555 Timer Calculator
Calculate frequency, duty cycle and pulse width for astable and monostable 555 circuits.
Component Values
Results
OUTPUT WAVEFORM
f = 1.44 / ((Ra + 2·Rb) × C)
How does the 555 timer work in astable mode?
In astable mode the 555 timer has no stable state — it oscillates continuously between high and low, producing a square wave at the output. The capacitor charges through Ra and Rb, then discharges through Rb only. This asymmetry means the high time is always longer than the low time, so the duty cycle is always above 50%.
Frequency and duty cycle depend on three components: Ra, Rb, and C. To lower the frequency, increase any of them. To get close to 50% duty cycle, make Rb much larger than Ra — the Ra contribution becomes negligible. For a true 50% duty cycle you need either a diode across Rb or a different circuit topology.
Frequency
f = 1.44 / ((Ra + 2·Rb) × C)Duty Cycle
D = (Ra + Rb) / (Ra + 2·Rb)Key Points
- Duty cycle always > 50% (Ra always in charge path)
- Frequency: f = 1.44 / ((Ra + 2·Rb) × C)
- To lower frequency: increase Ra, Rb, or C
- Minimum supply voltage: 4.5V (5V typical)
Applications
- LED blink circuits
- Clock signal generation
- PWM motor speed control
- Tone and sound generation
NE555 Timer Formulas
ASTABLE mode (oscillator):
f = 1.44 / ((R1 + 2×R2) × C)Duty cycle D = (R1 + R2) / (R1 + 2×R2)t_high = 0.693 × (R1 + R2) × Ct_low = 0.693 × R2 × CMONOSTABLE mode (one-shot):
t_pulse = 1.1 × R × CCommon Astable Configurations
| Frequency | R1 | R2 | C | Duty Cycle | Application |
|---|---|---|---|---|---|
| 1 Hz | 10 kΩ | 68 kΩ | 10 µF | 54% | LED blinker |
| 10 Hz | 10 kΩ | 68 kΩ | 1 µF | 54% | Slow PWM |
| 1 kHz | 1 kΩ | 10 kΩ | 47 nF | 52% | Tone generator |
| 10 kHz | 1 kΩ | 4.7 kΩ | 10 nF | 57% | PWM control |
| 38 kHz | 1 kΩ | 1 kΩ | 10 nF | 75% | IR remote carrier |
| 100 kHz | 1 kΩ | 1 kΩ | 4.7 nF | 75% | High-freq switching |
Worked Examples
R1 = 10 kΩ, R2 = 68 kΩ, C = 10 µF
f = 1.44 / ((10k + 136k) × 10µF) = 0.986 Hz ≈ 1 Hz. LED blinks ~60 times/minute.
Target: 100 ms pulse after button press. R = 100 kΩ, C = 1 µF
t = 1.1 × 100k × 1µF = 110 ms ✓
Design Tip
The NE555 can source/sink up to 200 mA — enough to drive LEDs, relays, and small motors directly. For 50% duty cycle astable, set R1 very small (1 kΩ) and use a diode bypass across R2. Supply voltage range: 4.5 V–16 V (CMOS TLC555: 2 V–15 V).
Did you know? The 555 timer IC was designed by Hans Camenzind in 1971 for Signetics. It is one of the best-selling ICs of all time with over 1 billion units sold per year. Its internal voltage divider uses three 5 kΩ resistors — giving it the "555" name.