ECE 304: Final Exam Fall '01 PROBLEM 1 (25 POINTS) 1. Define gain margin and phase margin using Bode plots for a three-pole amplifier as an example. 2. State the rate-of-closure rule and illustrate with a Bode plot. 3. Using a Bode gain plot, identify the midband region of an amplifier. Indicate in what regions you would use the short-circuit time constant method and the open-circuit time constant method to find the corner frequencies. Write a formula for each region expressing the corner frequencies as a function of the time constants. 4. Describe a situation where using a voltage follower could increase the bandwidth of an amplifier. Sketch the circuit. 5. Sketch the circuit for a cascode amplifier. Explain its advantages and how they are achieved. PROBLEM 2 (25 POINTS) 00+R4100Meg+CF{CF}+C1{C1}Sweep+-ACVS1V+RO{RO}PARAMETERS:CF = 100pFC1 = 10uFRO = 10kR1 = 1kVP VDB-++-E2GAIN = -1E5+R1{R1} VOV1VSFIGURE 1: Macromodel of two-pole amplifier using the VCVS PSPICE part E; gain is −105V/V. For the double-pole amplifier of Figure 1, • Derive a formula for the exact (no approximations) frequency dependence of the gain, VO/VS. Treat R4 as infinite. • Derive approximate formulas for the two pole frequencies assuming wide separation. • Using the given parameter values, make a Bode phase and Bode dB gain plot vs. frequency in Hz over the range 1Hz ≤ f ≤ 100 MHz. Mark values and frequencies for each break point and mark slopes/decade of frequency. • Define the Miller approximation and discuss its validity for the corner frequency of this example.John Brews Page 1 12/10/2001PROBLEM 3 (25 POINTS) VP-++-E3GAIN = -1E5+R1{R1}+C3{C2}Sweep+-ACVS1VVDB+R2{R1}+C1{C1}00PARAMETERS:C1 = 1nFR1 = 1kC2 = 1nF0+R3{R1}+C2{C2}-++-E2GAIN = 1 V2V1VS VOFIGURE 2: Macromodel of triple-pole amplifier using two VCVS PSPICE parts E; first-stage gain is 1V/V; second-stage gain is −105V/V. For the triple-pole amplifier of Figure 2, • Draw the open-loop Bode plots. Label all corners and slopes. • Move one pole to obtain zero phase margin if the amplifier is used in a non-inverting amplifier with feedback β=1 V/V. Use Bode plots for a first guess and use formulas to make any needed adjustments. • Draw the open-loop Bode plots with the new pole for frequencies from 1 mHz to the unity loop-gain frequency f1. Label all corners and slopes. • Find the value of capacitance C1 that will result in the new pole frequency. John Brews Page 2 12/10/2001PROBLEM 4 (35 POINTS) Parts for Problem 1 are shown in Figure 3 They are to be connected to form a feedback amplifier that provides a specified AC current to the load resistor RL. 0+R2{RC}+R4{RA}0T-SectionDifferential Amplifier FollowerLoad+C11_F0+R1{RN}+-VCCDC = 15VBiasResistor0+-VCCDC = 15VQ3Q2N2222Transient AnalysisI41mA1kHzQ2Q2N2222Capacitor+R6{RB}+R5{RV}+R3{RQ}+R3{RL}I3DC = {IVF}Q1Q2N2222I2DC = {IDA}Source FIGURE 3: Parts available are one each of diff amp, follower, source, load, bias resistor, capacitor, and T-section Given only the parts shown in Figure 3, make a schematic showing how to connect them to provide a specified current into load RL. The following specifications are given: 1. RL = 100Ω, RN = 1 kΩ 2. AC load current is 20 mA amplitude with no clipping for 1 mA AC input 3. Diff amp to have maximum difference-mode gain consistent with specifications 1 & 2. 4. Voltage follower to have minimum DC bias current consistent with specifications 1 & 2 5. T-section resistor values should be determined. Zero values are OK if desirable. Marks are based on your detailed answers to the following questions. Your design calculations should be keyed to these questions so I can locate where and how you executed your decisions. Explain the following parts of your design choices in words: 1. How did you select the type of feedback? 2. How did you determine the T-section resistor values (RA, RB, RV)? 3. How did you decide the follower DC bias current (IVF)? 4. How did you decide the diff amp DC bias current (IDA)? 5. How did you determine the value of RC? 6. Explain your use (or not) of the coupling capacitor C1. 7. Explain your use (or not) and sizing of the bias resistor RQ. John Brews Page 3 12/10/200115.00V-17.51mV+R4{RF}26.53uA+R9{RF}1.000mA2.801VI3DC = {IVF}-702.4mV+R10{rPI1+rPI2}12.96nA+R3{RL}0AI2DC = {IDA}I5DC = 1mA1.000mA+R6{(rPI3+RC)/BetaAC}20.00mAPARAMETERS:rPI1 = 907rPI2 = 1.34krPI3 = 277BetaAC = 20202.079V+-V1DC = 15V29.95mA0ACSweepI11A-627.1mV-29.12uVTransient AnalysisI41mA1kHz0V+R7{RL}20.00mA0+R5{RV}26.53uA+R1{RN}-2.627V00+R8{RV}19.00mA-17.38mVQ2Q2N22224.595mA-4.622mA+R2{RC}4.692mAQ3Q2N222296.83uA 19.90mA+R11{RN}44.68nA+C11_F-2.912V-875.5uV-875.5uV-++-E1GAIN = 1E5Q1Q2N222226.66uA5.352mA-5.378mAPARAMETERS:RN = {RF+1/(1/RV+1/RL)}RL = 100RF = {19*RV}RC = 2.6kIVF = 20mAIDA = 10mARV = 330 FIGURE 4: Solution for last problem 0-694.4mV-1.332mV+R1{RN}PARAMETERS:RN = {RF+1/(1/RV+1/RL)}RL = 100RF = {19*RV}RC = {(15-2*2.87)/(0.5*IDA+IVF/BetaDC)}IVF = 20mAIDA = 5.3mRV = 187BetaDC = 197RQ = {1/(1/RN+1/(RV+RF))}Q3Q2N2222100.8uA 19.90mAV-26.64mVI3DC = {IVF}6.249VQ1Q2N222214.49uA2.772mA+R7{RF}7.122uA5.527V+OUT{RL}Q2Q2N22222.499mA+-V1DC = 15V25.27mA+R3{RQ}14.49uA-1.332mV0+C11_FI2DC = {IDA}0+R2{RC}2.600mAACSweepI11ATransient AnalysisI41mA1kHz0I-26.64mV0+R17{RV}7.122uA OUTJohn Brews Page 4