The fact that many of the cheap key-fob lights use a button cell and LED with no sign of a series resistor or any current limiting often causes some confusion. If the LED doesn’t burn out in this case when connected to a 3 V battery then what’s all the fuss about current limiting and series resistors?
The answer is that there is a series resistor; we just can’t see it! All cells and batteries have some internal resistance. We usually model batteries as an ideal, constant voltage source with a series resistance.
- 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 2.2 V.
- (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
\( R = \frac {V}{I} = \frac {0.8}{0.03} = 26.7 Ω \)Note that the internal resistance may not be constant as it is due to complex chemical and physical action within the cell. If you repeat the measurements with two LEDs in parallel you will get another but close value. If you short out the cell with your ammeter you will get a different value again and usually higher as you are pulling much higher current.
The lesson?
Just because it works with a button cell, it doesn’t mean it’s a good idea with other power sources. Repeating the test with a battery of lower internal resistance will raise the current and may destroy the LED. Placing the LED across a regulated 3 V supply will definitely blow the LED. See the IV curves page for more details.
Other reading
To calculate the power dissipated in the internal resistance see the power calculation article.
