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!1"1LG#e"RIr!-G,!1")LG&K%9"<-HIr!!<6!1"0LG#e"bIr!-DF!1")LG&K%9"<-HIr!!>z!1"/LG#e"rIr!-A`!1")LG&K$W"<-HIr!(6j!1"!LK9Q)0Ir!'4&!A!6!(:;LG#e5HIr!&Ip-I!PLG#e2RIs!".[5]9P!1":LG#e10J\!#LV!1")LG&K$W"<-HIr!(6j!1"!LK9Q)0Ir!'4&!A!6!(:;LG#e5HIr!&Ip-I!PLG#e2RIs!".[ 5]9P!1":LG#e10J\!#LV!1")LG&K$W"<-HIr!(6j!1"!LK9Q)0Ir!'4&!A!6!(:;LG#e5HIr!&Ip-I!PLG#e2RIs!".[5]9P!1":LG#e10J\!#LV!1")LG&K$W"<-HIr!(6j!1"!LK9Q)0Ir!'4&!A!6!(:;LG#e5HIr!&Ip-I!PLG#e2RIs!".[5]9P!1":LG#e10J\!#LV!1")LG&K"s"<-HLV!&#e2RIs!!>{5hLV&ZJ\!*4&!A!,!(<6 LI#d-I"YLG&K"s"{5hLV&ZJ\!*4&!A!,!(<6LI#d-I"YLG&K"s"{5hLV&ZJ\!*4&!A!,!(<6LI#d-I"YLG&K"s"{5hLV&ZJ\!*4&!A!,!(<6LI#d-I"YLG&K"s"{5hLV &ZJ[!*6j!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!(!(<6LVLV!A!Q  .TXT 29 -5 5 0 0 Cg a72.000000,72.000000,1851 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}{\f1\fcharset2\fnil Symbol;}}\plain \cf1\fs24 \pard The operation of a fully-controlled bridge rectifier circuit with RL load is \par explained in this program. This bridge rectifier is fully-controlled, because \par all the devices used are SCRs. \par \par Let us assume that the circuit is switched on at wt = 0 and let the firing\par angle be {\f1 a}. Let the supply voltage v{ \dn s}(wt) = E sin (wt). When wt = {\f1 a}, the\par SCRs S{\dn 1} and S{\dn 4} get triggered and they start conducting since they are \par forward-biased. These two SCRs continue to conduct till wt = {\f1 p. }When \par wt = {\f1 p} radians, the supply voltage falls to zero, but the current through the\par \pard load does not become zero because the current lags voltage in an RL load. \par When the current through the load is higher than the holding level for SCRs \par S{\dn 1} and S{\dn 4}, they keep conducting by the inductor acting as a source. When the\par current through the inductor falls, the rate of change of current is negative and\par the voltage across the inductor is such that the terminal connected to the\par cathodes of SCRs S{\dn 2} and S{ \dn 4 }positive with respect to its other terminal and the\par inductor pumps energy back into the supply via the SCRs S{\dn 1} and S{\dn 4} and the \par load resistor. When{\f1 p} {\f1 f}, (firing angle > load angle) , the conduction is discontinuous. \par \par When {\f1 a} = { \f1 f}, (firing angle = load angle) , the conduction is about to \par become continuous and the value of A should satisfy both the equations.\par When {\f1 a} = {\f1 f}, A = 0 according to both equations. For this case, the current\par drawn from the source is sinusoidal and equals [E/Z]*sin (wt - {\f1 f) !}} .EQN 58 1 227 0 0 {0:A}NAME:{0:if}NAME({0:\a}NAME<{0:atan}NAME({0:\t}NAME),(2*{0:sin}NAME({0:atan}NAME({0:\t}NAME)-{0:\a}NAME))/((\(1+({0:\t}NAME)^(2)))*(1-({0:e}NAME)^(-(({0:\p}NAME)/({0:\t}NAME))))),({0:sin}NAME({0:atan}NAME({0:\t}NAME)-{0:\a}NAME))/(\(1+({0:\t}NAME)^(2) ))) .EQN 18 0 291 0 0 {0:\b}NAME:{0:atan}NAME({0:\t}NAME) .EQN 7 0 304 0 0 {0:n}NAME:0;360 .EQN 7 0 313 0 0 ({0:\q}NAME)[({0:n}NAME):({0:n}NAME)/(180)*{0:\p}NAME .EQN 7 -1 321 0 0 ({0:vs}NAME)[({0:n}NAME):{0:sin}NAME(({0:\q}NAME)[({0:n}NAME)) .EQN 7 0 328 0 0 ({0:v1}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<{0:deg}NAME,-({0:vs}NAME)[({0:n}NAME),({0:vs}NAME)[({0:n}NAME)) .EQN 7 0 334 0 0 ({0:v2}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<(180+{0:deg}NAME),({0:v1}NAME)[({0:n}NAME),-({0:vs}NAME)[({0:n}NAME)) .EQN 11 0 339 0 0 ({0:Cur1}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<{0:deg}NAME,(1)/({0:Z}NAME)*{0:sin}NAME(({0:\p}NAME+({0:\q}NAME)[({0:n}NAME))-{0:\b}NAME)+{0:A}NAME*({0:e}NAME)^(-(({0:\p}NAME-{0:\a}NAME+({0:\q}NAME)[({0:n}NAME))/({0:\t}NAME))),(1)/({0:Z}NAME)*{0:sin}NAME( ({0:\q}NAME)[({0:n}NAME)-{0:\b}NAME)+{0:A}NAME*({0:e}NAME)^(-((({0:\q}NAME)[({0:n}NAME)-{0:\a}NAME)/({0:\t}NAME)))) .EQN 9 0 343 0 0 ({0:Cur2}NAME)[({0:n}NAME):{0:if}NAME(({0:Cur1}NAME)[({0:n}NAME)>0.0,({0:Cur1}NAME)[({0:n}NAME),0.0) .EQN 8 0 344 0 0 ({0:CurLoad}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<(180+{0:deg}NAME),({0:Cur2}NAME)[({0:n}NAME),({0:Cur2}NAME)[({0:n}NAME-180)) .EQN 7 0 346 0 0 ({0:vBr}NAME)[({0:n}NAME):{0:if}NAME(({0:CurLoad}NAME)[({0:n}NAME)>0.0,({0:v2}NAME)[({0:n}NAME),0.0) .TXT 11 1 347 0 0 Cg a71.000000,71.000000,41 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard The plot of voltage of the bridge circuit} .EQN 24 2 348 0 0 &&(_n_u_l_l_&_n_u_l_l_)&({0:vBr}NAME)[({0:n}NAME)@&&(_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 25 16 0 3 .TXT 39 -2 349 0 0 Cg a71.000000,71.000000,40 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard The line current is obtained as follows.} .EQN 5 0 350 0 0 ({0:Cur3}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<{0:deg}NAME,-({0:CurLoad}NAME)[({0:n}NAME),({0:CurLoad}NAME)[({0:n}NAME)) .EQN 7 0 351 0 0 ({0:LineI}NAME)[({0:n}NAME):{0:if}NAME({0:n}NAME<(180+{0:deg}NAME),({0:Cur3}NAME)[({0:n}NAME),-({0:CurLoad}NAME)[({0:n}NAME)) .TXT 7 0 352 0 0 Cg a71.000000,71.000000,28 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard The waveform of load current} .EQN 45 1 354 0 0 &&(_n_u_l_l_&_n_u_l_l_)&({0:CurLoad}NAME)[({0:n}NAME)@&&(_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 47 26 16 0 3 .TXT 41 0 355 0 0 Cg a71.000000,71.000000,24 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard The plot of Line Current} .EQN 4 1 356 0 0 &&(_n_u_l_l_&_n_u_l_l_)&({0:LineI}NAME)[({0:n}NAME)@&&(_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 38 20 16 0 3 .TXT 35 -1 357 0 0 Cg a71.000000,71.000000,406 {\rtf\ansi \deff0{\colortbl;\red0\green128\blue128;}{\fonttbl{\f0 \fcharset0\fnil Times New Roman;}}\plain\cf1\fs24 \pard The output voltage of the bridge contains ripples. The inductor can be assumed to\par have been provided to reduce the ripple in the load current. The line current \par contains harmonics. We can obtain expressions for the ripple factor of the\par output voltage of the bridge, the ripple factor of the load current and the THD\par in the line current. A program written in C presents the calculations and the plots.}