Principle Functioning of the Circuit
To illuminate a white LED, we need a minimum of 3.6 volts and a maximum of 20 mA of current, which is the forward voltage drop of a white LED.
In the present circuit we use a 6 volt battery as the power source. So what's the big deal in illuminating a group of white LEDs with this battery? It's true, at first glance that the circuit is pretty ordinary, just connecting a group of LEDs in parallel to a battery. But something is special about this circuit. It's the inclusion of four diodes in series with the LEDs and, of course, the addition of S1.
We know that according to Ohms law, voltage is directly proportional to current, which implies that if the 6 volt supply is directly applied to the LEDs, they will start drawing excessive current, resulting in unnecessary dissipation of heat through the resistors and the LEDs and fast discharge of the battery, decreasing the efficiency of the circuit. This will happen because a potential difference of 6 volts is simply too high than the forward voltage of the LEDs (which is 3.6 volts as discussed above).
By adding 4 diodes we are able to drop the voltage to exactly the forward voltage of the LEDs (since each diode drops a voltage of approx. 0.6 volts).
Thus, when the battery is fully charged at about 6 volts, the LEDs will be receiving 3.6 volts, just enough to make them glow brightly without dissipating extra power.
Now suppose the battery voltage drops to an extent when you see the LEDs become weaker in intensity. You can simply adjust the switch S1 a step ahead by-passing one of the diodes. This will immediately restore the brightness of the LEDs, maybe for another couple of hours. After that the procedure may be repeated by by-passing the next diode.
Thus you are able to get multiple back-ups using the same battery- which wouldn’t be possible if the LEDs were connected directly.