Hello guys, I need to create db vs frequency graph of a low pass filter. I have capcitor and resistor values. How am I going to find the gain ? R1 is connected to - side of the op amp, + side is grounded. Rf and C is connected in paralel from r1 to vout.

 

Use circuit theory?

It works the same as if you just had resistors. You need to use either differential equations or Laplace. Probably Laplace...

It just becomes Zf/Zin instead of Rf/Rin for the gain.

 

I forgot it :/ i took the class 3 years ago :/ thats why I asked. For gain I have (Rf+Xc1)/R1, but i forgot what does Xc1mean and where should I plug the freguency:/

 

How did you manage to get that?

It's close but wrong if Rf and C are in parallel.

Xc is the reactance of the capacitor. It's the frequency dependent term:
\(X_c = \frac{1}{2\pi fC}\)

However your equation won't have Xc, it'll have Zc (the impedance):

\(Z_c = -jX_c\)

 

http://www.electronics-tutorials.ws/filter/filter_5.html , i have the same circuit in bottom with component values

 

How am i going to find the gain after this? (r2+xc)/r1?

 

Do you remember how to solve op-amp circuits?

I recommend you use the Laplace transform:

\(Z_c = \frac{1}{sC}\)

When you do your plot:
\(s = jw=j2\pi f\)

Zc is parallel with Rf. You treat it the same as a resistor when you write equations.

 

all i remember is current does not go inside of the op amp :/

 

can matlab plot imaginary numbers?

 

Matlab has a bode plot function.

http://www.mathworks.com/access/helpdesk/help/toolbox/control/ref/bode.html

Anyways, your only mistake is that Rf is parallel with Zc. You're adding them which implies series.
You're also missing a negative sign but that doesn't really matter in this case.

 

I have one more question, in my book it says that, K = -z2/z1 , Can i write z2 as (rf+c) , so the gain would be - (rf +c)/r1?

 

They're in parallel, not series:

\(Z_2 = R_f || Z_c = \frac{1}{\frac{1}{R_f} + sC}\)

 

I apologize i did correct on the paper , lol =)and yes i added the negative sign as well thank you. so is that final ( Zc//Rf)/R1 = gain is correct ?

 

Yep that's all it is.

There's standard form to put it into which is good practice to follow:

\(H(s) = \frac{K}{\frac{s}{w_o} + 1}\)

This way you can set s = 0 and see the low frequency gain instantly (K), as well as the cut off frequency (wo).

 

thank you very much my friend have a good night =)