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,

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.

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

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.