Engineering Calculating the size of a capacitor in a sample and hold circuit

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The discussion revolves around calculating the size of a capacitor in a sample and hold circuit by determining jitter power and signal power. The user encounters an issue with a negative value for thermal noise power while attempting to calculate the signal-to-noise ratio (SNR). It is clarified that the SNR is 60dB, translating to a power ratio of 1000, which suggests that the calculations for jitter power may have been misapplied. The conversation indicates that a simple mathematical error could be the reason for the negative thermal power. The user is encouraged to re-evaluate their calculations to resolve the issue.
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Homework Statement
In sampling system the sample rate is 500 Ms/s with an input signal of range of 1 V
peak−peak and a clock jitter of 1 ps rms . A performance of 60 dB is required for all
signals the Nyquist baseband. Calculate its size of the capacitor.
Relevant Equations
Analog-to-Digital Conversion by Marcel Pelgrom Book
First, I calculate the jitter power with an equation given in the book
1582481543814.png


Next, I calculate the signal power
1582481656848.png


Now, I know that SNR = Psignal/(Pjitter + PthermalNoise),
1582481688065.png


However, I seem to be getting a negative value for the thermal noise power?? My plan was to use this thermal noise power and equate it to kT/C to get my capacitor size. But the thermal power is negative??
What am I doing wrong?
 

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the thermal power is negative?? What do you calculate for the SNR?
 
sprog said:
the thermal power is negative?? What do you calculate for the SNR?

your SNR is stated as 60dB, so the factor 10^(60/20) yields 1000 fold power ratio, so your scaled Pjitter is around 0.0003

0.125 - 0.0003 is positive not negative simple math mistake?
 

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