Closed loop gain using small signal analysis for Op-amp

Thread Starter

u-will-neva-no

Joined Mar 22, 2011
230
Hi again everyone...I am really bad at this type of analysis. In particular, I have an issue forming equations (using KCL) from the small signal model.

The question and the small signal model is attached.

I want to form a KCL so I would have
\( I_i = I_o\) from my drawing. From the solution provided to me, I know that I am already wrong.

The solution I am trying to get to is:
\(\frac{v_o - A_0_l v_d}{R_o} + \frac{v_0 -v_i}{Ri} = 0 \)

and \( v_d = v_i - v_o\)

I may have 'i' wrong with 'I' subscript because it is not clear on the solution. This is the formula would like help with if possible. Thank you!
 

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Jony130

Joined Feb 17, 2009
5,183
For the current direction as show in your diagram we can write one nodal equation for Vo node.

\(\frac{Vin - Vo}{Ri} - \frac{Vo - Aol*Vd}{Ro} = 0 \)

And if we solve this for Vo/Vin we get this result

\(\frac{Vo}{Vin} = \frac{Aol*Ri+Ro}{Ri + Aol*Ri + Ro} =\frac{Aol*Ri+Ro}{(Aol+1)*Ri + Ro} \)

And if Aol = ∞ then Vo/Vin = 1
 
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Thread Starter

u-will-neva-no

Joined Mar 22, 2011
230
Hi Jony, am I correct to say that if the current flows in one direction then voltage across it points in the opposite direction so then I would subtract the voltage that the arrow point at. I will attach an image to show what I am on about. What confuses me is what voltage I subtract from it.

On my diagram I have the arrow head pointing opposite to the current flow so I would have \(\frac{v_I - v_o}R_i\) for \(I_i\) and \(\frac{v_o - A_ol*v_d}{R_o}\) for \(I_0\) I just want to see if my thinking is correct and I am never fully told how to analyse a circuit.

I generally look at an equation and try to make sense of it but I want to be able to go from a circuit and form the equation... Please let me know how you came to your equation if my thinking is incorrect!
 

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Jony130

Joined Feb 17, 2009
5,183
Firs when we apply the nodal analysis we write the KCL for the node .

In your diagram you assume that Iin is flow into the node and Iout flow out from the node. So we have
Iin - Iout = 0 (KCL current entering the node is equal to the current thats leaving the node)

Iin = (Vin - Vout)/Rin ---> becaues you assume that Iin is flow into the Vo node, so Vin voltage must be greater then Vout.

And we have a similar situation for Iout current.

Iout = (Vout - Aol*Vd) / Rout ----> we assume that Vout > Aol*Vd so the

current can flow out from the node.
http://www.youtube.com/watch?v=x6qmVkB_bLs&feature=channel_video_title
 
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