What does this calculator do?
In monostable mode, each trigger produces a single output pulse whose length depends on the timing resistor and capacitor.
t ≈ 1.1·R·C
Enter the values in the units shown. The calculator converts units automatically and displays the result in practical units.
Worked example
R = 100 kΩ and C = 10 µF produce a pulse of about 1.1 s.
Practical tips
- Use stable components if you need accurate timing.
- Electrolytic capacitors can leak enough to stretch long delays noticeably.
- Check the trigger and reset levels of the specific 555 you are using.
Common mistake to avoid
Capacitor tolerance usually dominates the timing error. Take leakage and dielectric type into account.
How the 555 monostable works
In monostable mode the 555 produces one pulse of fixed length every time it is triggered. While idle, the output is low and the capacitor is held discharged. When pin 2 (TRIGGER) drops below 1/3 of Vcc, the output goes high and the capacitor starts charging through R. When it reaches 2/3 of Vcc, the chip discharges the capacitor and the output returns low.
The pulse width is t = 1.1 · R · C. The 1.1 factor is ln 3 ≈ 1.0986: the time an RC network needs to charge from 0 to 2/3 of the supply voltage.
Typical values
| R | C | Pulse width | Typical use |
|---|---|---|---|
| 10 kΩ | 100 nF | ≈ 1.1 ms | Push-button debouncing |
| 10 kΩ | 10 µF | ≈ 0.11 s | Short beep |
| 100 kΩ | 10 µF | ≈ 1.1 s | Light an LED for one second |
| 1 MΩ | 10 µF | ≈ 11 s | Stair-light timer |
| 470 kΩ | 100 µF | ≈ 52 s | Long delay (sensitive to leakage) |
Step-by-step design
- Decide the pulse length you need, for example 5 s.
- Choose an available capacitor, for example 47 µF.
- Solve for R: R = t / (1.1 · C) = 5 / (1.1 × 47 µF) ≈ 96.7 kΩ, so use 100 kΩ.
- Check with the calculator: 1.1 × 100 kΩ × 47 µF ≈ 5.2 s.
- If you need precision, add a trimmer in series with R to fine-tune.
Frequently asked questions
What happens if the trigger pulse is longer than the output pulse?
The output can stay high for as long as pin 2 remains below 1/3 of Vcc. The trigger must be shorter than the pulse you want; if it isn’t, couple the input through a capacitor (a differentiator).
Why are long delays inaccurate?
With very large resistors and electrolytic capacitors, capacitor leakage and the chip’s input current disturb the charging. For delays of minutes, use a CMOS 555, low-leakage capacitors or a digital timer.
Can I restart the pulse before it ends?
The basic circuit is not retriggerable. You can cut the pulse short by pulling pin 4 (RESET) to ground.
Educational tool. Check tolerances, datasheets, standards and real operating conditions before using the results in critical applications.