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555 Timer Astable Frequency & Duty Cycle Calculator

Calculate the output frequency, period, and duty cycle of a 555 timer wired in astable (free-running oscillator) mode from its R1, R2, and C values.

Frequency
48.09 Hz
Duty cycle
66.67%
Output high time
13.8629 ms
Output low time
6.9315 ms
Period
20.7944 ms

How it works

In astable mode, the 555's timing capacitor repeatedly charges through R1+R2 from 1/3 to 2/3 of the supply voltage (output high), then discharges through R2 alone back down from 2/3 to 1/3 (output low) — the standard NE555 datasheet derivation. The time to cross that fixed 1/3-to-2/3 window of an RC charge/discharge curve is exactly ln(2) × R × C, so high time = ln(2)(R1+R2)C, low time = ln(2)(R2)C, period T = ln(2)(R1+2R2)C, and frequency f = 1/T.

Duty cycle is high time ÷ period = (R1+R2) ÷ (R1+2R2). Because the charge path (R1+R2) is always longer than the discharge path (R2 alone) whenever R1 is greater than zero, a plain 555 astable circuit can never quite reach a 50% duty cycle — it always runs somewhat 'on' more than 'off'.

R1 and R2 are entered in kilohms and C in microfarads (the sizes real 555 circuits typically use) and converted to ohms/farads internally before the formula runs.

FAQ

Why can't this circuit reach exactly 50% duty cycle?

Because the capacitor always charges through R1+R2 but discharges through R2 alone, so the high half of the cycle is inherently longer than the low half whenever R1 > 0 — duty cycle only approaches 50% as R1 shrinks toward zero relative to R2, never quite reaching it. A common workaround in real circuits is a diode across R1 that bypasses it during discharge, which this calculator doesn't model since that's a different circuit topology with its own formula.

Does making R1 and R2 equal give any particular duty cycle?

Yes — whenever R1 = R2 (any value), duty cycle works out to exactly 2/3, or about 66.67%, regardless of the actual resistance or capacitance chosen, since the ratio (R1+R2)/(R1+2R2) simplifies to 2R/3R.

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