Achieving 50 percent duty cycle with the 555 in astable mode Which combination of conditions best ensures tHI equals tLO (near 50 percent duty cycle) in a 555 astable configuration?

Introduction / Context:
In the classic 555 astable, the charge path includes RA + RB while the discharge path uses RB only. This asymmetry makes duty cycle greater than 50 percent unless modifications are made. Designers often add a steering diode across RB during charge so the capacitor charges through RA alone and discharges through RB alone, allowing duty cycle to be set by choosing RA and RB, including the special case of 50 percent when RA equals RB.

Given Data / Assumptions:

• Standard 555 astable topology with RA, RB, and timing capacitor C.
• Charge path normally goes through RA + RB; discharge path through RB.
• Diode across RB (anode to discharge pin side) conducts during charge only.

Concept / Approach:
With the diode, tHI ≈ 0.693 * RA * C and tLO ≈ 0.693 * RB * C. To achieve 50 percent, set RA = RB so tHI equals tLO. The extra statement tLO = tHI expresses the target, while the diode and equality of resistors define how to achieve it.

Step-by-Step Solution:

Verification / Alternative check:
Compute example with RA = RB = 10 k and C = 0.01 uF: tHI ≈ 0.693 * 10 k * 0.01 uF, tLO ≈ 0.693 * 10 k * 0.01 uF, clearly equal.

Why Other Options Are Wrong:

• tLO = tHI: states the goal but not the method to achieve it.
• RA = RB and diode only: implies 50 percent but does not explicitly assert equality of times.
• Capacitor rises above 1/3 VCC: true in all 555 astables and not specific to 50 percent.

Common Pitfalls:
Forgetting that without the diode, the charge path includes RB, making duty cycle greater than 50 percent even if RA = RB.

Final Answer:
tLO = tHI, RA = RB, and short RB with a diode during the capacitor charging cycle