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ECE361-PSPICE-Project1-F08 - RUTGERS UNIVERSITY DEPARTMENT...

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Unformatted text preview: RUTGERS UNIVERSITY DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING PSPICE SIMULATION 14530361: ELECTRONIC DEVICES 14:330:363: ELECTRONIC DEVICES LAB PROJECT 1: DIODE (EXAMPLES 8c PROBLEM ASSIGNMENT] INSTRUCTOR: DR. Y. LU Exl-l. Dicde. Plct the DC transfer characteristic of V0 1versus Vin, as Vin from -15 V to +15 V with increment cf 0.5 V. The juncticn potential of all the diodes is 0.7 V. R1=IO chins, R2=10 chms, R3210 01111151 Vcc=15 Vs and Vee=-15V . **** UBIlEIBE 10:43:22 ****'**'* NT Evaluation PSpiee {July 1997] wwwiwk-kfit-Irti *EXl-l.DIDDE APPLICATION **** CIRCUIT DESCRIPTION **tttit****tt******tt***i**fi**i**1**fi**fl**w*w*+***+*i++**+***+*+++***+****+** * iif“fl*1i*flf*iiitii****i***i*+*iii*******+**+****i***+t*** *eircuit description *DC input voltage of Gvipreset values} * +nede unpde model value VIN 1 0 DD 0 *Diede P nede N node model name D1 2 1 DM D2 2 4 DM D3 1 3 DM D4 4 3 DM *resistnr R1 is cnnncted tn node 2 and 5 with value pf lfiknhms R1 2 5 13K R2 3 1 10K R3 4 D lflK _ VCC 5 fl DC lSU VEE T G DC -15U *.Medel Cdmmand defines the parameters of diode with name DM .HDDEL DH DIUJ=U.7: **Analysis resquested *DC sweep from —lE.te 15v with increment of 6.5v .DC VIN -15 15 3.5 **Graphie nutput requested .PRDEE . - *All input files must be ended by .end command .END w ***' QBIIEIQS 1D:4E:22 ********* NT Evaluatign §Spice [July 1997: wi'flfii'l't'i'i'i'i *EXl—ltDIQDE APPLICATION **** Diode MDDEL PARAMETERS 1"-til-k1:i-1-1-1-i;*i-ir-kIk-W'i'fi******t***+*******i***i*‘I’*1’1’**************T**********W‘I*** DH IS 10.DDDUflflE-15 VJ .7 JOB CONCLUDED TOTAL JOE TIME .06 'EK‘]. . DIODE APPLICATION DatefTime run: flBHlEKSB 11:19:DB Temperature: 2T.D ____——.__—-_-.-.————--— tfl} Exl-l. Diode Application Date: August 12, 1993 Page 1 Time: 11:34:UT Exlwfl. Zener diode. Find the DC transfer eharaeteristie of V2 versus Vin, as Vin from -15 V to +15 V with increment of 0.5 V. The zener voltage of the diodes is Ve=6 V. R1=2 Kehnis, R222 Kehrns. The operating temperature is 2?“C. ® G) V (3 y Ext-2.29ner Diode **** DSflEIQS 11:13:fi9 ********* NT Evaluation PSpiCE {JU1Y 1997] tii-i-‘l-ieflti-fi-Ekir *Exl—Z. ZEHEE DIODE **+* CIRCUIT DESCRIPTION iri'i'*ti’*****t*‘k**+*‘k****itittt-kt'fi‘*Wi‘***+***+**‘k*1‘*******************w*w **+***** *** CIRCUIT DESCRIPTION *t'k‘k‘ki‘fi‘l‘ii-‘A-‘kI-‘kwwwi*************t*tflr**********‘l‘k*‘k‘k***** **Cireuit description VIN 1 0 DC E R1 1 2 2k R2 2 D 2k D1 2 3 DM D2 D 3 DH .HDDEL DH DIEV=61 **Anelyeie required .DC VIN -15 15 fl.5 **Dutput requested .PHINT DC ViE} .PRDBE ”END **** BEIlBFQB 11:13:09 ********* NT Evaluation PSpiCE {July 199?} *iri'ir'kirt'k'ki'i'ir *EXl-Z. ZENEE DIODE **** Diode MODEL EARAMETERS *1!-*****************************‘ka‘ki'ii-ifictiti-*1“!-*t*****************i**** *iriri-iri‘i-fi- DH .IS 1D.flDflDflflE-15 EU 6 **** UBI13EBB 11:18:G9 ********* NT Evaluation PEpiCE {July 199?} t**********T *EXl-E. ZENER DIODE 27.0fi0 DEG 1| **** DC TRANSFER CURVES TEMPERATURE *********‘k****‘k*“*******************‘I’i’i‘i’*‘k*********************t*****i *itiriirir‘ki' VIN W21 -1.5flflE+fll -T.G33E+DD ~1.45DE+D1 n?.flflDE+flD -1.4flDE+Dl —E.933E+DD “1.35flE+fll HE.T48E+DD -1.3flflE+fll -E.5flDE+DU -1.25DE+D1 —6.25DE+DD -1.2GflE+fll —6.DDDE+DU -1.15UE+Dl r5.?5DE+DD -1.lflDE+Ul F5.SDDE+DD -1.USDE+Dl -5.25flE+DD -1.GDGE+01 —5.flflflE+flD —9.5DUE+UD ~4.T5DE+DU -9.flDUE+UU #4.SDDE+DD HB.5DDE+DG —4.25DE+DD -3.flflDE+Ufl -4.BDDE+DD -T.SUUE+DD -3.T5flE+flD —7.DDDE+Dfl -3.5DDE+UD —E.500E+Ufl r3.25flE+flD -E.GDDE+Ufl —3.flflDE+flfl n5.SUUE+Ufi -2.T5DE+DD “5.DDGE+Ufl ~2.5flDE+flfl -4-500E+flfl -2.25DE+GU -4.DBUE+Dfl -2.DGUE+UD -3.5DDE+UD —1.T5flE+Ufl —3.DDDE+DU "1.5flflE+Dfl —2.SDUE+DD —1.25flE+Dfl n2.flGUE+UD —1.DDDE+DU -l.5flUE+flG ~?.SDflE—Dl -l.flDUE+DD -5.DUDE—Dl -5.flflDEFl -2.SUUE—Dl D.DflflE+flfl D.DDUE+DG 5.DflflE—fl1 2.5DDE—fl1' 1.UflflE+flD 5.0DDE—fl1 1.5flDE+DD T.5flDE—U1 2.DUUE+flfl 1.flflDE+flfl 2.5GUE+DD 1.25DE+DU 3.DUUE+UD 1.5flflE+flfl 3.50flE+flfl 1.?5DE+DD 4.DDDE+UD 2.DflflE+DD HI—‘I—H—Imxommdqmmmmh l. l. l. 1. l. 1. l- .50flE+flU .flDflE+DD .50fiE+UU .DUflE+GD .SDDE+UD .UDDE+OD .5fl0E+DU .DDDE+DU .SDfiE+flD .DDGE+UU .EDflE+DD .DUfiE+Dl .DSDE+D1 .IDGE+D1 15DE+D1 250E+Ul 253E+fll BDDE+D1 3SDE+D1 4UGE+01 450E+Dl EUUE+01 2 2 damn-.mmmmmmmbhnnwwwwm .25DE+DU .EDUE+UD .TSDE+UD .UflUE+UU .250E+DD .SDUE+Ufi .TSDE+UD .DUUE+flD .25flE+DD .EDUE+DU .?5DE+DU .DDDE+DD .25flE+UD .SUDE+DU .TSDE+DD .flDDE+Dfl .250E+fifl .SUDE+flfl .?4BE+DU .9333+flfl .flUUE+DD .fl33E+UD JOE CONCLUDED TOTAL JOE TIME .0? .D 7 2 11:34:22 amperature: Time: T ZENER DIODE Page 1 *EXl-E. 13!?3 11:18:09 1993 run: 05 August 13. DatefTime Date: ProbIFI. Rectifier. Plot the transient response of the voltage across the capacitor C and diode D2 within 60 ms with an increment of 0.2 ms. The zener voltage of D2 is 6 V. Parameters of D1 takes their default values. R1=10 ohrns, R2=100 ohms, C=100 nF, Vin=10sin(2n a 50 t). Prebl-Z. Find nut the transient response of V0 in 4 ms with an increment of 0.004 ms in this clipper eireuit. Parameters ef the diodes are of default values. Vin=5sin(2nx1000t), R1=2 ehrns. R2=15 ehtns, C=D.UI IJF. Prob1-3. Find out the output V8.3) and V(4.3) for the rectifier with a single phase center tapped transformer in 50 ms with a 0.1 ms increment. The ratio of the primary to the secondary is 5:1. Vin=100sin(2nx50t). R1=10 Gohms, R222 Kohms, C1=100 of, parameters for D1and D2 take their default values. (Hint: you might want to consider using inductors to represent the transformer] Prob. I -3 ...
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