How an astable 555 generates a square wave
In astable mode the 555 repeatedly charges the capacitor through R1+R2 and discharges it through R2 alone, toggling its output each time the cap crosses the internal 1/3 and 2/3 supply thresholds. Because charging always goes through both resistors but discharging only through R2, the output is HIGH longer than it's LOW — the duty cycle can approach 50% but, in this basic circuit, never drop below it.
- R1 — resistor from V+ to the discharge pin (Ω)
- R2 — resistor from discharge to threshold/trigger (Ω)
- C — timing capacitor (F)
- f — output frequency (Hz)
More detail
Why duty cycle can't go below 50% here
The output is HIGH while the capacitor charges through R1+R2, and LOW while it discharges through R2 alone. Since the charge path always includes an extra resistor (R1) that the discharge path skips, the HIGH portion is always at least as long as the LOW portion — duty = (R1+R2)/(R1+2R2) only reaches exactly 50% in the limit as R1 approaches 0, and rises toward 100% as R1 grows much larger than R2. To get duty cycles below 50%, real designs add a diode across R2 to give the charge and discharge paths separate resistors.
Picking components. Fix C first (common values: 0.01 µF–100 µF depending on target frequency), then solve for R1 and R2 to hit the frequency and duty cycle you need. Keep R1 and R2 in the kΩ range — under about 1 kΩ draws excess current through the chip, and above a few MΩ the timing becomes sensitive to leakage and noise pickup.
Frequently asked questions
R1 = 1kΩ, R2 = 1kΩ, C = 1µF — what frequency comes out?
f = 1.44 ÷ ((1000 + 2000) × 0.000001) = 1.44 ÷ 0.003 = 480 Hz, with a duty cycle of (1000+1000)/(1000+2000) = 66.7% — a classic textbook example of the astable 555.
How do I get close to a 50% duty cycle?
Make R1 as small as possible relative to R2 — the formula duty = (R1+R2)/(R1+2R2) only equals exactly 50% when R1 = 0, which isn't practical (it removes current limiting on the discharge pin). For a true 50% square wave, a common trick outside this basic circuit is adding a diode across R2 to separate the charge and discharge paths.
I need 1 kHz with R1 = R2 = 10kΩ — what capacitor?
Solve C = 1.44 / ((R1+2R2) × f) = 1.44 / (30,000 × 1000) = 48 nF (0.048 µF). Enter R1=10000, R2=10000 above and adjust C until frequency reads 1000 Hz to confirm.
Why is the high time always longer than the low time?
Because charging goes through R1+R2 while discharging only goes through R2 — the extra resistance in the charge path means it always takes longer, so tHigh > tLow whenever R1 > 0. Only as R1 shrinks toward 0 do the two times converge.