# Closed loop gain using small signal analysis for Op-amp

Discussion in 'Homework Help' started by u-will-neva-no, Nov 5, 2011.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
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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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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
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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

Last edited: Nov 6, 2011
3. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
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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4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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.