How to use it
- Choose Mode: Astable (repeating) for a clock or blinking LED, Monostable (one pulse) for a single timed pulse.
- Type Resistor R1 and, for astable, Resistor R2. In monostable mode R1 is the timing resistor R.
- Type the Timing capacitor C, for example 10µ or 100n.
- Read Frequency and Duty cycle in astable mode, or Pulse width in monostable mode. Open Working, step by step to check the numbers.
The formulas
The NE555 datasheet from Texas Instruments gives the timing of both modes. In astable mode the capacitor charges through R1 and R2 and discharges through R2 alone, so the high time is longer than the low time. The output is high for more than half of each cycle, because R1 is always in the charging path. In monostable mode the capacitor charges through R once per trigger, and the output stays high for 1.1 RC.
Astable:
f = 1.44 ÷ ((R1 + 2 R2) × C)
tH = 0.693 × (R1 + R2) × C
tL = 0.693 × R2 × C
Duty cycle = (R1 + R2) ÷ (R1 + 2 R2)
Monostable:
tw = 1.1 × R × C- Frequency and period are inverses (period = 1 ÷ f). The high and low times should add up to the same period; with the two constants above they agree to about 0.2 %.
- The duty cycle of a plain astable circuit is always more than 50 %. A diode across R2 changes that; this calculator does not include it.
- Constants: 1.44 and 0.693 (ln 2) are the values the datasheet uses; 1.1 is the monostable constant, about ln 3.
What this calculator does not cover
- The limits on R1, R2 and C given in the datasheet's recommended operating conditions. Check them for your values; very small resistors or very large capacitors move the timing away from these formulas.
- Timing tolerance: the resistors' and capacitor's tolerances and temperature changes move the frequency by several per cent.
- The CMOS versions (such as TLC555) and other timer ICs: their constants are different.
Worked example: astable with 10 kΩ, 10 kΩ and 10 µF
f = 1.44 ÷ ((10 k + 2 × 10 k) × 10 µF) = 1.44 ÷ 0.3 = 4.8 Hz, so the period is 208.3 ms.
The high time is tH = 0.693 × (10 k + 10 k) × 10 µF = 138.6 ms and the low time is tL = 0.693 × 10 k × 10 µF = 69.3 ms. Their sum, 207.9 ms, is 0.2 % shorter than 208.3 ms from the frequency, because the constants 0.693 and 1 ÷ 1.44 = 0.6944 differ slightly.
The duty cycle is (10 k + 10 k) ÷ (10 k + 2 × 10 k) = 66.67 %. For monostable mode with the same R and C, the pulse width is 1.1 × 10 k × 10 µF = 110 ms.
Frequently asked questions
How do I get a 50 % duty cycle from a 555 astable?
A plain astable cannot reach 50 %, because R1 is always in the charging path. A diode placed across R2 lets the capacitor charge through R1 alone, so with R1 close to R2 the duty cycle moves towards 50 %. The formulas here do not include the diode.
Why does the output frequency differ from the calculated one?
Resistor and capacitor tolerances, temperature and the supply voltage all shift the timing, so expect a difference of a few per cent in a real circuit.
What is the difference between R1 and R2 in the astable circuit?
R1 sits between the supply and the discharge pin, and R2 between the discharge pin and the threshold and trigger pins. R1 sets only the charging time, R2 sets both the charging and discharging times.
What does 1.1 RC mean in monostable mode?
It is the pulse width: the output stays high for about 1.1 times the product of the resistance and the capacitance after a trigger. Ten kilohms and 10 µF give a pulse of 110 ms.
Can I use this calculator for a 555 with a different supply voltage?
Yes, the timing formulas do not depend on the supply voltage, but the output levels and the limits on R and C in the datasheet do. Check the datasheet for your supply range.
Sources
- Texas Instruments: NE555, SA555, SE555 Precision Timers datasheet (SLFS022)
- Wikipedia: 555 timer IC (the same formulas, citing the NE555 datasheet)
Figures and text last checked on Tuesday, November 10, 2026.