1.5v LED FLASHERThe 1.5v LED Flasher circuit will flash a LED at approx 1 flash per second. The ZSCT1555 IC from Zetex is designed to operate from a single 1.5v cell and will continue to operate to a terminal voltage of 0.9v. The circuit was originally presented by A J De-Guerin.
The first thing we note is the 1.5v supply voltage. From our previous discussions we know that a LED will not illuminate until the voltage across it reaches 1.7v. This means the circuit will not work unless the output is generating a voltage higher than 1.7v.
And that is exactly what happens.
The circuit charges the 47u via the 330R resistor when the output of the chip is HIGH. This puts a voltage of 1.5v on the electrolytic.
We will now replace the electrolytic by a 1.5v battery to make the discussion easier to understand.
We have a 1.5v battery connected to the output of the chip with the positive of the battery connected to the output of the chip.
When the output goes LOW, the left-hand-side of the battery falls by 1.5v and this causes the right-hand-side to fall by the same amount.
The negative lead of the "battery" is actually 1.5v BELOW the 0v rail and this puts a total of 3v (in theory) across the LED.
We have already learnt that the voltage across a LED will never rise above 1.7v (for a red LED) and so the voltage on the electrolytic creates a voltage higher than 1.7v across the LED and the energy flows into the LED to make it flash. The circuit is called a VOLTAGE DOUBLER.