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#1 |
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diyAudio Member
Join Date: May 2010
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hey guys thought this might be a good place to ask...
im sure this is an easy one for most people here, but which way to run a LED is more efficient on a 12vdc battery running them at 3.3vdc or (requierd) voltage by way of a stepdown regulator say a" lm317t" or running them strait of a 12vdc line with the simple resistor attached to the anode??? im only asking because im converting the lighting on my boat to LED's to get the most from my battery life thanks guys here is the specs on the ones im useing to if it helps cheers and thankyou 0.5W 10MM HI POWER LENS Type:Water clear Color Temperature: -- Luminous Intensity-MCD: Typ: 200,000 mcd Reverse Voltage:5.0 V DC Forward Voltage: Typical: DC Forward Current:100mA DC Forward Current 100 mA Peak Pulse Forward Current (1) 150 mA Avg. Forward Current (Pulse Operation) 100 mA Operating Temperature -40 to +100
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#2 |
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diyAudio Member
Join Date: Mar 2007
Location: Canandaigua, NY USA
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Armchair thoughts- it makes no difference if you drop the voltage across a linear regulator or a resistor; both ways dissipate the power resistively. What you want to do is eliminate dissipation everywhere but the LEDs by using a PWM scheme (the eye integrates the pulses) or by building a switching supply that controls it's output *current* as needed. I'm pretty sure there are LED driver chips on the market that use these techniques.
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I used to be an audiophool like you but then I took an arrow to the knee. |
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#3 |
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diyAudio Member
Join Date: Mar 2007
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There are switching LED power supplies out there, some offering very high efficiency. Another solution is to use many LEDs in series, such that the nominal voltage drop is just below the minimum supply voltage and use a simple 2 transistor current regulator.
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"Fully on MOSFET = closed switch, Fully off MOSFET = open switch, Half on MOSFET = poor imitation of Tiffany Yep." - also applies to IGBTs! |
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#4 |
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diyAudio Member
Join Date: Sep 2009
Location: New York, the vampire state
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#5 |
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diyAudio Member
Join Date: Oct 2008
Location: Vancouver
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If there 3 volt LEDs and a twelve volt battery, make groups of 4 LEDs in series across the battery. That would use all the power in the LEDs and none in a voltage drop device. If the LED voltages are not a perfect multiple of the battery voltage than use as many LEDs as posssible and a small drop down resistor. (eg 3volt LEDs and a 14 volt battery = 4 LEDs and a resistor that drops the extra 2 volts.)
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#6 |
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diyAudio Member
Join Date: Mar 2008
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12V and 24V LED strips for automotive use are easy to find.
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#7 |
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diyAudio Member
Join Date: Sep 2002
Location: Lakewood, Ohio
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It sounds like a sailboat. The LEDs will do fine on the battery, but when re-charging the battery alternators can be hard on LEDs. Lots of the early auto LED tail lights didn't last long.
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Kevin |
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#8 | |
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diyAudio Member
Join Date: Aug 2008
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Quote:
You can use a LM317 as a constant current source to power series LEDs up to about 2V less than the minimum supply voltage (eg 3x 3V LEDs on a >11V supply).
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Tyler |
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#9 |
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diyAudio Member
Join Date: Jul 2009
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#10 |
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diyAudio Member
Join Date: Sep 2002
Location: Lakewood, Ohio
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I think that it was large voltage spikes caused by the alternator and voltage regulator combination.
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Kevin |
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