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555 timer calculator: frequency, duty cycle and pulse width

This 555 timer calculator uses the formulas from the Texas Instruments NE555 data sheet. In astable mode RA = 1 kΩ, RB = 10 kΩ and C = 10 µF give a period of 145.5 ms, i.e. 6.871 Hz with a duty cycle of 52.4 %; in monostable mode 100 kΩ and 10 µF give a 1.100 s pulse.

Example: Mode: astable (oscillator) · RA (monostable: R) 1 kΩ · RB 10 kΩ · Capacitor C 10 · Capacitor unit: µF → Result: 6.871 Hz – period 145.5 ms, duty cycle 52.4 %. Source: Texas Instruments – NA555, NE555, SA555, SE555 Precision Timers, data sheet SLFS022K (rev. March 2026), sections 6.3.1 (monostable) and 6.3.2 (astable), equations 1–6, and the note on the maximum RA + RB (3.4 MΩ at 5 V, 10 MΩ at 15 V) in the electrical characteristics (read 09.10.2026). Updated: .

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kΩ
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Result

Result
6.871 Hz – period 145.5 ms, duty cycle 52.4 %
Details
ItemValue
Frequency f = 1 ÷ T6.871 Hz
Period T = tH + tL145.5 ms
High time tH76.23 ms
Low time tL69.30 ms
Duty cycle (output high) tH ÷ T52.4 %
Output driver duty cycle tL ÷ T47.6 %
FormulatH = 0.693 × (RA + RB) × C, tL = 0.693 × RB × C
Calculated value from the data sheet approximations; resistor and capacitor tolerances (often ±5–20 %) shift the real timing.

How it is calculated

Astable: f = 1 ÷ (0.693 × (RA + 2RB) × C). The capacitor charges through RA and RB (high time tH = 0.693 × (RA + RB) × C) and discharges through RB only (low time tL = 0.693 × RB × C). The data sheet also writes the frequency as f ≈ 1.44 ÷ ((RA + 2RB) × C); for the default values that gives 6.857 Hz instead of 6.871 Hz – both are approximations of the same circuit.

Examples (astable)

RARBCFrequencyDuty cycle
1 kΩ10 kΩ10 µF6.871 Hz52.4 %
1 kΩ68 kΩ10 µF1.053 Hz50.4 %
10 kΩ100 kΩ10 nF687.1 Hz52.4 %
4.7 kΩ4.7 kΩ100 nF1.023 kHz66.7 %

In the first row tH is 76.23 ms and tL 69.30 ms. Because the capacitor charges through RA + RB but discharges through RB alone, the output is always high for more than half the period; a small RA and a large RB bring it close to 50 %.

Monostable (one-shot)

A trigger pulse starts a single output pulse of tw = 1.1 × R × C. 100 kΩ with 10 µF gives 1.100 s, 1 MΩ with 100 µF gives 110.0 s and 10 kΩ with 1 nF gives 11.00 µs – close to the minimum of about 10 µs. The timing does not depend on the supply voltage as long as it stays constant.

Limits from the data sheet

TI recommends a maximum of 100 kHz with the bipolar NE555 and gives about 3.4 MΩ for RA + RB at 5 V (10 MΩ at 15 V). For the RC time constant on its own use the RC filter cutoff frequency calculator; for capacitors in series or parallel the capacitor calculator.

Frequently asked questions

How do I calculate the frequency of a 555 timer?

f = 1 ÷ (0.693 × (RA + 2RB) × C). With 1 kΩ, 10 kΩ and 10 µF: 0.693 × 21 kΩ × 10 µF = 145.5 ms, so 6.871 Hz.

How do I get a 1 Hz blinker?

RA = 1 kΩ, RB = 68 kΩ and C = 10 µF give 1.053 Hz with a duty cycle of 50.4 % – close enough for an LED blinker.

What is the formula for monostable mode?

tw = 1.1 × R × C. 100 kΩ and 10 µF give a 1.100 s pulse.

Why is the duty cycle always above 50 %?

The capacitor charges through RA + RB but discharges through RB only, so tH is always longer than tL. With RA = RB the duty cycle is 66.7 %.

Is 1.44 ÷ ((RA + 2RB) × C) the same formula?

Yes, 1.44 is the rounded reciprocal of 0.693. The calculator uses 0.693 as in the period equations of the data sheet.

How exact is the result?

It is a calculated estimate. Component tolerances of 5–20 % and the chip itself shift the real frequency; trim with a potentiometer if you need an exact value.

Sources and legal basis

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