Why the offset of the buffer can be divided by the gain of input amp in this PIC?

[rainman.cn]

I read the paper which say the following pic, the offset of the buffer can be divided by the gain of input amp, I dont understand , can anyone show me why?

 

[brmadhukar]

Re: S&H problem

Simple -- The offset is amplified and fed back in the negative phase. This continues till the amplifier oputput is at the original ooffset leve.

hope it helps.

 

[A.Anand Srinivasan]

Re: S&H problem

the offset of the buffer is the same whether your signal is amplified or not.... so when your input is amplified the relative voltage level of your offset gets reduced by a factor of the gain of the input amplifier and hence it gets divided by the gain of the input amplifier...

hope you find this useful....

 

[rainman.cn]

Re: S&H problem

thanks brmadhukar and A.Anand Srinivasan,
my following analysis is right?
I assume input stage has no offset, the picc2 is the model,
A(Vin-Vout)+Vos=Vout, so
Vin-Vout+Vos/A=Vout/A, if A is large, then
Vout-Vin=Vos/A

 

[A.Anand Srinivasan]

Re: S&H problem

ya.... your derivation is right....

A direct theoretical way..... now consider the picture below... in the first case consider the output the signal to offset ratio is Vin/Vos.... consider the second case it is AVin/Vos and can be equivalently Vin/(Vos/A)... it can be seen that the contribution of the offset has been reduced by an factor given by the gain of the preamp....