RC Filter Behavior with a Realistic Source Impedance
A student builds a low-pass RC filter with R = 10 kΩ and C = 10 nF, expecting a cutoff near 1.6 kHz. The source is a function generator with 600 Ω output impedance, and the load is an oscilloscope with 1 MΩ input in parallel with 15 pF. The student measures a −3 dB point at about 1.5 kHz, slightly lower than predicted, and also notices that the output amplitude at 100 kHz is larger than the ideal filter would give. They must decide whether the discrepancy is due to component tolerance, source/load loading, or a wiring problem.
1. Compute the ideal cutoff and then estimate the effect of the 600 Ω source and the scope input on the measured cutoff.
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Ideal cutoff: f_c = 1/(2πRC) = 1/(2π × 10⁴ × 10⁻⁸) ≈ 1592 Hz. The source resistance adds in series with R, giving R_total ≈ 10.6 kΩ, which lowers the cutoff to about 1/(2π × 10.6 × 10³ × 10⁻⁸) ≈ 1502 Hz. The scope input is 1 MΩ in parallel with 15 pF; the 1 MΩ is negligible against 10 kΩ, but the 15 pF adds to C, giving C_total ≈ 10.015 nF, a 0.15% effect. So source loading explains most of the shift from 1592 Hz to about 1502 Hz, matching the observed ~1.5 kHz. Component tolerance (±5% on R and C) could shift it further but is not needed to explain the data.
2. Why is the 100 kHz output larger than the ideal filter predicts, and how would you test the cause?
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At 100 kHz, the ideal RC filter gives attenuation of about 1/(2πfRC) ≈ 1/(2π × 10⁵ × 10⁴ × 10⁻⁸) ≈ 1/63, or about −36 dB. A larger-than-ideal output suggests stray capacitance bridging the resistor, ground-loop coupling, or the scope probe's own capacitance and lead inductance forming a parallel path. The 15 pF scope capacitance is in parallel with C and actually helps the filter, not hurts it. More likely culprits are stray capacitance across R (a few pF can create a high-frequency bypass) or inductive pickup from long leads. Test by shortening leads, shielding, and measuring with the probe directly at the capacitor; if the high-frequency output drops, the cause was layout/stray coupling rather than component values.
3. The student wants the measured cutoff to match 1.6 kHz within 1%. What design change is needed?
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To hit 1592 Hz within 1%, account for the 600 Ω source: choose R such that R + 600 Ω gives the desired total. With C = 10 nF, required R_total = 1/(2π × 1592 × 10⁻⁸) ≈ 10.0 kΩ, so the resistor should be about 9.4 kΩ. Also account for the 15 pF scope capacitance by reducing C slightly, to about 9.985 nF, though that is a 0.15% correction. Use 1% resistors and capacitors, and verify with a calibrated measurement. The dominant correction is the source resistance; the scope capacitance is a minor trim.