Diode with an RLC Load

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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  = ω 0 “critically damped” s 1 = s 2 = - roots are equal i(t) = (A 1 + A 2 t)e s1t. 3 Cases (continued). Case 2 - PowerPoint PPT Presentation

Transcript of Diode with an RLC Load

  • Diode with an RLC LoadvL(t)vC(t)VCo

  • Close the switch at t = 0VCo

  • KVL around the loop

  • Characteristic Equation

  • 3 CasesCase 1 = 0critically dampeds1 = s2 = -roots are equali(t) = (A1 + A2t)es1t

  • 3 Cases (continued)Case 2 > 0 overdampedroots are real and distincti(t) = A1es2t + A2es2t

  • 3 Cases (continued)Case 3 < 0 underdampeds1,2 = - +/- jr r = the ringing frequency, or the damped resonant frequencyr = o2 2i(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 diodeThe conduction time will occur when the current goes through zero.

  • Conduction Time

  • Freewheeling DiodesFreewheelingDiode

  • Freewheeling DiodesD2 is reverse biased when the switch is closedWhen the switch opens, current in the inductor continues. D2 becomes forward biased, discharging the inductor.

  • Analyzing the circuitConsider 2 Modes of operation.

    Mode 1 is when the switch is closed.

    Mode 2 is when the switch is opened.

  • Circuit in Mode 1i1(t)

  • Mode 1 (continued)

  • Circuit in Mode 2I1i2

  • Mode 2 (continued)

  • Example 2.7

  • Inductor Current

  • Recovery of Trapped EnergyReturn Stored Energy to the Source

  • Add a second winding and a diodeFeedback windingThe inductor and feedback winding look like a transformer

  • Equivalent CircuitLm = Magnetizing Inductancev2/v1 = N2/N1 = i1/i2

  • Refer Secondary to Primary Side

  • Operational Mode 1Switch closed @ t = 0Diode D1 is reverse biased, ai2 = 0

  • Vs = vD/a Vs/avD = Vs(1+a) = reverse diode voltageprimary current i1 = isVs = Lm(di1/dt)i1(t) = (Vs/Lm)t for 0
  • Operational Mode 2Begins @ t = t1 when switch is openedi1(t = t1) = (Vs/Lm)t1 = initial current I0Lm(di1/dt) + Vs/a = 0i1(t) = -(Vs/aLm)t + I0 for 0
  • Find the conduction time t2Solve -(Vs/aLm)t2 + I0 = 0yields t2 = (aLmI0)/Vs

    I0 = (Vst1)/Lmt1 = (LmI0)Vs

    t2 = at1

  • Waveform Summary

  • Example 2.8Lm = 250HN1 = 10N2 = 100VS= 220VThere is no initial current.Switch is closed for a time t1 = 50s, then opened.Leakage inductances and resistances of the transformer = 0.

  • Determine the reverse voltage of D1The turns ratio is a = N2/N1 = 100/10 = 10vD = VS(1+a) = (220V)(1+10) = 2420 Volts

  • Calculate the peak value of the primary and secondary currentsFrom above, I0 = (Vs/Lm)t1I0 = (220V/250H)(50s) = 44 Amperes

    I0 =I0/a = 44A/10 = 4.4 Amperes

  • Determine the conduction time of the diode

    t2 = (aLmI0)/Vs t2 = (10)(250H)(44A)/220V t2 = 500s

    or, t2 = at1 t2 = (10)(50s) t2 = 500s

  • Determine the energy supplied by the SourceW = 0.5LmI02 = (0.5)(250x10-6)(44A)2W = 0.242J = 242mJW = (1/2)((220V)2/(250H))(50s)2W = 0.242J = 242mJ