555 Timer Calculator (astable & monostable)

Understanding the 555 timer calculator — concept and principle

The 555 timer is a classic integrated circuit that produces timed pulses and square waves using only a capacitor and one or two resistors. This calculator gives the frequency, duty cycle and high and low times in astable (free-running oscillator) mode, plus the pulse width in monostable (one-shot) mode, for blinkers, tone generators, debounce circuits and delays.

The capacitor charges through R1 + R2 and discharges through R2 alone, so high lasts a bit longer.

How it is calculated

Astable mode, with R1 between the supply and the discharge pin and R2 between discharge and the capacitor:

High time: t_H = ln 2 × (R1 + R2) C ≈ 0.693 (R1 + R2) C.
Low time: t_L = ln 2 × R2 C ≈ 0.693 R2 C.
Frequency: f = 1 / (t_H + t_L) = 1 / (ln 2 × (R1 + 2R2) C) ≈ 1.44 / ((R1 + 2R2) C).
Duty cycle: (R1 + R2) / (R1 + 2R2) × 100%.

Monostable mode, using R1 as the timing resistor: t = ln 3 × R1 C ≈ 1.1 R1 C. The calculator uses the exact ln 2 and ln 3, not the rounded constants.

Resistances are in ohms and C in microfarads (converted to farads by × 10⁻⁶), so times are in seconds.

The principle

Inside the 555, two comparators watch the capacitor voltage against thresholds of one third and two thirds of the supply. In astable mode the capacitor charges through R1 + R2 from 1/3 to 2/3 V_cc, then the discharge transistor drains it through R2 alone back to 1/3 V_cc. An RC circuit charging toward a fixed voltage covers that half-way gap in exactly ln 2 ≈ 0.693 time constants, which is where 0.693 comes from, and 1.44 is approximately 1 / 0.693. In monostable mode the capacitor charges from 0 to 2/3 V_cc, which takes ln 3 ≈ 1.1 time constants.

Worked example

C = 10 µF = 10⁻⁵ F; R1 + 2R2 = 1,000 + 20,000 = 21,000 Ω.
High time: ln 2 × 11,000 × 10⁻⁵ = 0.0762461899 s; low time: ln 2 × 10,000 × 10⁻⁵ = 0.0693147181 s.
Period 0.1455609 s, so frequency = 1 ÷ 0.1455609 = 6.869976385 Hz.
Duty cycle: 11,000 / 21,000 ≈ 52.38%.
Monostable pulse: ln 3 × 1,000 × 10⁻⁵ = 0.0109861229 s.

The shortcut 1.44 / 0.21 gives 6.857 Hz; the small gap comes only from rounding 1/ln 2 to 1.44.

Each point is the calculator re-run with only “C (µF)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

C (µF)	Astable frequency (Hz)
2.5	27.47990554
5	13.73995277
7.5	9.159968514
10	6.869976385
12.5	5.495981108
15	4.579984257
20	3.434988193

How to read the result

In the basic astable circuit the duty cycle is always above 50%, because the capacitor charges through both resistors but discharges through R2 only. Making R2 much larger than R1 brings it close to 50%; for lower duty cycles, add a diode across R2. Electrolytic capacitors are often ±20%, so the real frequency can differ as much.

Common mistakes

Entering C in farads or nanofarads instead of microfarads.
Making R1 very small, which overloads the discharge transistor; values of about 1 kΩ or more are common.
Expecting a duty cycle below 50% without the diode modification.

Keep in mind: The formulas describe the bipolar NE555 with ideal components. CMOS versions, capacitor leakage and tolerance, and temperature change the timing, so trim or measure where accuracy matters.

Frequently asked questions

How do I get exactly 50% duty cycle?

Add a diode across R2 so charging bypasses it, then make R1 equal to R2; diode drop makes it approximate.

Does supply voltage change the frequency?

Very little, since thresholds are fixed fractions of the supply.

Sources: Texas Instruments, NE555 / SA555 / SE555 Precision Timers datasheet · Horowitz and Hill, The Art of Electronics, 3rd edition

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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