555 Astable Frequency

f = \frac{1}{\ln 2 \, (R_{1} + 2R_{2}) C}

f =

R₁ =

R₂ =

C =

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Constant used — built into this formula, no need to enter

\ln 2 = 0.6931471805599453

Natural Logarithm of Two · exact

Learning zone

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 V_CC 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

f = \frac{1}{\ln 2 \, (R_{1} + 2R_{2}) C}

Where

$f$ = Output frequency
$R_{1}$ = Resistor 1 (supply to discharge)
$R_{2}$ = Resistor 2 (discharge to threshold)
$C$ = Timing capacitor

Missing one of these? Work it out first, then come back

Output frequency — Power-Factor Correction Capacitance, Synchronous Speed from Frequency and Poles
Resistor 1 (supply to discharge) — Conductor Resistance Temperature Correction, Voltage Divider
Resistor 2 (discharge to threshold) — Conductor Resistance Temperature Correction, Voltage Divider
Timing capacitor — Power-Factor Correction Capacitance, Q Factor of a Series Resonant Circuit