Shift registers , I don't know alot about them(applications and basic functionality) but I do know how to use them and in this instructable I will show you how you can use an 8 bit serial in parallel out shift register.

Supplies

For this circuit , you will need :
SN74HC595
LED (x9)
220 ohm (×9)
5V battery

Step 1: BASIC THEORY

As we all know,of course, there are about four types of shift registers but for this instructable I will only be focusing on the 8 bit Serial in , parallel out shift register . This register takes in data serially , stores it and releases it through 8 output pins.

It requires a clock input for its internal shift and storage registers in order for the data to be laoded serially. This will be provided by an A-stable clock circuit .

Step 2: FOLLOW THE SCHEMATIC

This circuit relies on the SN74HC595 IC , which is an 8 bit serial in parallel out shift register .Pins one to seven , 15 and 9 are output pins , while pins 16 , 10 and 8 connect to 5V and ground respectively . Pins 14 , 12 and 11 connect to to the A- stable clock circuit ( pin 3 ) that I made in one my previous instructables, (which you should definitely
go and check out ) "HOW TO MAKE YOUR OWN BINARY CLOCK"

Step 3: TEST THE CIRCUIT

By connecting the circuit to power and pressing the button in the A-stable clock circuit you should see the the LEDs connected to the shift register start lighting up in sequence . With the way the circuit is built now it may take a while before it works right , as it did with me , but once it does you should see the shift register taking in or shifting(no pun intended )the values one by one as you press the button .