Simple Alternate LED Flashing Using 555 Timer

About: electrical and electronics engineering student

here is the circuit diagram.!

Step 1: Components Required

1) 2 led of different colours
2) 100uf capacitor
3) 555 timer ic
4) wires
5)9v battery
6) battery connector
7) 1 k ohm resistor 1
8) 10 k ohm resistor 1
9) 220 ohm resistor 2
10) bread boaed

Step 2: Connecting the Components

1) mount the 555 ic in the middle of the bread board
2)connect pin no 7 and 8 with 1 k ohm resistor
3) connect pin 6and 7 with 10k ohm resistor
4)join 2nd and 6th pin
5) join 4th and 8th pin
6) connect pin 1 and 2 with the capacitor
7) take a separate line from the pin 3 and connect one of the 220 ohm resistor to it, and connect the anode(+) of one of the led to the other end of the resistor and connect the cathode(-) of led to pin no 2.
8) connect the anode of another led to pin 4, then connect the another 220 ohm resistor series with it
9)connect the another end of the resistor to pin no 3

Step 3: Testing

connect the positive end of the battery to pin no 8 and negative to pin no 1.
You will see the LEDs flashing alternatingly... plz try this @ home it will cost below 2$.

hope you guys like this..
comment your doubts below
thank you..! meet you all soon with another circuit..




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    6 Discussions


    4 years ago on Introduction

    Schematic as well as "assembly description" is wrong.

    I have attached correct schematic, PCB layout (1.5" x 1.0") and solderless breadboard layout, but you really ought to make an effort yourself.

    The two 'scope frames are with the LED to Pin 2 and the correct way to B- top trace (red) is pin 2 and bottom trace is pin 3. Timebase is 500ms/div, the pin 2 trace is 5V/div and the pin 3 trace is 10V/div (same settings on both frames. As you can see, you upset the timing by the wrong connection - if you experience a problem when using it right, maybe you have a faulty 555. It would be a good idea to put a 'scope to it and see where it goes wrong, but just a voltmeter over the battery terminals may reveal something. A 9V battery is considered flat at 5.4V (0.9V/cell) and the 555 is not able to reach the rails with its output, so perhaps that could be it - either way, always use a variable power supply, to find the lowest voltage it can work with and to make sure you have a known good voltage to work from.

    Also, please use common conventions, when you draw a schematic - like the more positive a supply line is, the higher up it should be drawn and circuit flow should go from left to right (whenever possible) i.e. inputs to the left and outputs to the right. Getting into the habit of doing it right will help you immensely in your education.

    3 replies

    Reply 4 years ago

    thanks omnivent.. sure I will follow your words.


    Reply 3 years ago

    Off topic a little, but what what program did you use to make the 2nd black and blue schematic? or if you didnt make it, do you know what was used?


    Reply 3 years ago

    Sorry for the late reply, I just found the announcement email when reviewing my spam folder.

    I assume you are referring to the PCB layout I posted 7 months ago (?) and if so... Both the schematic (pic #1) and the layout (pic #2) is made with Eagle CAD, the freeware version.


    4 years ago on Introduction

    Why isn't the cathode of the low-side diode connected to B-?

    Also, it would be easier to discuss your circuits if the schematics included reference designators. Like D1, D2, R1, R2, U1, etc.

    1 reply

    Reply 4 years ago

    if the cathode of lower side diode connect to B-, it will start to glow continuously. from next time onwards I will post the schematic with referring names. thanks for advice.