0
\$\begingroup\$

I'm building an LED lightbox but I really want to avoid using a power supply/cable so I'm trying to figure out a way to do it with batteries.

The strip I'm using is http://www.amazon.ca/dp/B00D76XBU0/ref=pe_386430_30332290_TE_3p_dp_1 I'll be using about half the strip. I've got it setup using 2 9v batteries in series with a switch but I'm sure I need a resistor I just don't know how to figure out which to use!

Also does this look right?

   + 9v - + 9v -
   |    |-|    |
  (res?)       |______ - 
   |                   STRIP
  (switch)____________ +
\$\endgroup\$
3

4 Answers 4

Reset to default
2
\$\begingroup\$

As Vladimir has stated, you'll definitely want much more power than a 9V battery can supply. You'll also want to ensure you don't waste lots of your energy through a resistor due to your supply voltage being much higher than your load voltage.

If you want a mobile light box, either find yourself a lead acid or lithium ion battery pack that will supply the amount of current you're after for extended periods of time. Your circuit itself isn't wrong besides the use of an 18V power supply to power a 12V load. At best, you'd have 66% efficiency (12/18).

AT 5 amps, you're going to need a heavy duty resistor. An ideal led supply would give it constant current, but that's likely too sophisticated for what you want to do. Given that these are power LEDs, their voltage-current curve isn't terribly steep and will likely be able to handle a bit more power so long as the power supply is relatively close to 12V. If it goes above 12.5-13 V on your supply. I would expect you to start decreasing the lifetime of the LEDs significantly. If you measure your supply (batteries) and it's below that range, then you would likely be able to eliminate the use of your resistor entirely. If your supply is above that range, then you'll want to determine how much you need to drop the voltage. Basically, you want it to drop 5 Amps down to 12 Volts.

For example: lets say your power supply gives 14 volts rather than 12 V. Go ahead and add a resistor that drops it to 12V. (14V-12V)/5A= 0.4 Ohms. That is the application of Ohm's Law. If you don't know or understand it, wikipedia is a great place to start.

\$\endgroup\$
5
  • \$\begingroup\$ that's perfect but I don't think that a 60W portable led strip is something feasible... I mean, it depends on what portable means of course. \$\endgroup\$
    – Vladimir Cravero
    Jul 3, 2014 at 9:20
  • \$\begingroup\$ Using resistors to regulate voltage is not an ideal solution. \$\endgroup\$
    – John U
    Jul 3, 2014 at 9:43
  • \$\begingroup\$ @JohnU Agreed, but I doubt the OP is looking for an ideal solution that would require a decent understanding of circuitry or an engineer to make it for him/her. A buck converter run with current control would be the ideal solution, but from the sounds of it, that's way beyond what the OP's capable of at this point in time. \$\endgroup\$
    – horta
    Jul 3, 2014 at 10:45
  • \$\begingroup\$ I realise there's a fair tolerance here, but with the OP's minimal understanding, using a resistor could result in a burned-out LED strip if he cuts it down by, say, 50% or the stated 5A draw is an "absolute max" and the reality turns out to be some smaller number. \$\endgroup\$
    – John U
    Jul 3, 2014 at 11:55
  • \$\begingroup\$ @JohnU In my experience with power LEDs they do state the "absolute max" rating you should run them at. Anything above that will limit the useful lifetime or need extra cooling. You'd have to go higher than 25% of the "max power" rating to start doing significant damage in a short time tho. You're right that if they're actually rated for 2Amps and he/she puts 5 amps into it, they'll definitely pop immediately. But I assume the OP posted correct info. If he/she didn't, it's their own fault. \$\endgroup\$
    – horta
    Jul 3, 2014 at 17:33
1
\$\begingroup\$

Cheap & easy portable 12v supply = car or motorbike battery. Used/scrap ones that can't crank a car engine will usually still have more than enough capacity to source 5-10A and are available for scrap prices. Chargers are cheap & widely available, and in-car cellphone chargers etc. offer cheap & easy sources of regulated 5v (for example) for powering other circuits.

\$\endgroup\$
1
  • \$\begingroup\$ This is the easiest and cleanest solution. Perfect success for a DIY beginner, and no hassle with too much details. \$\endgroup\$
    – Gee Bee
    Mar 14, 2016 at 13:36
0
\$\begingroup\$

Wise ones use a cheap silicone diode (rectifier) You can get one from Goodwill. By a used stereo for 5 bucks and open it up. Lots of diodes and transistors there for parts. Don't get a schotkey diode since they only drop the voltage like 0.4 volts and you need more of them to get the voltage down, but they can be used. Silicon is cheapest anyway. , just an old fashioned cheap silicon Diode. Just put it in line in Series on the positive lead so you don't float the whole LED strip. this will drop the voltage 1.2 volts so that you can run them off a car battery without all these gimmickry voltage regulators and such. Batteries are only able to make a maximum voltage so they are the best way to protect projects from high votages. They are self voltage limiting free of charge. they are the best voltage regulators. Get a cheap JUNK BATTERY worth nothing, and put a 2 or 5 amp trickle charger, and your LED will be powered cheap. Those little tiny 12 volt motorcycle batteries are the best. The batteries don't even have to work, they just buffer the voltage, but they should have water in them or they wont work. so a car battery, a diode or two or three or four or five or six or seven, in series, and a LED strip and you are good to go cost about 20 cents. Get one that handles about 20 amps.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ If the voltage while charger is on in a motorhome is like 14.7 volts, then you should put two diodes in series and the voltage will drop 2.4 volts and they wont burn out. \$\endgroup\$
    – Guestguy
    Aug 16, 2015 at 16:57
  • \$\begingroup\$ Measure the voltage across the LED terminal while the charger is on your battery system. Add diodes in series until the voltage is 12 volts. Simple Plugger electronics. We used to build radios that way, tweak things while they are on until they work. \$\endgroup\$
    – Guestguy
    Aug 16, 2015 at 17:02
0
\$\begingroup\$

For a quick solution you have to put a 1 ohm resistor. To be exact 1.2 ohm for safety. Or, you can use two silicon diodes in series as their individual voltage drop is 0.7 volts.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.