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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