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#1 |
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diyAudio Member
Join Date: Sep 2004
Location: UK
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Resisting the temptation to hijack the "brightness" thread, how do LED torches work?
I have one with 10 white LEDs, apparently in parallel, hooked across 3 series AAA cells. It works. Are they special LEDs with built in resistors? Is it a happy co-incidence that 3v to 3.8V is still on the V/I knee and does not s/c the cells? Can't find anything on this elsewhere, so I am looking in the wrong place |
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#2 |
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diyAudio Member
Join Date: Nov 2005
Location: San Antonio TX
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3 AAA cells in series is enough to light a white LED, so they may just be in parallel. A boost converter can also be used. 1 AAA cell can power a white LED. That's the method I like to use.
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It is error only, and not truth, that shrinks from enquiry. - Thomas Paine |
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#3 |
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diyAudio Member
Join Date: Sep 2004
Location: UK
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They are definitely in parallel, with no evident series resistor/element.
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#4 |
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diyAudio Member
Join Date: Nov 2005
Location: San Antonio TX
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If there's no PCB blob indicating a chip-on-board, then yes they're just connected direct to the battery. The inherent resistance and increased Vf at higher currents (as discussed in the "brightness" thread) combine to make a cheap working unit.
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It is error only, and not truth, that shrinks from enquiry. - Thomas Paine |
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#5 |
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diyAudio Member
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I had a cheap little LED desk light that used 3x AAA's,with no resistor or anything in between. It ran fine from the batteries for a good while,but one night when the batteries ran dead I connected it to my variable bench supply set to 4.5Vdc. After a few minutes the LED's started to change color,and a couple of them died. When I took it apart,the plastic reflector around the LED's was melted in spots.
It was relying on the internal resistance of the batteries to limit the current. |
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#6 |
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diyAudio Member
Join Date: Feb 2008
Location: Seattle
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This is certainly a cheap way to get something working.
A better idea is a boost regulator that can create the necessary voltage for the LED Vf even when the batteries drop well below their normal voltage. The obvious disadvantage here is that you're probably back to needing the current-limiting resistor, plus you need an inductor for the boost regulator as well as the silicon. Advanced boost regulators for LED drivers can output a target current rather than a target voltage, and those would not need the resistor, but they still need the inductor and FET(s). Somewhere online I saw a nice little circuit in a DIY context that worked with a hand-wound inductor. Of course, my second paragraph has little to do with the original question, but considering the disposable nature of most batteries, as well as the limited voltage of rechargeable batteries, I'd personally much prefer spending a little extra on the circuit to save many times the cost in batteries over the life of the flashlight. |
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#7 |
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diyAudio Member
Join Date: Sep 2004
Location: Minneapolis, Minnesota, USA
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A current regulator would make more sense in this case than a voltage regulator.
Mike |
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#8 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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LEDs discharge batteries.
A good LED circuit will be designed to discharge it's battery as little as possible. Good and cheap cannot be combined in this situation. Three AAAs in series gives ~4.5Vdc at low current. Use this to multiply up to a high voltage current regulator. 10 white LEDs in series require ~25 to 40V. A few mA will bring them on and 20mA will probably be near the limit of their brightness. I think all LEDs have their most efficient light output at well below their maximum current. If this is the case then reducing the brightness to below maximum by limiting the current to well below maximum will extract the most life and lumens from the batteries.
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regards Andrew T. |
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#9 |
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diyAudio Member
Join Date: Mar 2004
Location: Budapest, Hungary
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Are there LEDs with built-in current sources? If not, I will patent such device
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#10 |
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diyAudio Member
Join Date: Feb 2008
Location: Seattle
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