FCC Exam Question: 3-30D6

In the op-amp oscillator circuit shown in Figure 3D6, what would be the most noticeable effect if the capacitance of C were suddenly doubled?

A. Frequency would be lower.
B. Frequency would be higher.
C. There would be no change. The inputs are reversed, therefore the circuit cannot function.
D. None of the above.
Correct Answer: A

Explanation: In an op-amp oscillator circuit, the frequency of oscillation is typically determined by the time constants of its reactive components, usually resistors (R) and capacitors (C). For most common RC oscillator configurations (like Wien bridge or phase-shift oscillators), the oscillation frequency ($f$) is inversely proportional to the RC time constant ($RC$). Therefore, if the capacitance of C were suddenly doubled, the RC time constant would also double. Since frequency is inversely proportional to this time constant ($f \propto 1/(RC)$), doubling the capacitance would cause the frequency to halve, resulting in a **lower frequency**. Option B is incorrect because frequency is inversely related to capacitance. Option C is incorrect; op-amp oscillators are designed to function and reversing inputs in a way that prevents oscillation isn't implied for a *functioning* oscillator, nor does it address the effect of changing capacitance.

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Includes Elements 1, 3, 6, 7R, 8, and 9.