calculate closed loop gain with non ideal op-amp

Thread Starter

sterstar

Joined Aug 14, 2017
2
Hi.I am currently stuck on the follwing question (as attached).
i know the formula to calculate the closed loop gain is

Av=Ao/(1+Ao(R1/R1+R2))+s/so and for AD711 , Ao is 70db while so is 1kHz
However i am stuck at the resistor part. there is only one resistor. does it mean that the part R1/R1+R2 will become 1?
 

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WBahn

Joined Mar 31, 2012
24,674
Go back and review what R1 and R2 are in that equation. Is the resistor in your circuit either of those?

Are you sure you are using the s/s0 term correctly?

Are you aware that R1/R1+R2 is equal to 1+R2 ?

Which non-ideal properties are you supposed to be taking into account? The presence of that 10 kΩ resistor implies that you might be looking at more than just finite gain and limited bandwidth.
 

Thread Starter

sterstar

Joined Aug 14, 2017
2
Go back and review what R1 and R2 are in that equation. Is the resistor in your circuit either of those?

Are you sure you are using the s/s0 term correctly?

Are you aware that R1/R1+R2 is equal to 1+R2 ?

Which non-ideal properties are you supposed to be taking into account? The presence of that 10 kΩ resistor implies that you might be looking at more than just finite gain and limited bandwidth.
the R1 and R2 refers to the resistors present at inverting input, however in this case there is no resistor at inverting input as it is a buffer. so, i didnt know which R to account for in the formula.

i know that s=j(omega) = j(2*pi*f), but i am not sure whether s0 is equal to 1kHz.

yes, i am aware that R1/R1+R2 is equal to 1+R2, but still, there is no resistor at the inverting input, perhaps R2 is the 10k resistor at non inverting input?
the questions asked me to calculate closed loop gain and of the buffer at frequency 1kHz and also the buffer closed loop 3db bandwidth
 

WBahn

Joined Mar 31, 2012
24,674
the R1 and R2 refers to the resistors present at inverting input, however in this case there is no resistor at inverting input as it is a buffer. so, i didnt know which R to account for in the formula.
And, hence, my question. You are trying to take a formula and force things into it. You see a formula with an R in it and you see a circuit with an R in it, and by gosh you're going to cram that R into that formula somehow.

i know that s=j(omega) = j(2*pi*f), but i am not sure whether s0 is equal to 1kHz.
Then perhaps you need to spend more time understanding where that formula comes from, what it tells you, and what it doesn't tell you.

yes, i am aware that R1/R1+R2 is equal to 1+R2,
Actually, you've just demonstrate very, very effectively the point I was trying to make.

You did not mean to use R1/R1+R2, you meant to use R1/(R1+R2). You (like many other people) got sloppy with your order or operations (remember, multiplication before addition?) and rely on people "knowing what you mean". But here you clearly didn't even remember what you meant when you read it. You need to be explicit with your expressions and respect the proper order of operations, otherwise you WILL come to grief on a regular basis.

but still, there is no resistor at the inverting input, perhaps R2 is the 10k resistor at non inverting input?
Do you see why I say that you see a formula with an R in it and you see a circuit with an R in it, and by gosh you're going to cram that R into that formula somehow.
 

MrAl

Joined Jun 17, 2014
6,454
Hi,

I also see "non ideal op amp" in the subject title but i dont see any non idealities in the little schematic. Maybe you are supposed to find those from the data sheet?
 
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