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555 Timer Calculator

A 555 timer makes a steady blinking signal (astable mode) or a single timed pulse (monostable mode). In astable mode the frequency is 1 ÷ (ln 2 × (R1 + 2 × R2) × C), and in monostable mode the pulse lasts ln 3 × R × C, about 1.1 × R × C.

Result

Output frequency

6.87 Hz

Time HIGH per cycle: 76.25 ms
Time LOW per cycle: 69.31 ms
Period (one full cycle): 145.6 ms
Duty cycle (share of time HIGH)always above 50 % for a basic 555 astable: 52.38 %

The formula

Astable: time the output is HIGH

\htmlData{sym=t_HIGH}{t_{\mathrm{HIGH}}} = \ln 2\,(\htmlData{sym=R_1}{R_1} + \htmlData{sym=R_2}{R_2})\,\htmlData{sym=C}{C}

Astable: time the output is LOW

\htmlData{sym=t_LOW}{t_{\mathrm{LOW}}} = \ln 2\,\htmlData{sym=R_2}{R_2}\,\htmlData{sym=C}{C}

Astable: frequency

\htmlData{sym=f}{f} = \dfrac{1}{\htmlData{sym=t_HIGH}{t_{\mathrm{HIGH}}} + \htmlData{sym=t_LOW}{t_{\mathrm{LOW}}}} = \dfrac{1}{\ln 2\,(\htmlData{sym=R_1}{R_1} + 2\htmlData{sym=R_2}{R_2})\,\htmlData{sym=C}{C}}

Astable: duty cycle

\htmlData{sym=D}{D} = \dfrac{\htmlData{sym=R_1}{R_1} + \htmlData{sym=R_2}{R_2}}{\htmlData{sym=R_1}{R_1} + 2\htmlData{sym=R_2}{R_2}}

Monostable: pulse length

\htmlData{sym=T:pulse}{T} = \ln 3\,\htmlData{sym=R}{R}\,\htmlData{sym=C}{C}

What each letter means

$t_{\mathrm{HIGH}}$: Time the output is HIGH (s · second)
$R_1$: Resistance of R1 (Ω · ohm) : How strongly something opposes current.
$R_2$: Resistance of R2 (Ω · ohm) : How strongly something opposes current.
$C$: Capacitance (F · farad) : How much charge a capacitor stores per volt.
$t_{\mathrm{LOW}}$: Time the output is LOW (s · second)
$f$: Frequency (Hz · hertz) : How many times per second something repeats.
$D$: Duty cycle (no unit) : The fraction of the time a switch is on, from 0 to 1 (0 % to 100 %). A ratio, so it has no unit.
$T$: Pulse length (s · second) : How long the single output pulse lasts.
$R$: Resistance (Ω · ohm) : How strongly something opposes current.

How it works

The capacitor charges and discharges between one third and two thirds of the supply voltage. In astable mode it charges through R1 and R2 (the output is HIGH) and discharges through R2 alone (the output is LOW). Charging takes the longer path, so the output is always HIGH for longer than it is LOW: the duty cycle is always above 50 %, and it gets close to 50 % only when R1 is much smaller than R2.

In monostable mode a trigger starts the capacitor charging through a single resistor, and the output goes LOW again when it reaches two thirds of the supply. The time it takes is ln 3 × R × C. The 555 switches at one third and two thirds of its own supply, so the supply voltage cancels out of the timing. Real capacitors and resistors have tolerances, so expect the real frequency to be a few percent away.

Worked example

Astable with R1 = 1 kΩ, R2 = 10 kΩ and C = 10 µF.
HIGH time: ln 2 × (1 kΩ + 10 kΩ) × 10 µF = 76.25 ms. LOW time: ln 2 × 10 kΩ × 10 µF = 69.31 ms.
The period is 145.6 ms, so the frequency is 6.87 Hz and the duty cycle is 76.25 ms ÷ 145.6 ms = 52.38 %.
Monostable with R = 100 kΩ and C = 10 µF: the pulse lasts ln 3 × 100 kΩ × 10 µF = 1.099 s.

Common questions

Why can the duty cycle never go below 50 %?

In the basic circuit the capacitor charges through R1 + R2 but discharges only through R2, so the HIGH time is always longer than the LOW time. To get close to 50 % make R1 much smaller than R2 (but not below about 1 kΩ).

How do I get exactly 50 %?

Run the 555 at twice the frequency you want and divide it by two with a toggle flip-flop, which always gives an exact 50 % square wave. Another common trick is to add a diode across R2 so the capacitor charges through R1 only.

Does the supply voltage change the frequency?

Only slightly. The 555 switches at one third and two thirds of its supply, so the supply cancels out of the formula. The part’s tolerances and the capacitor’s leakage matter more.

Try it live

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