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Why  noise power is proportional to the mean square of the short-circuit current (Read 2256 times)
baab
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Why  noise power is proportional to the mean square of the short-circuit current
Dec 22nd, 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).
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RFICDUDE
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Re: Why  noise power is proportional to the mean square of the short-circuit current
Reply #1 - Dec 30th, 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

is , in and vnxYs

is and in are two completely independent noise sources that are uncorrelated.

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

Just ask if there is something confusing about this.

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baab
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Re: Why  noise power is proportional to the mean square of the short-circuit current
Reply #2 - Jan 7th, 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.
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