How we worked it out
Updated as you change the numbers.
- High = 0.693(R1 + R2)C. Low = 0.693 R2 C. f = 1.44 ÷ ((R1 + 2R2)C).
Method and assumptions
The 0.693 factor is ln 2: the capacitor swings between 1/3 and 2/3 of the supply. The duty cycle is always above 50% in this circuit unless a diode bypasses R2.
Questions
What frequency do 1 kΩ, 10 kΩ and 10 µF give?
About 6.86 Hz with a 52.4% duty cycle.
Sources
Method and figures checked on . Every worked answer on this page is covered by an automated test. Found a mistake? Tell us.