3V 25mAh button cell battery, a single 3-3.4v 20mA UV LED.

If I put a 1/8W 1 ohm resistor in the circuit will the battery last longer?

  • \$\begingroup\$ Not noticeably. Also, you need to rethink your LED selection. \$\endgroup\$ – Blair Fonville Apr 20 at 6:14
  • \$\begingroup\$ The resistor dissipates power, hence wasted energy. \$\endgroup\$ – Chu Apr 20 at 9:46
  • \$\begingroup\$ Most button cells won't provide 20mA \$\endgroup\$ – Scott Seidman Apr 21 at 23:23

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Figure 1. The equivalent circuit for a button cell and LED. Source: Battery + LED without resistor.

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Figure 2. Measuring the no-load cell voltage and the loaded cell voltage allows us to measure the internal cell resistance.

You can check this out yourself.

  • In Figure 2a we measure the open-circuit voltage of the battery and read 3 V.
  • In Figure 2b we measure the voltage again with the LED connected. We find that it is, say, 2.2 V. Yours will be higher due to the colour.
  • (c) We switch the meter to mA and connect it in series with the LED. We measure 30 mA.

With these three measurements we can calculate the internal resistance of the cell. The voltage drop is 3 – 2.2 = 0.8 V at 30 mA so, using Ohm’s law we can calculate the internal resistance as

$$ R=\frac {V}{I} = \frac {0.8}{0.03}=26.7\ Ω $$

It should be clear that, in this case at least, adding 1 Ω in series will cause a slight drop in voltage and slight reduction in current. You probably won't notice. Actually, with a UV LED you definitely won't see any difference unless you are observing some fluorescence caused by the UV LED.

See the linked article for more on the topic.

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Figure 3. Current versus voltage curves for various colours of LED. Source: IV-curves.

I based the graph above on various LED datasheets so the curves are approximate and will vary by device and supplier. The main point to note is that the UV LED has a higher forward voltage than the others and that, according to my graph, it would require about 3.6 V to get 20 mA. Note that all the LEDs show the characteristic steep curve once we get past the turn-on point and this is why we need to control the current through the LED rather than the voltage across it. See Variations is Vf and "binning" for more on this.


Your small resistor will cause a small reduction in current draw which will make the battery last slightly longer. But your LED is slightly dimmer so there is no free lunch with the proposed resistor.


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