Diode with an RLC Load vL(t) vC(t) VCo

Close the switch at t = 0 VCo

KVL around the loop

Characteristic Equation

3 Cases Case 1  = ω0 “critically damped” s1 = s2 = - roots are equal i(t) = (A1 + A2t)es1t

3 Cases (continued) Case 2  > ω0 “overdamped” roots are real and distinct i(t) = A1es2t + A2es2t

3 Cases (continued) Case 3  < ω0 “underdamped” s1,2 = - +/- jωr ωr = the “ringing” frequency, or the damped resonant frequency ωr = √ωo2 – α2 i(t) = e-t(A1cosωrt + A2sinωrt) 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 i1(t)

Mode 1 (continued)

Circuit in Mode 2 I1 i2

Mode 2 (continued)

Example 2.7

Inductor Current

Recovery of Trapped Energy Return 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 Lm = Magnetizing Inductance v2/v1 = N2/N1 = i1/i2

Refer Secondary to Primary Side

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

vD = Vs(1+a) = reverse diode voltage primary current i1 = is Vs = vD/a – Vs/a vD = Vs(1+a) = reverse diode voltage primary current i1 = is Vs = Lm(di1/dt) i1(t) = (Vs/Lm)t for 0<=t<=t1

ai2 = 0, D1 is reverse biased is = i1 v1 = Vs

v2 = av1 = aVs -v1 + vD/a – Vs/a = 0  vD = Vs(1+a)

Operational Mode 2 Begins @ t = t1 when switch is opened i1(t = t1) = (Vs/Lm)t1 = initial current I0 Lm(di1/dt) + Vs/a = 0 i1(t) = -(Vs/aLm)t + I0 for 0 <= t <= t2

i1 = 0 (is = 0) i1 becomes ai2 is = -ai2 (into source Vs)

vD = 0, D1 is forward biased

Waveform Summary

Find the conduction time t2 Solve -(Vs/aLm)t2 + I0 = 0 yields t2 = (aLmI0)/Vs I0 = (Vst1)/Lm t1 = (LmI0)/Vs t2 = at1

Example 2.8 Lm = 250μH N1 = 10 N2 = 100 VS= 220V There is no initial current. Switch is closed for a time t1 = 50μs, then opened. Leakage inductances and resistances of the transformer = 0.

Determine the reverse voltage of D1 The turns ratio is a = N2/N1 = 100/10 = 10 vD = VS(1+a) = (220V)(1+10) = 2420 Volts

Calculate the peak value of the primary and secondary currents From above, I0 = (Vs/Lm)t1 I0 = (220V/250μH)(50μs) = 44 Amperes I’0 =I0/a = 44A/10 = 4.4 Amperes

Determine the conduction time of the diode t2 = (aLmI0)/Vs t2 = (10)(250μH)(44A)/220V t2 = 500μs or, t2 = at1 t2 = (10)(50μs)

Determine the energy supplied by the Source W = (1/2)((220V)2/(250μH))(50μs)2 W = 0.242J = 242mJ W = 0.5LmI02 = (0.5)(250x10-6)(44A)2 W = 0.242J = 242mJ