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Diode with an RLC Load D1 DIODE_VIRTUAL R 160 Ohm L 2mH C 0.05uF Vs 220 V J1 Key = Space v L ( t) v C (t ) V Co
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### Transcript of Diode with an RLC Load v L (t) v C (t) V Co. Close the switch at t = 0 V Co.

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

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