Solution of Tutorial 2 - Uncontrolled Rectifier Circuits

 

ELEC4240/9240

Power Electronics

The University of New South Wales School of Electrical Engineering & Telecommunications

Solution for Tutorial 2 Question 1(a). 

v = Vmax sin ω t = Ri + L

di dt 

+ Eb   

The solution for i is i=

Vmax  R + ( ω L ) 2

2

 At   θ b = sin −1

 E    b V max 2

 R

∴i = At

  ωt

 

=

∴0 =

 R

 R

  e +A

− t  L

, i=0

Vmax

∴0 = i = ∴ A = − [

sin( ω t − φ  ) −

E b

+

2

( ω L ) Vmax

 R + ( ω L ) 2

Vmax  R 2 + ( ω L )2

2

sin( θ b − φ ) − sin( θ b − φ  ) −

sin( ω t − φ ) −

Eb

E b  R E b  R

+ Ae  

 R   ω L

]e

θ b

 R ω L

θ b

   R

Vmax

( θ b − ω t ) E sin( θ b − φ ) − b  ]  e ω L R R 2 + ( ω L )2

−[

 R

−

β ,   i = 0 Vmax  R 2 + ( ω L )2

Solution of Tutorial 2

sin( β − φ ) −

Eb  R

−[

Vmax

 

R 2 + ( ω L )2

1

sin( θ b − φ ) −

Eb R

 R

]  e ω L

 

( θ b − β  )

F. Rahman

 

ELEC4240/9240

Power Electronics

If  E   b = 0,

then θ b = 0

and  R β 

 sin( β − φ ) = − sin φ  e  

−

ω  L

i  

(b) 

 sin( β − φ  )

 β    Degree

Radian

200 220 240 230 225

3.491 3.840 4.188 4.014 3.927

  0.422 0.087 -0.259 -0.087 0

 R β 

 R β 

− sin φ  •  e

−

ω  L

  φ  • e    sin(  β  − φ  ) − sin

-0.022 -0.0156 -0.109 -0.0127 -0.014

−

ω  L

 

>0 >0 0 >0