Thermal (Johnson–Nyquist) noise of a resistor: How does the generated noise power depend on temperature?

Difficulty: Easy

Correct Answer: Directly proportional to absolute temperature (T)

Explanation:


Introduction / Context:
Every resistor generates random thermal noise due to the agitation of charge carriers. Understanding how this noise scales with temperature is essential for low-noise RF and instrumentation design.



Given Data / Assumptions:

  • Resistor at thermodynamic equilibrium (no shot noise source).
  • Thermal noise model (Johnson–Nyquist).
  • Bandwidth B is considered when converting spectral density to total noise.


Concept / Approach:

Thermal noise power spectral density at the terminals of a matched resistor is kT per hertz. The mean-square noise voltage over bandwidth B is 4kTRB. Hence noise power and voltage grow linearly with absolute temperature T (in kelvin).



Step-by-Step Solution:

Noise spectral density: N0 = kT (W/Hz).Mean square noise voltage: v_n^2 = 4kTRB.Therefore v_n ∝ sqrt(T) and noise power ∝ T.


Verification / Alternative check:

Cooling a low-noise amplifier or front-end (e.g., cryogenic LNAs) reduces T and thus reduces noise floor proportionally.



Why Other Options Are Wrong:

  • T^2: incorrect scaling; it is linear with T.
  • Independent of T: contradicted by k*T dependence.
  • Inversely proportional to T: opposite of true behavior.
  • Depends only on DC bias: thermal noise exists without bias.


Common Pitfalls:

  • Confusing noise voltage (proportional to sqrt(T)) with noise power (proportional to T).


Final Answer:

Directly proportional to absolute temperature (T)

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