Diode with an RLC Load
D1
DIODE_VIRTUAL
R
160 Ohm
L2mH
C0.05uF
Vs220 V
J1Key = Space
vL(t)
vC(t)VCo
Close the switch at t = 0
D1
DIODE_VIRTUAL
R
160 Ohm
L2mH
C0.05uF
Vs220 V
J1Key = Space
VCo
KVL around the loop
2
2
2
2
1(0)
10
0
s C
diV Ri L idt v
dt Cdi di
L R idt dt C
di R di idt L dt LC
Characteristic Equation
2
2
1,2
0
2 2
1,2 0
10
12 2
21
Rs s
L LC
R Rs
L L LCR
dampingfactorL
resonantfrequencyLC
s w
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
V1220V
D1
DIODE_VIRTUAL
R
160ohm
L2mH
C0.05uF
Determine an expression for the current
-3
50 -3 6
0
2 2 10 8r 0
-αt1 r 2 r
1
-αt2 r
-αtr r 2 r
R 160α = = = 40,000rad/s
2L (2)(2×10 )
1 1ω = = =1×10 rad/s
LC (2×10 )(0.05×10 )
α < ω
ω = ω -α = 1×10 -16×10 = 91,652rad/s
i(t) = e (A cosω t +A sinω t)
@t = 0,i(t = 0) = 0
A = 0
i(t) = e (A sinω t)
di= ω cosω tA e -αsinω tA
dt-αt
2e
sr 2
t=0
s2 -3
r
-40,000t
@t = 0,
Vdi= ω A =
dt LV 220V
A = = =1.2Aω L (91,652rad/s)(2×10 )
i(t) =1.2sin(91,652t)e
Determine an expression for the current
Determine the conduction time of the diode
• The conduction time will occur when the current goes through zero.
•
-40,000t1 1
1
1
i(t ) =1.2sin(91,652t )e A = 0
91,652t = π
πt = = 34.27μs
91,652
Conduction Time
Freewheeling Diodes
J1
100usec 200usec
D1
DIODE_VIRTUAL
D2DIODE_VIRTUAL
L200uH
R0.001 Ohm
XSC1
A B
G
T
V1200 V
Rf0.001 Ohm
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.
J1
100usec 200usec
D1
DIODE_VIRTUAL
D2DIODE_VIRTUAL
L200uH
R0.001 Ohm
V1200 V
Rf0.001 Ohm
Analyzing the circuit
• Consider 2 “Modes” of operation.
• Mode 1 is when the switch is closed.
• Mode 2 is when the switch is opened.
J1
100usec 200usec
D1
DIODE_VIRTUAL
D2DIODE_VIRTUAL
L200uH
R0.001 Ohm
V1200 V
Rf0.001 Ohm
Circuit in Mode 1
Vs220V
L20uH
R1 Ohm
i1(t)
Mode 1 (continued)
1
1
1
1 1 1
( ) (1 )
@
( ) (1 )
Rt
s L
Rt
s L
ss
Vi t e
ROpen t t
VI i t e
Rsteady state
VI
R
Circuit in Mode 2
L20uH
R1 Ohm
I1
i2
22
2 1
2 1
0
(0)
( )RtL
diL Ridt
i I
i t I e
Mode 2 (continued)
Example 2.7
J1
100usec 200usec
D1
DIODE_VIRTUAL
D2DIODE_VIRTUAL
L200uH
R0.001 Ohm
XSC1
A B
G
T
V1200 V
Inductor Current
SV
i(t) = tL
i(t) = 100A
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
Lm = Magnetizing Inductance
v2/v1 = N2/N1 = i1/i2
Refer Secondary to Primary Side
Operational Mode 1Switch closed @ t = 0
Diode D1 is reverse biased, ai2 = 0
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
Operational Mode 2Begins @ 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
Find the conduction time t2
Solve
-(Vs/aLm)t2 + I0 = 0
yields
t2 = (aLmI0)/Vs
I0 = (Vst1)/Lm
t1 = (LmI0)Vs
t2 = at1
Waveform Summary
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)
• t2 = 500μs
Determine the energy supplied by the Source
1 1t t 2
2s ss 1
m m0 0
V V1W = vidt = V tdt = t
L 2 L
W = 0.5LmI02 = (0.5)(250x10-6)(44A)2
W = 0.242J = 242mJ
W = (1/2)((220V)2/(250μH))(50μs)2
W = 0.242J = 242mJ