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Design >> RF Design >> Why noise power is proportional to the mean square of the short-circuit current
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Message started by baab on Dec 22^nd, 2013, 11:57am

Title: Why noise power is proportional to the mean square of the short-circuit current
Post by baab on Dec 22^nd, 2013, 11:57am

Hi,
I get stuck with noise calculation. Now, I am reading an application note about LNA.
Here is the link:
http://www.maximintegrated.com/app-notes/index.mvp/id/3169
Could you please help me explain this part in more detail?
It is an excerpt from that note.

It is assumed that noise from the source is uncorrelated with noise from the two-port network. Thus, noise power is proportional to the mean square of the short-circuit current (denoted by active-low Isc²) at the input port of the noise-free amplifier; and noise power due to the source alone is proportional to the mean square of the source current (active-low Is²).

Title: Re: Why noise power is proportional to the mean square of the short-circuit current
Post by RFICDUDE on Dec 30^th, 2013, 6:35pm

Hi Baab,

Note that when they talk about the short circuit current they are referring to the terminals of the noiseless circuit. When the input of the noiseless circuit is shorted then the current that flows through the short is due to three sources

i_s , i_n and v_nxY_s

i_s and i_n are two completely independent noise sources that are uncorrelated.

v_n is uncorrelated from the thermal source noise, but it can be partially correlated to the circuit input noise current i_n, so a correlation admittance Y_c is used to represent the correlation between i_n and v_n.

Just ask if there is something confusing about this.

Title: Re: Why noise power is proportional to the mean square of the short-circuit current
Post by baab on Jan 7^th, 2014, 1:55am

Thank you. I get it now. I misunderstood where the short refers to.
I have just re-read the note and this is the simplest explanation that I got.