Introduction
— three-phase diode bridge rectifier —
what is this all about?
vA
D1
D3
D5
+
iOU T
D2
D4
D6
vB
i1
i2
+
i3
+
v1
+
v2
v3
vOU T
−
input voltages
v1 = Vm cos (ω0 t)
2π
v2 = Vm cos ω0 t −
3
4π
v3 = Vm cos ω0 t −
3
2π
vk = Vm cos ω0 t − (k − 1)
,
3
k ∈ {1, 2, 3}
input voltages, waveforms
normalization of voltages
mX ,
vX
Vm
m1 = cos (ω0 t)
2π
m2 = cos ω0 t −
3
4π
m3 = cos ω0 t −
3
voltages?
vk = Vm cos ω0 t − (k − 1) 2π
3 ,
k ∈ {1, 2, 3}
voltages?
vk = Vm cos ω0 t − (k − 1) 2π
3 ,
k ∈ {1, 2, 3}
voltages?
vk = Vm cos ω0 t − (k − 1) 2π
3 ,
k ∈ {1, 2, 3}
voltages?
vk = Vm cos ω0 t − (k − 1) 2π
3 ,
k ∈ {1, 2, 3}
voltages?
vk = Vm cos ω0 t − (k − 1) 2π
3 ,
k ∈ {1, 2, 3}
voltages?
vk = Vm cos ω0 t − (k − 1) 2π
3 ,
k ∈ {1, 2, 3}
v1 , spectrum
v2 , spectrum
v3 , spectrum
voltages, quantitative characterization
k
1
2
3
Vk RM S
103.83 V
103.70 V
105.12 V
all graphs and data PyLab processed
T HD(vk )
3.34 %
2.77 %
3.06 %
T HD
And what is THD?
qP
∞
2
k=2 Ik RM S
T HD ,
I1 RM S
Parseval’s identity:
2
IRM
S =
∞
X
Ik2 RM S
assumed I0 = 0
k=1
results in
q
T HD ,
2
2
IRM
S − I1 RM S
I1 RM S
simple, but important
computational issues, finite sums . . .
normalization of currents and time
jX ,
iX
IOU T
unless otherwise noted
ϕ , ω0 t
good: physical dimensions lost, reduced number of
variables, results are generalized, core of the
problem focused
bad: physical dimensions lost, perfect double-check
tool is lost
how does it work? part 1: theory
vA
D1
D3
D5
+
iOU T
D2
D4
D6
vB
i1
i2
+
i3
+
v1
+
v2
v3
vOU T
−
one of the three: D1, D3, D5
vA
vA , analytical
mA = max (m1 , m2 , m3 )
vA , spectrum
√
3 3
mA =
2π
1+2
∞
X
(−1)k+1
k=1
9k 2 − 1
!
cos (3kω0 t)
what about vB ?
vA
D1
D3
D5
+
iOU T
D2
D4
D6
vB
i1
i2
+
i3
+
v1
+
v2
v3
vOU T
−
one of the three, again: D2, D4, D6
vB
vB , analytical
mB = min (m1 , m2 , m3 )
vB , spectrum
√
3 3
mB =
2π
−1 + 2
∞
X
k=1
!
1
cos (3kω0 t)
9k 2 − 1
the output voltage, vOU T
mOU T = mA − mB = max (m1 , m2 , m3 ) − min (m1 , m2 , m3 )
vOU T , spectrum
mOU T
√
3 3
=
π
1−2
∞
X
1
k=1
36k 2 − 1
!
cos (6kω0 t)
currents?
i1 (t) = (d1 (t) − d2 (t)) IOU T
i2 (t) = (d3 (t) − d4 (t)) IOU T
i3 (t) = (d5 (t) − d6 (t)) IOU T
states of the diodes
the input currents
consider i1
spectra of the input currents
spectra of the input currents, analytical
j1 (t) =
+∞
X
J1C, k cos (kω0 t)
k=1
J1C, k


1


− , k = 6n − 1


√  k
2 3 1
=
, k = 6n + 1
π 
k






0, otherwise
for n ∈ N0 , k > 0
double-check:
PIN
obtained using wxMaxima
√
√
3
2 3
3 3
= ×1×
=
= POU T
2
π
π
numerical verification, Gibbs phenomenon
THD of the input currents
r
2
IOU T
3
√
6
=
IOU T
π
Ik RM S =
Ik RM S, 1
T HD ,
q
Ik2 RM S − Ik2 RM S, 1
Ik RM S, 1
r
π2
− 1 ≈ 31.08 %
9
Parseval’s identity based formula turned out to be useful
T HD =
voltages and currents
some more parameters
s
XRM S ,
1
2π
2π
Z
(x(ω0 t))2 d(ω0 t),
0
x ∈ {i, v}
already used for the THD
S , IRM S VRM S
P ,
1
2π
Z
2π
v(ω0 t) i(ω0 t) d(ω0 t)
0
PF ,
P
S
DP F , cos φ1
and φ1 is . . .
and if the voltages are sinusoidal . . .
S = VRM S IRM S
P = VRM S I1, RM S cos φ1
PF =
I1, RM S
I1, RM S
P
=
cos φ1 =
DP F
S
IRM S
IRM S
DP F = cos ϕ1
T HD =
q
2
2
IRM
S − I1, RM S
I1, RM S
s
=
IRM S
I1, RM S
2
−1
i.e. everything depends on the current waveform and its position
some more parameters, plain rectifier
r
Ik RM S =
2
IOU T
3
1
Vk RM S = √ Vm
2
√
2
1
IOU T × √ Vm = 3 Vm IOU T
3
2
√
3 3
P = VOU T IOU T =
Vm IOU T
π
r
S =3×
PF =
3
≈ 95.5%
π
DP F = 1
actually, not so bad; T HD is the problem
back to the rectifier:
how does it work? part 2: experiment
vA
D1
D3
D5
+
iOU T
D2
D4
D6
vB
i1
i2
+
i3
+
v1
+
v2
v3
vOU T
−
input, at IOU T = 3 A
input, at IOU T = 3 A
input, at IOU T = 3 A
input, at IOU T = 3 A
input, at IOU T = 6 A
input, at IOU T = 6 A
input, at IOU T = 6 A
input, at IOU T = 6 A
input, at IOU T = 9 A
input, at IOU T = 9 A
input, at IOU T = 9 A
input, at IOU T = 9 A
output, at IOU T = 3 A
output, at IOU T = 3 A
output, at IOU T = 6 A
output, at IOU T = 6 A
output, at IOU T = 9 A
output, at IOU T = 9 A
in quantitative terms, input, 1st
IOU T
0A
3A
6A
9A
k
1
2
3
1
2
3
1
2
3
1
2
3
Ik RM S [A]
—
—
—
2.60
2.61
2.63
5.12
5.12
5.13
7.59
7.64
7.66
Vk RM S [V]
101.29
100.63
102.40
98.23
97.73
98.82
94.87
94.34
96.80
92.38
91.95
94.04
Sk [VA]
—
—
—
255.01
254.60
259.71
485.41
482.67
496.95
701.53
702.47
720.00
Pk [W]
—
—
—
245.16
244.37
251.00
466.87
464.25
477.08
673.86
675.16
692.30
in quantitative terms, input, 2nd
IOU T
0A
3A
6A
9A
k
1
2
3
1
2
3
1
2
3
1
2
3
P Fk
—
—
—
0.9614
0.9598
0.9665
0.9618
0.9618
0.9600
0.9605
0.9611
0.9615
T HD(ik ) [%]
—
—
—
30.50
29.57
29.97
29.26
28.37
28.31
28.00
27.21
27.06
T HD(vk ) [%]
4.33
3.75
4.75
4.17
3.86
5.38
3.87
3.66
3.87
4.01
3.92
4.19
in quantitative terms, output
IOU T [A]
0.00
3.21
6.27
9.41
VOU T [V]
239.79
229.51
221.23
212.91
POU T [W]
1.07
736.72
1386.56
2004.12
PIN [W]
−0.81
740.53
1408.20
2041.32
η [%]
—
99.49
98.46
98.18
overall impressions
I
pretty good rectifier
I
simple, robust, cheap
I
good symmetry
I
excellent DP F
I
acceptable P F
I
poor T HD (but not that poor)
I
up to this point:
I
I
I
diode bridge rectifier analyzed
measurement tools developed
is there a way to do something with the T HD?
fruitless effort #1: shaping the output current
vA
iOU T
D1
D3
D5
+
ROU T
D2
D4
D6
vB
i1
i2
+
i3
+
v1
+
v2
v3
vOU T
−
fruitless effort #1: waveforms
fruitless effort #1: quantitative
I
T HD = 30.79%
I
not a big deal of an improvement
I
only one degree of freedom, iOU T
I
shaping i1 , i2 , and i3 is the goal
I
two degrees of freedom needed, since i1 + i2 + i3 = 0
fruitless effort #2: additional deegree of freedom
vA
+
iA
D1
D3
D5
ROU T
2
vOU T
D2
D4
ROU T
2
D6
−
iB
vB
i1
i2
+
i3
+
v1
iN
+
v2
v3
fruitless effort #2: waveforms
fruitless effort #2: neutral current
fruitless effort #2: quantitative
I
T HD = 24.76%
I
somewhat better
I
all of i1 , i2 , and i3 cannot be fixed by programming iA and
iB in this circuit
I
example: i1 = iA , i2 = −iB , no way to fix i3
I
gaps in the input currents in both of the “patches”
I
the additional degree of freedom is taken by iN
I
which is a disaster of itself
I
we would need another degree of freedom to fix iN
I
but this is a wrong approach, iN was not an issue before
conclusions
I
three-phase diode bridge rectifier analyzed
I
quantitative measures of rectifier performance introduced
I
measurement tools developed
I
theoretical predictions related to experiments
I
gaps in the input currents identified as a problem
I
how to fill in the gaps?
I
an answer is current injection . . .