In this lab, we took a look at op amps by applying them to a real world scenario. The idea was that in the real world, different sections of your circuit may require different signal voltages/currents. In order to provide the correct signal to these sections, you have to pass them through an intermediate conditioning circuit, and op amps are an excellent way to produce the conditioning needed. Our problem involved a conditioning circuit where the input needed to be 0 to 1V while the output need to be 0 to -10V. So we created a voltage divider to provide the 0 to 1V input and then set up an op amp(LM741) as an inverting amplifier to get the 0 to -10V output. The calculated values for the input and feedback resistors needed to produce a gain of -10 were Ri=1kohm and Rf=10kohm. We used a variable resistor in our voltage divider to vary the input voltage at .25V increments from 0 to 1V and measured the output voltage at each reading. The gains at each of these readings was then calculated and most came out to be right around -10.
| Component | Nominal | Measured | Power Rating |
| Ri | 1000 ohm | 990 ohm | 1/8 W |
| Rf | 10000 ohm | 9870 ohm | 1/8 W |
| Rx | 360 ohm | 354 ohm | 1/8 W |
| Ry | 68 ohm | 67.7 ohm | 1/8 W |
| V1 | 6 V | 6.01 V | 36 W |
| V2 | 6 V | 6.05 V | 24 W |
| V in | V out | Gain | V ri | I ri | V rf |
| 0.00 V | 0.00 V | 0.00 | 0.00 V | 0.00 mA | 0.00 V |
| 0.25 V | -2.31 V | -9.91 | 0.233 V | 0.235 mA | 2.31 V |
| 0.50 V | -4.99 V | -10.00 | 0.499 V | 0.504 mA | 4.99 V |
| 0.75 V | -6.99 V | -9.98 | 0.700 V | 0.707 mA | 6.99 V |
| 1.00 V | -9.96 V | -9.98 | 0.997 V | 1.00 mA | 9.95 V |
Ideally the circuit would have been set up best with 3 power supplies, but we used only 2 which caused some undesired consequences. Since our + op amp supply was also being used to power the voltage divider, the current drawn from our 2 op amp power supplies came out to be 75.5mA and -1.8mA which is not what we desired.
