0
\$\begingroup\$

A resistor's Johnson (Inherent) noise can be modelled as an RMS voltage in series with the resistor: $$ e_{rms}\triangleq\sqrt{4kRT\Delta f} $$ where k is Boltzmann's constant, R is resistance, T is absolute temperature in Kelvin, and delta-f is the frequency bandwidth of the circuit.

Mathematically speaking, decreasing the frequency bandwidth decreases e_rms and effectively reduces noise. However I question what the tradeoff of this operation could be. In other words, there has to be some disadvantage or issue if we decrease the frequency bandwidth.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ What do you mean? You can't just reduce the bandwidth on a real resistor. You can reduce the bandwidth of the measurement, but that's not reducing the inherent noise. If you are talking about reducing measured noise by reducing the bandwidth a circuit operates at, then the obvious disadvantage is reduced bandwidth. \$\endgroup\$
    – DKNguyen
    May 6, 2022 at 14:58
  • \$\begingroup\$ can't we use some bandpass filter? @DKNguyen \$\endgroup\$
    – SPARSE
    May 6, 2022 at 15:00
  • 2
    \$\begingroup\$ Then you are asking about the disadvantage of reducing noised measured by reducing circuit bandwidth. See second half of my previous response. \$\endgroup\$
    – DKNguyen
    May 6, 2022 at 15:01

3 Answers 3

6
\$\begingroup\$

It depends on your application. You need to provide the bandwidth that your application requires. If you reduce the bandwidth below that value, you will reduce the resistor noise at the expense of reduced performance of your circuit. This is, in general, the tradeoff you must make in the design of any system.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ You beat me by 21 seconds, LOL! \$\endgroup\$
    – John D
    May 6, 2022 at 15:01
3
\$\begingroup\$

It all depends on the application. If you're building an audio amplifier and want to minimize thermal noise, you can decrease the bandwidth of your amplifier. The tradeoff of course is that you can't reproduce higher frequencies.

\$\endgroup\$
2
\$\begingroup\$

Halving bandwidth is the same as taking two measurements and averaging them together. The trade off is that it takes twice as long and you'll potentially average away any rapidly changing component of your signal. The benefit of averaging is that you reduce noise. Usually you set the bandwidth as high as it needs to be for your signal of interest and no higher in order to reduce noise.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.