[quote author=pancho_hideboo link=1266231079/0#13 date=1266482886]
Manas wrote on Feb 17th, 2010, 9:52pm:I MEAN TO SAY THAT ANY SPURIOUS CHARGE CAN BE REMOVED BECAUSE OF OPAMP ACTION
AND IN THE STEADY STATE IT WILL BE SAME AS IF YOU DON'T HAVE ANY CHARGE IN THE CAP.
Not correct.First, Vx is a floating node. So Qx can not be removed.
Second, it could result in stagnated bias points. So OP Amp can't work correctly.
Manas wrote on Feb 17th, 2010, 9:52pm:THE VOLTAGE AT THE ASKED NODE IS --> = Vdd * [C2/(C1+CX+C2)].
AS CHARGE REDISTRIBUTION HAPPENS.
PLEASE NOTE THAT IF C1>>CX (AS MENTIONED IN MY DIAGRAM 2.5pf),
THE EFFECT OF Cx CAN BE NEGLECTED AND ONE CAN GET THE DESIRED VOLTAGE DIVISION.
Not correct.C1, C2 and Cx share Qx in initial state.
Qx is not charged in only small Cx.
So after restribution of charge, I can't expect Vx correctly.
If feedback networks are composed of only capacitor,
- It is difficult to fix DC bias points properly.
It could results in extreme bias points, e.g. all devices are saturated or cut off.
This is very true for CMOS OP Amp.
- No feedback effect at DC and low frequency. It could cause unstability.
If feedback networks are composed of only resistor,
- It could be unstable for high frequency.SORRY,
I AM WRONG. AS IT IS A FLOATING NODE, THE Qx CHARGE CAN'T BE REMOVED. HENCE THAT POINT CAN'T HAVE DESIRED VOLTAGE DUE TO OPAMP ACTION.
SORRY AGAIN FOR THE WRONG INTERPRETATION.
AGAIN THANKS A LOT !!