555 Timer Calculator: Astable Frequency and Duty Cycle, Resistor Picker, and Monostable Pulse Width
The 555 timer is still the quickest way to blink an LED, make a square wave, or stretch a button press into a fixed-length pulse. This calculator covers the two circuits almost everyone uses: astable (free-running oscillator) and monostable (one pulse per trigger). In astable mode it can also work backwards: give it a target frequency and duty cycle, and it picks standard E24 resistors for you.
Astable formulas
In the standard astable circuit, the capacitor charges through R1 + R2 and discharges through R2 only:
thigh = 0.693 × (R1 + R2) × C
tlow = 0.693 × R2 × C
f = 1.44 ÷ ((R1 + 2·R2) × C)
duty = (R1 + R2) ÷ (R1 + 2·R2)
The 0.693 is ln 2: the capacitor swings between 1/3 and 2/3 of the supply, and that ratio doesn't depend on the supply voltage. That's why a 555's frequency stays put when the battery sags.
Example: R1 = 1 kΩ, R2 = 68 kΩ, C = 10 nF gives about 1.05 kHz at 50.4% duty, a decent near-square wave.
Why the duty cycle is always above 50%
Charging goes through R1 + R2 and discharging through R2 alone, so the high time is always longer than the low time. Making R1 small relative to R2 gets you close to 50%, but never below it. That's why the calculator refuses a target of 50% or less.
For duty cycles of 50% or lower, put a diode (a 1N4148 is fine) in parallel with R2, anode on pin 7 and cathode on pins 2/6. The capacitor then charges through R1 and the diode, and discharges through R2, so roughly thigh ≈ 0.693 × R1 × C and tlow ≈ 0.693 × R2 × C. The diode's forward drop makes the real high time a bit longer, more so at 5 V than at 12 V, so check it on a scope if the exact value matters.
Picking values
- Keep R1 at 1 kΩ or more. When the output goes low, pin 7 shorts the R1/R2 junction to ground, so R1 sits straight across the supply. Small R1 just burns current.
- Choose C first, then the resistors. Capacitors come in far fewer values than resistors (and with worse tolerance, often ±10–20%), so fix C and let the resistors do the fine-tuning. If the calculator says the resistors come out under 1 kΩ, use a smaller C. If they come out in the megohms, use a bigger C.
- Expect a few percent of error. Resistor tolerance, capacitor tolerance and the chip itself all add up. If you need an exact frequency, put a trimmer in series with R2, or use a microcontroller.
- Long delays: for periods of many seconds, electrolytic capacitor leakage starts to throw the timing off. Film capacitors, or a CMOS 555 such as the TLC555 or 7555 with bigger resistors, behave better.
- Add a 10 nF capacitor from pin 5 (control voltage) to ground, and a 0.1 µF bypass capacitor across the supply pins. The bipolar NE555 pulls a big current spike every time the output switches.
Monostable (one-shot)
Pull pin 2 below 1/3 of the supply and the output goes high for
t = 1.1 × R × C
then drops back low until the next trigger. The 1.1 is ln 3: the capacitor charges from 0 to 2/3 of the supply. A 100 kΩ resistor with a 10 µF capacitor gives about 1.1 seconds, which is handy for debouncing a button or keeping a light on for a moment after a sensor fires.
Driving an LED from the 555 output? Use the LED resistor calculator for the series resistor. More calculators are on the Tools page.
Worked examples
An LED blinker near 1 Hz
Suppose the target is 1 Hz at 60% duty cycle, using a 10 µF timing capacitor. Select the target-frequency mode and enter C as 10 µF. The target period is T = 1 / 1 = 1 s.
The widget uses ln(2), rather than the rounded constant 0.693. Calculate R2 = (1 − 0.60) × 1 / (ln(2) × 10 × 10⁻⁶) = 57,707.8 Ω. Then calculate R1 = (2 × 0.60 − 1) × 1 / (ln(2) × 10 × 10⁻⁶) = 28,853.9 Ω.
The widget selects the nearest E24 values by logarithmic distance: R1 = 30 kΩ and R2 = 56 kΩ. Substituting these gives t_high = ln(2) × (30,000 + 56,000) × 10⁻⁵ ≈ 0.596107 s and t_low = ln(2) × 56,000 × 10⁻⁵ ≈ 0.388162 s.
Thus f = 1 / (t_high + t_low) ≈ 1.015982 Hz, displayed as 1.016 Hz. Frequency error is approximately +1.60%. Actual nominal duty is 86,000 / 142,000 × 100 ≈ 60.5634%, displayed as 60.6%, or approximately 0.563 percentage points above the target.
Practical caveat: these times describe the output's high and low states. An LED connected to light while the output sinks current will illuminate during the low interval.
A quarter-second sensor pulse
For a sensor indicator, choose a target pulse of 250 ms and C = 1 µF. The widget's monostable calculation uses ln(3), so solve R = 0.250 / (ln(3) × 10⁻⁶) ≈ 227,559.8 Ω.
Manually choose the nearby E24 value 220 kΩ; monostable mode does not automatically select a resistor. Enter R as 220 kΩ and C as 1 µF.
The resulting pulse is t = ln(3) × 220,000 × 10⁻⁶ ≈ 0.241695 s. The widget displays 242 ms. Using the unrounded result, timing error is (0.241695 / 0.250 − 1) × 100 ≈ −3.32%.
Practical caveat: a one-shot does not queue separate output pulses for triggers arriving while its output is already high.
Common mistakes
- Leaving the reset input floating instead of holding it high during normal operation.
- Entering a capacitor's nanofarad number into the widget's microfarad field.
- Holding the trigger low throughout the intended monostable pulse; the trigger must return high for normal timing behavior.
- Placing the timing capacitor across breadboard contacts that already share the same electrical connection.
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