Op Amp circuit and RC circuit transient analysis.

Sep 2017
I have unable to figure out this 2 problem.

Please help me understand them.

For the first one #1, (Op Amp), I know that all ideal opAmp has V+ = V- but the problem states that Vout = A ( V+ - V-). If there were equal, Vout will be 0 then.

For the second one #2 ( I am really have a hard with how to redraw the circuit and derive the differential equation.

I have been out of school for some time so I am a bit rusty.

Please help


Nov 2013
New Zealand
The first looks like a controlled gain circuit (So A shouldn't figure in final analysis). I think you should be able to solve that with Kirchoff's laws. https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws. The voltage \(\displaystyle V^{+} = V_{in}\) as there is no voltage drop across RS as no current is flowing through the inputs of the op amp. R1 and R2 comprise a voltage divider which you can work out the voltage drop between them from ohms law. That should give you V- and from there you should be able to calculate \(\displaystyle V_{out}\) and the gain \(\displaystyle \frac{V_{out}}{V_{in}}\).

The second just looks like a capacitor resistance network. From google the current formula for a capacitor is \(\displaystyle I = C \frac{dV}{dt}\). I would assume the two \(\displaystyle V_{dd}\) points are connected, as they are at the same potential, even though they are not drawn that way. You should be able to solve this using Kirchoff's current laws.
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May 2017
The first circuit you uploaded is a negative feedback circuit for Op Amp. This configuration is named as the Non-Inverting config. The voltage at terminal 6 depends on the voltage coming from the output of Op Amp (terminal 1). You can use a voltage divider calc to find the voltage that will be in at the 6. Here he mentioned no current goes in, that's probably due to the high input impedance of op-amp. The attenuation formula for your circuit is B = R_i/(R_i + R_f).

For second question you will require watching some tutorials since it is not possible to answer it without revising the topic.