I have a couple of Harbor Freight free-when-you-buy-something-else flashlights.
I want to modify one of them into a light source for a microscope. I've taken (10) of the (24) white LEDs off of one of the circuit boards and resoldered the remaining (14) (most of them were really poorly soldered). The original current limiting resistor was 1.5 ohms,5% for 24 LEDs and they were powered by 3 AAA batteries.
I would like to power the remaining (14) LEDs by tapping into the USB cable used for the digital microscope I'm using.
The LED microscope has a circle of LEDs built in for looking at opaque objects, but now I've got it mounted on a traditional microscope base with an adjustable stage (this thing is cool). I want the (14) LEDs to be mounted in the base of the microscope so it will illuminate translucent objects (traditional glass microscope slides). My question is, what resistor value should I use to power the standard (14) white LEDs with 5V from the USB cable?
I've attached a photo of HF flashlight circuit boards, one with all the LEDs still attached and the other that I'm modifying, including sanding the tops.
I'm not real experienced with electronics. I know enough to get in trouble.
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2\$\begingroup\$ Knowing a bit about the reliability of things you pay for at Harbor Freight, I can't imagine how bad something would be that they give away for free. \$\endgroup\$– The PhotonCommented May 18, 2017 at 18:38
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\$\begingroup\$ Pretty much a duplicate of: electronics.stackexchange.com/questions/21243/… or any of the hundreds of LED questions here. \$\endgroup\$– winnyCommented May 18, 2017 at 18:46
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\$\begingroup\$ hobby-hour.com/electronics/ledcalc.php \$\endgroup\$– Bruce AbbottCommented May 18, 2017 at 18:53
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\$\begingroup\$ 3.2V@20mA =64mW nom x14 LEDs = 900mW approx or 280mA with a 1W R @ 6.4 Ohm 1.8V/0.28A or you can replace with a single 3W LED for <$2 on a DIY heatsink about 2"x2" digikey.com/product-detail/en/cree-inc/XPGDWT-01-0000-00ME3/… assuming your USB port supports it \$\endgroup\$– D.A.S.Commented May 18, 2017 at 18:59
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\$\begingroup\$ you probably want around 5000K and you will get a lot more lumens/W out with the Cree part (about double) \$\endgroup\$– D.A.S.Commented May 18, 2017 at 19:02
1 Answer
Short answer: 18 ohms.
Your USB may be one that complies with the spec that limits current to 100 mA. If your LED circuit wants more, it may be cut off entirely. While USB may provide 500 mA, you must request the extra current via enumeration, which involves USB message transactions (something you don't want to get involved with). So to be safe, a 18 ohm quarter-watt resistor should limit current to your 14 parallel LEDs to 100 mA.
Tony Stewart provides a similar approach, suggesting a 6.4 ohm (one watt) resistor, which would provide about 280 mA. Your USB port may not comply with USB specifications and provide any current in which case, you can get more light with his method. But your USB source may balk and shut down because the load is beyond the 100 mA initial limit.
However, you have not said how much current your microscope-to-USB interface is flowing. Your extra LED illuminator will add to this current.
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\$\begingroup\$ Very rare for a usb source to limit to 100 without enumeration. Most don't even limit to 500 ma either. \$\endgroup\$– PasserbyCommented Dec 24, 2017 at 5:18