DOC PREVIEW
UA ECE 304 - Study Notes

This preview shows page 1 out of 2 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Problem 1: Feedback network designProblem 2: Feedback compensationProblem 3: Corner frequency comparisonClosed book, calculators required; Friday, May 5; 11AM-1 PM ECE 304: Final Exam Spring ’06 Answers NOTE: IN ALL CASES 1. Solve the problem on scratch paper. Then, once you understand your answer, compose your answer sheet as follows: 2. Put your answer first, and 3. Follow your answer with an outline of your solution. Each major step in the outline should 3.1. Begin with a heading that describes the objective of that step, and should 3.2. Have a body where actual work is done, not just hand waving, and should 3.3. Conclude with a quantitative statement of the major result for that step (a number or formula or both). A mish-mash of calculations is not an outline! No marks without an outline. For all problems take the thermal voltage as VTH= 25.864 mV. Problem 1: Feedback network design A signal current source with an output resistance RN = 10 kΩ is to be used to drive a load RL = 50 Ω. A current gain of 10 A/A is required. This goal is achieved using a negative feedback amplifier. We have available a T-section of resistors for a feedback network and a voltage amplifier with a gain of Aυ0 = 105 V/V, input resistance RI = 1kΩ and output resistance of RO = 10 Ω. 1. Sketch the case of ideal feedback first and find the ideal βFB. Ideal means a single controlled current or voltage source and no feedback resistors. 2. Set up a two-port for the real-world feedback network using a T-section of resistors. 3. Select the T-section resistor values for the maximum loaded open-loop current gain. In maximizing the loaded gain, assume βFB maintains its ideal value from Part 1. 4. Sketch your final design with all components labeled. Show your work in your outline. Answers: Ideal βFB = 0.1A/A, resistor values: RA = 9RC =700.7 Ω, RC = vertical resistor, RA = left horizontal resistor, RB = 0 Ω. Problem 2: Feedback compensation A compensated open-loop amplifier has a gain expression given by EQ. 1 +α+α+=υυ3210ffj1ffj1ffj1A)f(A where Aυ0 = 105 V/V, f1 = 105 Hz, f2 = 5 x 106Hz and f3 = 107 Hz. The value of α is decided by the size of a compensation capacitor that shifts two of the pole frequencies (pole f1/α moves down and pole αf2 moves up in frequency as α increases). The amplifier is used in a feedback voltage amplifier with βFB = 10 mV/V. 1. Find the value of α for a closed-loop Butterworth two-pole step response. 2. Sketch the Bode phase and magnitude plots of your final design for the open loop and closed loop amplifiers, and label all breakpoints with (frequency, value) coordinates. 3. Mark the open-loop phase flip frequency f180 and the 1/βFB magnitude frequency f1/β. Show how you got them in your outline 4. Determine the gain and phase margin of the feedback amplifier. Answers: α=20, gain margin 26 dB, phase margin 58.5° Copyright 5/7/2006 by John Brews Page 1 5/7/2006Closed book, calculators required; Friday, May 5; 11AM-1 PM Copyright 5/7/2006 by John Brews Page 2 5/7/2006 Problem 3: Corner frequency comparison AMP TWO+R5{R_L}1.000mA0015.00V-1.479V2.000VI4{I_VF}-2.000VVDBPARAMETERS:R_S = 500V_CC = 15VR_L = 2kC_L = 10pFC_C = 5pF00+-V3{V_CC}51.10mA0015.00V+C7{C_C}-50.00mVQ_pQ02-100.0uA-10.00mA10.10mA0Sweep+-ACV21VI2{I_C}AMP ONE15.00V15.00V+C1{C_L}I6{I_CE}0Q_nQ0110.00mA-10.10mA+C2{C_L}714.7mV.model Q_p PNP (IS={I_S} BF={B_F}Cje={C_JE} Cjc={C_JC} Tf={T_F} Vaf={V_AF})-50.00mV15.00VI5{I_CB}00-1.285V+R1{R_S}VDBQ_nQ0310.00mA-10.10mADOT-MODEL:B_F = 100I_S = 10fAT_F = 0C_JE = 0C_JC = 0pFV_AF = 1E12+C6{C_C}I3{I_C5}-764.7mV0.model Q_n NPN (IS={I_S} BF={B_F}Cje={C_JE} Cjc={C_JC} Tf={T_F} Vaf={V_AF})0+R3{R_L}1.000mAI1{I_E1}Sweep+-ACV11V+C410_FI7{I_C3}Q_nQ2100.0uA10.00mA-10.10mA15.00V0I8{I_CE}0Q_nQ110.00mA-10.10mA+C310_F-764.7mV0+R2{R_S} VPVPOUT_2VPVPVP VPOUT_1Two amplifiers for comparison; AMP ONE (top) and AMP TWO (bottom) FIGURE 1 The transistors in Figure 1 have no internal capacitances, and do not exhibit Early effect (VAF = 1012 V). 1. Determine a formula for the upper 3dB frequencies f3dB of both amplifiers. 2. Evaluate your formulas. 3. Explain the effect (if any) of each stage in each amplifier upon the bandwidth. 4. Explain the effect (if any) of each stage in each amplifier upon the small-signal midband gain. 5. By simple inspection without calculation, when used in a feedback amplifier, which amplifier exhibits the largest overshoot in step response? Explain. Answers: f3dB(Amp 1) = 2.73 MHz, f3dB(Amp 2) = 7.63 MHz, Amp 2 shows most overshoot because CC doesn’t cause pole splitting (cascode), while Amp 1 does show pole


View Full Document
Download Study Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Study Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Study Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?