555 timer in astable mode (square-wave generator) Given the standard frequency formula for a 555 in astable mode, f = 1.44 / ((R1 + 2R2) * C1). What value of C1 is required if R1 = 1 kΩ, R2 = 1 kΩ, and the desired output frequency f = 1 kHz?

Introduction / Context:
The 555 timer integrated circuit configured in astable mode is a classic way to generate a continuous square wave. Its oscillation frequency depends on two resistors (R1 and R2) and one timing capacitor (C1). This problem checks your ability to apply the standard astable frequency formula and solve for the unknown capacitance to hit a target frequency.

Given Data / Assumptions:

• Astable 555 configuration.
• R1 = 1 kΩ, R2 = 1 kΩ.
• Target frequency f = 1 kHz.
• Use the ideal formula f = 1.44 / ((R1 + 2R2) * C1).

Concept / Approach:

The 555 charges C1 through R1 + R2 and discharges through R2, which leads to a duty cycle > 50% for equal resistors and the well-known frequency expression. Rearranging the formula allows direct computation of C1 for any chosen R1, R2, and f. Always keep units consistent (ohms, farads, hertz).

Step-by-Step Solution:

Verification / Alternative check:

If C1 = 0.48 µF, then (R1 + 2R2) * C1 = 3000 * 4.8 * 10^-7 = 1.44 * 10^-3. Finally f = 1.44 / 1.44 * 10^-3 ≈ 1000 Hz, confirming the result.

Why Other Options Are Wrong:

• 0.33 µF: Yields f ≈ 1.45 kHz, higher than required.
• 480 µF: Off by a factor of 10^6; would produce millihertz frequency.
• 33 nF: Gives f ≈ 14.5 kHz, far too high.
• 4.8 µF: Gives f ≈ 100 Hz, too low by a factor of 10.

Common Pitfalls:

• Mixing µF and F; always convert µF to 10^-6 F for calculations.
• Forgetting the 2*R2 term in the formula.
• Rounding too early; keep sufficient precision until the final step.

Final Answer:

0.48 µF