Single 6 V, 600 µs pulse into an RC integrator (R = 150 kΩ, C = 0.002 µF): To what voltage will the capacitor charge by the end of the pulse?

Introduction / Context:Here we calculate the charged voltage of a capacitor at the end of a finite pulse—key to understanding RC integrator performance for edge shaping and pulse area control.

Given Data / Assumptions:

• R = 150 kΩ.
• C = 0.002 µF = 2 nF.
• Pulse amplitude Vin = 6 V.
• Pulse width t = 600 µs.
• Initial capacitor voltage vC(0) = 0 V (uncharged).

Concept / Approach:Use the exponential charging law vC(t) = Vin * (1 − e^(−t/tau)). Compute the time constant tau = R * C, then evaluate at t = 600 µs.

Step-by-Step Solution:C = 0.002 µF = 0.002 × 10^−6 F = 2 × 10^−9 Ftau = R * C = 150,000 * 2e−9 = 3e−4 s = 300 µst/tau = 600 µs / 300 µs = 2vC = 6 * (1 − e^(−2))e^(−2) ≈ 0.1353 ⇒ vC ≈ 6 * 0.8647 ≈ 5.19 V ≈ 5.16 V

Verification / Alternative check:At t = 2 tau, a capacitor reaches about 86.5% of its final value. 0.865 * 6 ≈ 5.19 V, matching our computation and the provided option.

Why Other Options Are Wrong:

• 6 V: Requires t → ∞ for full charge; not at 600 µs.
• 3.78 V: Around 63% of 6 V (one tau) rather than two tau.
• 0 V: Only at the instant before the pulse begins.

Common Pitfalls:Misconverting 0.002 µF to farads, or forgetting that the capacitor does not reach 100% in finite time. Always compute t/tau first.

Final Answer:5.16 V