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210 J. W. Dixon
i
G
i
sc
V
f-f
L
S
L
S
L
D
I
D
T1 T2
T3
T5
T6
T4
V
D
v
D
v
A
v
B
v
C
L
S
N
ON
OFF
FIGURE 12.16 Commutation process.
12.2.6 Commutation
The description of the converters in the previous sections was
based upon assumption that the commutation was instanta-
neous. In practice this is not possible, because the transfer
of current between two consecutive valves in a commuta-
tion group takes a finite time. This time, called overlap time,
depends on the phase-to-phase voltage between the valves par-
ticipating in the commutation process, and the line inductance
L
S
between the converter and power supply. During the over-
lap time, two valves conduct, and the phase-to-phase voltage
drops entirely on the inductances L
S
. Assuming the dc current
I
D
to be smooth and with the help of Fig. 12.16, the following
relation is deduced
2L
S
·
di
sc
dt
=
√
2 ·V
f −f
sin ωt = v
A
−v
B
(12.16)
where i
sc
is the current in the valve being fired during the
commutation process (thyristor T2 in Fig. 12.16). This current
can be evaluated, and it yields
i
sc
=−
√
2
2L
S
·V
f −f
cos ωt
ω
+C (12.17)
The constant “C” is evaluated through initial conditions at
the instant when T2 is ignited. In terms of angle, when ωt = α
when ωt = α, i
sc
= 0
∴ C =
V
sec
f −f
√
2 ·ωL
S
cos α (12.18)
Replacing Eq. (12.18) in (12.17):
i
sc
=
V
f −f
√
2 ·ωL
S
·
(
cos α −cos ωt
)
(12.19)
Before commutation, the current I
D
was carried by thyristor
T1 (see Fig. 12.16). During the commutation time, the load
current I
D
remains constant, i
sc
returns through T1, and T1
is automatically switched off when the current i
sc
reaches the
value of I
D
. This happens because thyristors cannot conduct
in reverse direction. At this moment, the overlap time lasts
and the current I
D
is then conducted by T2. In terms of angle,
when ωt = α +µ, i
sc
= I
D
, where µ is defined as the “overlap
angle.” Replacing this final condition in Eq. (12.19) yields
I
D
=
V
sec
f −f
√
2 ·ωL
S
·
[
cos α −cos
(
α +µ
)
]
(12.20)
To avoid confusion in a real analysis, it has to be remem-
bered that V
f −f
corresponds to the secondary voltage in case of
transformer utilization. For this reason, the abbreviation “sec”
has been added to the phase-to-phase voltage in Eq. (12.20).
During commutation, two valves conduct at a time, which
means that there is an instantaneous short circuit between
the two voltages participating in the process. As the induc-
tances of each phase are the same, the current i
sc
produces the
same voltage drop in each L
S
, but with opposite sign because
this current flows in reverse direction and with opposite slope
in each inductance. The phase with the higher instantaneous
voltage suffers a voltage drop −v and the phase with the
lower voltage suffers a voltage increase +v. This situation
affects the dc voltage V
C
, reducing its value an amount V
med
.
Figure 12.17 shows the meanings of v, V
med
, µ, and i
sc
.
The area V
med
showed in Fig. 12.17, represents the loss of
voltage that affects the average voltage V
C
, and can be evalu-
ated through the integration of v during the overlap angle
µ. The voltage drop v can be expressed as
v =
v
A
−v
B
2
=
√
2 ·V
sec
f −f
sin ωt
2
(12.21)