306L1

.pdf

School

Purdue University *

*We aren’t endorsed by this school

Course

306

Subject

Electrical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

6

Uploaded by ProfessorSnow18757

Report
EXPERIMENT 1: Analog Operations 4.1.1 Rf = 22000 Ri = 1000 4.1.2 4.1.3 Compare the input and output signals shown in Screen Capture 4.1.2. The output is inverted output. 4.1.4 Measure the experimental gain Vo / Vi of your inverting amplifier. Discuss the difference between the theoretical and experimental gain. The gain is 1.97/0.117 = 16.837 < 22
4.1.5 4.1.6 Compare and comment on the input and output signals shown in Screen Capture 4.1.5. The output has higher peak to peak value and the system have gain = 20.975, may be due to the saturation. 4.1.7 Discuss the difference in the output signal seen in Screen Capture 4.1.2 and Screen Capture 4.1.5. Why are the output signals different? The output is different is because the input amplitude is 2 times 0.2cos(2π1000t) as the first experiment 0.1cos(2π1000t). 4.1.8 What is meant by “op-amp saturation”? Op-amp saturation occurs when the amplifier output stops increasing even as the input increases. 4.1.9 What determines the saturation voltage? An op-amp's voltage cannot exceed that of the power supply. The output at saturation does not reach the full input voltage due to minor internal voltage drops in the op amp output circuit. 4.1.10 How can one determine whether a given input signal will cause saturation without testing it experimentally? The amplitude and gain of the input signal can be used to determine whether saturation will occur. If the output signal exceeds the range of the supplied power voltage for the given input signal and gain, saturation will occur.
4.2.1 What bandwidth do you expect for your amplifier with a gain of -22? (Hint: convert 22 to decibels, and refer to Figure 24 of the datasheet) 20 log 22 = 26.8485 in LF357 between 100k and 1M(Hz). 4.2.2 Create a log-log frequency vs gain plot using Excel, MATLAB, or similar. Include a curve to indicate ideal performance. F/gain 10 Hz = 24.3062 100 Hz = 22.5837 1k Hz = 22.5359 10k Hz = 22.2966 100K Hz = 20.622 1M Hz = 6.8421 4.2.3 Measure and report A, the low frequency gain, and fC, the cutoff frequency from your plot. 0.707 * 22 = 15.554 A = 100k(Hz) From observing the oscilloscope, I found that when gain = 15.554, the frequency roughly = 27.9kHz. fc = 27.9kHz 4.2.4 Compute the gain-bandwidth product: A * fC A * fc = 100k * 2.79k = 279M
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help