555 astable: R1=1kΩ, R2=1kΩ, C=100nF → 4.809 kHz
Worked answer for a 555 timer free-running (astable) with R1=1kΩ, R2=1kΩ and C=100nF.
Oscillates at 4.809 kHz period 207.944 µs · duty 66.7%
| R1 | 1kΩ |
| R2 | 1kΩ |
| C | 100nF |
| Frequency f = 1/(ln2·(R1+2R2)·C) | 4.809 kHz |
| Period T = 1/f | 207.944 µs |
| t_high = ln2·(R1+R2)·C | 138.629 µs |
| t_low = ln2·R2·C | 69.315 µs |
| Duty cycle = t_high / T | 66.7% |
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.