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Spectre RF noise analysis

Northfork
Northfork over 10 years ago

I'm trying to simulate the thermal noise spectral density of a resistor in ic6.1.5-64b using the noise analysis in the ADE.
I am using the port component (50 ohm) along with a noiseless resistor (50 ohm).
I would expect the thermal noise spectral density to follow 4kTR, but I am getting something that is approximately 1/4th that when I simulate it.
Would expect 8.28e-19 V^2/Hz (@300K), but the plot shows 2.0019e-19 V^2/Hz.

Is there a reason for this, or is my setup incorrect?

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  • ShawnLogan
    ShawnLogan over 10 years ago

    Hi Northfolk,

    Just off-hand, if the two resistors are in series, I was wondering if you are observing the total noise of a 100 ohm resistor. Half of the resistor is not contributing any noise. Does this make sense?

    Shawn

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  • Northfork
    Northfork over 10 years ago

    I think I may understand what you are saying. Does this mean that if I take the output across the noiseless resistor for the noise analysis I will NOT get the variance of the noise source required to deliver the total noise power, RATHER I will obtain kTR/delta_f ( which is power density*R or Vout_rms^2/delta_f)? I guess I dont understand what the noise analysis is plotting. Maybe I'm just confused with the terminology.

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  • Andrew Beckett
    Andrew Beckett over 10 years ago

    A few years ago (nearly 10!) I wrote a solution on Cadence Online Support which describes precisely this and how both the input referred and output referred noise are calculated. The examples are not exactly the same, because I have a 50 series resistor and a 50 ohm load as well as the port, but it should be straightforward enough to figure out:


    There are two key points. First, the port is not at the simulator temperature, but at 290K (16.848°C) - this is controlled by the noisetemp parameter on the port (the reason is that the port is part of the test equipment and not on-chip, and is typically at a "standard" temperature. The second is that the noise you are measuring is the output noise, not the noise at the source. Because of this, there's a potential divider, dividing the noise voltage by 2, and thus the noise power by 4. So since 4kTR for a 50 ohm resistor at 290K is 8.0076e-19V^2/Hz, the output referred noise is 2.0019e-19V^2/Hz which agrees exactly with what you've observed.

    Regards,

    Andrew.

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  • Northfork
    Northfork over 10 years ago

    Ok that makes sense. Thanks!

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