We began the class discussing unity gain buffer amplifiers which simply output the same voltage that is put in because the difference between them is so negligibly small.
Continuing with class, we are asked to look at the image below. The sketch on the left can be transformed into the sketch on the right and some simple equations may be written.
It is a an amplified version of the original that is shifted upward.
Exercise 2:
Continuing , we are asked to look at a circuit, sketch below. It is an inverting op amp circuit powered by positive 9 volts and the negative power supply connected directly to ground. What may we predict will happen?
At the bottom of the above image is a sketch of what the graph may look like had it been powered by both positive and negative terminals. Now, due to the fact that the negative terminal has a zero supply delivered, the actual sketch is depicted to the right of the above image. The function only exists above the horizontal axis, zero.
Exercise 3:
Using nodal analysis, we are asked to solve for the voltages Va and Vo. Once the 2 KCL equations are written in terms of the variables, a few of them may be eliminated because they are equal to zero (i.e. ideal Op-Amp). At this point we end up with 2 equations and 2 unknowns, making this a much simpler problem.
Exercise 4:
If there is a parallel rule for solving circuits, does that imply there is a perpendicular rule? No, not really... But, while we know of inverting op-amps, does that imply that there are non-inverting op-amps? Yes! and to solve the gain we must once again do some nodal analysis.
Below is our analysis of the nodes for the above circuits. Since it is an ideal op-amp, two of the currents are equal and a single KCL equation may be written. Below, the output voltage may be expressed as a function of the gain and the input voltage.
Something that we may notice with non-inverting op-amps is that the gain may never be less that 1... this makes them less useful because of that limitation.
Let us now jump on over to analyzing summing op-amps. If we recall, op-amps are amplifiers that perform mathematical operations, this one will do just that by summing, currents in this case. Nodal analysis is performed and something interesting may be seen. If all the resistors are the same, then the voltage out is simply the sum of the voltages going in.
It may also be noted that by playing with the resistors, we can turn a sum amp into an averaging filter.
Exercise 6:
Below is the sketch of the circuit we are to design and build. We must chose resistors that will not result in a saturation of our data.
Once we calculate and design the circuit we may build and test it. Below is the simple summing circuit we built.
Data was taken and entered into MATLAB in order to graph and analyze. In a perfect world there are calculated sets of data that are equal to any measured data. In the real world there is wear and tear and that leaves room for error. As shown in the graph below, although we calculated for resistor values that would not lead to saturation, our sad little operational amplifier did so anyways. Still, we initially had a line almost parallel to that of the predicted result... and for a couple of simple engineering students and Mt. SAC, that is good enough.
Exercise 7:
If there are summing amplifiers then there must also be difference amplifiers, right? Right! Let us first looks at the sketches below and see what is happening. If you solve for the difference between the upper left sketch and the lower left sketch, the result is the lower right sketch. It is the upper right sketch shifted down 1 unit.
Exercise 8:
Continuing with difference amps, below we see a circuit, and by calling it an ideal op-amp and using nodal analysis we may solve for the voltage out in terms of some gain value and the voltage sources. The result is the product of the gain and the difference of the voltage inputs.
One more exercise to practice solving circuits with difference op-amps.
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