Circuito Rl Con Diodo

UNIVERSIDAD NACIONAL AUTÓNOMA DE MÉXICO FACULTAD DE INGENIERÍA

Electrónica de potencia Tarea #2 Circuitos RL con diodo ideal

Juárez Murillo Samuel Gonzalo Ing. Macías Pérez Roberto Grupo: 4Tarea #2, Juárez Murillo Samuel Gonzalo

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

Linductor:= 100mH = 100 × 10 Rcarga := 10Ω = 10 × 10 Ω
0

−3

H

Vimax := 110V⋅ 2 = 155.563 × 10 V0

s −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

f := 60Hz = 60 × 10

01

Análisis del problema por el método exacto
w := 2 ⋅π ⋅f = 376.991 × 10
01

s

Vi := Vimaxsin( w⋅t ) ⋅Diodo ideal entonces Vd=0 [V]
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

Diagrama

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Vi := Vdiodo + VResistor+ VLinductor
Vi := VLinductor+VResistor

2/20

Tarea #2, Juárez Murillo Samuel Gonzalo

Vi := L⋅

d  IL  + R ⋅Ir dt  

Vi L  d  ( R ⋅Ir) := ⋅ Il  + L L  dt  L Vi  d  ( R ⋅Ir) :=  Il  + L L  dt 

u := wtdu := wdt

dt :=

dwt w

Vimaxsen ( wt) ⋅ L

:= 

 d  ( R ⋅Ir) IL  + L  dt 

( R ⋅Ir) Vimaxsen ( wt) ⋅ d := ( IL⋅w) + L L dwt Vimaxsen ( wt) ⋅ d  w  ( R ⋅Ir) :=  IL⋅  + w ⋅L w ⋅Ldwt  w 

i ( wt) := In ( wt) + If ( wt)
01

w := 2 ⋅π ⋅f = 376.991 × 10

s

L := 100mH = 100 × 10

−3

H
0

Vimax := 110 ⋅ 2 V = 155.563 × 10 V R1 := 10Ω = 10 × 10 Ω
0

3/20Tarea #2, Juárez Murillo Samuel Gonzalo

 R  ⋅wt   w ⋅L  In ( wt) := A ⋅e
−

Z :=
φ

R1 + ( w⋅L) = 39.003 × 10 Ω

2

2

0

:= atan

 w⋅L  = 1.312 × 100   R1 
 Vimax⋅sin( wt − φ )   Z 
− 
R1

If ( wt) := 

 ⋅wt  w ⋅L   Vimax ⋅sin( wt − φ ) i ( wt) := A ⋅e  + 

 Z 

Z = 39.003 × 10 Ω
φ

0

= 1.312 × 10

0

 radianes    segundoUn dato a considerar es que tenemos un diodo ideal por lo tanto al tener un valor de voltaje V>0 queda polarizado en directa y conduce hasta que se polariza en inversa. Es por ello que se sabe...