We will examine the op-amp version of the differential amplifier. You will appreciate the relative simplicity of the op-amp version. You may also observe that this amplifier is actually a combination of the inverting and non-inverting amplifiers.

**An image of your Tinkercad circuit is required (please insert in the correct position in the report) for each measurement section.**

**Note: Some parts produce a perfect result, as expected from a simulation software. In practice, a difference is expected.**

THE EXPERIMENT

1. Setup the power supply to have ±10V dual voltage outputs.

2. Construct the circuit shown below; pay special attention to the ±10V voltage and the ground connections from power supply. Label the pin numbers to make your connection process clear.

3. By the Superposition Theorem, we can measure the voltage output produced by a single input signal at a time; and the total output voltage will be the algebraic sum of each individual output voltage (or gain).

4. Measure the differential voltage gain A_{va }using superposition as follows:

__a. Ground V___{b}__;__

b. Use the signal generator to generate the following signal and connect it to V_{a}:

- Type of the signal:
__Sine waveform__ - Frequency:
__f = 1kHz__; - Peak amplitude (Peak value):
__5V__ - Offset DC voltage:
__0V__.

Use the oscilloscope to observe input V_{a }on channel 1 and output V_{out }on channel

Sketch the signal voltages observed on the oscilloscope in the following:

c. From the waveform on scope, find the peak-to-peak values of the signals on input and output, respectively:

V_{a(PP) }=

V_{out(PP) }=

d. Please circle the correct answer: the phase relationship between V_{a }and V_{out }is:

**Out of phase or****In phase**

e. Calculate the voltage gain A_{VA }based on your measurement (show your calculation procedures):

A_{VA }=

5. Don’t change the setting of the signal generator, repeat step 4 but this time with __V__**_{a }grounded **and apply the input signal to V

V_{b(PP) }=

Vout(PP)

A_{VB }=

6. Calculate the Average of the ** absolute value **of the voltage gain A

A_{AVG(D) }=

7. Now we can measure the __common mode gain A__**_{cm }**as follows:

__Short V___{a }and V._{b }together- Apply a large 1kHz sine wave to both V
_{a }and V_{b }(e.g., 5V_{peak}) together. - Measure V
_{OUT }carefully and determine the common mode gain A_{cm }(show your calculations), it should be <<<1. (You may see some noise in the output, just record the RMS value of the input/output and calculate the A_{cm})

V_{a(RMS) }=

Vout(RMS)

A_{cm }=

8. Calculate the COMMON MODE REJECTION RATIO (CMRR) in dB:

CMRR = 20log(A_{AVG(D)}/A_{CM})

9. The highest CMRR results will be reached when the resistors are perfectly matched. For the poorly matched resistors, the results could be very different. To illustrate the effect of poorly matched resistors, change your resistors R_{3 }and R_{4 }as follows (leave R_{1 }and R_{2 }untouched):

**R****3****= 12K****Ω ****R****4****= 82K****Ω **

10. Repeat the measurements above and then determine the CMRR in dB:

Average A_{AVG(D) }=

A_{cm }=

CMRR in dB =

11. Compare results in step 10 and step 7, did you see any difference in A_{cm }and CMRR?

**QUESTIONS**

- Use the theory you have learned from class,
the output voltage V__derive___{out }in terms of V_{a }and V_{b}, assume R_{1}=R_{3}=10kΩ and R_{2}=R_{4}=100kΩ (Hint: You need to find the equation based on “virtual short”). - Repeat question a) when R1= 10KΩ, R2= 100KΩ, R3= 18KΩ and R4= 82KΩ.

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