Electrical – Calculation of transfer function in OP-AMP circuits

operational-amplifiertransfer function

In a simple circuit like the following:

schematic

simulate this circuit – Schematic created using CircuitLab

I calculate the transfer function using I1 = I2:

  1. Vin = I1*R1 => I1 = Vin / R1
  2. Vo = – I2*Ztotal => I2 = – Vo / ( (Zc * R2) / (Zc + R2) )
  3. H = Vo / Vi = …

In 2. in my mind I always simplify it like this (and then use KVL):

schematic

simulate this circuit

Now I have another circuit and I am a bit confused. This is the circuit and again how I simplify it in my mind like the previous circuit:

schematic

simulate this circuit

Now just like the previous circuit I used I1 = I2 and:

  1. Vin = I1*2R => I1 = Vin / 2R
  2. Vo = – I2*(Ztotal + R) => I2 = – Vo / ( R + ( (Zc * R2) / (Zc + R2) ) )
  3. H = Vo / Vi = …

My questions:

  1. I know the way I try to solve the second circuit is wrong. How should I do it?
  2. In the first circuit (which is an example and I know it is correct) why doesn't I2 split into I3 and I4? What I know so far is that before you get into a new loop the current splits.

Best Answer

After a while when you recognize the patterns of impedance ratios determine negative feedback gain inverts the transfer function of the feedback, We see a Low Pass filter with a load R suppressed the feedback so it now amplifies as a HPF. I have also included the low pass response due internal Gain Bandwidth product of a simple 300kHz Op Amp (OA)

Since +in is ground then -ve is treated as a virtual ground just like the OA output, so both feedback R's for ac signals becomes a parallel Norton equivalent cct.

schematic

simulate this circuit – Schematic created using CircuitLab

You start by learning the math, then with experience use these asymptotic approximation methods to get a fast solution.

It looks like this. With experience you can figure this out in your head in a few seconds to understand it.

enter image description here

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