I am used to thinking about the effect of an attenuator on a system's noise in the following way:

  • there is some system noise (and signal) at the attenuator's input, coming from the preceding part of the circuit
  • this system noise is attenuated by the attenuator by the same amount as the signal
  • the attenuator is resistive, so it adds some noise power at its output as a function of its temperature

I can then calculate the noise power at the attenuator's output as the total of the attenuated system noise and the attenuator's Johnson noise. The signal to noise ratio is the ratio of the attenuated signal power to this noise power total.

I wondered, can I cheat this by using a perfect capacitive (or inductive) divider instead? These idealized components do not create Johnson noise, so do I win on the signal to noise ratio that I acquire at the output of my attenuator, compared to the resistive case? If so, why would I not always do this, instead of using resistive attenuators?

This answer implies that I cannot do this: a capactive divider will still increase the noise in the system. It mentions statistical fluctuations in the division of currents between the signal path and ground as the reason.

Could you help me understand this, or provide some more detail on the physical processes here?

  • \$\begingroup\$ Why not use a bandpass filter to remove noise? \$\endgroup\$ Commented Dec 21, 2020 at 18:48
  • \$\begingroup\$ I think unless the noise of concern is at a different frequency than the signal, filtering won't help me. \$\endgroup\$
    – janizer
    Commented Dec 21, 2020 at 18:58
  • \$\begingroup\$ If it is at same frequency I don't think you can do something. \$\endgroup\$ Commented Dec 21, 2020 at 19:02

1 Answer 1


A reactive attenuator will not have a constant input or output impedance, which is a serious disadvantage in many cases. Especially if you need to measure 'DC' or close to that, the capacitors become open circuits and the inductors shorts, neither of which is very useful. Large value inductors and capacitors are often significantly less ideal.

Offhand I can't think of a situation where it would be useful. And if you're noise-limited, why are you attenuating?

  • \$\begingroup\$ That's a good point. This is mostly a thought experiment: reading the linked answer, I realized I did not understand noise in these reactive divider circuits. \$\endgroup\$
    – janizer
    Commented Dec 21, 2020 at 19:01

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