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I am thinking of making my own LED lamp - That is, have an array of LEDs and power them off, maybe, a battery pack? The power source I am lenient, I'll solve that issue when I get to it. My main question is regarding the LEDs

Is it more efficient to individually solder, say, 20 or 50 individual 5mm LEDs like these:

Individual LEDs

By soldering them, I mean taking two pieces of long wire, and soldering the all the positive ends of the LED strip to one wire and all the negative ends to another wire, and power it on from the two ends of the wires.

Or should I simply buy a LED Strip and break it up into little strips and slap them onto a template?

My main concern is energy consumption - the LED strip requires 12v, whereas the individual LEDs require 3.2-3.4v. Since I will be powering them off lithium ion batteries, each battery has a range between 4.2v and 3.2v. I will be regulating the voltage with a buck boost converter. TO power the individual LEDs, I'll need only one battery, whereas for the 12v I will need multiple batteries so that they are not stressed.

As well, are LED strips generally brighter than the individual LEDs, which are 15,000mCD each?

Each individual LED is, say, 3.3v and draws 20mA. for 20, that's 1.3Whrs.

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    \$\begingroup\$ If you put any LED's in parallel, they should have their own resistor to limit their current. Not all LED's got the exact same forward voltage, and when LED's get warmer they conduct better, so eventually some LED's might fry, because of no resistor balancing. \$\endgroup\$ – Harry Svensson Jul 17 '17 at 10:09
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The answer is "it depends".

If you want a flashlight, with a nice beam, then you need optics, thus it is much easier to use one high-power LED so only have to purchase one optic (lens or reflector).

If you want diffused light, then having many light sources (many LEDs) is easier to work with.

Now, if you use batteries, you'll be interested in the efficiency of your LEDs (ie, lumens/W) to maximize battery life. So you need to be aware of that. There is a compromise though, as LEDs with pleasant warm light and good color rendition tend to be less efficient. Very high lumens/W LEDs are usually "cool white" and low-CRI which isn't that good for stuff like reading or ambient light.

As well, are LED strips generally brighter than the individual LEDs, which are 15,000mCD each?

mcd (tmillicandela) has nothing to do with light output. A candela is a lumen per steradian, the latter being a unit of ANGLE. This means the same LED chip, which puts out the same amount of light (lumens) can be 100 mcd or 10000 mcd depending on how the optic in front concentrates the light into a wide or tight beam. If you want to make a lamp for diffuse ambient light, you need low-mcd, high lumen LEDs.

Each individual LED is, say, 3.3v and draws 20mA. for 20, that's 1.3Whrs.

W is power, not Whr which is energy.

Now, DO NOT wires your LEDs in parallel. Since all LEDs have a bit different voltage drops, the one with the lowest voltage drop will hog all the current, then burn. Then the next one will hog the current... etc.

If you got many LEDs, you need to put them in strings with resistors to equalize the current. Or use only 1-3 high power LED.

Losing less power on the resistor means putting more LEDs in series (ie, using higher voltage) but if you start from one 3.6V LiIon cell, boosting above 12V will also be less efficient. So 12V is OK.

12V LED strips will lose some power on the resistor. 3 high power LEDs in series will not (but may be more cumbersome to use, your choice).

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  • \$\begingroup\$ Thank you so much for your detailed response - I was about to go with wiring about 20 LEDs in parallel haha, thank you for warning me. I think I'll go to the high power LED option - I'll rip a heatsink out of my old computer and stick it on the back of the LED to reduce temperature. \$\endgroup\$ – Che0063 Jul 18 '17 at 21:55
  • \$\begingroup\$ Good! Now you know why wiring LEDs in parallel is a bad idea... Also the threshold voltage has a negative temperature coefficient (like all diodes), this means that the one which hogs the current heats up, its threshold goes down, so it hogs even more current (and then burns). Choice between a few power LEDs and lots of small SMD ones depends on how you want to diffuse your light. What do you want to do? \$\endgroup\$ – peufeu Jul 18 '17 at 23:12
  • \$\begingroup\$ I think I may simply go with one high power 3W LED. I have seen many on eBay, so I believe the best choice for me is to get one that runs off 3.0-4.2v, wire it up to a buck boost converter, with the power source being either an 18650 or 26650 Lithium Battery. As for the optics, I may go all out cheap and make a cardboard shaped cone covered with aluminium foil. I know it is conductive, but I'll work something else. Or maybe I'll buy an actual optic later. Again, thank you for your help \$\endgroup\$ – Che0063 Jul 20 '17 at 6:28
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if I were you I would make something like this:

enter image description here

Here's a link if you want to play around


EDIT

I reread your question and saw that you said the LED strips needed 12V. That it was not your battery that delivered that. Your battery delivered 4.2V - 3.2V.

In that case this schematic would be much more viable:

enter image description here

Here's a link if you want to play around


Let's compare how bad it is to use current limiting resistors instead of using a buck converter or boost converter. Assume that a X converter is 80% efficient. Let's also assume that you will use 4 LED's like in the schematic above.

\$P = VI\$
\$P_{LED} = 4(3.3×0.02) = 264 mW\$ the 4 comes from 4 LED's

With a boost converter you'd get this equation:
\$P_{w/converter}=\frac{P_{LED}}{0.8} = \frac{0.264}{0.8} = 330 mW\$

With 50 ohm resistor per lane like in the schematic it would become like this: \$P_{w/resistor} = I^2R + P_{LED} = 4(0.02^2×50)+0.264 = 344mW\$

\$\frac{344}{330}=104\%\$ So if you use a constant current boost converter that is 80% efficient instead of 50 ohm resistors, then you'll be 4% more efficient. It will be the same story if you use more LED's, the 4% that is. But if you however get a constant current boost converter that is more efficient than 80%.. then that one will turn the 4% to like 20% and then it might be more viable.

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  • \$\begingroup\$ OK, thank you for your detailed response. I understand the efficiency between the buck boost converter but wouldn't the 50ohm resister setup reduce the voltage drop? Wouldn't this increase the lifespan of the LEDs \$\endgroup\$ – Che0063 Jul 18 '17 at 22:06

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