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summi Community Member Offline Posts: 38	S/H ciruit noise Oct 13^th, 2011, 6:51am hi, i am new to analog design, i am just reading S/H noise, i came to know that irrespective of relation between switching time period and o/p time constant integrated noise is KT/C, i want to know why it is independt of the this relation? Thanks, Summi.
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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: S/H ciruit noise Reply #1 - Oct 13^th, 2011, 7:25am hello, Integrated noise is the area under the PSD, and so even if your switching frequency lower than ckt BW or otherwise you get same area, but only diference is if SW frequency is lower tahn the BW you get flat PSD because of uncorrelation in the othercase you get low pass PSD due to correlation. Hope this helps, raj.
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HdrChopper

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Re: S/H ciruit noise
Reply #2 - Oct 18^th, 2011, 6:52am

Noise PSD is set by the switch resistor. Thus there is no switching involved.
Now, depending on the switching rate, the output noise PSD might look flat or colored (low pass like) as raja suggested, but the total power has to be the same as the input one, since no noise power was dissipated in any load (the cap stores such energy and does not dissipates it).
Thus when integrating the resulting output noise, the total power will be the same and equal to KT/C.

Best
Tosei

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Keep it simple

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RobG

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Re: S/H ciruit noise
Reply #3 - Oct 18^th, 2011, 2:46pm

I found the answers a bit confusing so let me say it another way. The noise source is the resistance of the switch. When the resistance is put in series with the cap the whole thing acts like a low-pass filter with a cut-off of 1/(2*pi*R*C) since you are taking the voltage across the capacitor. The resistance value falls out of the equation when you sum up the noise over the entire frequency band.

The reason it falls out of the equation is that a larger resistance will generate more noise, but the capacitor interacts with the larger resistance, rolling off the noise at a proportionately lower frequency. So a large resistance will generate a lot of noise, but the bandwidth will be small. Conversely, a small resistance will generate less noise, but the bandwidth will be proportionately larger. Either way, the total noise stays the same.

A derivation of the result is here and in most decent textbooks.

Note that this is just the noise put onto the capacitance by the switch. The amplifier also adds noise. Also, a larger switch resistance filters off the noise from the voltage you are attempting to sample, so kT/C is a bit of a lower limit on the total noise.

rg

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« Last Edit: Oct 18^th, 2011, 5:05pm by Ken Kundert »

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HdrChopper Community Fellow Offline Posts: 493	Re: S/H ciruit noise Reply #4 - Oct 19^th, 2011, 5:29am Rob, Thanks for a clearer explanation. I found mine confusing too (it was too late for trying to explain it when I wrote it!) Best Tosei
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summi Community Member Offline Posts: 38	Re: S/H ciruit noise Reply #5 - Oct 23^rd, 2011, 3:38am dear all, thanks for all replies, but i am not clear how smapling frequency and rc time constant makes PSD white or colored shape? BR, Summi.
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raja.cedt

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Re: S/H ciruit noise
Reply #6 - Oct 23^rd, 2011, 4:59am

hello summi,
please read my previous post, when fs>1/rc (of course rare case for real S/H ), whatever noise noise sampled from the resister will be stored on the cap for more time (due to high time constant) means there is lot correlation hence lowpass PSD. Where as in the other-case noise will stored stored on cap for less time, so very less correlation hence white PSD.

please read 'Design-Oriented Estimation of Thermal Noise in
Switched-Capacitor Circuits'. It has very clear explanation.

Thanks,
Raj.

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