03/29/2016: Op ams and its application

I. Introduction:

Professor Mason briefly introduced us to the op ams, which was internally complicated, but contained only resistors, capacitors, and transistors

After showed us how op ams works, Professor Mason asked us to restate the definition of close loop feedback and open loop feedback, close loop simply means received the op am with external feedback, and open  loop means gain of op am without external feedback

 II. More example of op ams:

We were asked to redraw an non-ieal op ams. Notice that there is a resistors between Vn and Vp. Vin is from power supply, Vout is voltage obtained from the other side. The ratio of Vout/Vin is also equal to 20k/10K resistor

We were asked to redraw this op am equivalent. We found the ratio Vs/Vo is roughly -50

 II. Experiment:
After doing some op ams practice, Professor Mason let us to do the experiment to test our results

We found the ratio of Vout/Vin =2. We obtained two resistors, R1=1.8kohms and R2=3.6kOhms to do this experiement

We had to look back the information of this op am to know how to setup the wire correctly, The setup is not hard, but understand instruction of the lab is most time consuming

Here is our table of data. Notice that at Vin=-3 and Vin=3 V we got the saturated circuit.

Here is our graph of Vin vs Vout, which is matched with what we learned about op ams today

Here is the ideal op ams. Basically, the resistance at negative terminal is too high to that no current can go through. The current at 10k and 20k is the same. Professor Mason redrew the circuit for us to make it easier to analyze

 III. Summary:
Today, we learned about the op ams circuit and the real-life application of it. There were two types of circuits we learned, the ideal one and non-ideal one. We can make op am as an ideal op am. The input resistance is very large and there is no output resistance, we can use the constraint equation Vp=Vn to solve for ideal op and i_p=i_n=0