Case 1
- Resistor R1 (Ω)
- 1000Ω
- Resistor R2 (Ω)
- 10000Ω
- Timing capacitor C (µF)
- 0.1µF
687.1435Hz
Open with these valuesNE555 Astable Calculator
687.1435Hz
Result: 687.1435 HzIn astable mode the 555 never settles: the capacitor charges through R1 + R2 and discharges through R2 alone, over and over. The frequency is 1 / (0.693 × (R1 + 2 × R2) × C). Because the charge path is longer than the discharge path, the output is high more than half the time.
Worked examples
Case 2
- Resistor R1 (Ω)
- 4700Ω
- Resistor R2 (Ω)
- 4700Ω
- Timing capacitor C (µF)
- 0.01µF
10,234.0528Hz
Open with these valuesCase 3
- Resistor R1 (Ω)
- 2200Ω
- Resistor R2 (Ω)
- 10000Ω
- Timing capacitor C (µF)
- 0.047µF
1,382.9801Hz
Open with these valuesHow it's calculated
f = 1 / (0.693 × (R1 + 2 × R2) × C)
- StepEnter R1, the resistor from V+ to the discharge pin, in ohms.
- StepEnter R2, the resistor from discharge to threshold, in ohms.
- StepEnter the timing capacitor in microfarads — 100 nF is 0.1.
- ResultRead the frequency; the duty cycle is (R1 + R2) / (R1 + 2 × R2).
Reference table
| R1, R2, C (Ω, Ω, µF) | Duty cycle | Frequency (Hz) |
|---|---|---|
| 100000, 100000, 10 | 66.67 % | 0.4810 |
| 10000, 100000, 1 | 52.38 % | 6.8714 |
| 1000, 10000, 0.1 | 52.38 % | 687.1435 |
| 2200, 10000, 0.047 | 54.95 % | 1382.9801 |
| 4700, 4700, 0.01 | 66.67 % | 10234.0528 |
Questions
How do I calculate the frequency of a 555 astable circuit?
Add the high time and the low time and take the reciprocal: tH = 0.693 × (R1 + R2) × C, tL = 0.693 × R2 × C. For R1 = 1 kΩ, R2 = 10 kΩ and C = 0.1 µF that gives about 687.14 Hz. The familiar short form f = 1.44 / ((R1 + 2 × R2) × C) is the same equation with the reciprocal moved to the front.
Where does the 0.693 come from?
It is ln(2) rounded to three digits, and the rounding is the datasheet's, not ours. Between one third and two thirds of the supply the capacitor voltage crosses exactly ln(2) time constants, so the number is derived rather than measured. This calculator keeps 0.693 because a more precise value shifts the low-frequency examples in the third digit.
What is the duty cycle of a 555 astable circuit?
It is the share of each cycle the output spends high: (R1 + R2) / (R1 + 2 × R2). For R1 = 1 kΩ and R2 = 10 kΩ that is 11/21, about 52.38 %. The reference table lists it beside every frequency.
Why is the duty cycle always more than 50 %?
The timing capacitor charges through R1 + R2 but discharges through R2 alone, so the high time is always the longer one. Making R1 small against R2 pushes the duty cycle towards 50 % without ever reaching it. A diode across R2 gives charge and discharge separate paths and breaks the limit.
Do I enter the capacitance in farads or microfarads?
In microfarads (µF). A 100 nF capacitor is 0.1 µF, a 10 nF capacitor is 0.01 µF and a 1 nF capacitor is 0.001 µF. Entering raw farads gives a frequency a million times too low, and the wrong answer still looks plausible.
Sources and last check
Information, not professional advice.