555 Timer Calculator: Astable Frequency and Duty Cycle

Pick astable or monostable, enter the timing resistors and capacitor, and get the frequency and duty cycle or the pulse width.

Calculator

Frequency f
6.87 Hz
Period T
145.6 ms
High time tH
76.25 ms
Low time tL
69.31 ms
Duty cycle
52.38%

Type numbers like 4.7, 0.002 or 2e-3. Results are rounded to 4 significant figures.

The formulas

tH=ln⁡2 (R1+R2) CtL=ln⁡2 R2 C\begin{gathered} t_H = \ln 2\,(R_1 + R_2)\,C \\ t_L = \ln 2\,R_2\,C \end{gathered}

Astable: the capacitor charges through R1 and R2 (output high) and discharges through R2 alone (output low). ln 2 ≈ 0.693.

f=1ln⁡2 (R1+2R2) C≈1.44(R1+2R2) Cf = \dfrac{1}{\ln 2\,(R_1 + 2R_2)\,C} \approx \dfrac{1.44}{(R_1 + 2R_2)\,C}

The duty cycle tH / (tH + tL) = (R1 + R2) / (R1 + 2R2), always above 50% in this basic circuit.

tw=ln⁡3 R C≈1.1 RCt_w = \ln 3\,R\,C \approx 1.1\,RC

Monostable: the output stays high while the capacitor charges from 0 to 2/3 of VCC.

  • The 1/3 and 2/3 VCC thresholds make the timing independent of the supply voltage.
  • For a duty cycle below 50%, add a diode across R2 so the capacitor charges through R1 only.
  • Keep R1 at 1 kΩ or more so the discharge transistor isn't overloaded.

Worked example: R1 = 1 kΩ, R2 = 10 kΩ, C = 10 µF

  1. R1 + 2R2 = 1 kΩ + 20 kΩ = 21 kΩ.
  2. Period T = 0.6931 × 21 000 Ω × 10 µF = 0.1456 s, so f = 1 / 0.1456 s = 6.870 Hz.
  3. High time tH = 0.6931 × 11 kΩ × 10 µF = 76.25 ms; low time tL = 0.6931 × 10 kΩ × 10 µF = 69.31 ms.
  4. Duty cycle = 76.25 / 145.6 = 52.38%. An LED on the output blinks about 7 times a second.
  5. Monostable with R = 100 kΩ and C = 10 µF: tw = 1.0986 × 100 kΩ × 10 µF = 1.099 s.

Learn the topic

Sources