I've got a 555-timer circuit. When I examine the output-pin on an oscilloscope, I get a nice square wave.

I want to turn an LED on and off, so I've attached an LED and resistor to the output pin (see image, below). However, the LED stays on. When I examine the output pin again, I find a solid signal, rather than a square wave (see images).

If I remove the LED and resistor, and attach the timer-output to GND, I still see the square-wave (I thought perhaps creating the circuit from output -> GND was the issue). I'm not sure how to determine exactly what is happening, or why.

My questions:

  1. Why is this occurring?
  2. How can I modify my circuit to turn the LED on and off?
  3. What steps could I have taken (before asking here) to further 'debug' my circuit? Using the oscilloscope showed what my issue was, but not why it was occurring.

The circuit: My circuit

My wave, before I attach the LED/resistor [sorry for the blurriness]: Before the LED is attached

When the LED is attached, this is what I get: When the LED is attached

Here are some pictures of my breadboard:

My breadboard. My breadboard, angle #2

When I insert a capacitor (1uF) over the power supply pins:

Noisier signal with a cap over the power supply pins.

  • \$\begingroup\$ Do you have a picture of your circuit setup? \$\endgroup\$
    – jippie
    Apr 23, 2013 at 5:45
  • \$\begingroup\$ @jippie You mean a photo of my breadboard? \$\endgroup\$
    – simont
    Apr 23, 2013 at 5:46
  • \$\begingroup\$ yes (padding..) \$\endgroup\$
    – jippie
    Apr 23, 2013 at 5:47
  • \$\begingroup\$ @jippie Yes. Added two, can add more if these don't show adequate information. \$\endgroup\$
    – simont
    Apr 23, 2013 at 5:53
  • \$\begingroup\$ Could it be something as simple of you not adjusting your trigger level? I apologize if you know what I'm talking about but you never know ... \$\endgroup\$ Apr 23, 2013 at 5:56

2 Answers 2


You were right in using an oscilloscope to check the output, but many of them have an "auto ranging" feature meaning the screen is scaled. You may need to adjust how many seconds of the signal are displayed at a time. Also, the scope might have features that allow you to measure the frequency of the signal on the screen as well as the max and min voltages. Double check to ensure your screen is displaying the right amount of information. It's OK to ask this type of question here, just so long as you have done a bit of work on your own. That's what we're here for!

In your pictures, you are not using 1.8 Mohm resistors as noted in the schematic, which would be brown, grey, green. The resistors in the picture are orange, orange, green which are 3.3Mohm. With these values, the output would be 31 Hz with a duty cycle of 66.67%. This pulse is too fast to see with the naked eye. To actually see the LED flash, the pulse frequency needs to be be less than 20Hz, but this would be a fast strobe light. Something like twice per second, or 2Hz, is more appropriate. Of course, all of that is assuming you are actually using a 4.7nF capacitor.

That type of capacitor is rated in pF (10^-12). 4.7nF = 4.7(10^-9) = 47(10^-10). A 4.7nF capacitor should have "472" printed on it, meaning 47(10^2)(10^-12). Double check that value and report back.

Increasing the capacitor value from 4.7nF to 47nF (473) would increase the cycle time, decreasing the frequency by a factor of 10. The new pulse frequency would be 3.1Hz instead of 31Hz, meaning the LED would flash 3 times per second. Remember, you can increase a capacitor value by putting them in parallel. For example, three parallel 100nF caps = one 300nF cap.

Here is a resistor color code chart and calculator.

Here is a note on reading capacitor values and a value calculator.

Here is a great calculator for 555 timer circuits.

Hopefully that gets you on the right path!


I'm assuming your supply voltage isn't really 1V as drawn in the schematic, 1V is a little low for a 555 (can't make it out from the scope screenshot). If this is in fact the case increase it to at least 5V and redo your test. Also, the resistors values are fairly high, it might work this way but I wouldn't go over 1M, rather increase the capacitor to increase the timing if need be.

  • 1
    \$\begingroup\$ I was going to point that out although the top trace appears to be 12V, I can just make out the 5V/div the same as the second. \$\endgroup\$
    – PeterJ
    Apr 23, 2013 at 6:17
  • \$\begingroup\$ That's an oopsie in the schematic. I'm measuring the supply voltage at around 11-12V - I'll fix that in the question. \$\endgroup\$
    – simont
    Apr 23, 2013 at 6:20
  • 1
    \$\begingroup\$ Of course, a 555 won't work on 1V. The recommended minimum Vdd is 4.5V. It will work somewhat lower as well, but not on 1V. \$\endgroup\$
    – user17592
    Apr 23, 2013 at 6:22

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