EECS 145L Final Examination Solutions (Fall 2013)

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Bernard Ellis
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1 UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering, Electrical Engineering and Computer Sciences Department 1.1 Instrumentation amplifier (1) differential amplification (2) very high input impedance (3) low output impedance (4) constant gain over a bandwidth (5) high common mod rejection (6) constructed from three op-amps [2 points off for low input impedance] 1.2 Thermistor semiconducting material exponential increase in conductivity with temperature temperature promotes electrons into the conduction band [3 points off if semiconducting material not mentioned] [3 points off for saying that R increases with T] [2 points for not describing that temperature promotes electrons into conduction band] 1.3 PIN photodiode pn diode with an intrinsic layer between the p-doped and n-doped layers absorbed photons create electron-hole pairs (photocurrent) in I layer photocurrent is proportional to the intensity of light received [3 points off if P, I, and N layers not described] [3 points off if conduction described as electrons promoted into conduction band and no mention of a diode junction or n- and p-type doping] [3 points off for not describing that absorbed photons create electron-hole pairs] 1.4 Peltier heat pump series of elements with alternating high and low electrical mobility (or n- and p-doped with high and low electron concentrations) electrons are pumped by a dc voltage into elements of low mobility (or high electron concentration, p- type), causing heating electrons expand into elements of high mobility (or low electron concentration, n-type) to cause cooling [3 points off for polarity rather than mobility or concentration differences] [3 points off for not describing compression and expansion of electrons] [8 points off for describing Joule heating, which cannot pump heat] [8 points off for explaining that electron-hole pairs are created and then combine; this will not cause cooling] December 18, 2013 page 1 S. Derenzo

2 1.5 Metal film strain gauge long, thin metal layer bonded to an insulating sheet that changes its geometry with strain tension increases length, decreases area => resistance increases compression decreases length, increases area => resistance decreases [3 points off for not describing that electrical resistance changes due to geometrical changes] [3 points off for not describing shape and composition (metal on insulating sheet)] 2.1 1, Hz mv 1 p-p 0.1 Electrode 0.01 drift EEG signal EMG background ,000 10,000 Electrode drift 1.0 mv p-p from 0 to 0.1 Hz EEG signal mv p-p from 0.5 to 30 Hz 5mV at60hz EMG background 0.10 mvp-p from 100 to 3000 Hz [2 points off for each incorrect section] ,000 Frequency (Hz) 2.2 V 2 V 1 Gain 500 Notch filter LPF HPF Gain 200 December 18, 2013 page 2 S. Derenzo

3 [3 points off if differential gain >1000 before the notch filter- the 10 mv differential will saturate] [2 points off for gain error 10x] 2.3 EEG output 5 V from 0.5 to 30 Hz- system gain = 5 V/50 µv = 100,000 (100 db) Need to reduce all other backgrounds to below 2% of 50 µv p-p or 1 µv p-p in 0.5 to 30 Hz band. This corresponds to 100 mv at the output (2% of 5V) Electrode drift is 1 mv p-p at 0.1 Hz, need to reduce this by to below 1 µv p-p. => gain at 0.1 Hz = 100,000 x = 100 (40 db) EMG is 100 µv p-p at 100 Hz, need to reduce this by 0.01 to below 1 µv p-p. => gain at 100 Hz = 100,000 x 0.01 = 1,000 (60 db) 60 Hz differential is 10 mv p-p, need to reduce this by to below 1 µv p-p. ð gain at 60 Hz = 100,000 x = 10 (20 db) ð [2 points off for using a 12-pole LPF to reduce 60 Hz differential; this is not good engineering when at 60 Hz notch filter can be used] These critical points are shown by the filled circles below: Filter gain (db) 100 Filter gain 100, , , ,000 10,000 Frequency (Hz) [10 points off if system gain not shown outside the 0.1 to 30 Hz frequency band] 2.4 Analog filter must have the following gains 0.9 to 1.0 from 0.5 to 30 Hz HPF: 0.90 at 0.5 Hz and at 0.1 Hz LPF: 0.90 at 30 Hz and 0.01 at 100 Hz HPF: frequency ratio between 0.90 and is 5 December 18, 2013 page 3 S. Derenzo

4 n = 4 frequency ratio = 1.199/0.178 = 6.74 (n too low) n = 6 frequency ratio = 1.128/0.316 = 3.74 (n okay) fc between 0.5/1.128 = 0.44 Hz and 0.1/0.316 = 0.32 Hz. LPF: frequency ratio between 0.9 and 0.01 is 3.33 n = 4 frequency ratio = 3.162/0.834 = 3.79 (n too low) n = 6 frequency ratio = 2.154/0.886 = 2.43 (n okay) fc between 100/2.154 = 46.4 Hz and 30/0.886 = 33.9 Hz. [4 points off each for wrong LPF or HPF order] [2 points off each for wrong LPF or HPF corner frequency] 2.5 The common mode background is 105 mv at 60 Hz. To keep below the 100 mv output background limit, the overall system common mode gain must be < 1. [5 points off if background voltage given but no gain number] 3.1 The diameter d of the beam at a distance D is given by d = 10 3 D. For D = 1000 m, d = 1 m. The fraction of the beam subtended by a photodiode of diameter p is (p/d) 2. For p = 1 mm, the fraction is (1 mm/1 m) 2 = [3 points off for 10 4 ] [4 points off for using ratio of diameters rather than ratio of areas] [4 points off for ] 3.2 If the amplifier output is 1V, the input is 100 mv across the 100 MΩ resistor, and the photodiode current is 1 na. Since the wavelength is 1240 nm, each photon has an energy of 1 ev. With a quantum efficiency of 80% this means that a power of 1.25 nw is received by the photodiode. This is 1.25 x 10 9 of the laser power of 1 watt. From problem 3.1 we have (p/d) 2 = 1.25 x d/p = sqrt(0.8 x 10 9 ), and d = 28,000 p = 28,000 x 1 mm = 28 m. d = 10 3 D; D = 28 km. [4 points off for 28 m] [4 points off for 10 6 km and the answer to 3.1 used the ratio of diameters] [5 points off for > 10 6 km] In summary a 1W laser beam at a distance of 28 km will have a diameter d = 28 m, and only 1.25 nw will reach the 1 mm diameter photodiode to produce a photocurrent of 1 na, The resistor will produce a voltage of 100 mv which is amplified to 1 volt. December 18, 2013 page 4 S. Derenzo

5 3.3 From the equation sheet, the rms Johnson noise = 1.287x10 10 sqrt(r f). The amplifier bandwidth is 100 khz (gain-bandwidth/gain) The resistor Johnson noise = x sqrt(10 13 ) = mv, which produces 4.07 mv at the output of the amplifier. [2 points off for using a bandwidth of 1 MHz rather than 100 khz] 3.4 At 28 km the signal to noise is 1 V/4.07 mv = 245. This larger than S/N = 6 by a factor of Increasing the distance by a factor of sqrt(40.95) = 6.40 will reduce the S/N to 6. The distance is 28 km x 6.40 = 179 km. Check: At 179 km the beam diameter us 179 m. The fraction of 1 W received by the 1 mm photodiode is (1/179,000)2 = 3.12 x = nw. This will produce na photocurrent, and 2.5 mv across the 100 MΩ resistor. This is 6 imes larger than the Johnson noise rms of [6 points off if Johnson noise not considered] [answer accepted if statistical S/N used to get 6.4 times the answer to part 3.2] 3.5 The lens effectively increased p by a factor of 100 and the light collection efficiency by a factor of Distance would increase by a factor of 100 to 17,900 km. Signal depends on p/d. Both increase by 100. [2 points off for 10x increase] [4 points off for answering increase without a number] [6 points off for 2x increase] [7 points off for answering decrease ] 4.1 Want a sensitivity of 0.1 V/K and a shift of 273K. Inverting amplifier sums (1 mv/k) T V and amplifies by a factor of 100. December 18, 2013 page 5 S. Derenzo

6 10 V V 1 k! 1 µa/k Solid-state temperature sensor 1 k! 1 mv/k 100 k! [4 points off for not converting K to C] [4 points off for using a 100 kω series resistor and not using a bias voltage The 10 V bias cannot provide (100 kω)(273+50)(1 µa/k) = 32.2 volts] [4 points off for assuming that the solid state sensor produces a current without a bias voltage] [3 points off for 10x gain error] 4.2 Mount to strain gauges on a membrane cemented to a vacuum tank. The external pressure will cause the membrane to stretch. Place the two resistive strain gauges in a bridge and connect the bridge to an instrumentation amplifier. Bridge output is # R + "R V +!V! = V b 2R + "R! R & % $ 2R + "R ( ' = V # "R & b % $ 2R + "R ( ' ) V "R b 2R Strain is proportional to pressure: L/L = kp, where k depends on Young s modulus and the membrane geometry R/R = 2 L/L = 2kP Instrumentation amplifier output V 0 = G (V + V ) = V b kpg Output voltage per mm = 0.1V/mm = GVb10 4 Choosing Vb = 1 V, G = 10 3 Choose R so that V + V = 0 at 700 mm Hg. December 18, 2013 page 6 S. Derenzo

7 R S1 R S2 V b R S2 Fixed pressure below 700 mm Hg V+ V R S1 R [The membrane is under tension on both sides. However, no points off for incorrectly assuming tension on one side, compression on the other side.] [3 points off for a correct bridge circuit but gain error 10x] [2 points off for not providing a fixed pressure on one side of the membrane] 4.3 Attach the shaft of a circular resistor to a weathervane or wind sock that rotates to point in the direction of the wind. Contact point Shaft [4 points off for no bias voltage] 0 V 10 V 4.4 Attach four strain gauges to a flexible sheet and measure the force of the wind, similar to measuring the force of the weights in the strain lab. Connect the strain gauge in a bridge and connect the bridge output to an instrumentation amplifier with a gain G. L/L = kw, where W is wind speed, and k depends on Young s modulus and the sheet geometry Bridge output is # R + "R V +! V! = V b 2R! R! "R & % $ 2R ( ' = V b "R R December 18, 2013 page 7 S. Derenzo

8 Instrumentation amplifier output V 0 = G (V + V ) = 2 V b G k W Want to choose V b k G = 1 V per 40 km/hr Choosing Vb = 1V, at 200 km/hr L/L = 2 x 10 3, R/R = 4 x 10 3, need gain = In general gain = 2500/Vb V b Wind b 2 t 1 t 1 b 1 V + V t 2 b 2 t 2 b1 [2 points off for 2x error] [3 points off for 10x error] December 18, 2013 page 8 S. Derenzo

9 145L FINAL EXAM GRADE STATISTICS Problem Total Average rms Maximum Total final exam score distribution: L COURSE GRADE STATISTICS Grade Undergraduate Graduate Scores Scores A A 897, 901, 906, A 888, 889 B+ 860, 861, 870, B B 833 C+ 800, 807 C 776 C D+ D D F Maximum 1000 Average rms 42.2 December 18, 2013 page 9 S. Derenzo

UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences

UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences EECS 145L: Electronic Transducer Laboratory FINAL EXAMINATION Fall 2013 You have three hours to