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I am working on a project where I am needing to power 13-20 LEDs of varying sizes and types from a single battery / battery pack. I am generally a novice when it comes to circuitry, however, I have successfully made similar simple set ups before using resistors and a fixed 12v power supply and they still work after years of use.

My concern now relates to the voltage variability of batteries based on their charge and burning out the LEDs. Should I use something like a linear voltage regulator (L7912 ?) in combination with the resistors to protect the LEDs?

Thank you.

Edit: Would a simple circuit with a voltage regulator like the below work? enter image description here

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  • \$\begingroup\$ What is the forward voltage range of the LEDs and what is the nominal voltage range of the battery. Are all the LEDS on at the same time or are they individually controlled? \$\endgroup\$
    – Trevor_G
    Commented Mar 8, 2018 at 20:01
  • \$\begingroup\$ forward voltage varies between 2.8v and 3.2v depending on which LED. My general thought was to group like LEDs in series up to the source voltage (use resistor for remainder) and put the groups in parallel. As for battery I am flexible as to what can be used, considering a 12v battery pack made up of AA batteries. All LEDs in this setup will be on at the same time with a single toggle switch. \$\endgroup\$
    – TAllen
    Commented Mar 8, 2018 at 20:05
  • \$\begingroup\$ Yes, that schematic should work with the 9V version. - You would get a constant brightness until the battery discharges to a little more than 9V (not sure of the voltage drop across that device). \$\endgroup\$
    – Norm
    Commented Mar 12, 2018 at 17:16

2 Answers 2

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Your series resistor(s) should be sized to limit current to their LED (to probably 10-20 mA depending on the LED) and of appropriate wattage to be able to drop the extra voltage without getting too hot. The resistor is what's going to do all the work here. You want it big enough (wattage) that it won't overheat when the battery is at full charge. As the battery discharges the resistor will drop less voltage, until the battery voltage drops close to the LED's working voltage, and the LED's will dim.

Are these LED's used as some type of indicators?

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  • \$\begingroup\$ the LEDs are being used to light a scale model (like a diorama) so once I have the preferred brightness dialed in i would prefer to try and maintain that as much as possible. \$\endgroup\$
    – TAllen
    Commented Mar 8, 2018 at 20:16
  • \$\begingroup\$ If you want to maintain brightness as the batteries discharge then you would want a regulator in there to keep the voltage constant across the LED's \$\endgroup\$
    – Norm
    Commented Mar 8, 2018 at 20:27
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It depends on how much change in brightness you can tolerate over the battery voltage range. If you don't care much, design it with series resistors based on the highest battery voltage and lowest forward voltages.

Otherwise, you would need some form of current regulators to fix the forward current of the LEDs to whatever value you choose. Since you have a variety of LEDs each string would need to be equivalent forward currents and the total forward voltage would need to be less than the minimum battery voltage minus a bit to let the regulator do it's job.

See this cross-post

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  • \$\begingroup\$ The variation in brightness would not be desirable. As for the current regulator, I don't have any experience here. Not sure what i would be looking for, would I be trying to match up the current regulator rating with the exact current rating on the LEDs? \$\endgroup\$
    – TAllen
    Commented Mar 8, 2018 at 20:23
  • \$\begingroup\$ @TAllen yes. See my edit \$\endgroup\$
    – Trevor_G
    Commented Mar 8, 2018 at 20:25
  • \$\begingroup\$ i see how the current regulator would work, seems like i would have to do that for each group of LEDs. Would it also work to use a voltage regulator to regulate the battery voltage to a fixed amount (see diagram in updated question above). Or would that not work \$\endgroup\$
    – TAllen
    Commented Mar 9, 2018 at 19:14
  • \$\begingroup\$ @TAllen yup that works too, though watch your power dissipation on the regulator. \$\endgroup\$
    – Trevor_G
    Commented Mar 9, 2018 at 19:17

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