Per the TI datasheet, f = 1 / T and T = 0.693 x (RA + 2RB) x C, giving f ≈ 1.44 / ((RA + 2RB) x C). With RA = 1 kΩ, RB = 10 kΩ and C = 1 µF, T is 14.56 ms and f is 68.7 Hz.
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Work out the frequency, duty cycle and timing of a 555 timer from RA, RB and C, or find standard resistor and capacitor values for a target frequency and duty cycle.
The 555 is a precision timer that its maker describes for delays from 10 µs to hours, or frequencies from under 1 mHz to 100 kHz. This calculator uses the relations in the Texas Instruments datasheet (NA555, NE555, SA555 and SE555, document SLFS022K). In astable mode the output stays high for tH = 0.693 x (RA + RB) x C and low for tL = 0.693 x RB x C, so the period is 0.693 x (RA + 2RB) x C and the frequency is about 1.44 / ((RA + 2RB) x C). The capacitor swings between 1/3 and 2/3 of the supply voltage, which is why the timing does not depend on it as long as it is steady. In monostable mode the pulse lasts about 1.1 x R x C. The calculator uses the exact constants, ln 2 = 0.6931 and ln 3 = 1.0986, which the datasheet rounds to 0.693 and 1.1, and the difference is under 0.1%. Be careful with the term duty cycle: the datasheet gives two definitions, the driver one (RB / (RA + 2RB), the low fraction) and the waveform one ((RA + RB) / (RA + 2RB), the high fraction). Here both are shown apart. Because RA is always positive, the high duty cycle stays above 50% in the standard circuit. For 50% or less, the variant with a diode across RB works, but it is an ideal-diode estimate that the datasheet does not give. Components are treated as ideal: the tolerance fields show the possible range of frequency and duty cycle with the parts at their limits.
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Per the TI datasheet, f = 1 / T and T = 0.693 x (RA + 2RB) x C, giving f ≈ 1.44 / ((RA + 2RB) x C). With RA = 1 kΩ, RB = 10 kΩ and C = 1 µF, T is 14.56 ms and f is 68.7 Hz.
With 5% on the resistors and 10% on the capacitor: frequency from 911.7 mHz to 1.232 Hz and high duty cycle from 50.3% to 50.4%.
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