Prelim_exam_sol_f05

Prelim_exam_sol_f05 - ECE Prelim – Fall 2005 Problem...

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Unformatted text preview: ECE Prelim – Fall 2005 Problem Solutions Problem 1 (Core) Code Number:__________ 2 Problem 2 (Core) Code Number:__________ (10pts) Consider the following binary function F = (a + b)(cd ) . a. (3pts) Draw the CMOS transistor-level schematic of F. b. (4pts) Draw the stick diagram of F such that there is no break on the diffusion strips. c. (3pts) Assuming that the width of all transistors used in F is 2w, compute the worst-case delay in terms of τ when F is driving a load of 10Cinv. The meaning of the symbols is: w is the width of the minimum-size nFET, Cinv is the input capacitance of a minimum-sized inverter, R is the resistance of the minimum-size nFET, and τ = R·Cinv. Assume that the mobility of electrons is twice as high as that of holes. Solution to ECE3060 problem (Fall’05): Answer: F = (a + b)(cd ) = ab + c + d (c) pull-up: worse delay path is Vdd-a’-b-F. So, delay = (R+R)·10Cinv = 20τ. pull-down: worse delay path is F-a’(or b)-c-d-Gnd. So, delay = 3·0.5R·10Cinv = 15τ. Thus, the worst-case delay is 20τ. 3 Problem 3 (Core) Code Number:__________ 4 Problem 4 (Core) Code Number:__________ 5 Problem 5 (Core) Code Number:__________ 6 Problem 5 (continued) Code Number:__________ 7 Problem 6 (Core) Code Number:__________ 8 Problem 6 (Core) Code Number:__________ 9 Problem 6 (Core) Code Number:__________ 10 Problem 7 (Core) Code Number:__________ 11 Problem 8 (Core) Code Number:__________ 12 Problem 8 (continued) Code Number:__________ 13 Problem 9 (Core) a) Output power = 1.732 x 13,800 x I x 0.8 Line current = I = 784.46 Amperes Apparent power = 1.732 x 13,800 x 784.46 = 18,750 kVA Code Number:__________ Reactive power = ( Apparent power 2 – Load power 2 ) 0.5 = 11,250 kVAR b) At full current, the Apparent Power can be calculated to be Apparent Power = 1.732 x 13,800 x 1000 = 23,901 kVA Real Power = 22,500 kW Net Reactive Power = ( 23,901 2 – 22,500 2 ) 0.5 = 8,062 kVAR Additional Load Reactive Power = 7500 / 0.7 x ( 1 – 0.7 2) = 7,651 kVAR Total Load Reactive Power = 7,651 + 11,250 = 18,901 kVAR Total Capacitive kVAR Needed = 8,062 – 18,901 = - 10,839 kVAR Capacitive Current Needed = 10,839,000 / (3 x 13,800) = 261.8 A/capacitor Capacitance value = 13,800 / 261.8 A = 52.7 ohms = 50.3 uF/capacitor c) Power factor at new operating point = 22,500 / (1.732 x 13,800 x 1000) = 0.941 14 Problem 10 (Core) SOLUTION Code Number:__________ (a) Van Ia -j Ia Xs Ean Phasor diagram (b) Van = 208/1.732 = 120 volts Ia = 7200/(3 x 120x0.8) = 25 A (c) power factor is 0.8 lagging, so Cos Θ = 0.8 so Θ = - 36.87 Ean = Vt – jIa Xs = 120 /_0o - j4(25/_-36.87o) = 100/_-53.13o (d) Ia-new Θnew δ Van -j Ia-new Xs Ean-new 15 Problem 10 (continued) Code Number:__________ (e) Now Ean-new magnitude is simply 1.5 x 100 = 150 volts due to increased field current. But power remains the same, So power P = {(120 x 150)/4} sin δ = 7200/3 So sin δ = 0.533 and thus δ = 32.23o (f) Now 120/_0o - Ian-new jXs = Ean-new /_-32.23o 16 Problem 11 (Core) Code Number:__________ 17 Problem 11 (Core) Code Number:__________ 18 Problem 12 (Core) Code Number:__________ Solution: (a) Without light illumination, there will be no excess carriers. Δpn = 0 (b) At steady state, the excess minority carrier Δpn = GLightτp = 1015·10-6=109 cm-3 (c) Δpn = 109 << ND Low-level injection FN = Ei + kTln(ND/ni) FP= Ei - kTln(p/ni) ≅ kTln(Δpn /ni) FN – FP = kT ln(NDΔpn/ni2)=0.026ln(105)=0.30 (eV) (d) Under low-level injection, the electric field in the uniformly doped semiconductor is assumed to be zero. The time-dependent light illumination will incur the time-dependent carrier generation. Since the EHP are uniformly generated inside of the semiconductor, there is no spatial distribution for excess carriers and hence ∇p = 0 ∂Δp n 1 Δp = ∇ ⋅ J p − n + G Light τp ∂t q J P = qpμ p ε − qDP ∇p ε ≈ 0, ∇p = 0 ∂Δp n Δp = − n + G Light ∂t τp B.C.' s Δp n (t = 0) = 0, Δp n (t → ∞) = G Lightτ p (e) Solve for 1st order differential equation in (d): Δp n (t ) = G Lightτ p [1 − exp(−t / τ p )]....t ≥ 0 Δpn(t) GLightτp t=0 1 ` 2 3 4 5 t/τp 19 Problem 13 (Core) Code Number:__________ 20 Problem 14 (Core) A. Code Number:__________ 21 Problem 14 (continued) B. Code Number:__________ 22 Problem 15 (Core) Solution: (a) The closed loop characteristic polynomial is: Code Number:__________ 0 = s 4 + 9 s 3 + 26 s 2 + (24 + K ) s + (−K ) Routh Array is: s4 : 1 s3 : 9 s 2 : b1 s1 : c1 s 0 : −K b1 = c1 = For stability: 26 24 + K −K 0 0 −K 0 0 0 0 9 * 26 − (24 + K ) 210 − K = 9 9 b1 (24 + K ) − 9(−K ) 5040 + 267K − K 2 = 210 − K b1 210 − K > 0 2 (1) 5040 + 267K − K > 0 (2) −K > 0 (3) From (1): K<210, (2) K < 0, (3) K 2 − 267K − 5040 < 0 ⇒ −17.7 < K < 284.7 −17.7 < K < 0 Therefore, limits on K for closed-loop stability is: (b) Auxiliary equation s2 : 210 − K 2 s + (−K ) = 0 9 When K = 284.7, roots are at When K = -17.7, roots are at s = ± j 5.86 s = ± j .84 When K = 0, coefficient in the s0 row is 0 therefore root at origin of the s-plane Open-loop Poles: 0, -2, -3, -4 Zeros: 1 23 Problem 15 (continued) Code Number:__________ K>0 10 8 6 4 2 0 -2 -4 -6 -8 -10 -10 -8 -6 -4 -2 0 Real Axis 2 4 6 8 10 K=284.7 K=0 K<0 10 8 6 4 2 0 -2 -4 -6 -8 -10 -10 -8 -6 -4 -2 0 Real Axis 2 4 6 8 10 K=-17.7 24 Problem 16 (Core) Code Number:__________ 25 % ‡ ‹ Ž ¤q h ! %  Ž Wq h ! 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(a) What is the oxide thickness after the dry oxidation? (a) What is the final oxide thickness after the wet oxidation? Dry Oxidation: B=0.01042, A=0.233, tau=0.618 Wet Oxidation: B=0.529, A=0.183 33 Problem 23 (Specialized: Bioengineering) Code Number:__________ 34 Problem 24 (Specialized: Bioengineering) Code Number:__________ 1. Show a differential amplifier design using 3 Op-amps which give a total gain of 10,000 with very high input impedance. Lots of designs will be acceptable. To get high input impedance, the + and – EEG electrodes could be each connected to the + input of a non-inverting amplifier. The output of each of these two amplifier circuits could be then connected to the + and – inputs of the third Op-amp. The gain of the two non-inverting amplifiers could be 100, and the gain of the third amplifier circuit could be also 100. 2. Describe 2 reasons why it is important for the differential amplifier to have high input impedance. It is always safer to have less current passing through the patient, and the smaller current levels also reduces the polarization and resulting capacitance of the electrodes. 3. Why is it unsafe to connect the earlobe electrode to line ground? Any short in a building circuit could allow large amounts of current to pass through the line ground, and a portion of this current could then pass through the ground electrode attached to the patient. 4. Describe 2 simple methods for increasing the RMS Signal-to-Noise levels that do not involve filtering the signal. You should shorten the leads between the electrodes and the amplifier as much as possible, which might entail moving the amplifier circuit closer to the patient. It is also advisable to twist the two electrode wires and to shield the twisted wires using a wire mesh that’s connected to the earlobe electrode. 5. How do you expect the EEG signal to change when the level of anesthesia increases? The dominant bandwidth will shift to Delta, which are the lowest frequencies in the EEG power spectrum. 6. How would you display the EEG data to give the surgeon a simple indicator of the level of anesthesia? Again, many answers will be acceptable, e.g. plots of the Power Spectral Density. 35 Problem 25 (Specialized: Bioengineering) Code Number:__________ 36 ...
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