around next year.
 
  #25  
Old 07-14-2011, 05:14 PM
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Originally Posted by abadv8
I have to chime in here...

Load equalizers are in parallel, and increasing resistance decreases the load.

P=IV & V=IR
An LED is a diode and acts like a small resistor but does not change or effect current. Each LED creates a voltage drop. A resistor in series creates a voltage drop and reduces current supplied to the LED's. (The LED's are the load and when a resistor is placed in series it is now an added part of the overall load on the supply)

http://www.hebeiltd.com.cn/?p=zz.led...tor.calculator

It is a resistive cicuit in application but a silcone device is also non-ohmic so technically ohms law does not apply. Resistance in regards to the applied resistor (whether parallel or in series) is constant but the resistance involved with LED's in the circuit can curve so it becomes non-linear in application. These swings are what cause problems.

So with a resistor you have a contsant R value and as a result I and V are affected. When you couple in LED's R is affected and V is affected but I is not affected. So a resistor in parallel with an LED or group of LED's is not the same as having other OHM type devices in parallel.

The resistor is more or less a bootstrap device or dummy load used not to regulate or control but rather to absorb current because LED's don't. In effect the resistor is the load and is now replacing the original lamp filament and effectively not gaining any advantage in regards to strain on the source/supply. Without a resistor you don't really have a load with LED's technically speaking.

Placing a resistor in series does limit voltage and current supplied to the LED's depending on it's value. It's rating only applies to how much it can tolerate before opening in the circuit (switching it off). Of course besides opening it could short and thus loose any resistive effect at all. Ohm's Law would apply to the supply voltage and the resistor in the circuit at this point only.

30 years as a bench level technician in consumer electronics. First question Electronic Engineers ask me? "How do you tell the difference between an NPN transistor and PNP? Then they quit about a day to two weeks later because Electronics Engineering and troubleshooting electronics are not the same fields.

If you want theory? Ask an engineer. If you want help ask a tech.
 
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