How to use the 555 Timer Calculator
- 1. Pick a modeAstable analysis, astable design or monostable.
- 2. Enter R and C valuesUse SI prefixes like 10k and 10u.
- 3. See the timingsHigh time, low time, frequency, duty cycle and nearest E24 values.
555 timer formulas
In astable mode the capacitor charges through R1 + R2 and discharges through R2 only:
- tH = 0.693 (R1 + R2) C
- tL = 0.693 R2 C
- f = 1 / (tH + tL) = 1.44 / ((R1 + 2R2) C)
- Duty cycle D = (R1 + R2) / (R1 + 2R2)
In monostable mode, one trigger gives one pulse of width T = 1.1 RC. The constant 0.693 is ln 2 and 1.1 is close to ln 3.
Worked example: R1 = 1 kΩ, R2 = 10 kΩ, C = 10 µF. tH = 0.693 × 11k × 10µ = 76.2 ms, tL = 0.693 × 10k × 10µ = 69.3 ms, T = 145.5 ms, f = 6.87 Hz and duty cycle = 11/21 = 52.4%.
Design tips and exam questions
With the basic astable circuit, the duty cycle is always above 50% because tH includes R1. For 50% or less, put a diode across R2 so the capacitor charges through R1 only; then tH = 0.693 R1 C and tL = 0.693 R2 C. The design mode handles both cases.
- Design an astable at 1 kHz with 60% duty cycle using C = 10 nF.
- Find the pulse width of a monostable with R = 47 kΩ, C = 10 µF (about 0.52 s).
- Why can the duty cycle of the basic astable not go below 50%?
Keep R1 at 1 kΩ or more so the discharge transistor is not overloaded.
Frequently asked questions
Why is it 0.693 and 1.1?
The capacitor swings between 1/3 and 2/3 of VCC in astable mode, which takes ln 2 ≈ 0.693 time constants. In monostable mode it charges from 0 to 2/3 VCC, which takes ln 3 ≈ 1.1 time constants.
How do I get exactly 50% duty cycle?
Add a diode across R2 and make R1 equal to R2. Then charge and discharge both happen through equal resistors.
Will my circuit match the result exactly?
Close, but not exactly. Capacitor tolerance (often ±10% to ±20%) and the chip's own variation shift the frequency by a few percent.
Is this tool free?
Yes. You enter your name and WhatsApp number once to see your result, and all other Uncodemy tools then open without the form.