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555 Astable Frequency

f=1ln⁡2 (R1+2R2)Cf = \frac{1}{\ln 2 \, (R_{1} + 2R_{2}) C}

Worked example: 10 kΩ, 10 kΩ, 10 nF → 4.81 kHz — press Try an example to run it live, then adjust anything.

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Constant used — built into this formula, no need to enter
ln⁡2=0.6931471805599453\ln 2 = 0.6931471805599453Natural Logarithm of Two · exact
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555 Astable Frequency explained

R₁R₂C555f

The 555 charges C through R₁ + R₂ from ⅓ to ⅔ of the supply, then discharges it through R₂ alone back to ⅓. Each leg takes ln 2 ≈ 0.693 time constants, so the period is ln2·(R₁ + 2R₂)C and the frequency its reciprocal. Datasheets print the constant as 1.44 (that is 1/ln 2 rounded), which is where the familiar f = 1.44/((R₁ + 2R₂)C) comes from. With R₁ = R₂ = 10 kΩ and C = 10 nF the output runs at about 4.8 kHz.

Notice what is missing: the supply voltage. The thresholds are ratios of VCC set by an internal three-resistor divider, so the timing is immune to supply changes — the design insight that made Hans Camenzind's 1971 chip the best-selling IC of all time, with billions shipped. The catch is duty cycle: the charge path is always longer than the discharge path, so plain astable output is always above 50%. Making R₂ ≫ R₁ approaches it, and a diode across R₂ splits the paths for true symmetry.

555 Astable Frequency formula

f=1ln⁡2 (R1+2R2)Cf = \frac{1}{\ln 2 \, (R_{1} + 2R_{2}) C}
Where
  • ff= Output frequency (Hz)
  • R1R_{1}= Resistor 1 (supply to discharge) (Ω)
  • R2R_{2}= Resistor 2 (discharge to threshold) (Ω)
  • CC= Timing capacitor (μF)

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