How can I calculate how many LEDs in parallel I can light with three 1.2V batteries in series?

How long will 3.000mAh batteries last in that case? Do I have to put a resistor at only one of the parallel lines or in every line?

  • \$\begingroup\$ Yes, you can make these calculations, provided that you know the LED current. \$\endgroup\$ Jun 17, 2014 at 0:33
  • 2
    \$\begingroup\$ s/random/arbitrary/ \$\endgroup\$ Jun 17, 2014 at 1:21

3 Answers 3


You'll always want a single resistor per LED. If you don't, only one LED will really be lit to its full brightness. The rest won't be at a similar brightness because their V-I curve will be slightly different due to process differences.

To calculate how many LEDs you can supply in parallel, all you need to do is determine how much current you want each LED to consume. Let's say you want each LED to consume 10mA.

You'd have 3mAh/10mA = 0.3 hours or 18 minutes for one LED. If you are from Europe and actually mean 3000mAh rather than 3mAh, then one LED would last 300 hours. Then you can divide the amount of hours by how many LEDs you want. So 2 would be 150 hours total. Here should be your formula:

$$Total\;time\;LED's\;will\;burn\;(hours) = \frac{Battery\;capacity\;(mAh)}{number\;of\;LEDs \times current\;each\;LED\;consumes}$$

Lastly, notice that it doesn't matter how much voltage you have so long as it's enough to light the LEDs with as much current as you desire. Beyond that, any voltage will simply be dropped across the series resistors you need to add to each LED.


Your first step would be to look at the LED's datasheet to determine the current. Common LED currents are around 25mA, though they vary widely depending on the construction, age, brightness, and so on.

Once you find the current, you should be able to determine the total current drawn by each series string. The number of strings in parallel will determine how long your battery will last.

  • \$\begingroup\$ If batteries work for example 5 hours and the voltage drops as the time passes LEDs will stop lighting up? \$\endgroup\$
    – konsalex
    Jun 17, 2014 at 2:25
  • \$\begingroup\$ They will dim as the voltage decreases over time, and eventually the battery voltage will no longer be able to keep them lit. It is also important to remember that the mAH rating does not tell you how long it will last, since different batteries are considered "dead" at different points, and take different amounts of time to drain to said point. There really is no real way of answering your question, since it's practically impossible to get all of the information necessary \$\endgroup\$
    – DerStrom8
    Jun 17, 2014 at 2:35

LED's are low resistance, current driven devices, thus the LED current draw, as was previously mentioned is an important factor in calculating the power draw.

In a bank of 2mAH LEDs say 10 to a bank you would multiply the number of LEDs times .002 (or 2mAH) so 10 LEDs times .002AH per LED would be .020AH per bank. The resistor you would need for a bank is about 3.6V DC / (10 LEDs times .002) or .020 or 20mAH per bank. See equation #3 below

  1. Current times Voltage = Power available .... 3.6vdc times 3AH = ~ 11 Watts Available
  2. Current times Voltage = Power in Watts. 3.6V DC times .002AH = ~.0072 Watts per LED / Hr.
  3. Voltage / Current (amps) = Limit resistor size. .. 3.6/.020mAH (for string of 10 LEDs) Limit resistor for 10 LEDs drawing .002 amps is: 3.6 / (10 times .002) or .020AH which gives 180 ohms limiting resistor per string of 10 LEDs.

11WH/.0072WH gives you how many LEDs you can run for an hour.

About 1,500 LEDs ideally. Keep in mind the LEDs need voltage behind that current to operate. So realistically at 3.6 V DC of battery voltage, and an 2.0V cut-off voltage for the LEDs you have less time or LEDs than stated. The battery calculations are out of my zone of patience.


Remember the mAH in equation 2 increments by the mfg. datasheets estimate of their operational current draw. In this case 2mAH increments. Batteries being low resistance devices lose Voltage over time.

Keep in mind that when your batteries run down they will not meet the forward voltage drop across an array of LEDs once the batteries fall below 2 volts output. The voltage drives the current much like water pressure drives water out of your garden hose. Current being Volume, and Voltage being pressure.


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.