ENGR-4300 Electronic Instrumentation Quiz 4 Fall 2010 Name Section. Question Value Grade I 20 II 20 III 20 IV 20 V 20. Total (100 points)

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1 ENGR-43 Electronc Instrumentaton Quz 4 Fall 21 Name Secton Queston Value Grade I 2 II 2 III 2 IV 2 V 2 Total (1 ponts) On all questons: SHOW LL WORK. EGIN WITH FORMULS, THEN SUSTITUTE VLUES ND UNITS. No credt wll be gven or numbers that appear wthout justcaton. Provde as much explanaton as possble so that partal credt can be gven approprate. K.. Connor 1 Troy, NY

2 Queston I Dode Recter Crcuts (2 ponts) V1 Transormer V2 - + V3 R1 Load Resstor n:1 DIODE RIDGE The dagram above shows the applcaton o a dode brdge or perormng rectcaton o the voltage rom the output o the transormer. The snusodal source voltage V1 = 12V RMS and R1 = 5kΩ. 1. (3pt) Knowng that the voltage ampltude s 2 larger than the RMS voltage, what transormer turns rato n:1 wll gve as close as possble to a 6V ampltude at V2? (n should be rounded to an nteger.) n = (12 x 1.414)/6 = => 28 n:1 = 28:1 2. (3pt) What wll the actual peak voltage be on the output o the ull wave brdge (across R1). Let the dealzed dodes have V on =.6V and V2 s the voltage rom the turns rato n queston 1? V = (12 x 1.414)/28 2x(.6) = 4.86V 3. (3pt) Gven R1 above, what s the peak current that wll low through any o the 4 dodes? I = V/R = 4.86/5k =.972m K.. Connor 2 Troy, NY

3 Queston I Dode Recter Crcuts (contnued) 4. (3pt) For a 6Hz nput voltage V1 a capactor s added n parallel wth R1 to reduce the rpple n the voltage across the load resstance so that the droop s less than.25v. Whch o the ollowng values s the mnmum capactance necessary to acheve ths? a) 1μF b) 17μF c) 33μF d) 1μF Droop =.25/4.86 =.5 T = 1/12Hz = 8.3ms τ = RC =.83/.5 =.166 C =.166/5 = 33μF 5. (3pt) It s decded to use a 68μF capactor to lter the supply voltage. What 3 dgt code wll be wrtten on ths capactor to ndcate ts value? 687 => 68 x 1 7 pf = 68 x 1 7 x 1-12 = 68 x 1-5 =.68F 6. (5pt) For a quck calculaton o the voltage droop wth the 68μF capactor and 5kΩ load resstance, assume a 5V ampltude 5Hz sne wave has been deally ull wave rected (V on = V). Use the perod between adjacent peaks as the maxmum droop tme and assume the exponental decay can stll be modeled as a straght lne n ths nterval. Wth these smplcatons, how much wll the voltage droop rom ts 5V maxmum value? 5 T= 1/(2x5Hz) = 1/1Hz =.1s τ = RC =3.4s droop/.1 = 5/3.4 droop = 5x.1/3.4 = 14.7mV.1 RC = 5x68u = 3.4 K.. Connor 3 Troy, NY

4 Queston II Zener Dode Crcuts (2 ponts) R1 C + C - 5 V1 = -1 V2 = 1 TD = TR =.5ms TF =.5ms PW = 1e-1 PER = 1ms V1 Most o the source voltage wll appear across the dode due to 1/15 dvder. Input should be trangular wave wth p- p ampltude o 2V R2 1k D1 D1N75 The crcut above s a Zener dode voltage regulator. Zener dode s used to regulate the voltage across the load resstor R2 n the crcut. Shown below are two gures, one o whch shows the correct voltages or ths conguraton. 1V a) (5pts) Identy whch o the two gures s correct by crossng out the one that s ncorrect. Explan your answer. Most o 1V s enough to turn on the Zener. b) (5pts) On the correct gure, label whch plot corresponds to the voltages at ponts and and across resstor R1 (C + to C - ) 5V C V -5V -1V s.5ms 1.ms 1.5ms 2.ms 2.5ms 3.ms V(V1:+) V(R1:2) V(V1:+,R1:2) Tme Zener clamps voltage at about.7v orward and about 5V reverse (Zener) K.. Connor 4 Troy, NY

5 1V 5V V -5V -1V s.5ms 1.ms 1.5ms 2.ms 2.5ms 3.ms V(R1:1) V(R1:2) V(R1:1,R1:2) Tme c) (6pts) From the gven normaton, determne the Zener voltage or ths dode V Z and the orward bas voltage necessary to turn the dode on n the orward drecton. Exact answers are not requred. V Z s a lttle less than 5V, say about 4.7V, V orward s a lttle less than 1V, say about.8v wth a range o answers accepted. It s not necessary to be able to read the graphs precsely but the answers should be reasonable. d) (4pts) Sketch the V-I plot or ths dode. Your sketch does not have to be perect, but t should show the man characterstcs o the dode. I V Should have more curvature at the corners, but ths s also acceptable. K.. Connor 5 Troy, NY

6 Queston III LEDs and Phototransstor Crcuts (2 ponts) +5V R U2 NC COM NO 1 2 U1 3 E Relay _SPDT_b C D bove s a typcal optcal solaton crcut wth an LED/phototransstor par. The logc gate and nputs may be n a cage whose reerence voltage s 5kV hgher than the phototransstor and relay crcut, but the optcal solaton removes the danger o hgh voltage gettng through. 1. (9pt) Fll n the ollowng table: Relay (on or o?) LED C (on or o?) LED D (on or o?) OFF OFF ON 1 ON ON OFF 1 ON ON OFF 1 1 OFF OFF ON 2. (6pt) ssumng the resstance o the relay col s neglgble but 2m s needed to turn t on, what s the maxmum resstance value R can be the on-resstance o the phototransstor s 15Ω? nswer: 5V/2m = 15 + R R = 5/.2 15 = 1Ω 3. (5pt) Usng the phototransstor on-resstance o 15Ω, settng R = 5Ω and assumng the phototransstor behaves lke an deal swtch when t s turned o, what are the mnmum and maxmum voltages at E, the collector o the transstor? nswer: Mnmum = 5 = 5 = 3.75V Maxmum = 5V K.. Connor 6 Troy, NY

7 Queston IV - Dode Lmter Crcuts (2 ponts) VOFF = V R1 C + C - 1k D4 D1 D1N42 D1N42 VMPL = 1 D5 D2 R2 FREQ = 1k D1N42 D1N42 1k D6 D3 D1N42 D1N42 Sx 1N42 dodes are used to protect a load (n ths case R2) rom havng too large a voltage across t. The 42 dode has somewhat derent propertes than the 4148 dode we have studed n class. 1V 5V a) (9pts) Ths conguraton s tested wth the snusodal voltage source shown n the crcut above. The voltages measured are shown below. Label whch plot s the voltage at pt, at pt and across resstor R1 (C + to C - ). Explan your answer. C V -5V -1V s.5ms 1.ms 1.5ms 2.ms 2.5ms 3.ms V(V:+,D1:2) V(D1:2) V(V:+) Tme s 1V snusod at the nput, cannot be larger than the orward bas o three dodes whch s about 2V, C s what s let. K.. Connor 7 Troy, NY

8 2.V b) (6 pts) The source voltage ampltude s changed to 2V. Sketch the resultng voltage across resstor R1 (C + to C - ) and at the load () on the plot below. Explan your answer. 1.5V 1.V.5V.V -.5V -1.V -1.5V -2.V s.2ms.4ms.6ms.8ms 1.ms 1.2ms 1.4ms 1.6ms 1.8ms 2.ms 2.2ms 2.4ms 2.6ms 2.8ms 3.ms V(R1:1) V(R2:1) Tme t 2V, the nput voltage s not sucent to turn on the dodes so they reman open and out o the crcut. ll that remans s a voltage dvder wth two 1k resstors, so the voltage observed s a snusod at hal the nput. c) (5pts) Usng the gven normaton, determne the orward voltage necessary to turn the 1N42 dode on. Explan your answer. 3 dodes at about 2V means about.67v. good range or the answer would be.6v <.67V <.7V or the usual range or smple dodes. K.. Connor 8 Troy, NY

9 Queston V Sgnal Modulaton and Flterng (2 ponts) modulated sgnal s to be ltered to remove the eects o the modulaton and other nose. The desred sgnal s rom 1Hz 8kHz. The undesred parts o the sgnal are below 5Hz and above 1kHz. Desgn a combnaton o lter types (low pass and/or hgh pass) that wll remove everythng except the desred sgnal. It has been decded that the lters should be n seres (cascaded) as shown below. The Zs n the op-amp crcuts represent complex mpedances and may be combnatons o resstors and capactors. (Hnt: one o these lters was used n Project 4.) FILTER 1 FILTER 2 Z2 Z4 Z1 - OPMP Z3 - OPMP OUT OUT + U3 + U4 1. (4pt) For each lter determne the approprate type. Flter 1: LOW PSS Flter 2: HIGH PSS Order o lters may be swtched around 2. (4pt) For each lter determne the corner requency. Flter 1 c : 9kHz 8kHz 1kHz OK Flter 2 c : 7Hz 5Hz 1Hz OK (5pts?) For one o the rst our problems on ths quz, you can choose one o the ollowng optons. Wrte do not grade on t n really huge and clear letters and you wll receve the ull 2 ponts, regardless o your actual answer, or Solve the problem and, your answer s completely correct, you wll receve 25 ponts Note that you have ths opton or only one o the rst our problems. You must solve the other three. K.. Connor 9 Troy, NY

10 Queston V Sgnal Modulaton and Flterng (contnued) 3. (6pt) Usng 2kΩ resstors and whatever capactor values are necessary, draw the crcuts or Flter 1 and Flter 2 wth the correct components replacng the Zs above. LPF Flter 1: ω = 2 π = C = = =.884nF R C R 2π 2, 2π 9k HPF Flter 2: ω = 2 π = C = = =. 114μF R C R 2π 2, 2π 7 FILTER 1 FILTER 2 2k 2k 2k -.884n OUT.114u 2k - OUT + U5 + U6 K.. Connor 1 Troy, NY

11 4. (4pt) Wrte down the transer uncton, H(jω), or each lter wth numercal values or the coecents. LPF Flter 1: R H ( jω) = R (1 + jωr C = 1+ 1 jω17.7x1 ) 6 HPF Flter 2: jωr C H ( jω) = 1+ jωr C jω2.27x1 = 1+ jω2.27x (2pt) Whch type o lters, Mller ntegrators or practcal derentators, have problems due to nherent nose n sgnals? Practcal Derentators: they greatly amply hgh requency nose They are also nherently unstable and some nose at the nput can cause them to oscllate. Practcally speakng, t s best to lter out ths part o the response so the derentator can do ts job. Ether descrpton s ne or ths queston. K.. Connor 11 Troy, NY