555 timer calculator

Enter the timing resistors and capacitor of an NE555 circuit to get its output frequency, duty cycle and high and low times in astable mode, or the length of the output pulse in monostable mode.

Mode
kΩ
kΩ
µF
Frequency1.3872 Hz
Duty cycle (output high)
54.81 %
Period
720.8731 ms
High time
395.0939 ms
Low time
325.7792 ms

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How to use

  1. Choose Astable for a free-running oscillator (blinkers, tone generators, clock signals) or Monostable for a single pulse after a trigger.
  2. Enter the resistors in kΩ and the capacitor in µF. For nanofarads type a decimal: 10 nF = 0.01 µF, 100 nF = 0.1 µF.
  3. Read the frequency, duty cycle and high and low times, or the pulse width.

Formulas

The 555 charges the timing capacitor between one third and two thirds of the supply voltage, so the times do not depend on the supply.

Astable

  • high time = ln 2 × (R1 + R2) × C
  • low time = ln 2 × R2 × C
  • frequency = 1 / (high + low) = 1 / (0.693 × (R1 + 2R2) × C) ≈ 1.44 / ((R1 + 2R2) × C)
  • duty cycle = (R1 + R2) / (R1 + 2R2)

Monostable: pulse width = 1.1 × R × C.

With kΩ and µF the product R × C comes out directly in milliseconds.

Worked example

R1 = 10 kΩ, R2 = 47 kΩ, C = 10 µF (the default values):

  • high time = ln 2 × 57 × 10 = 395.0939 ms (ln 2 = 0.693147)
  • low time = ln 2 × 47 × 10 = 325.7792 ms
  • period = 720.8731 ms, so the frequency is 1.3872 Hz — an LED blinks a bit more than once a second;
  • duty cycle = 57 / 104 = 54.81 %.

Change C to 0.01 µF (10 nF) with R1 = 1 kΩ and R2 = 4.7 kΩ and the same circuit runs at 13,872.0677 Hz, an audible tone.

Tips

  • Keep R1 at 1 kΩ or more: when the discharge transistor switches on, R1 alone limits the current from the supply.
  • To get close to 50 % duty cycle, make R2 much larger than R1.
  • Use film or ceramic capacitors for stable timing; electrolytic capacitors drift with temperature and age.

Frequently asked questions

What formula does the calculator use for the astable 555?

High time = ln 2 × (R1 + R2) × C, low time = ln 2 × R2 × C, so the frequency is 1 / (0.693 × (R1 + 2R2) × C), the familiar 1.44 / ((R1 + 2R2) × C). The duty cycle is (R1 + R2) / (R1 + 2R2).

Why can the duty cycle not go below 50 %?

In the standard astable circuit the capacitor charges through R1 and R2 but discharges only through R2, so the high time is always longer than the low time. For a duty cycle under 50 % add a diode across R2 or use a different circuit.

How long is the pulse in monostable mode?

About 1.1 × R × C. The exact factor is ln 3 = 1.0986, which datasheets round to 1.1. With R = 100 kΩ and C = 10 µF the pulse lasts 1.1 s.

Why does my circuit run at a slightly different frequency?

Capacitors often have a tolerance of 10 % or 20 %, electrolytic ones even more, and the 555 itself adds a small error. The formula gives the ideal value; measure and trim R2 if you need an exact frequency.

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