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I have two models by Chip Carter, one is the Enterprise, and the other the Galileo Shuttle, as shown below. Both models use three tiny LR41 button batteries, 1.5v each. That adds up to 4.5V. USB power is 5V. I've successfully custom wired dioramas with LEDs using resistors and USB power, but converting these models for installation into dioramas using USB power makes me nervous considering that they are sealed items in which I cannot replace the LEDs if I burn them out. Has anyone done this with these models, or similar? What is a safe resistor that I should use to power these models from USB?

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  • \$\begingroup\$ Good Question, see the answer I just posted. You can also purchase a 4.5V wall wart (phone charger) and use that. \$\endgroup\$
    – Gil
    Jun 3, 2021 at 2:54
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    \$\begingroup\$ One thing you want to be careful about is whether or not they are depending on the internal resistance of the battery to limit current. For instance, have a look at this tear down of a LEGO Light Brick: youtube.com/watch?v=Rhcr5wX0-SE \$\endgroup\$
    – ErikR
    Jun 3, 2021 at 2:59
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    \$\begingroup\$ @ErikR Those are 3V LEDs attached to two 1.5V LR41 batteries... so it doesn't need a current limiting resistor. It's nothing to do with the battery's internal resistance, it's just that the voltage is already matched to the LED. An LR41 battery has a short circuit current of about 220mA which is much higher than what the LED will consume. \$\endgroup\$
    – J...
    Jun 3, 2021 at 11:28
  • \$\begingroup\$ Thanks for all the responses and advice, will check into it. \$\endgroup\$ Jul 16, 2021 at 7:20
  • \$\begingroup\$ LEDs are current devices, not voltage devices. They have a forward voltage drop when operating, which changes with current. If you look at the data sheets it tells you the current requirements and may give you a range of forward voltage which as we all know changes with the color and process. \$\endgroup\$
    – Gil
    Jul 16, 2021 at 16:39

3 Answers 3

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Part of the problem is that you don't know if there are even any resistors in the models. It is quite possible they rely on the internal resistance if the batteries to limit current.

Some things about modern LEDs and LR41 batteries...

  • they don't require a lot of current... most are quite visible at just a couple of milliamps
  • 5 - 10mA continuous current is no problem
  • LR41 batteries have a small capacity -- like under 50mAh

To get an idea of how much the LEDs are drawing you can use a multimeter if you have one or just install a fresh set of LR41 batteries and see how long it takes for them to run down. Take 50 mAh divided by the run-down time and that will give you a ballpark figure for the current draw.

Once you have a current draw figure you can pick a resistor to match it. Start at 5K and work your way down. You can measure the current draw by taking the voltage across the resistor divided by the resistor value.

As I mentioned before, LEDs should be able to handle a 10mA continuous current, but you'll probably find they are bright enough at just a couple of milliamps or even just 1mA.

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  • \$\begingroup\$ Thanks for your response, will check into it. \$\endgroup\$ Jul 16, 2021 at 7:19
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You can do it several ways,the simplest is to add a diode in series with the 5V power, that will drop it by about 0.7V. This is what I would recommend. You can measure the current then calculate a series resistor, however if it makes noises or starts and stops thing this would be a poor choice. You can calculate the resistor by using Ohm's law There many calculators online free to use.

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  • \$\begingroup\$ Thanks for your response, will check into it. \$\endgroup\$ Jul 16, 2021 at 7:19
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First, use a multimeter to measure the current being drawn from a fresh set of the LR41 batteries. Next, make a current regulator set to this value; the LM317 is an ancient but trusty device for this and there are sample schematics/calculators online. You just need two resistors and a small capacitor.

With the current regulator in place you can use (within reason) any supply voltage. It's really current the LEDs are concerned about anyway.

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    \$\begingroup\$ For currents as low as in this situation, it could be prudent to take into account the burden voltage of the multimeter. More information, from Fluke: Can you live with the burden? \$\endgroup\$ Jun 3, 2021 at 20:26
  • \$\begingroup\$ Thanks for your response, will check into it. \$\endgroup\$ Jul 16, 2021 at 7:19

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