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Mar 25th, 2023, 7:25pm
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 Summarized phrasing of below topic "Outp.ref kT/C noise combined with an in (Read 1304 times)
 bernd2700 Community Member Offline Posts: 34 Summarized phrasing of below topic "Outp.ref kT/C noise combined with an in Nov 12th, 2021, 2:31am   I indeed found a much simpler and clearer formulation of my core question, here it is (but therefore without all the Matlab equation proofes, these can be found in my detailled post some days ago). Root-spectral density @ low frequencies obtained with Cadence PNoise analysis for the following:Track-and-Hold circuit stand-alone:T&H stand-alone:               (Node “Vcs”):          ~ 45.4 nV/sqrt(Hz)Sampled with Ideal S&H:    (Node “Vcs_ext”):    ~ 90.3 nV/sqrt(Hz)     ((See attached picture 1))Complete integrator circuit:Node “Vo” :         903 uVrms/sqrt(Hz)Node “Vo_ext” :   903 uVrms/sqrt(Hz)     (here exactly the same as the result for node “Vo”)     ((See attached picture 2))If we divide this result by the open loop gain of 10000, we get again the 90.3 nVrms/sqrt(Hz). So with just now adding the (lossy, open-loop) integrator to the stand-alone T&H, the whole circuit now suddenly resembles a Sample & Hold (S&H), not anymore a T&H. Why?? The first part is still a T&H circuit! What is a correct theoretical model for this complete stuff?Thank you very much for any answer, any hint, anything in advance!Bernd2700 Back to top IP Logged
 bernd2700 Community Member Offline Posts: 34 Re: Summarized phrasing of my core question "kT/C noise combined with an integrator" Reply #1 - Nov 12th, 2021, 2:32am   Complete integrator Back to top IP Logged
 Ken Kundert Global Moderator Online Posts: 2358 Silicon Valley Re: Summarized phrasing of below topic "Outp.ref kT/C noise combined with an in Reply #2 - Nov 12th, 2021, 12:29pm   Quote:the whole circuit now suddenly resembles a Sample & Hold (S&H), not anymore a T&H. Why?? The difference between your T&H and your S&H is that the duty cycle of the output of the T&H is roughly 50% whereas the duty cycle of the S&H is roughly 100%.  That explains why the noise the output of the T&H is roughly half that of the S&H.Your integrator is acting as a S&H.  The duty cycle at the output the integrator is roughly 100%.You do not need pnoise to see this phenomenon. You can also see it with a PAC analysis.-Ken Back to top IP Logged