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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.
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  • \$\begingroup\$ Within operating region of the opamp, the node at the inverted input is at 0V (same as ground). But it is not ground. The currents going through C into ground is not going into that node at the inverted input. Therefore you cannot combine R and C as if they are connected in parallel. \$\endgroup\$ – rioraxe Aug 22 '16 at 1:02
  • \$\begingroup\$ @rioraxe Well, yeah I know, that's why I said "I know the way I try to solve the second circuit is wrong." What I don't know is how to do it the right way. \$\endgroup\$ – DimChtz Aug 22 '16 at 1:30
  • \$\begingroup\$ I do not understand the philosophy behind your "simplification". When you are interested in the transfer function (gain) - where is the input signal? \$\endgroup\$ – LvW Aug 22 '16 at 7:39
  • \$\begingroup\$ @LvW The "simplification" is for the second step (Vo) \$\endgroup\$ – DimChtz Aug 22 '16 at 13:07
  • \$\begingroup\$ @DimChtz, OK now it is clear. See my detailed answer. \$\endgroup\$ – LvW Aug 22 '16 at 14:13
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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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Dimchtz - both currents (I1 and I2) meet at the inverting node.

  • For the first circuit, you must therefore consider that I2 is the total current driven by output. With other words: The current through the parallel connection. That is what you have done - hence correct.

  • For the second circuit, I2 is only the current through R (connected to the inv. input) which is to be used. My recommendation: use the voltage divider formula for finding the voltage Vx across R||C - and as a next step you will find I2 by applying Ohms law for the resistor R (I2=Vx/R)

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You now have a system that has two unknown voltages rather than just one. You need to write out a system of equations for all the unknown voltages and currents and then solve the resulting system.

Lets define the following...
Vc = the voltage across the capacitor.
Ic = the current through the capcacitor. Let positive values point into node Vc.
Zc = the impedance of the capacitor at some frequency.
R1 = the resistor of value "2R" at the input.
R2 = the resistor of value "R" connected to the op-amp minus terminal.
R3 = the resistor of value "R" connected to the output.
I1 = the current through resistor R1.
I2 = the current through resistor R2. Let positive values point into node Vc.
I3 = the current through resistor R3. Let positive values point into node Vc.
Vin = the input voltage.
Vo = the output voltage.

I1 = Vin / R1, by ohms law
I2 = I1, by ohms law
Ic = -Vc / Zc, by ohms law
Vc = - I2 * R2, by ohms law
I3 = (Vout - Vc) / R3, by ohms law
I2 + I3 + Ic = 0, by Kirchhoff's Current Law

Now all that is left is to solve the set of five equations.

substitute -Vc = I2 * R2 in each equation to get...
I1 = Vin / R1
I2 = I1
Ic = I2 * R2 / Zc
I3 = (Vout + I2 * R2) / R3
I2 + I3 + Ic = 0

Substitute I1 = I2 in each equation to get...
I2 = Vin / R1
Ic = I2 * R2 / Zc
I3 = (Vout + I2 * R2) / R3
I2 + I3 + Ic = 0

Substitute Ic = -(I2 + I3) in each equation to get...
I2 = Vin / R1
-I2 - I3 = I2 * R2 / Zc
I3 = (Vout + I2 * R2) / R3

Substitute I2 = Vin/R1 in each equation to get...
-Vin / R1 - I3 = Vin / R1 * R2 / Zc
I3 = (Vout + Vin / R1 * R2) / R3

Substitute I3 = (Vout + Vin / R1 * R2) / R3 into the other remaining equation to get...
-Vin / R1 - (Vout + Vin / R1 * R2) / R3 = Vin / R1 * R2 / Zc

Solving for Vout gives...
Vout = -Vin / R1 * (R3 * R2 / Zc + R3 + R2)

A quick sanity check shows that the equation appears right for the DC case as Zc approaches infinity.

This approach generalizes to linear circuits with any number of nodes.

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