555 timer in astable mode — adjust duty cycle to 0.5 without changing frequency Given a standard astable 555 configuration, how should R1 and R2 be adjusted to correct the duty cycle to 0.5 (50%) while keeping the output frequency unchanged?

Introduction / Context:
In the classic astable 555 (without diodes), the timing is set by R1, R2, and C. The frequency is f ≈ 1.44 / ((R1 + 2R2) * C), and the duty cycle D ≈ (R1 + R2) / (R1 + 2R2). Designers often need to move D toward 0.5 while holding frequency constant. This requires coordinated changes in R1 and R2 rather than altering the capacitor or supply conditions.

Given Data / Assumptions:

• Standard 555 astable (single timing capacitor C; charge through R1 + R2, discharge through R2).
• Target duty cycle D_target = 0.5.
• Frequency constraint: keep (R1 + 2R2)C constant.

Concept / Approach:
From D = (R1 + R2) / (R1 + 2R2), moving toward 0.5 requires reducing the numerator relative to the denominator. Decreasing R1 lowers the numerator more strongly than the denominator, while increasing R2 can compensate in the denominator so that (R1 + 2R2) remains unchanged to preserve frequency. Note: with the basic circuit, exact 50% ideally requires R1 → 0; in practice, we reduce R1 substantially and increase R2 appropriately (or add a steering diode for perfect 50%).

Step-by-Step Solution:

Verification / Alternative check:
Compute a sample: suppose initially R1 = 10 kΩ, R2 = 10 kΩ, C fixed. K = (10 + 20)kΩC = 30kC. To move closer to 0.5, choose R1 = 1 kΩ; then to keep K, solve 1 + 2R2 = 30 ⇒ R2 ≈ 14.5 kΩ. New D ≈ (1 + 14.5)/(1 + 29) ≈ 15.5/30 ≈ 0.517 (closer to 0.5).

Why Other Options Are Wrong:

• Increase C: changes frequency directly.
• Change Vcc or RL: does not control duty cycle in the ideal 555 timing equations.
• Decrease both R1 and R2: frequency will change unless tuned precisely; not the directed move toward D = 0.5.
• Increase R1 and decrease R2: pushes D away from 0.5 for a fixed K.

Common Pitfalls:

• Expecting exact 50% without adding a charge/discharge steering diode; the standard topology cannot achieve perfect D = 0.5 unless R1 approaches zero.

Final Answer:
Decrease R1 and increase R2.