I have a question from Razavi's book.

[twonsr]

Hi All :

In page 428, line 8,
it said "...raising the charge on CH to (CHVo + CinVx)."

Why ?

I don't understand why the charge on CH is to be raised.

how to explain and get this formula ?

Thanks

 

[samuel]

u'd better describe your problem in details.

 

[twonsr]

in sample mode :
the voltage across Cin is Vx.
the voltage across CH is Vo.


in amplification mode :
the voltage across Cin is still Vx.
but the voltage across CH is (CHVo + CinVx)/CH

Why is the CinVx added ?

 

[pd]

Hi,
I feel u r missing one thing here.Let me explain the complete scenario 2 u.
First only the feed back wire was connected between i/p and o/p.So Vo=Vx=V+=0
(virtual gnd concept )b'cos of -ve feed back(with no attunuation).Hence charge across Cin=0 in sampling mode.
In the amplification mode the feed back is hindered with a capacitor.Hence now there might b some diff. betn 2 i/p nodes.And Vx is no longer =0(line 6).
So where from this new charge across Cin come from?Definitely it has 2 come from the node Vin.So charge across Cin = Cin*Vx.But the basic poperty of cap is it doesn't allow the voltage 2 change across it.Hence the charge stored in Cin will also appear on the +ve terminal of Ch.Hence the new charge across Ch is now Ch*Vo+Cin*Vx.
I think this helps.
regards,

 

[pugongying]

Yes,pd did a good job,the key point is that the voltage at node X changes.
i.e. in sample mode,Vx=0.
in amplification mode,Vx<>0