Which statement correctly describes how source impedance affects op-amp noise performance?

Which statement correctly describes how source impedance affects op-amp noise performance?

• A. Voltage noise is a spectral density at the input.
• B. Current noise times source impedance adds to the total input-referred noise; high source impedance makes current noise more significant.
• C. Source impedance has no effect on noise.
• D. Noise is determined solely by the op-amp's gain.

Answer: (B)

The main idea is that op-amp noise comes from two sources at the input: the amplifier’s own voltage noise and the current noise flowing through the source path. The voltage noise is characterized as a spectral density at the input, often denoted en, and it adds to the total noise regardless of what the source impedance is. But the input current noise, when it flows through the source impedance Rs, creates an additional noise voltage across the input equal to in multiplied by Rs. This contribution grows with higher source impedance, so a larger Rs makes the current-noise term more significant. The total input-referred noise density is effectively sqrt(en^2 + (in*Rs)^2). That shows why the statement in question is correct: the current noise times the source impedance adds to the total noise, and increasing Rs amplifies that part of the noise. The other options miss this interplay. Noise isn’t solely determined by the op-amp’s gain, and source impedance does affect noise through the current-noise contribution. Also, while voltage noise density is a real characteristic, relying on it alone ignores how Rs makes the current-noise term matter.

The main idea is that op-amp noise comes from two sources at the input: the amplifier’s own voltage noise and the current noise flowing through the source path. The voltage noise is characterized as a spectral density at the input, often denoted en, and it adds to the total noise regardless of what the source impedance is. But the input current noise, when it flows through the source impedance Rs, creates an additional noise voltage across the input equal to in multiplied by Rs. This contribution grows with higher source impedance, so a larger Rs makes the current-noise term more significant.

The total input-referred noise density is effectively sqrt(en^2 + (in*Rs)^2). That shows why the statement in question is correct: the current noise times the source impedance adds to the total noise, and increasing Rs amplifies that part of the noise. The other options miss this interplay. Noise isn’t solely determined by the op-amp’s gain, and source impedance does affect noise through the current-noise contribution. Also, while voltage noise density is a real characteristic, relying on it alone ignores how Rs makes the current-noise term matter.