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A closed loop expression of OP

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xidian123

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An OP as fig1. The Gclosed=Vo/Vi=-Rf/R1

Now I want to write the transfer function in form of fig2.
So ,the forward transfer function and feedback loop transfer function can be writed as below

Gforward=-A (A is the voltage gain of OP)
β=R1/(R1+Rf)

But the result is wrong~~Who could kindly tell me why? Thanks~~
 

LvW

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The formula according to Fig. 2 is false. Where does it come from ?
The correct formula is

Gclosed=Hfor*A/1+Hback*A

with Hfor=-Rf/(R1+Rf) and Hback=-R1/(R1+Rf) and A positive.
(Another choice of signs is possible with A negative and both H expressions positive)
 

xidian123

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As the pic below,when calculate the Gforward , should the point A be disconnected ?Or how ?

Pls kindly express the process in detail. Ths ~~~
 

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xidian123 said:
As the pic below,when calculate the Gforward , should the point A be disconnected ?Or how ?
Pls kindly express the process in detail. Ths ~~~

No, it´s not necessary as the neg. input resistance is considered to be infinite (in reality Rin>>R1 parallel to Rf).

In words, based on the superposition theorem:
Hfor=part of Vi appearing at the opamp input (if Vo=0)
Hback=part of Vo appearing at the opamp input (if Vi=0)
 

    xidian123

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xidian123

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Thank you very much ,I get it ~~~

Hfor=part of Vi appearing at the opamp input (if Vo=0)
Hback=part of Vo appearing at the opamp input (if Vi=0)
 

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xidian123 said:
Thank you very much ,I get it ~~~

Hfor=part of Vi appearing at the opamp input (if Vo=0)
Hback=part of Vo appearing at the opamp input (if Vi=0)

Yes, thus you can calculate both voltages appearing at point A (add them together according to the superposition theorem),resulting in Vin.
As a 3rd equation you use Vo=-A•Vin and you arrive at the final expression.
 

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