555 astable: R1=1kΩ, R2=10kΩ, C=1µF → 68.7 Hz
Worked answer for a 555 timer free-running (astable) with R1=1kΩ, R2=10kΩ and C=1µF.
Oscillates at 68.7 Hz period 14.556 ms · duty 52.4%
| R1 | 1kΩ |
| R2 | 10kΩ |
| C | 1µF |
| Frequency f = 1/(ln2·(R1+2R2)·C) | 68.7 Hz |
| Period T = 1/f | 14.556 ms |
| t_high = ln2·(R1+R2)·C | 7.625 ms |
| t_low = ln2·R2·C | 6.931 ms |
| Duty cycle = t_high / T | 52.4% |
In the classic two-resistor astable, the cap charges through R1+R2 but discharges only through R2, so the duty cycle is always > 50%. To approach a 50% square wave, add a diode across R2 so charging bypasses it.
Different values? Change R1, R2 or C in the interactive tool:
Open the 555 Timer Calculator →Read more: NE555 pinout, specs & circuits · NE555 pinout reference
Disclaimer: These are ideal values from the textbook astable formula. Real capacitor tolerance and leakage shift the actual frequency — trim by measurement for anything time-critical.