Hi,

I'm trying to figure out the transfer function of the following Op Amp Intergrator:

Can you please advice on that?

Thank you!

 

It is a integrator. Not sure what the biasing network on the + input is about, it could be replaced with either power supply or a resistor network and a single power supply.

 

Replace the feedback network with a single impedance that is the parallel combination of the R and C. Then analyze it like any other opamp circuit, noting that the voltage on the inverting input is not at virtual ground, but rather at Vplus.

 

Hi guys.
Thanks!

I analyzed it and reach its transfer function:
* R1 - Resistor at (-) Input
* Zf - Impedance of Feedback

How can you tell what is the pole of this transfer function?

Thanks again!

 

Back to page 1 if I may

 

\(Z_f=\frac{R_ f\frac{ 1}{ C_f s}} {R_f+\frac{ 1}{C_f s } }\)

etc.

No negative sign.

 

Well, I do not see R1 in your schematic so the transfer function must be:

Vo = -Zf/R2 * Vi + V(+)*(1+Zf/R2)

Since V(+) is DC, it has no dynamic influence. It just moves the output DC level up or down depending on the combination of VSET and Vdc DC voltage level. The signal rides on top of this DC level. So, the transfer function can be rewritten as:

Vo = -Zf/R2 * Vi + V(+)*(1+Rf/R2)

The signal amplitude is given by

Vo_signal = -Zf/R2 * Vi, and can be written as

Vo_signal = -1/(1+ s Cf Rf)* Rf/R2 * Vi

So the pole is at

1/(2 pi Cf Rf) = 159.15 kHz. It is only determined by Cf Rf time constant. Since both resistors Rf and R2 are 1kohm, the signal amplitude is identical with the input amplitude (inverted), but rides on top of the DC level we discussed. The unity gain starts dropping after 159.15kHz.