1
\$\begingroup\$

I had a question about finding the input common mode of a differential switched capacitor integrator (shown in the figure below). The integrator is a conventional delaying integrator with two non-overlapping phases (\$\phi _1 \$ and \$\phi _2\$). In \$\phi _1 \$ the capacitors (\$C _{1+/-} \$) sample \$vin+\$ and \$vin-\$ respectively against the common mode voltage \$v_{cm}\$. In \$\phi _2 \$, the charge sampled is integrated onto the capacitors (\$C _{2+/-} \$). For simplicity we can assume that all the capacitors are equal \$(C _{2+/-}=C _{1+/-})\$

My question has to do with the final voltages at \$v_{x},v_{y} \$ at the the end of \$\phi _2 \$, after the amplifier hasenter image description here completely settled (we can assume infinite gain). What will these voltages be at the end of \$\phi _2 \$? and how will they depend on \$v_{cm}\$?

Thanks.

\$\endgroup\$

0

Reset to default

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.