Op Amp Capacitance multiplier

[madelineinpar]

Homework Statement

Please, I would like to know how to find the factor by which the capacitance is multiplied
A picture of the Op Amp is here: http://en.wikipedia.org/wiki/File:Cap-mult-op.svg
Zc is complex impedance of C

Homework Equations

none given

The Attempt at a Solution

choose,
I2 through R2 left to right
I1 through R1 top to bottom

so Iin = I1+I2 = (Vin-Vout)/R2 + Vin/R1 +Zc
Vout =V- = V+ = (ZcVin)/(R2+Zc)
...
Iin=Vin((R2+R1)/(R1(R2+Zc)))

And then I'm not sure how to calculate the effective capacitance. I think it is supposed to be increased by a factor of R2/R1 but I don't know how to get that.

 

[mark726]

To find the factor by which the capacitance is multiplied, you need to first understand the circuit and its components. The op amp in the picture is a non-inverting amplifier, which means that the output voltage is equal to the input voltage multiplied by a factor, known as the gain (A). In this case, the gain is given by the equation A = 1 + (R2/R1). This means that the output voltage will be amplified by a factor of (1 + (R2/R1)).

Now, to find the effective capacitance, we can use the formula for the impedance of a capacitor, which is Zc = 1/(jωC), where j is the imaginary unit, ω is the frequency, and C is the capacitance. Since we are dealing with a complex impedance, we can use the parallel impedance formula to find the effective capacitance (Ceff). This formula is given by Ceff = 1/((1/Zc) + (1/R2)). Substituting Zc = 1/(jωC) and solving for Ceff, we get Ceff = (R2C)/(1 + (R2/R1)). This shows that the effective capacitance is indeed increased by a factor of (1 + (R2/R1)).

I hope this helps you understand how to find the factor by which the capacitance is multiplied in this circuit. Please let me know if you have any further questions. Good luck with your studies!