Page 154 - Rashid, Power Electronics Handbook

10 Diode Recti®ers 143

where V is the effective (rms) value of the ac component of The poor TUF of a half-wave recti®er signi®es that the
ac
load voltage v , transformer employed must have a 3.496 (1=0.286) VA
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rating in order to deliver 1 W dc output power to the load.
q 
2
2
V ¼ V ÿ V dc ð10:24Þ In addition, the transformer secondary winding has to carry a
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ac
dc current that may cause magnetic core saturation. As a
result, half-wave recti®ers are used only when the current
Substituting Eq. (10.24) into Eq. (10.23), the ripple factor can
requirement is small.
be expressed as
In the case of a full-wave recti®er with a center-tapped
transformer, the circuit can be treated as two half-wave
s 
2
V L p  recti®ers operating together. Therefore, the transformer
2
RF ¼ ÿ1 ¼ FF ÿ 1 ð10:25Þ secondary VA rating V I is double that of a half-wave recti®er,
V dc s s
but the output dc power is increased by a factor of four due to
the higher recti®cation ratio as indicated by Eqs. (10.5) and
In the case of a half-wave recti®er,
(10.15). Therefore, the TUF of a full-wave recti®er with center-
p  tapped transformer can be found from Eq. (10.32)
2
Half-wave RF ¼ 1:57 ÿ 1 ¼ 1:21 ð10:26Þ
4 0:318 2
In the case of a full-wave recti®er, Full-wave TUF ¼
2 0:707 0:5
p 
2
Full-wave RF ¼ 1:11 ÿ 1 ¼ 0:482 ð10:27Þ ¼ 0:572 ð10:33Þ
10.2.3.6 Transformer Utilization Factor The bridge recti®er has the highest TUF in single-phase
The transformer utilization factor (TUF), which is a measure recti®er circuits because the currents ¯owing in both the
of the merit of a recti®er circuit, is de®ned as the ratio of the primary and secondary windings are continuous sinewaves.
dc output power to the transformer volt-ampere (VA) rating By substituting Eqs. (10.5), (10.15), (10.29), and (10.31) into
required by the secondary winding, Eq. (10.28), the TUF of a bridge recti®er can be found
P dc V I 2
dc dc
TUF ¼ ¼ ð10:28Þ 0:636
V I V I Bridge TUF ¼ ¼ 0:81 ð10:34Þ
s s s s 0:707 0:707
where V and I are the rms voltage and rms current ratings of
s
s
the transformer secondary The transformer primary VA rating of a full-wave recti®er is
equal to that of a bridge recti®er because the current ¯owing
V m in the primary winding is also a continuous sinewave.
V ¼ p ¼ 0:707 V m ð10:29Þ

s
2
10.2.3.7 Harmonics
The rms value of the transformer secondary current I is the Full-wave recti®er circuits with resistive load do not produce
s
same as that for the load current I . For a half-wave recti®er, I s harmonic currents in their transformers but they are produced
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can be found from Eq. (10.14) in half-wave recti®ers. The amplitudes of the harmonic
currents of a half-wave recti®er with resistive load, relative
0:5 V m to the fundamental, are given in Table 10.1. The extra loss
Half-wave I ¼ ð10:30Þ
s
R caused by the harmonics in the resistively loaded recti®er
circuits is often neglected because it is not high compared with
For a full-wave recti®er, I is found from Eq. (10.16). other losses. However, with nonlinear loads, harmonics can
s
cause appreciable loss and other problems such as poor power
0:707 V m
Full-wave I ¼ ð10:31Þ factor and interference.
s
R
Therefore, the TUF of a half-wave recti®er can be obtained by
TABLE 10.1 Harmonic percentages of a half-wave recti®er with resistive
substituting Eqs. (10.3), (10.13), (10.29), and (10.30) into Eq.
load
(10.28).
Harmonic 2nd 3rd 4th 5th 6th 7th 8th
0:318 2
Half-wave TUF ¼ ¼ 0:286 ð10:32Þ % 21.2 0 4.2 0 1.8 0 1.01
0:707 0:5