Rectifier Concepts1. Basic Concepts

a. Pure resistive loadb. Inductive load

c. Load with an internal DC voltage

Single-phase bridge rectifiers1. Idealised circuit with 0sL

)0()cos1(2

sin2

sin2

00

LiLItV

dtdt

diLdttV

dt

diLtV

dus

tt

s

s

uu

When the source voltage is zero and increasing, we call it t = 0.

Just prior to that instant, diode D3 is conducting and the voltage at the top of the diode is approximately zero volts.

Let i(t) be the inductor current; load current Id.

When D3 is conducting.

Note that i(0) is zero and when D2 turns off the inductor current is Id.

Single-phase bridge rectifiers

1. Idealised circuit with

2. Effect of on current commutation

sL

0sL

sL

Single-phase bridge rectifiers

sL

sL

f

b

f

bt

t

t

t

ds

s

dbds

dtdt

diLdtVtV

V

Vt

dt

diLVtV

)sin2(

2sin

1;sin2 1

This equation gives the final time.

Three-phase, full-bridge rectifiers

Idealised circuit with

sL

0sL

Three-phase, full-bridge rectifiers

1. Effect of on current commutationsL

Tutorial problems – 5-1, 5-3, 5-4, 5-11, 5-23, 5-25.

When D1 starts to conduct, D5 is already in conduction but doesn't switcho®becauseof the inductors.

va ¡ L sdiadt

+L sdicdt

¡ vc =0

Since I d = ia +ic and I d is constant, wehave:

va ¡ 2L sdiadt

¡ vc = 0

va ¡ vc =p(2)Vs (sin! t ¡ sin(! t ¡ 240)) =

p(3)

p(2)Vs sin(! t ¡ 30) giving

2L s

Z u+¼=6

¼=6

diadtd! t =

p(3)

p(2)Vs

Z u+¼=6

¼=6sin(! t ¡ 30)d! t

This means (with ia(¼=3) = 0 and ia(¼=3+u) = I d):

2! L sp(3)

p(2)Vs

=1¡ cosu:

Theaverageoutput voltagecan begiven by:

1¼=3

ÃZ u+¼=6

¼=6(va ¡ L s

diadt

¡ vb)d! t +Z ¼=6+¼=3

u+¼=6(va ¡ vb)d! t

!

=

=1¼=3

ÃZ ¼=6+¼=3

¼=6(va ¡ vb)d! t ¡

Z u+¼=6

¼=6L sdiadtd! t

!

=1¼=3

ÃZ ¼=6+¼=3

¼=6(va ¡ vb)d! t ¡ ! L sI d

!

Z ®+u

®212Vs sin! td(! t) =

Z ®+u

®L sdisdtd(! t)

= ! L s

Z I d

¡ I d

dis

) cos(®+u) = cos®¡2! L sI dp2Vs

Maple Script is: restart; currentdir("C:\\wattle\\courses\\PowerElectronics\\maple\\");

read "scrPracticalPage133.ma";

scrfw1phase-2.cir has the pSpice script to simulate it.

Tutorial Problems

• 6-1

• 6-3 • 6-5 (b) DPF = 0.707, PF = 0.636, THD = 48.55%. (c) DPF = 0.5, PF = 0.45,

THD = 48.43%.

• 6-9

• 6-12

• 6-18