Frequency of 555 timer astable multivibrator

In astable operation the 555 timer works as a free-running oscillator giving a rectangular wave. There is no stable state.

Circuit

          +VCC
     +-----+-----+
     |     |     |
   [RA]   (4)(8)  |
     |            |
     +--- (7) discharge
     |
   [RB]
     |      (6) threshold
     +------(2) trigger
     |
   [C]       (3) output
     |
    GND      (5) via 0.01 µF to GND

Pins 2 and 6 are tied together at the capacitor, and the capacitor charges through RA + RB and discharges through RB into pin 7.

Operation

The comparators have thresholds of 2VCC/3 (upper) and VCC/3 (lower).

1. Capacitor charges from VCC/3 to 2VCC/3 through (RA + RB). Output is HIGH, and the discharge transistor is OFF.
2. At 2VCC/3, the upper comparator resets the flip-flop; output goes LOW and the discharge transistor turns ON.
3. The capacitor discharges through RB to VCC/3. At VCC/3, the lower comparator sets the flip-flop and the cycle repeats.

Derivation

Charging (HIGH) time: the capacitor voltage is vC(t) = VCC − (VCC − VCC/3) e^(−t/(RA+RB)C). It reaches 2VCC/3 at time

2VCC/3 = VCC − (2VCC/3) e^(−tH/τ1) → e^(−tH/τ1) = 1/2

tH = 0.693 (RA + RB) C

Discharging (LOW) time: vC(t) = (2VCC/3) e^(−t/RB C), reaching VCC/3:

1/2 = e^(−tL/RB C)

tL = 0.693 RB C

Period and frequency:

T = tH + tL = 0.693 (RA + 2RB) C

f = 1/T = 1.44 / [(RA + 2RB) C]

Duty cycle: D = tH/T = (RA + RB)/(RA + 2RB), which is always above 50 %. D approaches 50 % when RA << RB.

Example: RA = 1 kΩ, RB = 10 kΩ, C = 0.1 µF gives f = 1.44/(21 k × 0.1 µ) = 686 Hz.