Download Class notes on controlled rectifiers

ELEC 431 Class notes AC/DC ‐ Controlled Rectifiers P.K. Jain Controlled rectifier
™ Unlike the uncontrolled rectifiers, by replacing the diodes with thyristors, the average output
voltage can be controlled by adjusting the turn-on time or the control angle, denoted by α, of the
thyristors (Q1, Q2, Q3 and Q4):
™ The turn-on instant of the thyristor is controlled by the gate signal applied to the thyristor
Figure 1
™ One example is to use the controlled rectifier to control the terminal voltage (vo) in DC machines
with armature resistance Ra and inductance La
DC machine Figure 2
1 ELEC 431 Class notes AC/DC ‐ Controlled Rectifiers v s = 2V sin (ωt ) Figure 3
The DC component of vo is given by:
Vo =
1
π
=
=
π +α
∫
2V sin (ωt )dωt
α
2V
π
[− cos(ωt )] απ +α
2 2V
π
cos(α ) ≈ 0.9V cos(α )
Hence, Vo can be controlled by α.
2 P.K. Jain ELEC 431 Class notes AC/DC ‐ Controlled Rectifiers P.K. Jain DC power at the terminal of the rectifier:
Pt = Vo I o
DC voltage across La = 0 Vo = Ra I o + Eo
Eo = Vo − Ra I o
Hence,
= 0.9V cos(α ) − Ra I o
Po = Eo I o
Output power of the DC machine:
Now, take a look on the input current again:
Figure 4
Recall from Fourier series, the fundamental input current of is is given as: is1 =
The RMS value of is1 is then: I s1 =
4I o
π
4I o
2π
sin (ω t − α )
=
2 2I o
π
≈ 0.9 I o
The total harmonic distortion (THD) in the input current can be calculated to be:
=
Recall:
input power factor is:
I s2 − I s21
I s1
=
I o2 − 0.81I o2
0.9 I o
active or real power Pin
=
=
apparent power
S
=
Vo I o 0.9V cos(α )I o
=
Vs I s
VI o
= 0.9 cos(α )
3 ≈ 0.48 I o
Assume lossless, so Pt = Pin ELEC 431 Class notes AC/DC ‐ Controlled Rectifiers ™ When α < π/2
→
Po = Vo I o
Power flows from AC to DC (rectifying mode/motoring
mode)
™ When α > π/2
→
Po = −Vo I o
Power flows from DC to AC (inverting mode/generating
mode)
4 P.K. Jain ELEC 431 Class notes AC/DC ‐ Controlled Rectifiers P.K. Jain Effect of source inductance (Ls) on current commutation in controlled rectifier
La
Ls
Q1
Q2
is
vo
vs +_
Q4
Q3
Figure 5
v s = 2V sin (ωt ) Figure 6
5 +
_
Ra
ia
+
Eo
_
ELEC 431 Class notes AC/DC ‐ Controlled Rectifiers P.K. Jain Average output voltage:
Vo =
1
π +α
∫ vo dωt =
π α +μ
2V
=
π +α
π α ∫+ μ
2V sin (ωt )dωt
[− cos(ωt )] απ ++αμ
π
2V
Vo =
Hence,
1
π
[cosα + cos(α + μ )]
------------------ (1)
Using similar approach in the uncontrolled rectifier case, voltage across Ls is equal to vs for α ≤ ωt ≤ α+µ
Ls
di Ls
= 2V sin (ωt )
dt
ωt
2V
sin (ωt )d (ωt )
iLs =
ωLs α∫
Hence,
=
At ωt = α,
is = -Io which means k = -Io
→
is =
Hence,
At ωt = α+µ,
→
2V
[− cos(ωt ) + cosα ] + k
ωLs
2V
[− cos(ωt ) + cosα ] − I o
ωLs
i s = Io
Io =
2 I o ωL s
2V
2V
[− cos(α + μ ) + cosα ] − I o
ωLs
= cos α − cos(α + μ )
------------------------- (2)
Substitute (2) into (1)
Vo =
2 I ωL ⎞
2V ⎛
⎜ cosα + cosα − o s ⎟
π ⎝
2V ⎠
=
2 2V
π
6 cos α −
2 I oωLs
π
ELEC 431 Class notes AC/DC ‐ Controlled Rectifiers P.K. Jain Controlled rectifier with a parallel diode
Figure 7
™ By placing a diode (DF) across the output of the controlled rectifier, during the negative cycles of
vo, DF becomes forward biased and vo is clamped to zero
v s = 2V sin (ωt ) vs
ωt
ig1 ig3
α
ωt
ig2 ig4
α
ωt
io
Io
ωt
vo
Vo
ωt
is
Io
Q2 Q4
turn on
0
α
Q1 Q3
turn on
ωt
π π+α
Figure 8
7 2π
ELEC 431 Class notes AC/DC ‐ Controlled Rectifiers P.K. Jain Average output voltage:
Vo =
1
π
∫ vo dωt =
πα
2V
=
π
Power factor =
Recall:
2V
π
π α∫
→
When α = π
Po = Vo I o =
π
→
Vo = 0
(1 + cosα )I o = Pin
1
Is =
2π
= Io
2V
power factor =
Vo = 2 2V / π
Assume lossless thyristors P Pin
=
S VI s
RMS input current:
Hence,
2V sin (ωt )dωt
(1 + cosα )
When α = 0
Since
1
π
(1 + cosα )I o
VI o
(π − α )
2π
π
I o2
∫0 i dt = 2π 2α∫ dt
2
s
(π − α )
π
=
2
(1 + cos α )
π (π − α )
π
Alternative circuit configuration:
Figure 9
™ By replacing a pair of the thyristors with a pair of diodes (D1 and D2) across the output of the
controlled rectifier, vo is clamped during the negative cycles
8