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Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,*,*,Diode with an RLC Load,v,L,(t),v,C,(t),V,Co,Close the switch at t=0,V,Co,KVL around the loop,Characteristic Equation,3 Cases,Case 1,=0“critically damped,s1=s2=-,roots are equal,i(t)=(A1+A2t)es1t,3 Cases(continued),Case 2,0“overdamped,roots are real and distinct,i(t)=A1es2t+A2es2t,3 Cases(continued),Case 3,0“underdamped,s1,2=-+/-jr,r=the“ringing frequency,or the damped resonant frequency,r=o2 2,i(t)=e-t(A1cosrt+A2sinrt),exponentially damped sinusoid,Example 2.6,Determine an expression for the current,Determine an expression for the current,Determine the conduction time of the diode,The conduction time will occur when the current goes through zero.,Conduction Time,Freewheeling Diodes,Freewheeling,Diode,Freewheeling Diodes,D2 is reverse biased when the switch is closed,When the switch opens,current in the inductor continues.D2 becomes forward biased,“discharging the inductor.,Analyzing the circuit,Consider 2“Modes of operation.,Mode 1 is when the switch is closed.,Mode 2 is when the switch is opened.,Circuit in Mode 1,i,1,(t),Mode 1(continued),Circuit in Mode 2,I,1,i,2,Mode 2(continued),Example 2.7,Inductor Current,Recovery of Trapped EnergyReturn Stored Energy to the Source,Add a second winding and a diode,“Feedback winding,The inductor and feedback winding look like a transformer,Equivalent Circuit,L,m,=Magnetizing Inductance,v,2,/v,1,=N,2,/N,1,=i,1,/i,2,Refer Secondary to Primary Side,Operational Mode 1Switch closed t=0,Diode D,1,is reverse biased,ai,2,=0,V,s,=v,D,/a V,s,/a,v,D,=V,s,(1+a)=reverse diode voltage,primary current i,1,=i,s,V,s,=L,m,(di,1,/dt),i,1,(t)=(V,s,/L,m,)t for 0=t=t,1,Operational Mode 2Begins t=t,1,when switch is opened,i,1,(t=t,1,)=(V,s,/L,m,)t,1,=initial current I,0,L,m,(di,1,/dt)+V,s,/a=0,i,1,(t)=-(V,s,/aL,m,)t+I,0,for 0=t=t,2,Find the conduction time t,2,Solve,-(V,s,/aL,m,)t,2,+I,0,=0,yields,t,2,=(aL,m,I,0,)/V,s,I,0,=(V,s,t,1,)/L,m,t,1,=(L,m,I,0,)V,s,t,2,=at,1,Waveform Summary,Example 2.8,L,m,=250,HN,1,=10N,2,=100V,S,=220V,There is no initial current.,Switch is closed for a time t,1,=50,s,then opened.,Leakage inductances and resistances of the transformer=0.,Determine the reverse voltage of D,1,The turns ratio is a=N,2,/N,1,=100/10=10,v,D,=V,S,(1+a)=(220V)(1+10)=2420 Volts,Calculate the peak value of the primary and secondary currents,From above,I,0,=(V,s,/L,m,)t,1,I,0,=(220V/250,H)(50s)=44 Amperes,I,0,=I,0,/a=44A/10=4.4 Amperes,Determine the conduction time of the diode,t,2,=(aL,m,I,0,)/V,s,t,2,=(10)(250,H)(44A)/220V,t,2,=500,s,or,t,2,=at,1,t,2,=(10)(50,s),t,2,=500,s,Determine the energy supplied by the Source,W=0.5L,m,I,0,2,=(0.5)(250 x10,-6,)(44A),2,W=0.242J=242mJ,W=(1/2)(220V),2,/(250,H),(50,s),2,W=0.242J=242mJ,
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