NE555 Astable Calculator
From two resistors and a capacitor, get the frequency, duty cycle, and high/low times of a 555 timer running free as an oscillator.
Three inputs, four answers
Enter R1, R2, and the timing capacitor (in microfarads) and the calculator returns the frequency, the duty cycle, and the output-high and output-low times at once.
Duty cycle is always above 50 %
In the classic astable circuit the capacitor charges through R1 + R2 but discharges through R2 alone, so the output spends longer high than low — the duty cycle never quite reaches 50 %.
What is an NE555 astable calculator?
Two resistors and a cap in, a square wave out
An NE555 astable calculator turns three component values — two timing resistors (R1 and R2) and a timing capacitor — into the numbers that describe the square wave a 555 timer produces when it free-runs: how often it oscillates (frequency), what fraction of each cycle the output is high (duty cycle), and how long each high and low phase lasts. Astable mode is the 555's most popular trick: it is the heart of blinking LEDs, tone generators, clock pulses, PWM dimmers, and countless hobby projects. Pick R1, R2, and C and the timer does the rest.
Enter R1, R2, and the timing capacitor (in microfarads) to get the frequency, duty cycle, and high/low times instantly.
A few short formulas, all built from R1, R2, the capacitance C (converted from microfarads to farads), and the constant 0.693 — that is ln(2), the natural log of two.
f = 1.44 / ((R1 + 2·R2) × C)The output-high time is tH = 0.693 × (R1 + R2) × C, because the capacitor charges through both resistors. The output-low time is tL = 0.693 × R2 × C, since it discharges through R2 only. The period is their sum, so the frequency is f = 1 / (tH + tL). The duty cycle — the fraction of each cycle spent high — is (R1 + R2) / (R1 + 2·R2), which is always above 50 %. Because you enter the capacitance in microfarads, the calculator first multiplies by one-millionth to get farads.
Suppose you build an astable oscillator with R1 = 1 kΩ, R2 = 10 kΩ, and C = 0.1 µF.
High time
0.693 × (1000 + 10000) × 0.0000001 = 0.000762 s — how long the output stays high.
Low time
0.693 × 10000 × 0.0000001 = 0.000693 s — how long it stays low.
Frequency and duty cycle
1 / (0.000762 + 0.000693) ≈ 687.14 Hz, and (11000 / 21000) × 100 ≈ 52.38 % duty cycle.
The four outputs together describe the square wave. The frequency (about 687.14 Hz for the example) is how many full cycles happen each second — drop it into the audible range for a tone, or down to a few hertz to blink an LED. The duty cycle (about 52.38 %) is the fraction of each cycle the output is high; in the classic two-resistor circuit it is always above 50 % because the capacitor charges through R1 + R2 but discharges through R2 alone. The high and low times (here roughly 0.762 ms and 0.693 ms) are the same information in seconds, handy when you care about pulse width rather than rate. To raise the frequency, shrink any of R1, R2, or C; to push the duty cycle toward 50 %, make R1 small relative to R2 (or add a diode across R2 so charge and discharge use separate paths). For a true 50 % square wave the diode trick or a different topology is needed — the bare circuit cannot reach it.
The formulas are the standard datasheet approximations, but a couple of practical points are worth keeping in mind.
Real timers, tolerances, and the 0.693 constant
These formulas use the idealised 0.693 (ln 2) timing constant and assume an ideal 555 with no leakage and instantaneous switching. Real resistor and capacitor tolerances (often ±5 % to ±20 %, and electrolytics worse) shift the measured frequency from the computed value, so treat the result as a starting point and trim with a trimmer pot if precision matters. Very small resistors draw large discharge currents and may exceed the 555's ratings, while very large resistors make timing sensitive to leakage. The duty cycle of the bare two-resistor circuit is always greater than 50 % and cannot reach it without a diode or a modified topology.