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Design >> Mixed-Signal Design >> switched capacitor question
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Message started by nxing on Nov 19^th, 2010, 9:02am

Title: switched capacitor question
Post by nxing on Nov 19^th, 2010, 9:02am

Hello
I have a basic question regarding switched-capacitor integrator circuit, look at the attached drawing (from Paul Gray's book, chapter 6). on the top phase (sample phase), I am wondering what's the charge at right plate of C1? I am thinking because it's connected to ground (or any DC voltage), there should be no Charge, if that is the case, why we say at integrating phase (low drawing), the charge will be dumped to capacitor C2?

Thanks,

Title: Re: switched capacitor question
Post by raja.cedt on Nov 19^th, 2010, 7:39pm

hi,
The charge on the plate is -cs*v1. Connected to ground doesn't mean zero charge, here more or less ground will provide this charge to cap plate.

Thanks.

Title: Re: switched capacitor question
Post by buddypoor on Nov 20^th, 2010, 1:57am

Hi nxing,

the answer is simple: During the sample phase the feedback capacitor must be discharged to be prepared to the next period (with charge transfer).
However, this assumes that the charge from the last period has been transferred before to the next stage which acts as a sample-and-hold stage (sample phase of the second stage).
That means, you never can treat a S/C integrator as a separat and isolated stage because, in reality, it always is operated in conjunction with some other circuitry.

Title: Re: switched capacitor question
Post by nxing on Nov 22^nd, 2010, 9:46am

Hi Guys,

Thanks for the reply. I think now I understand it. The ground actually is a AC ground but a DC one. So there is still charge on the left side of C1. Is that correct?

Regards,

Title: Re: switched capacitor question
Post by Julian18 on Dec 17^th, 2010, 4:57am

Hi nxing:
at zero voltage doesn't mean no charge is stored. e.g. the earth can be thought of as always at voltage 0, but of course it has a lot of charges.

Best regards

Julian :)

Title: Re: switched capacitor question
Post by Julian18 on Dec 17^th, 2010, 4:59am

nxing wrote on Nov 22^nd, 2010, 9:46am:

Hi Guys,

Thanks for the reply. I think now I understand it. The ground actually is a AC ground but a DC one. So there is still charge on the left side of C1. Is that correct?

Regards,

IMHO. whether it is a AC ground or DC one is irrelevant.

Title: Re: switched capacitor question
Post by adesign on Jan 4^th, 2011, 3:16am

Hi,

I'm referring the Figure6.8 on page 413.

Here is the sequence of operation:
During phase Φ1, LHP(Left Hand Plate) of C1 is connected to V_s whereas RHP(Right Hand Plate) of C1 is connected to ground. This means the net charge on C1 is C1.V_s during phase Φ1. During this phase only, the capacitor C2 is fully discharged. In this phase, RHP of C1 and LHP of C2 are hard-grounded.

During phase Φ2, switches S1, S3 and S4 opens, whereas switches S2 and S5 closes. The hard-grounding from RHP of C1 and LHP of C2 is removed. Then the virtual-ground functionality of opamp starts, such that this node acquires virtual-ground. During this operation, the charge transfers occurs and this leads to the net voltage on V_o as

V_o = (C1/C2)*V_s

Thanks.

Title: Re: switched capacitor question
Post by Nandish Mehta on Feb 22^nd, 2011, 9:34pm

Just a note.

Also during phase when C1 bottom plate is grounded it gets charged to Vs i.e. the input. In the next phase that charge gets transferred by forcing virtual ground at the negative plate.

I feel this is a very simple scheme which is parasitic insensitive (parasitic of sampling capacitor & switches which actual matters the most).