What does this calculator do?
In astable mode the 555 oscillates continuously: the timing capacitor charges and discharges between two internal thresholds, and the output toggles each time a threshold is crossed.
f ≈ 1.44 / ((R1 + 2R2)·C)
Enter the values in the units shown. The calculator converts units automatically and displays the result in practical units.
Worked example
R1 = 1 kΩ, R2 = 10 kΩ and C = 0.1 µF give about 686 Hz with a 52 % duty cycle.
Practical tips
- The formula is approximate: component tolerances shift the real frequency.
- For a duty cycle close to 50 %, add a diode so the capacitor charges and discharges through different resistors.
- Respect the current limits of the DISCHARGE pin and of the capacitor.
Common mistake to avoid
Real 555 parts and extreme R/C values limit accuracy. Treat the result as a starting point and measure the final circuit.
Complete 555 astable formulas
In the astable circuit the capacitor charges through R1 + R2 but discharges through R2 only, so the high time is always longer than the low time:
- High time: tH = 0.693 · (R1 + R2) · C
- Low time: tL = 0.693 · R2 · C
- Period: T = tH + tL = 0.693 · (R1 + 2R2) · C
- Frequency: f = 1 / T ≈ 1.44 / ((R1 + 2R2) · C)
- Duty cycle: D = (R1 + R2) / (R1 + 2R2) × 100 %
The 0.693 factor is ln 2: the capacitor swings between 1/3 and 2/3 of Vcc, and each charge or discharge leg lasts ln 2 time constants. The 1.44 constant is simply 1 / 0.693.
Typical values
| R1 | R2 | C | Frequency | Duty cycle | Typical use |
|---|---|---|---|---|---|
| 10 kΩ | 68 kΩ | 10 µF | ≈ 0.99 Hz | 53 % | 1-second LED blinker |
| 4.7 kΩ | 47 kΩ | 1 µF | ≈ 14.6 Hz | 52 % | Fast blink, testing |
| 1 kΩ | 1 kΩ | 10 µF | ≈ 48 Hz | 67 % | Low-frequency signal |
| 1 kΩ | 10 kΩ | 0.1 µF | ≈ 686 Hz | 52 % | Audible tone, buzzer |
| 1 kΩ | 6.8 kΩ | 10 nF | ≈ 9.9 kHz | 53 % | Clock, basic PWM |
Values come from the ideal formula. In a real circuit they vary with capacitor tolerance, which is often ±10 % or worse for electrolytics.
How to design a 555 astable step by step
- Pick the capacitor C first (for example 10 nF, 100 nF or 10 µF): there are fewer standard capacitor values than resistor values.
- Make R1 small, but not below about 1 kΩ, so the DISCHARGE pin is not overloaded.
- Solve for R2: R2 = (1.44 / (f · C) − R1) / 2.
- Round R2 to the nearest standard value and check the new frequency with this calculator.
- Check the duty cycle. If you need less than 50 %, use the diode variant below.
How do I get a duty cycle of 50 % or less?
With the classic circuit the duty cycle can never drop below 50 %, because charging goes through R1 + R2 and discharging only through R2. The usual fix is a diode in parallel with R2 (anode towards the DISCHARGE pin): the capacitor then charges through R1 only and discharges through R2. In that case tH ≈ 0.693 · R1 · C and tL ≈ 0.693 · R2 · C, so R1 = R2 gives roughly 50 % (slightly more because of the diode drop).
Frequently asked questions
Why doesn’t my measured frequency match the calculated one?
Because of capacitor and resistor tolerances, electrolytic leakage and differences between 555 versions. A stable film or C0G ceramic capacitor improves accuracy.
What should I do with pin 5 (CONTROL)?
If you don’t use it, connect it to ground through a capacitor of about 10 nF. This reduces noise and keeps the thresholds stable.
What is the maximum frequency of an NE555?
The bipolar NE555 works well up to a few hundred kHz. For higher frequencies or low power, use a CMOS version such as the TLC555 or LMC555, and always check the datasheet.
Can I connect an LED directly to the output?
Yes, with a series resistor. Work out its value with the LED resistor calculator.
Educational tool. Check tolerances, datasheets, standards and real operating conditions before using the results in critical applications.