555 timer in basic astable mode: For a 555 timer configured as a standard astable multivibrator (using two resistors and one capacitor), which element(s) set the duty cycle?

Introduction / Context:
The ubiquitous 555 timer can generate square or rectangular waves in astable mode. Designers frequently need to set both frequency and duty cycle. In the classic configuration, two resistors (RA and RB) and one capacitor (C) define the waveform characteristics. This question pinpoints which components control duty cycle specifically.

Given Data / Assumptions:

• Standard 555 astable using RA, RB, and C.
• No diode modifications or symmetry-correction tricks are applied.
• Thresholds at 1/3 Vcc and 2/3 Vcc apply as usual.

Concept / Approach:
In the basic astable, the capacitor charges through RA + RB and discharges through RB only. Thus, the high-time T_H depends on (RA + RB) * C, while the low-time T_L depends on RB * C. Because T_H and T_L depend differently on RA and RB, the duty cycle D = T_H / (T_H + T_L) is determined by the ratio of these two resistances, not by the capacitor alone. The capacitor scales both times equally and sets frequency in combination with the resistors, but it does not independently set duty cycle.

Step-by-Step Solution:

Verification / Alternative check:
Perform a parametric sweep: vary C while keeping RA and RB constant; the duty cycle remains unchanged while frequency shifts. Vary RA or RB with C fixed; duty cycle changes accordingly.

Why Other Options Are Wrong:

• One resistor or one capacitor: Cannot independently control both T_H and T_L.
• A resistor and a capacitor: C affects timing, but duty cycle depends on the resistor ratio.
• Control-voltage pin only: CV can modulate thresholds for special effects, not the basic duty formula.

Common Pitfalls:
Assuming duty cycle equals 50% by default; neglecting that without a steering diode, perfect symmetry is not achieved; misattributing duty variation to C rather than RA/RB ratio.

Final Answer:
two resistors