AC Circuit Problem - Notes on an Operational Amplifier Circuit - with General Impedance Values

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AC Circuit Problem - Notes on an Operational Amplifier Circuit - with General Impedance Values

       You have seen the circuit below with two different sets of impedances.  Here we will do a general analysis of this circuit.  We have these ends in mind.

• To give you some guided practice at the analysis of circuits like these.

        Writing those equations, we get:

• At the inverting node (the "-" input to the op-amp), there are two currents flowing towards that node, once through the C₁ from V₂, the other through R₂ from the output node, V_o.  Using KCL, you should get:
• V₂/Z₂ + V_o/Z₃ = 0
• You can solve this fairly easily for V₂.  If you can getV₂ in terms of the output voltage, then when you write the equation at node "2" you will be able to get the output voltage there instead of V₂.
• V₂ = -V_oZ₂/Z₃  - - - Use this expression below!!!
• Now, you can also write the node equation at node "2", and you should get:
• (V_in - V₂)/Z₁ - V₂/Z₂ + (V_o - V₂)/Z₄ = 0
• Then, you can take this equation and insert the expression above for V₂.
• First - rearrange the equation at node "2" to get:  (V_in)/Z₁- (V₂)(1/Z₁ + 1/Z₂ + 1/Z₄) = -V_o/Z₄
• That isolates V₂ so that you can insert the expression for V₂ in terms of the output voltage.  That should give you the expression below.  Here we left in all of the minus signs so you can see how things work out.  Do that one step at a time.
• (V_in)/Z₁- (-V_oZ₂/Z₃)(1/Z₁ + 1/Z₂ + 1/Z₄) = -V_o/Z₄
• Put all of the output voltage terms on the LHS:
• [V_oZ₂/(Z₁Z₃)] + [V_o/Z₃] + [V_oZ₂/(Z₃Z₄)] + [V_o/Z₄]= -V_in/Z₁
• V_o= -[V_in/Z₁]/[Z₂/(Z₁Z₃)] + [1/Z₃] + [1Z₂/(Z₃Z₄)] + [1/Z₄]
• V_o= -[V_inZ₃Z₄]/[Z₁Z₃ + Z₂Z₄ + Z₁Z₄ + Z₁Z₂]
• You can take this last as a result, and it is in a standard form, i.e. a ratio of polynomials, after you put in the specific forms for the various impedances.