find the transfer function of op amp

Thread Starter

squirby

Joined Aug 21, 2009
15
hey guys. wondering if anyone can help me determine the transfer function v2/v1. im a bit buggered on this, the placement of resistor one has stumped me.

much appreciated~
 

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steveb

Joined Jul 3, 2008
2,436
hey guys. wondering if anyone can help me determine the transfer function v2/v1. im a bit buggered on this, the placement of resistor one has stumped me.

much appreciated~
You would need to write out the equations and derive it. However, you need to decide what OPAMP model you will use. If you use the simplest OPAMP model assuming infinite gain and bandwidth, you will get the following:

\( T={{s^2}\over{s^2+s({\frac{1}{R_2C_2}}+{{\frac{1}{R_2C_1}})+{\frac{1}{R_1R_2C_1C_2}} }}\)

If you need help in deriving this (or a more complicated version with a more advanced OPAMP model), you should post your attempt so we can see where your trouble is.
 

mik3

Joined Feb 4, 2008
4,843
Find the current for the node between the capacitors and then add them algebraically to find the transfer function.
 

hgmjr

Joined Jan 28, 2005
9,029
The results of my analysis agreed with steveb's equation.

What you need to do is give the analysis your best shot and then post it here. There are many members who can assist you in understanding where you may have taken a wrong turn.

hgmjr
 

Thread Starter

squirby

Joined Aug 21, 2009
15
thx for such a quick response guys. sorry i forgot to mention i assumed ideal op-amp. i have attached my working. i know it is plain wrong but hopefully someone can tell me where i went wrong.

ps. i was also wondering if anyone has any texts or sites they can recommend for understanding op-amps. i have read the text on this site and other texts and it has helped me, i understand basic set ups with divided resistors and negative feedback, but i get stumped when there are more complex setups like this one (placement of resistor in between middle of 2 capacitors etc).

thx for all ur help!
 

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steveb

Joined Jul 3, 2008
2,436
thx for such a quick response guys. sorry i forgot to mention i assumed ideal op-amp. i have attached my working. i know it is plain wrong but hopefully someone can tell me where i went wrong.
The major mistake that I see is that the equations for I1 and I3 are not right. For current through an impedance device, you need to take the current equal to the voltage difference, divided by the impedance. Instead, you took the voltage at one end only. You should consider the node voltage (Vn) at the connection of C1, C2 and R1.

The equations could be written as follows:

V2=i2*R2
I2=(Vn-V2)*s*C2
Vn=V1-I1/s/C1
I1=I2-I3
I3=(V2-Vn)/R1
 
Last edited:

hgmjr

Joined Jan 28, 2005
9,029
Below is the beginnings of an analysis based on Millman's Theorem:

Simplifying equalities. Near the end of the analysis I replace the Xc terms with their s-domain equivalents.

\(X_{C1} = \frac{1}{sC_1}\)

\(X_{C2} = \frac{1}{sC_2}\)

\(V_x={\frac{{\frac{V_2}{R_1}}+{\frac{V_1}{X_{C1}}}}{\frac{1}{R_1}+{\frac{1}{X_{C1}}}+{\frac{1}{X_{C2}+R_2}}}}\)

\(V_+=\frac{{\frac{V_x}{X_{C2}}}} {{\frac{1}{X_{C2}}}+{\frac{1}{R_2}}}\)

The above equation can be solved for Vx:

\(V_+=\frac{{V_x}{R_2}} {{R_2}+{X_{C2}}\)

\(V_x=\frac{V_+(R_2+X_{C2})}{R_2}\)

Simplify the first equation at the top of this box and then set the two different expressions for Vx equal to each other and solve for V2/V1.

Oh yeah, I almost forgot. V+ = V_ = V2 due to ideal opamp operation.
Look Ma, no loops.

hgmjr
 
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