I drew this small circuit from a design a friend gave me to blink an LED, but I am missing a certain component. I am missing a 12KΩ resistor. Can I get away with a 10KΩ resistor in its place? enter image description here

  • \$\begingroup\$ What do the equations for designing 555 circuits tell you? You should be able to find this on the wikipedia entry for the 555. \$\endgroup\$
    – Daniel
    Commented Nov 23, 2015 at 22:51
  • \$\begingroup\$ You "designed" this circuit and you cannot figure out what the difference from 12K to 10K would cause?? From your question it would seem that it is more likely that you COPIED this circuit from some place. \$\endgroup\$ Commented Nov 24, 2015 at 0:31
  • \$\begingroup\$ that 20% difference is sometimes within specs for, say, the capacitor.. so if precision was really an issue you'd have to worry about other things too.. \$\endgroup\$
    – Wesley Lee
    Commented Nov 24, 2015 at 3:18
  • \$\begingroup\$ @MichaelKaras Here I edited it to make you happy. Happy? \$\endgroup\$
    – user92562
    Commented Dec 9, 2015 at 6:03
  • \$\begingroup\$ It's ok, everyone, when I assembled the circuit, it didn't work! why? may you ask? because... I HAVE NO CLUE, I DUNNO IF IT IS A BATTERY ISSUE OR WHAT, BUT MY STUPID LED WON"T CLICK \$\endgroup\$
    – user92562
    Commented Dec 15, 2015 at 0:45

2 Answers 2


Yeah, it will work, just with slightly different timing characteristics. See https://en.wikipedia.org/wiki/555_timer_IC#Astable for how to calculate the frequency, on time, and off time.


What values do you have? If you can add a resistor put your 10k + a 2k to achive your original design. (or 10k + 1k + 1k, etc)

  • \$\begingroup\$ I was aiming for small on this circut, i was planning on putting it in a tyny cylinder about an inch in diameter, it was really a test to see if I could do a soldering joint strong enough to hold up to... let's say 20gs at a small point in time (about 1 nanosecond) and average about 15g over 8.6 seconds (I put it in a model rocket) (yes it survived) \$\endgroup\$
    – user92562
    Commented Dec 18, 2015 at 6:31

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