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, Vo.
Using KCL, you should get:
-
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 =
-VoZ2/Z3 -
- - Use this expression below!!!
-
Now, you can also write
the node equation at node "2", and you should get:
-
(Vin
- V2)/Z1 - V2/Z2
+ (Vo - 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: (Vin)/Z1-
(V2)(1/Z1 + 1/Z2
+ 1/Z4) = -Vo/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.
-
(Vin)/Z1-
(-VoZ2/Z3)(1/Z1
+ 1/Z2 + 1/Z4) = -Vo/Z4
-
Put all of the output
voltage terms on the LHS:
-
[VoZ2/(Z1Z3)]
+ [Vo/Z3] + [VoZ2/(Z3Z4)]
+ [Vo/Z4]= -Vin/Z1
-
Vo=
-[Vin/Z1]/[Z2/(Z1Z3)]
+ [1/Z3] + [1Z2/(Z3Z4)]
+ [1/Z4]
-
Vo=
-[VinZ3Z4]/[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.