.MCAD 304020000 1 76 447 0 .CMD PLOTFORMAT 0 0 1 1 1 0 0 1 1 0 0 1 1 1 0 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 21 15 0 0 3 .CMD FORMAT rd=d ct=10 im=i et=3 zt=15 pr=3 mass length time charge temperature tr=0 vm=0 .CMD SET ORIGIN 0 .CMD SET TOL 0.001000000000000 .CMD SET PRNCOLWIDTH 8 .CMD SET PRNPRECISION 4 .CMD PRINT_SETUP 1.200000 0.983333 1.200000 1.200000 0 .CMD HEADER_FOOTER 1 1 *empty* *empty* *empty* 0 1 *empty* *empty* *empty* .CMD HEADER_FOOTER_FONT fontID=14 family=Arial points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD HEADER_FOOTER_FONT fontID=15 family=Arial points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFAULT_TEXT_PARPROPS 0 0 0 .CMD DEFINE_FONTSTYLE_NAME fontID=0 name=Variables .CMD DEFINE_FONTSTYLE_NAME fontID=1 name=Constants .CMD DEFINE_FONTSTYLE_NAME fontID=2 name=Text .CMD DEFINE_FONTSTYLE_NAME fontID=4 name=User^1 .CMD DEFINE_FONTSTYLE_NAME fontID=5 name=User^2 .CMD DEFINE_FONTSTYLE_NAME fontID=6 name=User^3 .CMD DEFINE_FONTSTYLE_NAME fontID=7 name=User^4 .CMD DEFINE_FONTSTYLE_NAME fontID=8 name=User^5 .CMD DEFINE_FONTSTYLE_NAME fontID=9 name=User^6 .CMD DEFINE_FONTSTYLE_NAME fontID=10 name=User^7 .CMD DEFINE_FONTSTYLE fontID=0 family=Times^New^Roman points=14 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=1 family=Times^New^Roman points=14 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=2 family=Times^New^Roman points=14 bold=0 italic=0 underline=0 colrid=7 .CMD DEFINE_FONTSTYLE fontID=4 family=Arial points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=5 family=Courier^New points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=6 family=System points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=7 family=Script points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=8 family=Roman points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=9 family=Modern points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD DEFINE_FONTSTYLE fontID=10 family=Times^New^Roman points=10 bold=0 italic=0 underline=0 colrid=-1 .CMD UNITS U=1 .CMD DIMENSIONS_ANALYSIS 0 0 .CMD COLORTAB_ENTRY 0 0 0 .CMD COLORTAB_ENTRY 128 0 0 .CMD COLORTAB_ENTRY 0 128 0 .CMD COLORTAB_ENTRY 128 128 0 .CMD COLORTAB_ENTRY 0 0 128 .CMD COLORTAB_ENTRY 128 0 128 .CMD COLORTAB_ENTRY 0 128 128 .CMD COLORTAB_ENTRY 128 128 128 .CMD COLORTAB_ENTRY 192 192 192 .CMD COLORTAB_ENTRY 255 0 0 .CMD COLORTAB_ENTRY 0 255 0 .CMD COLORTAB_ENTRY 255 255 0 .CMD COLORTAB_ENTRY 0 0 255 .CMD COLORTAB_ENTRY 255 0 255 .CMD COLORTAB_ENTRY 0 255 255 .CMD COLORTAB_ENTRY 255 255 255 .TXT 5 3 2 0 0 Cg a71.000000,71.000000,54 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard HALF-WAVE CONTROLLED RECTIFIER CIRCUIT WITH AN RL LOAD} .TXT 9 0 90 0 0 Cg a73.000000,73.000000,1059 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard The operation of the RL circuit supplied by a half-wave controlled rectifier \par is illustrated in this program. Let v{\dn s}(t) be defined to E * sin (wt). At \par wt = 0, the current through the circuit is zero. As wt becomes >0, v{\dn s} becomes positive. If a diode instead of an SCR has been used, the diode would start conduction at wt = 0. With an SCR, the conduction does not start till the SCR is triggered. Let the SCR be triggered when wt = {\f1 a}. Then {\f1 a} is called the firing angle and the SCR continues to conduct. \par When wt = {\f1 p}, the source becomes zero, but at this instant, the current through the circuit is not zero and there is some energy stored in the inductor. When v{\dn s} becomes negative, the current through the circuit \par would not become zero suddenly because of the inductor. The inductor \par acts as a source and keeps the SCR forward-biased till the \par energy stored in the inductor becomes zero. Let the current through the circuit become zero at wt = {\f1 b} and the value of {\f1 b} > {\f1 p} . For {\f1 b} < wt < 2{\f1 p}, the current through the circuit is zero. 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!1!PLG#e$VIs!!&J!1!OLG#e+4Ir!"&J!1!]LG&K#u"<-HJD!"4&)A!\LH)1+DIs!".[5`9P!1!6LG)1(~Ir!$A`!1!oLG&K%9"<5XIr!"+t!A!#LG#e(~Ir!$A`!1!oLG&K%9"<5XIr!"&J!A!%LG#e(~Ir!$A`!1!oLG&K%9"<5XIr!!LV!A!'LG#e(~Ir!$A`!1!oLG&K%9"<5XIr!!G,!A!)LG4''|Ir!$A`!1!oLG&K%y"<5XIr!!A` !A!+LG#e"2Ir!#6j!1!\LG#e.JIs!"DG5`9P!1!*LG&K#4Ir!!6j!1!GLG#e+4Ir!%Ip!A!>!(:kLG#e"BIs!!Ip!1!)LG#e'LIr!$A`!1!oLG&K%y"<5XIr!!1@!A!1LG#e"bIr!#.Z!1!\LG#e.JIs!"DG5`9P!1!$LG&K$6Ir!!>z!1!DLG#e+4Ir!%Ip!A!:!(:kLG#e!@Is!".Z!1!OLG#e+4Ir!%Ip!A!.!(:kLG&K,vJP!%Ip!A!( !(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!(!(<6LKA`!A!( !(<6LKA`!A!(!(<6LKA`!A!Q  .TXT 61 -8 373 0 0 Cg b75.800000,75.800000,240 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard L (di/dt) + Ri = E * sin (wt ) for {\f1 a} < wt < { \f1 b},\par i({\f1 a}) = i({\f1 b}) = 0 , i(wt) = 0 for 0 < wt < { \f1 a}. and i(wt) = 0 for {\f1 b} < wt < 2{\f1 p .}\par \pard The complementary solution is;\par i(wt) = A *exp [- (R(wt - {\f1 a}) / wL ) ], where A is a constant to be evaluated.} .TXT 19 0 375 0 0 Cg b75.800000,75.800000,275 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard The particular solution is:\par \par i(wt) = [E/Z] sin (wt - {\f1 f} ), where {\f1 f} = tan {\up -1} (wL/R) , Z {\up 2} = R{ \up 2} + (wL){\up 2} .\par \par The total solution is then:\par \par i(wt) = [E/Z] sin (wt - {\f1 f} ) + A *exp [- (R/wL *(wt - {\f1 a}) ]. \par \par Since i({\f1 a}) = 0 , A = [E/Z * sin ({\f1 f} - {\f1 a})]. The solution is illustrated below.} .TXT 29 0 376 0 0 Cg a75.800000,75.800000,54 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard First specify the values of parameters as shown below.} .EQN 9 2 377 0 0 {0:E}NAME:340 .TXT 0 13 378 0 0 Cg a63.000000,63.000000,2 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard V.} .EQN 0 10 379 0 0 {0:R}NAME:10 .TXT 0 12 380 0 0 Cg a64.000000,64.000000,1 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard {\f1 W}} .EQN 0 11 381 0 0 {0:wL}NAME:10 .TXT 0 12 382 0 0 Cg a63.000000,63.000000,2 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard {\f1 W}} .EQN 10 -58 383 0 0 {0:\f}NAME:{0:atan}NAME(({0:wL}NAME)/({0:R}NAME)) .EQN 0 22 384 0 0 {0:\f}NAME={0}?_n_u_l_l_ .TXT 0 14 385 0 0 Cg a61.000000,61.000000,4 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard rad.} .EQN 10 -36 386 0 0 {0:Z}NAME:\(({0:R}NAME)^(2)+({0:wL}NAME)^(2)) .EQN 0 24 387 0 0 {0:Z}NAME={0}?_n_u_l_l_ .TXT 0 13 388 0 0 Cg a60.000000,60.000000,2 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard {\f1 W}} .TXT 19 -37 389 0 0 Cg a72.000000,72.000000,22 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard Let firing angle {\f1 a} be:} .EQN 0 27 390 0 0 {0:\a}NAME:({0:\p}NAME)/(6) .TXT 0 11 391 0 0 Cg a44.000000,44.000000,3 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard rad} .TXT 7 -38 392 0 0 Cg a73.800000,73.800000,46 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Next compute the coefficient A as shown below.} .EQN 10 0 393 0 0 {0:A}NAME:({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME({0:\f}NAME-{0:\a}NAME) .EQN 0 34 394 0 0 {0:A}NAME={0}?_n_u_l_l_ .TXT 0 17 395 0 0 Cg a8.900000,8.900000,3 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Amp} .TXT 8 -51 396 0 0 Cg a73.800000,73.800000,73 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Define a range variable n to find solution at each degree within a cycle.} .EQN 7 0 397 0 0 {0:n}NAME:0;360 .TXT 6 0 398 0 0 Cg a72.800000,72.800000,44 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Express the angle within a cycle in radians.} .EQN 8 0 399 0 0 ({0:\q}NAME)[({0:n}NAME):({0:n}NAME)/(180)*{0:\p}NAME .TXT 8 0 400 0 0 Cg a72.800000,72.800000,90 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Find the load current. If the angle is less than the firing angle, assign \par it zero value.} .EQN 15 0 401 0 0 ({0:Cur}NAME)[({0:n}NAME):{0:if}NAME(({0:\q}NAME)[({0:n}NAME)<{0:\a}NAME,0,({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME(({0:\q}NAME)[({0:n}NAME)-{0:\f}NAME)+{0:A}NAME*({0:e}NAME)^(-((({0:R}NAME*(({0:\q}NAME)[({0:n}NAME)-{0:\a}NAME))/({0:wL}NAME))))) .TXT 10 0 403 0 0 Cg a72.800000,72.800000,326 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard The expression for current calculated thus would be negative if wt \par exceeds the extinction angle, {\f1 b}. The extinction angle is that\par angle at which load current becomes zero. Create an array for current,\par called I{\dn n}. This array has the same value as Cur{\dn n} if Cur{\dn n} >0.0. \par Otherwise the other elements of array I{\dn n} are set to zero.} .EQN 23 -1 404 0 0 ({0:I}NAME)[({0:n}NAME):{0:if}NAME(({0:Cur}NAME)[({0:n}NAME)>0.0,({0:Cur}NAME)[({0:n}NAME),0.0) .TXT 7 0 417 0 0 Cg a72.800000,72.800000,201 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs28 \pard The voltage across the SCR is zero when I{\dn n} > 0. Otherwise it is equal to\par the applied voltage. Note that when wt < {\f1 a} , the voltage across\par the SCR is positve. Then the SCR blocks forward conduction.} .EQN 14 0 418 0 0 ({0:VSCR}NAME)[({0:n}NAME):{0:if}NAME(({0:I}NAME)[({0:n}NAME)>0.0001,0.0,{0:E}NAME*{0:sin}NAME(({0:\q}NAME)[({0:n}NAME))) .TXT 8 0 419 0 0 Cg a73.800000,73.800000,60 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Next find the extinction angle. Assign a guess value first.} .EQN 9 0 420 0 0 {0:\b}NAME:(7*{0:\p}NAME)/(6) .EQN 7 0 421 0 0 {0:Given}NAME .EQN 10 0 422 0 0 ({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME({0:\b}NAME-{0:\f}NAME)+{0:A}NAME*({0:e}NAME)^(-((({0:R}NAME*({0:\b}NAME-{0:\a}NAME))/({0:wL}NAME))))÷0 .EQN 10 0 423 0 0 {0:\d}NAME:{0:Find}NAME({0:\b}NAME) .EQN 0 28 426 0 0 {0:\d}NAME={0}?_n_u_l_l_ .TXT 0 13 427 0 0 Cg a60.800000,60.800000,3 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard rad} .EQN 10 -41 428 0 0 {0:extinction_angle}NAME:({0:\d}NAME*180)/({0:\p}NAME) .EQN 9 0 429 0 0 {0:extinction_angle}NAME={0}?_n_u_l_l_ .TXT 0 29 430 0 0 Cg a44.800000,44.800000,3 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard deg} .TXT 19 -28 431 0 0 Cg a73.800000,73.800000,101 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard The extinction angle is computed to be 225.49{\up o}.\par \par Next the average value of load current is found out.} .EQN 22 0 432 0 0 {0:CurAvg}NAME:(1)/(2*{0:\p}NAME)*(({0:\a}NAME&{0:\b}NAME`({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME({0:y}NAME-{0:\f}NAME)+{0:A}NAME*({0:e}NAME)^(-((({0:R}NAME*({0:y}NAME-{0:\a}NAME))/({0:wL}NAME))))&{0:y}NAME)) .EQN 14 0 433 0 0 {0:CurAvg}NAME={0}?_n_u_l_l_ .TXT 0 19 434 0 0 Cg a54.800000,54.800000,3 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Amp} .TXT 8 -19 435 0 0 Cg a73.800000,73.800000,38 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard The RMS load current is evaluated next} .EQN 16 0 436 0 0 {0:CurRMS}NAME:\((1)/(2*{0:\p}NAME)*(({0:\a}NAME&{0:\b}NAME`((({0:E}NAME)/({0:Z}NAME)*{0:sin}NAME({0:y}NAME-{0:\f}NAME)+{0:A}NAME*({0:e}NAME)^(-((({0:R}NAME*({0:y}NAME-{0:\a}NAME))/({0:wL}NAME))))))^(2)&{0:y}NAME))) .EQN 15 0 437 0 0 {0:CurRMS}NAME={0}?_n_u_l_l_ .TXT 0 23 438 0 0 Cg a50.800000,50.800000,3 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Amp} .EQN 31 -20 441 0 0 &&(_n_u_l_l_&_n_u_l_l_)&({0:I}NAME)[({0:n}NAME)@&&(_n_u_l_l_&_n_u_l_l_)&{0:n}NAME 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 36 20 18 0 3 .TXT 37 11 442 0 0 Cg a61.800000,61.800000,25 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard The load current waveform} .EQN 12 -11 443 0 0 &&(_n_u_l_l_&_n_u_l_l_)&({0:VSCR}NAME)[({0:n}NAME)@&&(_n_u_l_l_&_n_u_l_l_)&{0:n}NAME 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 38 25 18 0 3 .TXT 39 17 444 0 0 Cg a51.800000,51.800000,26 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard The voltage across the SCR} .TXT 14 -19 445 0 0 Cg a74.800000,74.800000,35 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs28 \pard Next output voltage can be plotted.} .EQN 6 0 446 0 0 ({0:Vout}NAME)[({0:n}NAME):{0:if}NAME(({0:I}NAME)[({0:n}NAME)>0.001,{0:E}NAME*{0:sin}NAME(({0:\q}NAME)[({0:n}NAME)),0.0) .EQN 7 1 447 0 0 &&(_n_u_l_l_&_n_u_l_l_)&({0:Vout}NAME)[({0:n}NAME)@&&(_n_u_l_l_&_n_u_l_l_)&{0:n}NAME 0 1 1 1 0 4 0 1 1 0 1 1 1 0 4 0 1 1 0 1 0 0 1 1 NO-TRACE-STRING 0 2 1 0 1 1 NO-TRACE-STRING 0 3 2 0 1 1 NO-TRACE-STRING 0 4 3 0 1 1 NO-TRACE-STRING 0 1 4 0 1 1 NO-TRACE-STRING 0 2 5 0 1 1 NO-TRACE-STRING 0 3 6 0 1 1 NO-TRACE-STRING 0 4 0 0 1 1 NO-TRACE-STRING 0 1 1 0 1 1 NO-TRACE-STRING 0 2 2 0 1 1 NO-TRACE-STRING 0 3 3 0 1 1 NO-TRACE-STRING 0 4 4 0 1 1 NO-TRACE-STRING 0 1 5 0 1 1 NO-TRACE-STRING 0 2 6 0 1 1 NO-TRACE-STRING 0 3 0 0 1 1 NO-TRACE-STRING 0 4 1 0 1 1 NO-TRACE-STRING 0 1 1 45 29 18 0 3