Hello,

i was reading about one SC amplifier, which is called flip-around architecture. it is very simple circuit with just one cap around the loop.

i am trying to understand the hold phase operation.

the question is:

when we connect the left plate of the cap (which is charged to VIN) to the o/p of the amplifier (which is at Vcm) and if the o/p node moves from VCM to VIN i think there must some current (i mean just imagine the amplifier is made-up of a 5T OTA which has high rout), right?

if yes, who will supply the current since the cap can not because it is OPEN on the other side?.

Thanks,
A Kumar R

Hi Ken,

Thanks for the reply.

it is impossible to explain the operation w/o assuming a parasitic cap at the -ve i/p of the amp. other parasitics do exist, but w/o this parasitic the circuit doesn't work i believe.

after giving some thought, i kind of getting at the following.

if we assume some parasitic at that -ve i/p of the amp....we have a current path. but, since this node is the amp input and if any charge lost at that node the amp tries to correct the output towards that new VIN voltage, and it does so as long as the lost charge is put back (or taken out depending on the direction of VIN w.r.t VCM level) at that node. of course, it depends on the gain A, and the new voltage would be VCM +/- (VCM-VIN)/A.

Thanks,
A Kumar K