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#21
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The advantage of putting your emmiters in series (rather than
parallel) is that you pull less current. The mark to space ratio people have been discussing isn't that small. I don't think it would be a problem, but if you wanted to reduce the surge you could charge and discharge the gate of that big FET more slowly thus having a slower turn-on speed. You could do that by putting a high value resistor in series with the 555 output. The high voltage switching supplies that power those flash tubes can draw progressively more current as they degrade over time. anonymous coward wrote in message e... Something I've been pondering... 4 Luxeon star 1W emmitters in parallel will draw 1A or so. Your 555 based timer circuit will draw large(ish) currents for short periods of time, because of the small mark:space ratio. Might this interfere with other systems powered by the battery? The switching regulators for standard Xenon flash tubes draw a lower but much more constant current (though goodness knows they can generate radio noise if they're not shielded right). AC On Sat, 17 Apr 2004 23:25:24 -0700, Jay wrote: I think someone may have already pointed this out, and maybe I didn't make it as clear as I should have... I stacked the forward drop of MULTIPLE LEDs up until I got somewhere near the bottom end of the supply voltage. So for the example I gave, I got to 4 LEDS in series. Why waste all that power as long IR (heat) off a big resistor when we want red and green light right? Regarding 2.8V- The forward drop of these devices now-a-days is all over the place. The new chemistries seem to be making higher forward drops, plus the trend is to package multiple die into one larger device and this can effect the forward drop of the composite device. By the way, anyone building my circuit should try one instance of it (4 LEDS and resistor) on your bench supply before you go fly at night cross country. Jim Weir wrote in message . .. Before everybody in the Western Hemisphere blows a bucket full of light emitting diodes, would you care to calculate the resistor one more time? And perhaps post a retraction? (Jay) shared these priceless pearls of wisdom: -You can find examples on how to power the LEDs on the manufacturer web -site. - -Having said that... So lets say the recommended current for -the LED is 20mA. Ohms law is R=E/I, so that gives you a resistor -value of .3V/.02A=15 ohms. Um, no. Suppose the diode has a forward voltage drop of 2.8 volts (that's not a common value, but I'll give it to you for argument. Now the power supply (battery) is a 12 volt supply, but 14.2 volts at full charge with the alternator going, so the drop across the series resistor is going to be 14.2 minus 2.8 equals 11.4 volts, which is the voltage across the resistor. This current limiting resistor is going to have 20 mA flowing through it, so Ohm tells us that resistance equals voltage divided by current. In this case, 11.4 volts divided by 20 mA gives us a resistor of 570 ohms (560 is the nearest standard value). You put your calculated 15 ohm resistor in series with this diode and I guarantee you that the SNAP you hear is the gallium aluminum arsenide semiconductor of the diode being sacrificed on Ohm's altar. I'm serious. You owe the newsgroup a correction before somebody takes your error and blows up a whole bunch of LEDs. Jim Jim Weir (A&P/IA, CFI, & other good alphabet soup) VP Eng RST Pres. Cyberchapter EAA Tech. Counselor http://www.rst-engr.com |
#22
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#23
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#25
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"Blueskies" wrote in message . com...
Oh darn, you mean I'll only get about 50,000 hours???? or Nope ...does it mean that the light output at 100,000 will be 1/2 the original level? Ya that is right, thats how I understand it. Its the 3dB point. At that time it won't appear to be half as bright to the human eye because THAT sensor has a log response to give us all fantastic dynamic range. Measured with a calibrated instrument, you'd see half the output. Theoretically, if you exactly met the spec at time zero, in 1 hour you'd be under spec all other conditions being equal. |
#26
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(Jay)
shared these priceless pearls of wisdom: -I think someone may have already pointed this out, and maybe I didn't -make it as clear as I should have... I stacked the forward drop of -MULTIPLE LEDs up until I got somewhere near the bottom end of the -supply voltage. So for the example I gave, I got to 4 LEDS in series. - Why waste all that power as long IR (heat) off a big resistor when we -want red and green light right? And as somebody else has pointed out, running a series string close to the bottom limit of Vcc and hoping that a single resistor will provide constant current to these devices will cook them when Vcc rises to the charging voltage. No designer in his right mind would use a circuit in this manner. Jim Jim Weir (A&P/IA, CFI, & other good alphabet soup) VP Eng RST Pres. Cyberchapter EAA Tech. Counselor http://www.rst-engr.com |
#27
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Jim Weir wrote:
(Jay) shared these priceless pearls of wisdom: -I think someone may have already pointed this out, and maybe I didn't -make it as clear as I should have... I stacked the forward drop of -MULTIPLE LEDs up until I got somewhere near the bottom end of the -supply voltage. So for the example I gave, I got to 4 LEDS in series. - Why waste all that power as long IR (heat) off a big resistor when we -want red and green light right? And as somebody else has pointed out, running a series string close to the bottom limit of Vcc and hoping that a single resistor will provide constant current to these devices will cook them when Vcc rises to the charging voltage. No designer in his right mind would use a circuit in this manner. Jim The criteria I had when laying out my circuit was that the lights would operate down to 10V and still not exceed the max current limit at 15V. I'm working from memory here, but I believe the SuperBright LEDs I used had a forward voltage of 3.5V. Putting 2 in series with a 220ohm, 1/4W resistor maximized the efficiency while maintaining the range I wanted. -- http://www.ernest.isa-geek.org/ "Ignorance is mankinds normal state, alleviated by information and experience." Veeduber |
#28
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The criteria I had when laying out my circuit was that the lights would
operate down to 10V and still not exceed the max current limit at 15V. I'm working from memory here, but I believe the SuperBright LEDs I used had a forward voltage of 3.5V. Putting 2 in series with a 220ohm, 1/4W resistor maximized the efficiency while maintaining the range I wanted. Okay, thats not too bad because your using a resistor to cover HALF the voltage drop....of course your losing half your power in the resistor rather than having it produce light in the LED... My previous post was more concerned with folks both running the LED at absolute max power ratings,,,,and trying to use the LED voltage drop to cover something like 80 percent or more voltage range...thats when you have to be a little more careful... Your 1/4 watt resistor is NOT high powered enough....youll be running it close to or past its maximum power rating for normal voltages.... Get something more like 1/2 watt or 1 watt resistors....or use 2,3,or or even 4 of the 1/4 watt resistors with appropriate resistance values in series to spread the power load....resistors are cheap and there AINT no harm in using ones that can handle significantly more power than expected....while the risk to benefit ratio of trying to push their limits seems rather high.... Also make sure the resistors can cool effectively.....their power ratings dont mean diddly if they are well insulated or packed into a tight space and can warm up... take care Blll |
#29
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On Wed, 28 Apr 2004 09:18:57 -0700, Jay wrote:
"Blueskies" wrote in message . com... Oh darn, you mean I'll only get about 50,000 hours???? or Nope ...does it mean that the light output at 100,000 will be 1/2 the original level? Ya that is right, thats how I understand it. Its the 3dB point. At that time it won't appear to be half as bright to the human eye because THAT sensor has a log response to give us all fantastic dynamic range. That wouldn't be true at threshold though. For example, if you moved the LED away from the observer until it was only just visible then reduced the brightness by half, it would become invisible and you'd have to move it quite a long way back before the observer could see it again. Measured with a calibrated instrument, you'd see half the output. Theoretically, if you exactly met the spec at time zero, in 1 hour you'd be under spec all other conditions being equal. |
#30
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Have you considered using a constant current regulator, instead of a
resistor? I believe there is an example circuit given in the LM337/317 datasheet showing how to build one using only the LM337 (normally used as a voltage regulator) & one resistor. It would need to be bolted to a heatsink, like the Luxeon Star LEDs, but IIRC the LM337 and cousins also shut down if they overheat. AC |
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