I'm looking to convert a flashlight I have to LED, the flashlight uses a 6 volt "lantern battery but the LED's max voltage rating is 3.1 volts DC. What value resistor would I need for this? I also want to run it off a 9 volt supply, again I need to know the value of the resistor.
According to the website, the LED draws 350 mA. If the resistor drops
6 volts minus 3.1 volts = 2.9 volts, the required resistance is E/I or
2.9V/.350A = 8 ohms. The resistor will dissipate more than one watt so it will get hot. A 2 watt resistor would be appropriate.
The battery voltage will drop slightly under load but you might prefer to operate the LED at a lower current, say 300 mA or even 250 mA to give the LED longer life. It is not cheap!
A similar calculation can be done for the 9 volt supply.
Chuck
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A higher power resistor will work. For 6 volts it calculates to 10.9 ohms / 1.33 watts and for 9 volts 19.4 ohms / 2.38 watts
Where are you getting 16?
6 volts minus 2.2 volts across the led = 3.8 volts across the resistor, divided by .35 amps = 10.857 ohms
9 volts minus 2.2 volts = 6.8 volts across the resistor, divided by .35 amps = 19.4 ohms
The battery voltage may drop a little when you pull current - a lot if your intention is to use a 9V transistor radio battery.
A resistor will work but they say to use a constant current source in the ad. And the ad itself is suspect - you notice all the leds are rated at one watt and only the blue and white are actually one watt ? Actually 1.225 watts versus .77 for the red.
You may burn it out because they don't show the data. The led itself will be dissipating .77 watts and depending on the physical size, that heat has to be dissipated without raising the temperature of the die above ~100 degrees C.
You need to look at the data sheet to see if they will do what you want.
It is normal to derate leds to compensate for ambient temperature -
I use 4 one watt leds on my motorcycle. Each one is mounted to a 1" diameter aluminum slug to spread the heat, all four are mounted to 20 square inches of 1/8" aluminum. I run them at 10% power for dim and
100% for bright and calculated the ambient for around 37 degrees C.
Mine are called one watt leds made by Cree and dissipate 1.4 watts (650 milliamps) and cost $8 each in the US.
What Dick Smith is showing appears to be a bare small surface mount chip - it will take some work to make it dissipate a watt if that's the case. Unless you just turn it on briefly . . .
Yes. Again, I would view these resistance values as minimums. Unless you really neet to extract maximum output from the LED, you might go as high as 37 ohms with the 9 volt battery. Ideally, you would start with a high resistance and see what brightness results. You could then reduce the resistance by say 10% at a time and observe the differences. Yeah, it's much easier said than done. But you should make sure that the voltage and current to the LED do not exceed the maximums.
Chuck
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Do you have any interest in max light out, or max battery life?
This is basics, so perhaps just slapping in a resistor is as far as you want to go - but you could run 2 in series from the 6 volt battery, and
3 from the 9 volt battery, at no increase in power drawn - just making more into light, and less into heat from resistors.
Alternatively, you could get somewhat more complicated and build a small switch-mode driver circuit to get 2-3 times the battery life, with the same light out, by converting more voltage at less current into less voltage at more current, again reducing the power lost as heat from resistors.
If you need more power dissipation, you can use resistors in parallel and/or series to increase power without having to go track down a high-power resistor. i.e., if you need 20 ohms at 2 watts, you can parallel 2 1-watt 40 ohm resistors (39 in practice), or run 4 5 ohm (5.1 in practice) 1/2-watt resistors in series.
Don't use a resistor, that's going to suck, brightness will drop with the rapid fall in battery voltage. Use a proper DC-DC 350mA constant current converter designed for these LED's. A 6V converter is not easy to find, but 3V converters designed to run off two D cells are readily available, like this one:
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_id=3D320 If you used one of those D cell latern battery holders like this:
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then you can use the cheaper D cells and have room to mount a module.
For 9V use get another module like this one desing for the higher input voltage:
In that case you may be very disappointed indeed. These cheap asian copy 1W LED's aren't nearly as bright as the genuine Luxeon ones, and even the luxeon ones are old hat now compared to the Cree brand. The Cree are much more efficient than Luxeon, and Luxeon in turn are much more efficient than the cheap copies. All "1W" LEDs are not the same, they can have *vastly different* light outputs.
Also, unless you drive the LED with it's maximum current, light output will be very significantly reduced. That's why you should be using a constant current driver, so the current remains the same as the battery voltage drops.
I didn't see a constant current solution posted, probably because it is costs more than using just a series resistor. But constant current was mentioned, so here's a constant current circuit that uses 1 watt resistors and an LM317 chip, and works either 6 or 9 volts input. It provides about 347 mA to the LED.
Use Google, plenty of circuits available, here is one that uses a purpose designed chip:
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A constant current regulator puts out a constant current regardless of the input (battery) voltage. Yes, it is a regulator, but in this case it's a current regulator instead of a voltage regulator.
No need to build one, plenty of ones you can buy off the shelf.
That circuit is not entirely suitable for use with a 6V battery, as the constant current will not be maintained once the battery voltage drops low enough.
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