I'm a beginner working on a project which requires flashing high power LEDs, they will be supplied by a battery and flash at a frequency with a 555 timer. However, because I would like the battery to last longer I need to alternate between 3 LEDs with each output of the timer. Currently I have the 555 timer flashing a single LED but I was unsure of how to proceed with alternating between 3 LEDs.

Any help would be appreciated, thanks.

current circuit

  • \$\begingroup\$ Can you elaborate on how using three LEDs will preserve battery life? Assuming they require the same current, there should be no difference in battery life whether you use one or one hundred LEDs (assuming one is on at any time). \$\endgroup\$ – JYelton Feb 28 '20 at 0:12
  • \$\begingroup\$ Sorry I didn't phrase that well, I'm using 3 LEDs from the start, I was saying that I'm looking for a way to alternate between the 3. I wasn't trying to say that the number of LEDs was 3 to conserve power. The reason for why I'm flashing between 3 and not using one is for greater coverage in the implementation. \$\endgroup\$ – monocathron Feb 28 '20 at 0:26
  • \$\begingroup\$ @JYelton I think the flashing is necessary, but in order to prevent immense spikes in current they want the LEDs to be on at different times. A more consistent current draw will lead to better battery life over current spikes of 3X the current. \$\endgroup\$ – ambitiose_sed_ineptum Feb 28 '20 at 0:26
  • \$\begingroup\$ Thanks for the clarification. I was going to suggest using a decade counter as @AnalogKid has done. \$\endgroup\$ – JYelton Feb 28 '20 at 1:23

A very common way to do what you want is to have one 555 acting as the oscillator, and drive a CD4017 Johnson counter with it. The counter steps through 10 outputs in series, and you can make a connection such that after three steps it repeats. You will need one driver transistor per LED because the 4017 output is pretty weak.

This type of circuit is called a chaser, and there are many example schematics on the web.


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.