I would like to create this 8 led beacon with only a single mode, that is turn 2 leds on at a time and move to the next one in a loop like shown in the picture below.

I want to achieve this without a microcontroller in the long run. I can use led drivers or shift registers or timers. I also have no problem programming the intitial setup using arduino but the project should not depend on a microcntroller for regular operation.

leds will be of 8mm straw hat size with 150mA each at 2.4V (red). Power supply will be 2x AA NiMH cells. No need of any input button function, if I power the setup then it should go straight to this function and keep looping until I power off.

enter image description here

Can it be done using 74HC595 or similar 74 series or should I go for LM3914 or is there any other simpler way? I understand that 74HC595 can't handle the high current requirement, which is why asked for alternatives.

These are the available IC's I can buy. the site also has other drivers.

Thanks in advance.

  • \$\begingroup\$ Search for "chaser circuit" \$\endgroup\$ May 15 '17 at 13:08
  • \$\begingroup\$ @Scott Seidman Exactly, but the chaser circuits I saw was based on CD4017 IC, does that support 150mA per LED? \$\endgroup\$
    – Kokachi
    May 15 '17 at 16:25
  • 1
    \$\begingroup\$ if it doesn't, it will with a transistor on each output and a big enough power supply. \$\endgroup\$ May 15 '17 at 16:35
  • \$\begingroup\$ From the CD4017 datasheet, it can only source/sink less than 10 mA. \$\endgroup\$
    – Matthew R.
    May 15 '17 at 16:35

You should be able to achieve this using a 555 timer and CD4017 combined with some MOSFETS on the output to switch the higher currents. The two LEDs in parallel will draw 300mA so ensure that your battery is capable of providing this power. The only other problem I can see is that the voltage of the battery may not be high enough to run the electronics so you need to check all that too.


  • \$\begingroup\$ I'm making a slight modification. The two leds are going to be in series and the voltage source is gonna be 5v. \$\endgroup\$
    – Kokachi
    May 17 '17 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.