555 Timer Calculator
Frequency, duty cycle and pulse width for astable and monostable 555 timer circuits
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Works out the frequency and duty cycle of a 555 astable oscillator and the pulse width of a monostable one-shot from R and C, and suggests standard parts for a target.
How to use
- On "Astable (oscillator)", enter R1 (RA), R2 (RB) and C. Prefixes work, such as 4k7 or 10n.
- Read the frequency, high and low times and duty cycle under the circuit drawing and waveform.
- For 50 % duty or less, choose "With diode across R2".
- Under "Find parts for a frequency", enter a target and press "Use" on a row.
- On "Monostable (one pulse)", enter R and C, or find parts for a pulse width.
How it works
- TI xx555 datasheet (SLFS022K): tH = 0.693 × (R1 + R2) × C, tL = 0.693 × R2 × C, f = 1 ÷ (tH + tL); monostable tw = 1.1 × R × C.
- With the diode, tH = 0.693 × R1 × C; the diode drop is ignored, so the real high time is a little longer.
- Suggestions use E6 capacitors from 1 nF to 100 µF and resistors of at least 1 kΩ with R1 + R2 up to 3.4 MΩ. Part tolerances usually matter more than the listed error.
Questions and answers
- What frequency do R1 = 1 kΩ, R2 = 10 kΩ and C = 10 nF give?
- About 6.87 kHz with a duty cycle of 52.4 %: the high time is 76.2 µs and the low time 69.3 µs.
- Why can't the standard 555 astable reach 50 % duty?
- The capacitor charges through R1 and R2 but discharges through R2 only, so the high time is always longer. Add a diode across R2 or invert the output.
- How do I get a 1 second pulse from a 555?
- Make 1.1 × R × C = 1 s, for example 91 kΩ with 10 µF.
- What does pin 5 of the 555 do?
- Pin 5 (CTRL) reaches the internal two-thirds reference. Unused, it is usually decoupled to ground; the datasheet's example uses 0.01 µF.