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I was doing some reading on noise improvement and stumbled on an article by Scott Hunt at Analog Devices. Here is the reference link: https://www.analog.com/en/technical-articles/11-myths-about-analog-noise-analysis.html#author

According to the first subsection, due to the Johnson noise, reducing resistance helps improve SNR. However, for applications such as measuring resistance, as V=IR, the gain in voltage supersedes the increase in Johnson noise. This leads to an improvement of exactly 3dB for doubling the resistance.

Does anyone know how he obtained the value 3dB, if the resistance value itself is unknown?

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All resistors produce the same noise power.

If the measuring system has a higher input resistance than the resistor, which is often the case, then the noise is predominantly a voltage measurement, and reducing the resistor will lower the noise voltage.

The converse is true for current. And measured into a matched load, the received noise power is independent of the resistor value.

3dB is an approximation for half power (it's 3.010 to a few more decimal places).

Regardless of the initial resistor value, if changing the ratios produces half the noise power, you'll get 3dB.

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Your reference is talking about the SNR of a source, where the signal is the voltage developed across a resistor from a fixed current flowing through the resistor, and the noise is the resistor’s Johnson noise. The signal voltage squared S is proportional to R^2, while the mean square noise voltage N (in some bandwidth) is proportional to R. Thus the SNR = S/N is proportional to R, and doubling R increases the SNR by 2, or by 3 dB. The SNR of this source does not depend on anything outside the source. It is independent of the input resistance of whatever it is connected to. Of course, if the source is connected to a noisy amplifier, then there are other noise sources and the source SNR only tells us the best that we can possibly do.

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