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555 Timer Calculator

The 555 has outlasted almost every chip of its generation because two resistors and a capacitor are all it needs. Astable mode gives a continuous square wave; monostable gives one pulse per trigger.

555 Timer

Free-running oscillator: the output toggles continuously. Frequency and duty cycle follow from R1, R2 and C.

Frequency
Period
Time high
Time low
Duty cycle
  1. 1.Given: R1 = 1 kΩ, R2 = 10 kΩ, C = 10 µF
  2. 2.Formulat(high) = 0.693 × (R1 + R2) × CSubstitutet(high) = 0.693 × 11000 × 10e-6Resultt(high) = 76.2462 ms
  3. 3.Formulat(low) = 0.693 × R2 × CSubstitutet(low) = 0.693 × 10000 × 10e-6Resultt(low) = 69.3147 ms
  4. 4.FormulaT = t(high) + t(low), f = 1 ÷ TSubstituteT = 76.2462 ms + 69.3147 ms = 145.5609 msResultf = 6.87 Hz
  5. 5.FormulaDuty = t(high) ÷ T × 100SubstituteDuty = 76.2462 ms ÷ 145.5609 ms × 100ResultDuty = 52.381 %
Common trap: The basic astable circuit can't reach 50% duty — high time always exceeds low time because C charges through R1 + R2 but discharges through R2 alone. For a true square wave, put a diode across R2 or make R2 ≫ R1.

The formula

Astable: t_high = 0.693(R₁+R₂)C, t_low = 0.693 R₂C, f = 1/(t_high+t_low) | Monostable: t = 1.1 R C

R₁, R₂
timing resistors (ohms)
C
timing capacitor (farads)
f
output frequency (hertz)
duty
fraction of each cycle spent high (per cent)

Worked example

Astable with R₁ = 10 kΩ, R₂ = 10 kΩ, C = 100 nF.

  1. t_high = 0.693 × (10000 + 10000) × 100 × 10⁻⁹ = 1.386 ms
  2. t_low = 0.693 × 10000 × 100 × 10⁻⁹ = 0.693 ms
  3. Period = 1.386 + 0.693 = 2.079 ms, so f = 481 Hz
  4. Duty = 1.386 / 2.079 = 66.7 %

481 Hz at 66.7 % duty — the classic 555 astable, which can never quite reach 50 % without a diode.

Where you'll use it

Blinkers, tone generators, PWM and timing delays. The duty cycle is always above 50 % in the standard astable because the capacitor charges through R₁+R₂ but discharges through R₂ alone.

The laws behind it

Parts this applies to

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