graded_CHENG75-MP4

pdf

School

University of British Columbia *

*We aren’t endorsed by this school

Course

301

Subject

Electrical Engineering

Date

Feb 20, 2024

Type

pdf

Pages

Uploaded by DoctorPenguinMaster203

ELEC 301 Mini Project 4 Report By Marcus Cheng – 40407975 94

Part A) 1) First, we know that , R = 10k Ω and we get C = 1.592nF. Then, since it is a 2 nd order Butterworth filter, we know that the damping factor is . Knowing this information, we get A M = 1.586V/V from the calculation below. We know that A M = 1 + R 2 /R 1 and R 2 + R 1 = 10k Ω. Solving these two equations, we get R 2 = 3.69398k Ω and R 1 = 6.30602k Ω. The table below shows the values of C and A M that will turn the filter into a 2 nd order Butterworth filter with a 3dB frequency of 10kHz. A M C 1.586V/V 1.592nF Plugging in the values, we get the Butterworth filter circuit on the right: Below is the Bode plots for both the magnitude and phase of the circuit on the right. The solid line represents the magnitude, and the dotted line represents the phase. The figure on the right shows the location of the poles in the s-plane for this 2 nd order Butterworth filter.

To figure out what A M does my circuit start to oscillate, we can use the Routh array with the denominator of the H(s). Therefore, we expect A M = 3V/V when the circuit start to oscillate. Let’s see if that is the case when we simulate on LTspice. When I change R 2 = 6.7105k Ω and R 1 = 3.2895k Ω , A M ≈ 3. 05V/V. That’s w hen I start to see oscillation, which is quite close to the expected value. Below is the plot that shows the oscillation with the circuit on the right. The frequency of the oscillation is about 7.9kHz. I got this frequency by inverting the time different between two peaks. The graph on the right shows the root locus when A M = 3V/V. ° -2

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

The graph on the top right shows the root locus when A M = 2.5V/V, and the graph on the top left shows the root locus when A M = 3.5V/V. According to these plots, the poles will always stay on the left-hand side if A M < 3V/V, and the poles will always stay on the right-hand side if A M > 3V/V. Part B) Below is the time plot of the phase shift oscillator circuit on the right. The oscillation eventually decays to zero. After increase R to 29.1kΩ, we get the following stable oscillation.

Zooming in, we can see the oscillation in more detail. The measured frequency is about 64.957Hz. The calculated frequency is 64.975Hz using the formula from the course note handout: . They are quite close to the actual frequency. Below is the time plot of the phase shift oscillator circuit with half the value of R and C. The measured frequency is about 257.6Hz, and the calculated frequency is 259.898Hz. Below is the time plot of the phase shift oscillator circuit with double the value of R and C. The measured frequency is about 16.116Hz, and the calculated frequency is 16.244Hz. Overall, the calculated frequency and the simulated frequency are very close, any discrepancies should be due to human error, specifically not lining up the two peaks exactly when measuring the simulated frequency.

This phase-shift oscillator comprises of three RC filters, and each one is contributing to a 60° phase shift. These stages of filter made up of the feedback network of the overall circuit. The 180° phase shift of the feedback network cancel with the 180° phase shift of the inverting OpAmp circuit. Furthermore, the attenuation of the three RC filters is 1/29 at the operating frequency. If the gain of the inverting amplifier is 29, then there will be unity gain, which satisfy the Barkhausen criterion. If there is unity gain for this circuit, you get a pair of complex conjugate poles at the imaginary axis. This explains the oscillation. One reason why we might need to increase the 29R could be to move the poles was almost at the imaginary axis at the left half plane due to OpAmp non-idealities, so we need move them slightly to the right for them to be on the imaginary axis. Part C) To find the R B2 for the high gain at 1kHz, we can plot R B2 vs the gain at 1kHz. Plots below are exact that. Zooming in (2 nd plot), we can see the highest gain is about 42.12dB at R B2 = 20.13kΩ.

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

1) To find g m , r o , and h FE for each transistor on the right, we need to find the DC operating points Below is a table for that. V C V B V E I C I B I E Q 1 1.691V 0.6546V 0V 1.318mA 10.95uA 1.329mA Q 2 15.00V 1.691V 1.031V 1.828mA 13.13uA 1.841mA Now we can find the g m , r o , and h FE . 2) Below is the amplitude plot with R f = ∞. The 3dB frequencies are also include. The low 3dB point is 2.887Hz, the high 3dB point is 87.68kHz, and the midband gain is about 42.124dB or 127.7V/V. To measure the input impedance, we can put a 1mV AC source and measure the current through it. The input impedance is 1mV/132.66nA – 5kΩ = 2.538kΩ. Performing similar steps for finding the output impedance , we get 63.382Ω. By inspection, we know the feedback circuit uses shunt-shunt topology. To predict the frequency response, and input and output impedance for closed-loop, we need to first find the y-parameters shown below. 00 O O -4 -

I did not find the values of y 21 because it will not be used in the following calculation. After find the y-parameters, we can find A =V o /I s . Now we can find the parameters for predicting the closed-loop frequency response, which includes A f , f L3dB , and f H3dB , as well as the input and output impedance with feedback. Let’s verify these values via simulation. Below is the Bode plot with R f = 100kΩ. Below is a table comparing the calculated (predicted) values and the measured values. We can notice that the error for f L3dB and R if is higher than others. f L3dB f H3dB |A V | R if R of Calculated 0.3909Hz 647.5kHz 17.292V/V 343.66 Ω 8.582 Ω Simulated 0.5135Hz 651.1kHz 17.251V/V 248.43 Ω 8.706Ω % Error 24% 0.55% 0.24% 38% 1.4%

3) Below is the amplitude plot for R f = 1k Ω , 10k Ω , 100k Ω , 1M Ω , and 10M Ω. To find the measured βs, we can perform the following manipulations on A f : Below is a table comparing the calculated β s and measured βs. The error between the calculated β and measured β is almost non -existent. R f Measured A vf Calculated β Measured β % Error 1kΩ -14.02dB -10 -3 A/V -1.006*10 -3 A/V 0.60% 10kΩ 5.862dB -10 -4 A/V -1.003*10 -4 A/V 0.30% 100k 24.74dB -10 -5 A/V -1.002*10 -5 A/V 0.20% 1MΩ 37.83dB -10 -6 A/V -1.0015*10 -6 A/V 0.15% 10MΩ 41.58dB -10 -7 A/V -1.012*10 -7 A/V 1.2% 4) To measure the input and output resistance of the amplifier, we can follow the same steps for 2). To get the predicted amount of feedback, we can use these formulas: The table below shows the results of the procedures above. R f Measured R if Measured R of Estimate 1+Aβ using R if Estimate 1+Aβ using R of Predicted 1+Aβ 10kΩ 26.19 Ω 1.139 Ω 96.90 55.65 65.04 100k Ω 238.4 Ω 8.706 Ω 10.65 7.280 7.398 1MΩ 1.292k Ω 38.71 Ω 1.964 1.637 1.640 As seen in the table, the predicted amount of feedback is closer to the estimate amount of feedback using R of than the estimate amount of feedback using R if . Moreover, as R f increases, the difference between the estimate amount of feedback and predicted amount of feedback decreases.

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

5) If R f = ∞, then , so . For finding the estimated desensitivity factor 1+Aβ , we can find β using the formula in 3), then plug it into the equation. The table below shows values when R f = ∞ . R C A vf β Estimated 1+Aβ Predicted 1+Aβ 9.9kΩ -127.1V/V -7.426*10 -9 A/V 1.005 1 10kΩ -127.7V/V -45.09*10 -12 A/V 1 10.1kΩ -128.2V/V 6.076*10 -9 A/V 0.9961 The table below shows values when R f = 100kΩ . R C A vf β Estimated 1+Aβ Predicted 1+Aβ 9.9kΩ -17.24V/V -10.04*10 -6 A/V 7.407 7.385 10kΩ -17.25V/V -10.03*10 -6 A/V 7.403 10.1kΩ -17.26V/V -10.02*10 -6 A/V 7.399 As seen in the table, the estimated desensitivity factor is very close to the predicted desensitivity factor. Moreover, as we vary R C , the estimated desensitivity factor does not change much for both R f = 100kΩ and R f = ∞. Finally, for smaller values of Rf, the gain at lower frequency is larger at midband. This might be due to the frequency-dependent behavior of the amplifier and the feedback network. Specifically, at lower frequencies, the capacitive reactance is higher, which might lead to the increases of gain.