NE555 astable timer calculator
Frequency, period and duty cycle of an NE555 multivibrator from two resistors and a capacitor.
Fill in the fields and the result will appear here automatically.
Calculate nominal frequency, period and high/low times of the classic NE 555 astable circuit. The capacitor charges through R1+R2 and discharges through R2, so positive resistors give a high-level fraction strictly between 50% and 100%. The displayed percentage is tH/T; the reciprocal T/tH is a different quantity.
How it works
Formula and logic
tH=ln 2·(R1+R2)C, tL=ln 2·R2 C, T=tH+tL, f=1/T, D=100·tH/T. Enter R1, R2 in kΩ and C in nF; conversion to Ω and F is internal. Ideal thresholds are Vcc/3 and 2 Vcc/3. Frequency uses ln 2 rather than the rounded 1.44 coefficient. The rounded display can show 50% or 100% even though the exact model with positive R1 and R2 does not reach either boundary.
Example
10 kΩ, 47 kΩ and 100 nF give 138.72 Hz, 7.209 ms, tH=3.951 ms, tL=3.258 ms and D=54.808%. Doubling C doubles both times and halves frequency without changing D. R1=0 gives a formal 50% limit but is rejected because this selected circuit requires positive R1.
Fields and units
- Resistance R1 — kΩ
- Resistance R2 — kΩ
- Capacitance C — nF
How to use
- — R1 connects Vcc to pin 7; R2 connects pin 7 to the tied pins 2/6; C connects that node to ground.
- — Enter resistance in kΩ and capacitance in nF.
- — Compare frequency and loading with documentation for your specific 555 variant.
- — Diodes, flip-flops and other circuit modifications need new timing equations and are not modelled.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- TI NE555 RevK: classic circuit, nominal times and waveform high-level fraction
- Limitation
- Nominal classic NE 555 timing with ideal thresholds; no guaranteed frequency, load capability or temperature stability.
FAQ
Why is the high-level fraction greater than 50%?
Because the capacitor charges through R1 and R2 but discharges through R2 alone. The charging time is always longer, and exactly fifty per cent is only approached in the limit where R1 is far smaller than R2.
How can the high-level fraction be below half?
A different pulse-forming circuit is needed. Modified charging/discharging paths or frequency division can change D; these equations apply only to the classic circuit without a bypass diode.
Why does ln 2 appear in the formula?
With nominal thresholds at 1/3 and 2/3 supply, the exponential transition takes RC·ln 2. Supply cancels in the ideal formula; real thresholds, discharge voltage and timing depend on operating conditions.
Why does the real frequency differ?
R, C tolerances, leakage, actual thresholds, discharge-switch voltage, supply, temperature and loading matter. Component datasheets define their bounds; the calculation does not guarantee frequency or operation outside ratings.