Diode with an RLC Load
J1
Key = Space
D1
R
DIODE_VIRTUAL
160 Ohm
L
2m H
vL(t)
Vs
220 V
C
0.05uF
VCo
vC(t)
Close the switch at t = 0
J1
Key = Space
D1
R
DIODE_VIRTUAL
160 Ohm
L
2m H
Vs
220 V
C
0.05uF
VCo
KVL around the loop
di 1
V  R i  L   idt  v (0)
dt C
di
di 1
L
R  i  0
dt
dt C
di R di
i


0
dt
L dt LC
s
C
2
2
2
2
Characteristic Equation
R
1
s  s
0
L
LC
R
R
1
s 


2L
2L
LC
R
  dampingfactor 
2L
2
1,2
 
2
1
  resonantfrequency 
LC
s      w
0
2
1,2
2
0
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
D1
DIODE_VIRTUAL
V1
220V
R
160ohm
L
2mH
C
0.05uF
Determine an expression for the current
R
160
α=
=
= 40, 000rad/s
-3
2L (2)(2×10 )
1
1
ω0 =
=
= 1×105 rad/s
LC
(2×10-3 )(0.05×106 )
α < ω0
ωr = ω02 - α 2 = 1×1010 -16×108 = 91, 652rad/s
i(t) = e-αt (A1cosωr t + A 2sinωr t)
@t = 0,i(t = 0) = 0
A1 = 0
i(t) = e-αt (A 2sinωr t)
di
= ωr cosωr tA 2 e-αt - αsinωr tA 2 e-αt
dt
Determine an expression for the current
@t = 0,
Vs
di
= ωr A 2 =
dt t=0
L
Vs
220V
A2 =
=
= 1.2A
-3
ωr L (91, 652rad/s)(2×10 )
i(t) = 1.2sin(91, 652t)e-40,000t
Determine the conduction time of the diode
• The conduction time will occur when the current goes
through zero.
•
i(t1 ) = 1.2sin(91, 652t1 )e
91, 652t1 = π
π
t1 =
= 34.27μs
91, 652
-40,000t
A=0
Conduction Time
Freewheeling Diodes
XSC1
G
T
A
100usec 200usec
J1
D1
DIODE_VIRTUAL
Freewheeling
V1
200 V
D2
DIODE_VIRTUAL
L
200uH
Rf
0.001 Ohm
R
0.001 Ohm
Diode
B
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.
100usec 200usec
J1
D1
DIODE_VIRTUAL
D2
DIODE_VIRTUAL
L
200uH
Rf
0.001 Ohm
R
0.001 Ohm
V1
200 V
Analyzing the circuit
• Consider 2 “Modes”
of operation.
100usec 200usec
J1
D1
• Mode 1 is when the
switch is closed.
• Mode 2 is when the
switch is opened.
DIODE_VIRTUAL
D2
DIODE_VIRTUAL
L
200uH
Rf
0.001 Ohm
R
0.001 Ohm
V1
200 V
Circuit in Mode 1
i1(t)
L
20uH
Vs
220V
R
1 Ohm
Mode 1 (continued)
Vs
i1 (t )  (1  e
R
Open @ t  t1
R
 t
L
)
Vs
I1  i1 (t1 )  (1  e
R
steady  state
Vs
Is 
R
R
 t1
L
)
Circuit in Mode 2
I1
L
20uH
i2
R
1 Ohm
Mode 2 (continued)
di2
0L
 Ri2
dt
i2 (0)  I1
i2 (t )  I1e
R
 t
L
Example 2.7
XSC1
G
T
A
100usec 200usec
J1
D1
DIODE_VIRTUAL
L
200uH
V1
200 V
D2
DIODE_VIRTUAL
R
0.001 Ohm
B
Inductor Current
i(t) = 100A
i(t) =
V
t
L
S
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
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 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
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
t1
t1
2
s
Vs
1V 2
W =  vidt =  Vs tdt =
t1
Lm
2 Lm
0
0
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