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 –  > ω 0 “overdamped” –roots are real and distinct –i(t) = A 1 e s2t + A 2 e s2t

3 Cases (continued) Case 3 –  < ω 0 “underdamped” –s 1,2 = -  +/- jω r –ω r = the “ringing” frequency, or the damped resonant frequency –ω r = √ω o 2 – α 2 –i(t) = e -  t (A 1 cosω r t + A 2 sinω r t) –exponentially damped sinusoid

Example 2.6

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 I1I1 i2i2

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 L m = Magnetizing Inductance v 2 /v 1 = N 2 /N 1 = i 1 /i 2

Refer Secondary to Primary Side

Operational Mode 1 Switch 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 2 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 = 250μHN 1 = 10N 2 = 100V S = 220V There is no initial current. Switch is closed for a time t 1 = 50μs, 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/250μH)(50μs) = 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)(250μH)(44A)/220V t 2 = 500μs or, t 2 = at 1 t 2 = (10)(50μs) t 2 = 500μs

Determine the energy supplied by the Source W = 0.5L m I 0 2 = (0.5)(250x10 -6 )(44A) 2 W = 0.242J = 242mJ W = (1/2)((220V) 2 /(250μH))(50μs) 2 W = 0.242J = 242mJ