I am using this design:

The only exception that C2 is 150nF to change the timing slightly. When I plug in the circuit and attach an LED across GND and pin 3 output the LED is static on. However, on closer inspection with a multimeter frequency setting it says that it is 700Hz. Why though? This should provide a frequency of around 1Hz with these resistors and capacitors. I am using an LM555CN chip.

  • 2
    \$\begingroup\$ Do you have a series resistor to control the current through the LED? \$\endgroup\$ Jan 3, 2019 at 22:31

1 Answer 1


To start with C2 has nothing whatsoever to do with timing. It simply serves to decouple one of the internal thresholds.

Second, the output period for this configuration is

$$period = 0.693 \cdot 2 \cdot R3 \cdot C1$$

which works out to about 65 µs (15.35 kHz) for the component values shown. I don't know why you're measuring 700 Hz. If you changed C1 to 150 nF instead, then the period would be 977 µs (just over 1 kHz). Either way, this is too fast to see, so the LED will appear to be continuously lit.

If you want to see blinking, you need to increase the values of R3 and/or C1 by a couple of orders of magnitude. 10k and 10 µF will give you 140 ms (about 7 Hz).

  • \$\begingroup\$ But i used the calculator at ohmslawcalculator.com/555-astable-calculator. Is that wrong then? If so what components for roughly 1Hz? \$\endgroup\$ Jan 3, 2019 at 23:22
  • \$\begingroup\$ That's a different circuit configuration from the one you're using. Can you really not figure out what component values would give you 1 Hz from the information above? \$\endgroup\$
    – Dave Tweed
    Jan 3, 2019 at 23:25
  • \$\begingroup\$ i would use your formula but dont know which one to change for optimum efficiency and convention \$\endgroup\$ Jan 3, 2019 at 23:42
  • 2
    \$\begingroup\$ Energy efficiency? The only effect on power is that which is dissipated by R3. Make it large, but I wouldn't go above 1 Mohm. Then, to get a period of 1s, you calculate \$\frac{1}{0.693\cdot 2\cdot 1M} = 0.72 \mu F\$. That isn't a standard value, so make it 1 uF and change the resistor to 750 kohms, which will get you close. \$\endgroup\$
    – Dave Tweed
    Jan 3, 2019 at 23:51
  • \$\begingroup\$ And note that the circuit in the question will only have a symmetrical on/off time to the extent that it's output goes to the rails -- and the NE555 is a TTL output part. If you're going to use that topology, use a CMOS version of the 555. \$\endgroup\$
    – TimWescott
    Jan 4, 2019 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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