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LG#e#DIr!-+t!1!.LG#e#TIu!&LV!A!'LG#e!pIs!%LV!1!iLG&K'}"<(nIt!"&J!1!.LG#eBBIr!!DF!1!.LG+u1`Is!!A`!A!pLG#e-HIs!#9Q5[A`!a!4LG#e#$Ir!-1@!1!,LG#e#4It!'+t!A!,LG&K.ZIr!%9P!A!N!(9zLG+u$fIr!!A`!1#*LG#e"rIr!!DF!A!&LG#e1@Is!!A`!A!pLG#e-HIs!#DG5[9P!Q!9LG#e"bIr!-6j !1!*LG#e#4Is!!1@!1""LG&K#$Is!%LV!1!iLG&K)A"<'lIs!!1@!1!5LG#e"bIr!-6j!1!*LG#e#DIs!!+t!A""LG&K#$Is!%LV!1!iLG&K)A"<'lIs!!1@!1!6LG#e"BIr!-<6!1!(LG#e#dIu!!&J!1""LG&K#$Is!%LV!1!iLG&K'}"<'|Is!!+t!A!6LG#e"BIr!-<6!1!(LG#e4vIs!!<6!a!pLG#e-HIs!#9Q5[<6!a!"LG#e$vIr !!1@!1#0LG#e""Ir!(+t"#!*LG&K.ZIr!%9P!A!:!(:8LG#e""Ir!-A`!1!&LG#eE(Ir!%9P!A!:!(:9LG#e!`Ir!-G,!1!$LG#eE8Ir!%9P!A!:!(:9LG#e!`Ir!-G,!1!$LG#eE8Ir!%9P!A!:!(::LG#e!@Ir!-LV!1!"LG#eEHIr!%9P!A!:!(::LG#e!@Ir!-LV!1!"LG#eEHIr!%9P!A!2!(:;LG#eDfIr!.4&!1!iLG&K#u"<+DJ+ !,>z$w#1LG#e-HIs!"9Q5]9P!1!&LG#eCdIr!!.Z!1#6LG#e-HIs!"9Q5]9P!1!&LG#eCdIr!!.Z!1#6LG#e-HIs!"9Q5]9P!1!&LG#eCdIr!!.Z!1#6LG#e-HIs!"9Q5]9P!1!&LG#eCdIr!!.Z!1#6LG#e-HIs!"9Q5]9P!1!&LG#eCdIr!!.Z!1#6LG#e-HIs!"9Q5]9P!1!&LG#eCdIr!!.Z!1#6LG#e-HIs!"9Q5]9P!1!'LG#eCTIr !!1@!1#5LG#e-HIs!"9Q5]9P!1!(LG#eCDIr!!4&!1#4LG#e-HIs!"9Q5]9P!1!)LG#eC4Ir!!6j!1#3LG#e-HIs!"9Q5]9P!1!*LG#eC$Ir!!9P!1#2LG#e-HIs!"9Q5]9P!1!+LG#eBrIr!!<6!1#1LG#e-HIs!".[5]9P!1#6LG#e10J\!#LV!1!iLG&K$W"<-HIr!.&J!1"!LK9Q)0Ir!%9P!A!6!(:;LG#eDfIr!&Ip-I!PLG#e-HIs !".[5]9P!1#6LG#e10J\!#LV!1!iLG&K$W"<-HIr!.&J!1"!LK9Q)0Ir!%9P!A!6!(:;LG#eDfIr!&Ip-I!PLG#e-HIs!".[5]9P!1#6LG#e10J\!#LV!1!iLG&K$W"<-HIr!.&J!1"!LK9Q)0Ir!%9P!A!6!(:;LG#eDfIr!&Ip-I!PLG#e-HIs!".[5]9P!1#6LG#e10J\!#LV!1!iLG&K$W"<-HIr!.&J!1"!LK9Q)0Ir!%9P!A!6!(:; LG#eDfIr!&Ip-I!PLG#e-HIs!!>{5]9PLG"n!%9P!A!,!(<6LN>z-I";LG&K"s"{5hLV5xJ\!(9P!A!,!(<6LN>z-I";LG&K"s"{5hLV5xJ\!(9P!A!,!(<6LN>z-I";LG&K"s"{5hLV5xJ\!(9P!A!,!(<6LN>z-I";LG&K"s"{5hLV5xJ\!(9P!A!* !(<6LVLV+DIs!!9Q5hLVLVJOLG&K"S" }{\cf2\f1 f}{\cf2 , \par the load current is discontinuous and the behaviour of the circuit is \par similar to the circuit without source inductance operating in the \par discontinuous mode. The peak source voltage is assumed to be unity \par and the current E/R is also assumed to be unity.}} .EQN 38 0 256 0 0 {0:\t}NAME:2 .EQN 0 12 250 0 0 {0:\a}NAME:({0:\p}NAME)/(6) .EQN 0 14 251 0 0 {0:XL2}NAME:({0:\t}NAME)/(10) .EQN 9 -26 252 0 0 {0:\f}NAME:{0:atan}NAME({0:\t}NAME+{0:XL2}NAME) .EQN 0 51 253 0 0 {0:deg}NAME:({0:\a}NAME)/({0:\p}NAME)*180 .EQN 1 -26 254 0 0 {0:Z}NAME:\(1+(({0:\t}NAME+{0:XL2}NAME))^(2)) .TXT 22 -25 255 0 0 Cg a71.000000,71.000000,906 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}} \plain\cf1\fs24 \pard {\cf2 The solution is slightly difficult. Four unknown values are to be computed. They\par are:\par i. \tab the load current at wt = }{\cf2\f1 a}{\cf2 , called A ,\par ii. the instant }{\cf2\f1 b}{\cf2 when the commutation overlap ends,\par iii.\tab the load current at wt = }{\cf2\f1 b}{\cf2 , called B , and \par iv.\tab the coefficient for the exponential term, K.\par \par The equations are formed as follows.\par \par Let the load current be i(}{ \cf2\f1 a}{\cf2 ) at the instant of triggering and let it be }{\cf2\f1 i(b) }{\cf2 at wt = }{\cf2\f1 b}{\cf2 .\par Then the supply current changes from i(}{\cf2\f1 a}{\cf2 ) to -i(}{\cf2\f1 b}{\cf2 ) when wt varies from }{\cf2\f1 a}{\cf2 to }{\cf2\f1 b}{\cf2 .\par During this period, the load current varies from i(}{\cf2\f1 a}{\cf2 ) to i(}{\cf2 \f1 b}{\cf2 ) exponentially, with\par the time constant in radians being }{\cf2\f1 t}{\cf2 . The third equation is based on the periodic \par nature of the load current. Since the load current repeats itself every }{\cf2\f1 p}{\cf2 radians,\par i(}{\cf2\f1 a}{\cf2 ) = i(}{ \cf2\f1 p}{\cf2 +}{\cf2\f1 a}{\cf2 ). Another expression can be formed for the load current i(}{\cf2\f1 b}{\cf2 ) using\par the source voltage and the coefficient for the exponential term. }{ \cf2\dn }{\cf2 }} .TXT 41 0 266 0 0 Cg a73.800000,73.800000,849 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs24 \pard For solving the problem, the SOLVE BLOCK facility within MathCad is\par used. First, the guess values of the variables to be solved for are \par assigned. Then in the block below, the four constraints for solving\par are stated in the form of equations, with the equality sign created using\par by pressing CONTROL and = keys simultaneously. The program yields\par the solution as an array. The solution technique is elegant.\par \par The first equation equates the change in load current due to source voltage\par being applied across it directly from {\f1 a} to {\f1 b} . The second equation finds the \par load current at {\f1 b} from its value at {\f1 a}, based on the exponential decay. From\par the equation that would define the load current from {\f1 b} till ({\f1 p} + {\f1 a), }two equations\par can be obtained. At wt = {\f1 b} , the exponential part would equal K. At wt = {\f1 p} + {\f1 a},\par the current would have decayed. \par } .EQN 61 -1 257 0 0 {0:A}NAME:0.0 .EQN 0 12 258 0 0 {0:\b}NAME:{0:\a}NAME .EQN 0 11 259 0 0 {0:B}NAME:0.0 .EQN 0 13 260 0 0 {0:K}NAME:0.5 .EQN 6 -36 261 0 0 {0:Given}NAME .EQN 5 0 262 0 0 {0:A}NAME+{0:B}NAMEś(10)/({0:\t}NAME)*({0:cos}NAME({0:\a}NAME)-{0:cos}NAME({0:\b}NAME)) .EQN 9 0 263 0 0 {0:B}NAMEś{0:A}NAME*({0:e}NAME)^(-(({0:\b}NAME-{0:\a}NAME)/({0:\t}NAME))) .EQN 9 0 264 0 0 {0:A}NAMEś(1)/({0:Z}NAME)*{0:sin}NAME({0:\p}NAME+{0:\a}NAME-{0:\f}NAME)+{0:K}NAME*({0:e}NAME)^(-(({0:\p}NAME+{0:\a}NAME-{0:\b}NAME)/(({0:\t}NAME+{0:XL2}NAME)))) .EQN 10 0 265 0 0 {0:B}NAMEś(1)/({0:Z}NAME)*{0:sin}NAME({0:\b}NAME-{0:\f}NAME)+{0:K}NAME .EQN 13 0 267 0 0 {0:C}NAME:{0:Find}NAME({0:A}NAME,{0:\b}NAME,{0:B}NAME,{0:K}NAME) .EQN 5 0 268 0 0 {0:A}NAME:({0:C}NAME)[(0) .EQN 0 12 269 0 0 {0:\b}NAME:({0:C}NAME)[(1) .EQN 0 12 270 0 0 {0:B}NAME:({0:C}NAME)[(2) .EQN 0 15 271 0 0 {0:K}NAME:({0:C}NAME)[(3) .EQN 10 -39 276 0 0 {0:A}NAME={0}?_n_u_l_l_ .EQN 0 14 277 0 0 {0:\b}NAME={0}?_n_u_l_l_ .EQN 0 15 278 0 0 {0:B}NAME={0}?_n_u_l_l_ .EQN 0 16 279 0 0 {0:K}NAME={0}?_n_u_l_l_ .TXT 37 -45 280 0 0 Cg a71.000000,71.000000,220 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}} \plain\cf1\fs24 \pard {\cf2 Now the plots of load current, line current and output voltage of the bridge can\par be obtained. Define a range variable, n, to correspond to the degrees within a\par cycle of source voltage. Obtain the angle }{\cf2\f1 q}{\cf2\dn n }{\cf2 in radians.}} .EQN 14 0 295 0 0 {0:n}NAME:0;360 .EQN 6 0 297 0 0 ({0:\q}NAME)[({0:n}NAME):({0:n}NAME)/(180)*{0:\p}NAME .TXT 9 0 309 0 0 Cg a74.800000,74.800000,199 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs24 \pard The equation presented below computes the load current when wt < {\f1 a}. During this part,\par the solution is obtained by equating wt = {\f1 p} + {\f1 q}{\dn n}, and the elapsed angle for exponential\par decay is ({\f1 p} + {\f1 q}{\dn n }- {\f1 b})} .EQN 17 0 315 0 0 ({0:PartI}NAME)[({0:n}NAME):(1)/({0:Z}NAME)*{0:sin}NAME({0:\p}NAME+({0:\q}NAME)[({0:n}NAME)-{0:\f}NAME)+{0:K}NAME*({0:e}NAME)^(-(({0:\p}NAME-{0:\b}NAME+({0:\q}NAME)[({0:n}NAME))/({0:\t}NAME+{0:XL2}NAME))) .TXT 9 0 323 0 0 Cg a74.800000,74.800000,103 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs24 \pard When {\f1 a} < wt < {\f1 b}, the load current decays exponentially. Note that the load current at\par wt = {\f1 a} is A.} .EQN 13 0 337 0 0 ({0:PartOL}NAME)[({0:n}NAME):{0:A}NAME*({0:e}NAME)^(-((({0:\q}NAME)[({0:n}NAME)-{0:\a}NAME)/({0:\t}NAME))) .TXT 7 0 348 0 0 Cg a74.800000,74.800000,75 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs24 \pard The expression below computes the load current when { \f1 b} < wt < ( {\f1 p} + {\f1 a}).} .EQN 12 0 346 0 0 ({0:PartNorm}NAME)[({0:n}NAME):(1)/({0:Z}NAME)*{0:sin}NAME(({0:\q}NAME)[({0:n}NAME)-{0:\f}NAME)+{0:K}NAME*({0:e}NAME)^(-((({0:\q}NAME)[({0:n}NAME)-{0:\b}NAME)/({0:\t}NAME+{0:XL2}NAME))) .TXT 18 0 351 0 0 Cg a74.800000,74.800000,78 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Now a single expression for the load current for half-a-cycle can be obtained.} .EQN 9 0 349 0 0 ({0:LoadCur1}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<{0:deg}NAME,({0:PartI}NAME)[({0:n}NAME),({0:PartOL}NAME)[({0:n}NAME)) .EQN 8 0 350 0 0 ({0:LoadCur2}NAME)[({0:n}NAME):{0:if}NAME(({0:\q}NAME)[({0:n}NAME)<{0:\b}NAME,({0:LoadCur1}NAME)[({0:n}NAME),({0:PartNorm}NAME)[({0:n}NAME)) .TXT 8 0 398 0 0 Cg a74.800000,74.800000,122 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard From the expression for load current over half-a-cycle, an expression for the load\par current over a whole cycle is obtained.} .EQN 9 0 400 0 0 ({0:AmpLoad}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<180,({0:LoadCur2}NAME)[({0:n}NAME),({0:LoadCur2}NAME)[({0:n}NAME-180)) .TXT 8 0 405 0 0 Cg a74.800000,74.800000,336 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Next an expression for load current is developed. At instants outside the overlap\par region, the line current has the same amplitude as the load current. It has the same\par sign when SCRs S{\dn 1} and S{\dn 4} conduct and has the opposite sign when SCRs S{\dn 2} and S{\dn 3} \par conduct. During overlap period, its value is different from that of the load current.} .EQN 16 0 410 0 0 ({0:OLapI}NAME)[({0:n}NAME):-{0:A}NAME+(1)/({0:XL2}NAME)*({0:cos}NAME({0:\a}NAME)-{0:cos}NAME(({0:\q}NAME)[({0:n}NAME))) .EQN 10 0 414 0 0 ({0:LineCur1}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<{0:deg}NAME,-({0:AmpLoad}NAME)[({0:n}NAME),({0:OLapI}NAME)[({0:n}NAME)) .EQN 6 0 415 0 0 ({0:LineCur2}NAME)[({0:n}NAME):{0:if}NAME(({0:\q}NAME)[({0:n}NAME)<{0:\b}NAME,({0:LineCur1}NAME)[({0:n}NAME),({0:AmpLoad}NAME)[({0:n}NAME)) .EQN 6 0 416 0 0 ({0:CurLine}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<180,({0:LineCur2}NAME)[({0:n}NAME),-({0:LineCur2}NAME)[({0:n}NAME-180)) .TXT 23 1 417 0 0 Cg a71.000000,71.000000,24 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard { \cf2 The Plot of Load Current}} .EQN 3 1 418 0 0 _n_u_l_l_&_n_u_l_l_&(_n_u_l_l_&_n_u_l_l_)&({0:AmpLoad}NAME)[({0:n}NAME)@_n_u_l_l_&_n_u_l_l_&(_n_u_l_l_&_n_u_l_l_)&({0:n}NAME)/(180) 0 1 1 1 0 6 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 35 19 16 0 3 .TXT 33 -1 422 0 0 Cg a71.000000,71.000000,25 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard { \cf2 The Plot of Line Current}} .EQN 3 1 421 0 0 _n_u_l_l_&_n_u_l_l_&(_n_u_l_l_&_n_u_l_l_)&({0:CurLine}NAME)[({0:n}NAME)@_n_u_l_l_&_n_u_l_l_&(_n_u_l_l_&_n_u_l_l_)&({0:n}NAME)/(180) 0 1 1 1 0 6 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 44 24 16 0 3 .TXT 65 0 431 0 0 Cg a71.000000,71.000000,42 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard { \cf2 Getting an expression for the Load Voltage}} .EQN 6 0 432 0 0 ({0:vs}NAME)[({0:n}NAME):{0:sin}NAME(({0:\q}NAME)[({0:n}NAME)) .TXT 8 1 442 0 0 Cg a71.800000,71.800000,230 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs24 \pard At instants outside the overlap period, the bridge output voltage is the drop \par across the load resistor plus the voltage across the load inductance. Now two\par expressions are needed, one for wt < {\f1 a} and another for {\f1 b} < wt < {\f1 p} .} .EQN 16 -1 443 0 0 ({0:VP1}NAME)[({0:n}NAME):({0:AmpLoad}NAME)[({0:n}NAME)-((({0:AmpLoad}NAME)[({0:n}NAME)+({0:vs}NAME)[({0:n}NAME))*{0:\t}NAME)/({0:\t}NAME+{0:XL2}NAME) .EQN 14 0 444 0 0 ({0:VP2}NAME)[({0:n}NAME):((({0:vs}NAME)[({0:n}NAME)-({0:AmpLoad}NAME)[({0:n}NAME))*{0:\t}NAME)/({0:\t}NAME+{0:XL2}NAME)+({0:AmpLoad}NAME)[({0:n}NAME) .EQN 9 0 448 0 0 ({0:LoadVolt1}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<{0:deg}NAME,({0:VP1}NAME)[({0:n}NAME),0.0) .EQN 8 0 449 0 0 ({0:LoadVolt2}NAME)[({0:n}NAME):{0:if}NAME(({0:\q}NAME)[({0:n}NAME)<{0:\b}NAME,({0:LoadVolt1}NAME)[({0:n}NAME),({0:VP2}NAME)[({0:n}NAME)) .EQN 6 0 450 0 0 ({0:VoltLoad}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<180,({0:LoadVolt2}NAME)[({0:n}NAME),({0:LoadVolt2}NAME)[({0:n}NAME-180)) .TXT 36 -1 451 0 0 Cg a70.000000,70.000000,25 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard { \cf2 The Plot of Load Voltage}} .EQN 5 1 452 0 0 _n_u_l_l_&_n_u_l_l_&(_n_u_l_l_&_n_u_l_l_)&({0:VoltLoad}NAME)[({0:n}NAME)@_n_u_l_l_&_n_u_l_l_&(_n_u_l_l_&_n_u_l_l_)&({0:n}NAME)/(180) 0 1 1 1 0 6 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 45 26 16 0 3 .TXT 52 0 455 0 0 Cg a72.800000,72.800000,70 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Calculating the reduction in output voltage due to commutation overlap} .TXT 5 1 454 0 0 Cg a71.000000,71.000000,475 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}} \plain\cf1\fs24 \pard {\cf2 During commutation over lap, the output voltage is zero. All the four SCRs are\par in conduction, with the current through the incoming SCRs rising from zero\par to load current level and the current through the outgoing SCRs falling from\par the load current level to zero. There is a slight reduction in the output voltage.\par Let the commutation overlap in degrees be }{\cf2\f1 m}{ \cf2 . Let the average voltage \par with no source inductance be VavNoOL and the average voltage with overlap\par be VAvgWOL. }} .TXT 39 0 462 0 0 Cg a71.800000,71.800000,49 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Commutation overlap angle is obtained as follows.} .EQN 8 0 459 0 0 {0:\m}NAME:({0:\b}NAME-{0:\a}NAME)/({0:\p}NAME)*180 .EQN 7 0 460 0 0 {0:\m}NAME={0}?_n_u_l_l_ .TXT 0 11 461 0 0 Cg a59.000000,59.000000,3 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard { \cf2 deg}} .TXT 8 -11 466 0 0 Cg a71.800000,71.800000,105 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Next the output voltage without overlap is obtained. It is assumed that there is no\par source inductance. } .EQN 12 0 470 0 0 {0:VavNoOL}NAME:(1)/({0:\p}NAME)*({0:\a}NAME&{0:\p}NAME+{0:\a}NAME`{0:sin}NAME({0:x}NAME)&{0:x}NAME) .EQN 8 0 471 0 0 {0:VavNoOL}NAME={0}?_n_u_l_l_ .TXT 7 0 652 0 0 Cg a71.800000,71.800000,116 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard To obtain, the actual output voltage given a peak voltage, multiply the above value by\par the peak value of the source.} .TXT 7 0 651 0 0 Cg a71.800000,71.800000,190 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard From the bridge output voltage during the period when there is no overlap, the average\par value of bridge voltage can be obtained. During the overlap period, the bridge\par output voltage is zero.} .EQN 14 0 482 0 0 {0:Amp}NAME({0:x}NAME):(1)/({0:Z}NAME)*{0:sin}NAME({0:x}NAME-{0:\f}NAME)+{0:K}NAME*({0:e}NAME)^(-(({0:x}NAME-{0:\b}NAME)/({0:\t}NAME+{0:XL2}NAME))) .EQN 10 0 483 0 0 {0:VAvgWOL}NAME:(1)/({0:\p}NAME)*(({0:\b}NAME&{0:\p}NAME+{0:\a}NAME`(({0:\t}NAME)/({0:\t}NAME+{0:XL2}NAME)*({0:sin}NAME({0:x}NAME)-{0:Amp}NAME({0:x}NAME))+{0:Amp}NAME({0:x}NAME))&{0:x}NAME)) .EQN 21 -1 653 0 0 {0:VAvgWOL}NAME={0}?_n_u_l_l_ .TXT 6 0 659 0 0 Cg a71.800000,71.800000,116 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard To obtain, the actual output voltage given a peak voltage, multiply the above value by\par the peak value of the source.} .TXT 10 0 657 0 0 Cg a70.000000,70.000000,49 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard { \cf2 Fractional reduction in output Voltage, FRed, is }} .EQN 6 0 656 0 0 {0:FRed}NAME:({0:VavNoOL}NAME-{0:VAvgWOL}NAME)/({0:VavNoOL}NAME) .EQN 8 0 501 0 0 {0:FRed}NAME={0}?_n_u_l_l_ .TXT 7 0 507 0 0 Cg a70.000000,70.000000,66 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;\red0\green128\blue128;}{ \fonttbl{\f0\fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard { \cf2 The unit value corresponds to E, the peak value of source voltage.}} .TXT 8 0 517 0 0 Cg a72.800000,72.800000,163 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Verification \par \par The average load current should have the same value as the average bridge output\par voltage, since the average voltage across the load inductor is zero.} .EQN 18 0 521 0 0 {0:olapcur}NAME({0:x}NAME):{0:A}NAME*({0:e}NAME)^(-(({0:x}NAME-{0:\a}NAME)/({0:\t}NAME))) .EQN 11 0 522 0 0 {0:AvgLoadCur}NAME:(1)/({0:\p}NAME)*((({0:\b}NAME&{0:\p}NAME+{0:\a}NAME`{0:Amp}NAME({0:x}NAME)&{0:x}NAME))+(({0:\a}NAME&{0:\b}NAME`{0:olapcur}NAME({0:x}NAME)&{0:x}NAME))) .EQN 12 0 524 0 0 {0:AvgLoadCur}NAME={0}?_n_u_l_l_ .TXT 4 0 660 0 0 Cg a71.800000,71.800000,151 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard To obtain, the actual current given a peak voltage, multiply the above value by\par E/R, where E is the peak value of the source and R the load resistance.} .EQN 19 0 530 0 0 {0:RMSLoadCur}NAME:\((1)/({0:\p}NAME)*((({0:\b}NAME&{0:\p}NAME+{0:\a}NAME`({0:Amp}NAME({0:x}NAME))^(2)&{0:x}NAME))+(({0:\a}NAME&{0:\b}NAME`({0:olapcur}NAME({0:x}NAME))^(2)&{0:x}NAME)))) .EQN 14 0 531 0 0 {0:RMSLoadCur}NAME={0}?_n_u_l_l_ .EQN 11 0 533 0 0 {0:MaxAvgLoadCur}NAME:(2)/({0:\p}NAME) .EQN 10 0 534 0 0 {0:MaxAvgLoadCur}NAME={0}?_n_u_l_l_ .EQN 11 0 537 0 0 {0:RFCur}NAME:(\(({0:RMSLoadCur}NAME)^(2)-({0:AvgLoadCur}NAME)^(2)))/({0:MaxAvgLoadCur}NAME) .EQN 11 0 538 0 0 {0:RFCur}NAME={0}?_n_u_l_l_ .TXT 7 -1 608 0 0 Cg a73.800000,73.800000,224 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Next the ripple factor in output voltage of the bridge is computed. This ripple factor\par should be much greater than the ripple factor for the load current, since the\par load inductance acts as a filter for harmonic components. } .EQN 16 0 607 0 0 {0:RMSBrOutV}NAME:\((1)/({0:\p}NAME)*(({0:\b}NAME&{0:\p}NAME+{0:\a}NAME`((({0:\t}NAME)/({0:\t}NAME+{0:XL2}NAME)*({0:sin}NAME({0:x}NAME)-{0:Amp}NAME({0:x}NAME))+{0:Amp}NAME({0:x}NAME)))^(2)&{0:x}NAME))) .EQN 21 0 609 0 0 {0:RFBrOutV}NAME:(\(({0:RMSBrOutV}NAME)^(2)-({0:VAvgWOL}NAME)^(2)))/({0:MaxAvgLoadCur}NAME) .EQN 10 0 619 0 0 {0:RFBrOutV}NAME={0}?_n_u_l_l_ .TXT 9 0 620 0 0 Cg a44.800000,44.800000,40 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Function for line current during overlap} .EQN 7 0 621 0 0 {0:f2}NAME({0:x}NAME):-{0:A}NAME+(1)/({0:XL2}NAME)*({0:cos}NAME({0:\a}NAME)-{0:cos}NAME({0:x}NAME)) .TXT 10 0 622 0 0 Cg a73.800000,73.800000,12 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Verification} .EQN 7 0 623 0 0 {0:f2}NAME({0:\a}NAME)={0}?_n_u_l_l_ .EQN 0 15 635 0 0 {0:f2}NAME({0:\b}NAME)={0}?_n_u_l_l_ .EQN 0 18 634 0 0 {0:Amp}NAME({0:\p}NAME+{0:\a}NAME)={0}?_n_u_l_l_ .EQN 0 23 631 0 0 {0:Amp}NAME({0:\b}NAME)={0}?_n_u_l_l_ .EQN 13 -56 636 0 0 {0:RMSlineCur}NAME:\((1)/({0:\p}NAME)*((({0:\a}NAME&{0:\b}NAME`({0:f2}NAME({0:x}NAME))^(2)&{0:x}NAME))+(({0:\b}NAME&{0:\p}NAME+{0:\a}NAME`({0:Amp}NAME({0:x}NAME))^(2)&{0:x}NAME)))) .EQN 11 0 637 0 0 {0:RMSlineCur}NAME={0}?_n_u_l_l_ .TXT 11 0 638 0 0 Cg a73.800000,73.800000,100 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard Next THD in line current is found out. Find the Fundamental trignometric Fourier \par Series components.} .EQN 27 0 639 0 0 {0:fa1}NAME:(2)/({0:\p}NAME)*((({0:\a}NAME&{0:\b}NAME`({0:f2}NAME({0:x}NAME)*{0:cos}NAME({0:x}NAME))&{0:x}NAME))+(({0:\b}NAME&{0:\p}NAME+{0:\a}NAME`({0:Amp}NAME({0:x}NAME)*{0:cos}NAME({0:x}NAME))&{0:x}NAME))) .EQN 10 0 640 0 0 {0:fa1}NAME={0}?_n_u_l_l_ .EQN 12 0 641 0 0 {0:fb1}NAME:(2)/({0:\p}NAME)*((({0:\a}NAME&{0:\b}NAME`({0:f2}NAME({0:x}NAME)*{0:sin}NAME({0:x}NAME))&{0:x}NAME))+(({0:\b}NAME&{0:\p}NAME+{0:\a}NAME`({0:Amp}NAME({0:x}NAME)*{0:sin}NAME({0:x}NAME))&{0:x}NAME))) .EQN 11 0 642 0 0 {0:fb1}NAME={0}?_n_u_l_l_ .EQN 13 0 643 0 0 {0:FundLineCurRMS}NAME:\((({0:fa1}NAME)^(2)+({0:fb1}NAME)^(2))/(2)) .EQN 10 0 644 0 0 {0:FundLineCurRMS}NAME={0}?_n_u_l_l_ .EQN 9 0 645 0 0 {0:THD}NAME:\(((({0:RMSlineCur}NAME)/({0:FundLineCurRMS}NAME)))^(2)-1) .EQN 10 0 646 0 0 {0:THD}NAME={0}?_n_u_l_l_ .TXT 7 0 647 0 0 Cg a73.800000,73.800000,51 {\rtf\ansi \deff0{\colortbl;\red0\green0\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard The THD in line current is obtained as shown above.}