I've started to play with the 555 chip. There is a lot of documentation out there. So I want to create a strobe light project, with an adjustable frequency.

I've played with a lot of capacitors (1nf, 1uf, 10uf, 100uf, 1000uf) and resistances (330ohm, 1kohm, 10kohm, 100kohm, 1mohm). If I can change the frequency, I run into the problem where the light does not stay off for an equal amount of time than it stays on.

For example, right now I have this kind of output when I play with my potentiometer (numbers are not accurate):

  • 100ms on, 100ms off
  • 500ms on, 100ms off
  • 1000ms on, 100ms off

But I'm trying to achieve this kind of scale:

  • 100ms on, 100ms off
  • 500ms on, 500ms off
  • 1000ms on, 1000ms off

I have multiple 555 chips, but I can't find a way to have the on and off signal to have an equal duration.

  • \$\begingroup\$ I think it is not possible to get a 50% duty with a 555, At least I didn't find to do it when I was in the University, and related books didn't talk about it. It always was >50% or <50%. \$\endgroup\$ – Chirry Nov 28 '13 at 22:41
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    \$\begingroup\$ There are many ways to get 50% duty cycle with a 555. A simple Google search will turn up many examples. The simplest is to connect the timing resistor to the output (pin 3) instead of the usual connection to Vcc via pin 7. \$\endgroup\$ – Dave Tweed Nov 28 '13 at 23:01
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    \$\begingroup\$ @DaveTweed Maybe I just don't use the good keywords, but I've searched all day prior to writing up this question... could you point me to some links on the subject ? Thanks ! :) \$\endgroup\$ – FMaz008 Nov 29 '13 at 10:34
  • \$\begingroup\$ @Chirry: For my application, I don't need a very high precision, could be 45%+55% and it would do just fine. But right now my results are not anywhere close to 50%+50%, more like 95%/5% when at a low frequency. \$\endgroup\$ – FMaz008 Nov 29 '13 at 10:37
  • \$\begingroup\$ @FMazoo8 What I used to do in those cases, is to find a combination of capaticor-resistor that could work in the frequency i need, then with a potenciomenter as resistor I find the resistance it fix better. As you know, it's easier to find resistors than capacitors at a custom value. \$\endgroup\$ – Chirry Nov 29 '13 at 14:46

Dave Tweed has indicated a simple solution (+1 from me). The circuit for which is...

enter image description here

By charging and discharging the capacitor (C) through the resistance connected to the output (pin 3) the charge and discharge times are the same. By using a fixed resistor (R1) you can set the minimum time for the astable. By adding a variable resistor (VR1) you can easily alter the time without altering the mark/space ratio. The voltage across the capacitor (C) charges and discharges between 1/3 and 2/3rds of the supply voltage set by the internal comparators of the 555. The ON time = OFF time = T = 0.7CR (where R = R1 + VR1) - C in Farads, R in Ohms, T in Seconds.

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  • \$\begingroup\$ To be honest I don't understand all the explanation, but I did the circuit and it works way better than the traditional 555 circuit I see all over the Internet (for what I wanted to do). I randomly used a 10K for R1, and tried few Cap for C1 to have the speed that looked OK. (I know, very scientific approach here) The C2 cap seems to be useless, when I remove it, it's working too (?) My setup: C1 = 1uf, R1=10K, VR1=10K potentiometer. Get me a very quick strobe light for visual illusions. \$\endgroup\$ – FMaz008 Nov 29 '13 at 22:54
  • \$\begingroup\$ @FMaz008 C2 stabilizes the thresholds that defines at which voltage the output toggles. You don't want any noise on these voltage levels and that is exactly what C2 does. It filters noise from the reference voltage. \$\endgroup\$ – jippie Nov 30 '13 at 16:46
  • \$\begingroup\$ Of course this looks very similar to and can also be accomplished with an opamp based relaxation oscillator \$\endgroup\$ – jippie Nov 30 '13 at 17:06
  • \$\begingroup\$ Not that it was a prerequisite in my original question, but this diagram is impossible to do on a single layer board... Linking pin 6 and 2 is ... not fun. \$\endgroup\$ – FMaz008 Dec 4 '13 at 5:07
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    \$\begingroup\$ @FMaz008 Its certainly not impossible but it is an additional question - draw copper track from 6 and 2 and put in a wire link from 8 to 4 under the chip but on other side of board to copper track. \$\endgroup\$ – JIm Dearden Dec 4 '13 at 18:05

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