Introduction / Context:
In analog circuits we often need to block DC bias from reaching a following stage, meter, or load, while passing the AC information. The standard technique is capacitive coupling, which provides high impedance at DC and low impedance over the desired AC band, thereby eliminating DC return while preserving signal content.
Given Data / Assumptions:
- Goal: block DC (0 Hz), pass AC band of interest.
- Source and load impedances are finite and the coupling capacitor value can be chosen.
- We ignore transformer coupling here to keep the solution component-minimal and broadband around low audio/RF as needed.
Concept / Approach:
A capacitor's reactance Xc = 1 / (2 * π * f * C) is infinite at f = 0, so it perfectly blocks DC. At sufficiently high frequency, Xc becomes small and the capacitor behaves like a short for AC. Choosing C such that the high-pass cutoff f_c = 1 / (2 * π * R_total * C) lies well below the lowest signal frequency ensures negligible amplitude distortion.
Step-by-Step Solution:
Specify the lowest frequency f_min you need to pass.Estimate R_total seen by the capacitor (source resistance + load resistance as appropriate).Choose C so that f_c = 1 / (2 * π * R_total * C) ≤ f_min / 10.Insert capacitor in series; include a bleed resistor if necessary to set reference bias.
Verification / Alternative check:
Simulate a small-signal AC sweep; confirm flat response above f_c and DC isolation at 0 Hz.
Why Other Options Are Wrong:
Resistor: provides a DC path and drops DC per divider action, not full isolation.Inductor: passes DC (low reactance at 0 Hz is not correct—an ideal inductor has zero DC resistance, so it actually allows DC) and blocks high frequencies (opposite of the requirement).Transformer: can block DC but imposes bandwidth, impedance, and size constraints; the question asks what is 'necessary' and simplest—capacitive coupling.
Common Pitfalls:
Picking too small a capacitor causing low-frequency roll-off; forgetting biasing at the load side when AC-coupling into transistor/op-amp inputs.
Final Answer:
A capacitor between source and load
Discussion & Comments