555 timer calculator (astable and monostable)
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Result
In the standard astable circuit the duty cycle is always above 50%, because the capacitor charges through R₁ + R₂ but discharges through R₂ only. A diode across R₂ allows duty cycles below 50%.
Design a 555 timer circuit. In astable mode the calculator gives the frequency, the high and low times, the period and the duty cycle from R₁, R₂ and C; in monostable mode it gives the pulse width from R and C. Capacitors can be entered in µF, nF or pF, and every time is shown in a suitable unit.
Step-by-step solution
Charging and discharging times
Frequency and duty cycle
Key formulas
- Astable high and low time
- tH = 0.693·(R1 + R2)·C, tL = 0.693·R2·C
- Astable frequency
- f = 1/(tH + tL) ≈ 1.44/((R1 + 2R2)·C)
- Duty cycle
- D = tH/T = (R1 + R2)/(R1 + 2R2)
- Monostable pulse width
- T = ln 3·R·C ≈ 1.1·R·C
How to calculate the frequency of a 555 timer
- Astable: the capacitor charges through R₁ + R₂ and discharges through R₂, between ⅓ and ⅔ of the supply voltage.
- 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) ≈ 1.44/((R₁ + 2R₂) C). Duty cycle: t_H/(t_H + t_L).
- Monostable: the output stays high for T = ln 3 × R × C ≈ 1.1 RC after a trigger.
Questions and answers
Why is the duty cycle always above 50%?
In the standard circuit the capacitor charges through R₁ + R₂ but discharges through R₂ only, so the high time is always longer. A diode across R₂ (anode at pin 7) bypasses R₂ while charging and allows duty cycles below 50%.
Does the supply voltage change the frequency?
Ideally not: the thresholds are fixed fractions of the supply, so the timing depends only on R and C. In practice the frequency drifts slightly with voltage and temperature.
What values of R and C are sensible?
Keep R₁ and R₂ above about 1 kΩ and below about 1 MΩ, and C above about 100 pF. Large electrolytic capacitors have wide tolerances and leakage, which make long times inaccurate.
Where do 0.693 and 1.1 come from?
The capacitor voltage changes exponentially. Charging from ⅓ to ⅔ of the supply takes RC·ln 2 = 0.693 RC; charging from 0 to ⅔ takes RC·ln 3 = 1.0986 RC.
Practise it yourself
Practise Diodes & transistors
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