Operational Amplifier circuit analysis...help!!!?

Question

I have a test tomorrow on this. It kind of makes sense in my textbook (Introduction to Mechatronics and Measurement Systems by Alciatore and Histand) but it doesn't do a good job with explaining problems and stuff. Looking at the back of the book I am pulling out my hair. Our professor gave us practice problems and solutions in pdf format but I still really dont understand. Does anybody know any good resources for this? Like how to analyze inverting & noninverting amps along with summers, difference amps, differentiators and integrators??

Answer 1

Always go straight to the horse's mouth - in this case look to the people who helped develop op-amps in the first place.

Here's just one of many wonderful resources from the company.

http://www.philbrickarchive.org/r1_febru...

Answer 2

I found this site to be useful once upon a time.  Its very basic, not sure if it will match the level of detail that your professor wants.

http://www.eas.asu.edu/~holbert/ece201/o...

Answer 3

you need to be more specific.

The technique I use is to assume the opamp has zero offset and zero input current, and infinite gain. That works in a lot of cases, and almost all DC examples.

For example, an inverting amp (always start with the simple example and be sure you understand it before moving on), with a resistor (Rf) between the output and the negative input and a resistor (Ri) from there to the input. + input to ground.

Assume: Ri = 1k, Rf = 10k, Vin = 1 volt.

The opamp will work to make the negative input zero. So the current into R1 = 1v/1k = 1ma. That current has to flow into Rf, as there is no other place for it to flow. 1ma through 10k is 10 volts, but because of the polarities, it is -10 volts at the output.

So +1v at the input gives you -10 volts at the output, and the gain is -10.

Case 2, the non-inverting opamp, with Rf (from output to - input), and Rg (resistor from - input to ground), with input to the + input.

Assume Rf = 9k and Rg = 1k

Again, assume +1v input. Since there can be no voltage between the inputs, that means the - input is also at +1v. This is across the 1k, causing 1ma of current. This same current must flow through the 9k, producing 9 volts across it. So the output is at +1v + 9v = +10 volts, for a gain of +10.

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