3
\$\begingroup\$

I'm putting some battery-powered lights into a boat, and I have the following question:

I bought a bunch of cheapo LED lights, as pictured:

lights

They run on 3 AAA batteries. They work fine using 1.2V NiMH rechargeable batteries, and I've wired them up just to share one set of batteries, rather than each containing their own (still works fine). Rather than recharging batteries all the time, I figured it would be more convenient to get one of those USB LiPo power-packs for phones, and charge that when needed.

My question: Is anything bad going to happen if I hook up 5V USB power-pack in place of the (3.6V total) batteries? If I need a resister in series, how do I tell what resistance I need?

\$\endgroup\$
6
  • \$\begingroup\$ If I need a resister in series, how do I tell what resistance I need? If you cannot figure that out yourself then you should not even be thinking about hooking up that 5 V power pack. \$\endgroup\$ Commented Nov 15, 2016 at 12:52
  • \$\begingroup\$ When you tested with NiMH can you tell if they were any dimmer at 3.6V vs 4.5V? If not can you retest? \$\endgroup\$
    – D.A.S.
    Commented Nov 15, 2016 at 14:54
  • \$\begingroup\$ Not sure. Will respond if I do the test. \$\endgroup\$
    – Peter
    Commented Nov 15, 2016 at 17:02
  • \$\begingroup\$ I take it that if the answer is "no, the brightness doesn't change" it means the LEDs are hooked up in series, and with a boost converter, so that switching the power sources is ok, and if the answer "yes, it's brighter with 4.5V" then they're in parallel and swapping to a 5V power source may burn out the LEDs? \$\endgroup\$
    – Peter
    Commented Nov 15, 2016 at 17:19
  • \$\begingroup\$ not burn out going from a 2V drop from 4.5 to a 2.5V drop from 5V is like 20 to 25mA scaled, so I see no serious problem, 25% temp rise can be felt if too hot \$\endgroup\$
    – D.A.S.
    Commented Nov 16, 2016 at 3:54

4 Answers 4

5
\$\begingroup\$

My question: Is anything bad going to happen if I hook up 5V USB power-pack in place of the (3.6V total) batteries?

The LEDs will likely fail

If I need a resister in series, how do I tell what resistance I need?

Measure the voltage when operating on new batteries and note it down. Swap to the 5V supply but put 1 kohm in series and gradually lower that 1 kohm until you see the same voltage as measured previously. Use that modified value of resistor for each LED lamp you have i.e. they will all require their own individual dropper resistor.

\$\endgroup\$
1
\$\begingroup\$

If I'm not mistaken by the photo, it has 5 LEDs which are each around 3V and requires a 4.5V battery source but there is no indication of current or power or Lumens. So if 1.2V cells work, I conclude it uses an el cheapo boost regulator to drive the LEDs in series near 15V. So a battery pack from 3.3 to 5V will work as long as it has same Ah capacity or better than Alkaline for useage time.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ It is more "cheapo" to just switch the LEDs in parallel without any boos converter. I have a similar product myself, it also uses 3 x 1.2 V NiMh cells and it does not have a boost converter. \$\endgroup\$ Commented Nov 15, 2016 at 14:47
  • \$\begingroup\$ Good point , let's ask. \$\endgroup\$
    – D.A.S.
    Commented Nov 15, 2016 at 14:52
1
\$\begingroup\$

I tried it, and everything seems ok, LEDs have not burned out, and now I have a USB-power-pack-powered lighting system. Thanks Tony and Andy for your responses.

Notes:

  • If I open the light up, it does seem that the LEDs are wired in parallel. So I'm guessing Tony's el cheapo boost-regulator with LEDs in series hypothesis is not true.
  • I did not see a noticeable difference in LED brightness when comparing 3x 1.2V NiMH batteries to 3x 1.5V Alkaline batteries. Have not yet compared with 5V source (which leaves me a little confused, as I'd expect a change in brightness if there's no voltage regulator on this thing).
\$\endgroup\$
0
\$\begingroup\$

If the string of lights is on one 2 conductor wire, they are wired IN SERIES if only one conductor is going into the bulb socket. It would be unusual to be wired in parallel as that is more material and production cost. If the battery pack is outputting in series, it's going to be 3.6- 4.8 Volts. Approximately 150 milliamps (current flow / battery drain) assuming each battery can provide a steady 50 MA. What watt bulbs are used will determine how long they will remain lit / bright. LED's are preferred for low watt consumption per lumens, but not all LED's are as efficient as advertised. Display lights and decorations powered by 3 AAA batteries are going to eat tons of batteries. Either find the correct DC plug in transformer (which will already have a rectifier and resistor protection within) or do the research on the rechargeable battery pack to match your current needs. If you use a higher amperage power supply than necessary, you're going to need capacitors and heat sinks on one 'leg' at the front of the 'loop' and the other 'leg' on the end / last device. And know which ranges they operate under. If you're going to increase your lights, you can change those capacitors each time you add a string until your string consumption matches the string of lights' amperage draw (it becomes the 'capacitor'). If you're not adding to it, I'd stick with wanting about 4.5vdc at 100-150ma supply. So the bulbs won't fry. Another thing you can do is when the battery is on the lights: If you pull one bulb out and all the other bulbs go off/dark, then you know it's 'in series'. If it's in parallel, the other bulbs will remain 'on', and you'll need to design the capacitor/heat sinks at the beginning of the circuit on each 'leg' until your desired amperage is attained for the number of lights on the string required amps.

\$\endgroup\$

Your Answer

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

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