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555 astable multivibrator and duty cycle

Short answer

In 555 astable mode, the capacitor charges through RA+RB (output HIGH) and discharges through RB (output LOW). The duty cycle is D = (RA+RB)/(RA+2RB). Since RA>0, D always exceeds 50%.

Explain the operation of astable multivibrator using 555 timer IC square wave generator and derive its duty cycle.

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.

Duty cycle

From the charge and discharge analysis:

  • HIGH time: tH = 0.693 (RA + RB) C
  • LOW time: tL = 0.693 RB C
  • Period: T = tH + tL = 0.693 (RA + 2RB) C

Duty cycle D = tH/T = (RA + RB) / (RA + 2RB), usually written as a percentage.

Since RA > 0, D is always more than 50 %. When RA << RB, D approaches 50 %; for example, RA = 1 kΩ and RB = 100 kΩ gives D = 50.2 %. When RB << RA, D tends towards 100 %.

Example: RA = 4.7 kΩ, RB = 10 kΩ, C = 0.1 µF: D = 14.7/24.7 = 59.5 %, f = 1.44/(24.7 kΩ × 0.1 µF) = 583 Hz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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