what happens if R is omitted?
iA
D1
D3
vA
C/2
iIA
D5
+
IOU T
The Discontinuous Conduction Mode
L
D2
iR1
D4
iR2
D6
iR3
iB
iIB
C/2
vB
iX
iX
iX
i1
i2
+
+
v1
what really happens?
I
in the case VA, 1 and VB, 1 remain the same as in the CCM,
the amplitude of iY won’t be limited
I
something, though, limits the amplitude; try and see . . .
let’s look at the circuit and search for the answers:
I
i3
+
v2
v3
v1 , i1 , IOU T ≈ 3 A
1. what limits the amplitude of iY ?
2. is it safe to operate in this mode?
3. is there any use of this operating mode?
v1 , i1 , IOU T ≈ 3 A, spectra
v1 , IOU T ≈ 3 A
v2 , i2 , IOU T ≈ 3 A
v2 , i2 , IOU T ≈ 3 A, spectra
vN
iY
vOU T
−
v2 , IOU T ≈ 3 A
v3 , i3 , IOU T ≈ 3 A
v3 , i3 , IOU T ≈ 3 A, spectra
v3 , IOU T ≈ 3 A
v1 , i1 , IOU T ≈ 6 A
v1 , i1 , IOU T ≈ 6 A, spectra
v1 , IOU T ≈ 6 A
v2 , i2 , IOU T ≈ 6 A
v2 , i2 , IOU T ≈ 6 A, spectra
v2 , IOU T ≈ 6 A
v3 , i3 , IOU T ≈ 6 A
v3 , i3 , IOU T ≈ 6 A, spectra
v3 , IOU T ≈ 6 A
v1 , i1 , IOU T ≈ 9 A
v1 , i1 , IOU T ≈ 9 A, spectra
v1 , IOU T ≈ 9 A
v2 , i2 , IOU T ≈ 9 A
v2 , i2 , IOU T ≈ 9 A, spectra
v2 , IOU T ≈ 9 A
v3 , i3 , IOU T ≈ 9 A
v3 , i3 , IOU T ≈ 9 A, spectra
v3 , IOU T ≈ 9 A
vOU T , iOU T , IOU T ≈ 3 A
vOU T , iOU T , IOU T ≈ 3 A, spectra
vOU T , iOU T , IOU T ≈ 6 A
vOU T , iOU T , IOU T ≈ 6 A, spectra
vOU T , iOU T , IOU T ≈ 9 A
vOU T , iOU T , IOU T ≈ 9 A, spectra
does it worth?
does it worth?
IOU T [A]
≈ 3A
≈ 6A
≈ 9A
k
1
2
3
1
2
3
1
2
3
Ik RM S [A]
2.90
2.91
2.88
5.66
5.66
5.65
8.38
8.48
8.42
Vk RM S [V]
99.78
99.47
100.12
95.89
95.75
97.01
91.67
92.14
94.72
S [VA]
289.66
289.58
288.67
542.29
541.53
548.57
767.90
781.43
797.43
P [W]
282.36
282.42
281.42
533.10
532.38
539.83
752.36
767.26
785.45
does it worth?
IOU T [A]
3.08
6.06
9.41
IOU T [A]
≈ 3A
≈ 6A
≈ 9A
k
1
2
3
1
2
3
1
2
3
PF
0.9748
0.9753
0.9749
0.9830
0.9831
0.9841
0.9798
0.9819
0.9850
T HD(ik ) [%]
17.76
17.86
17.74
14.69
14.19
14.19
11.98
11.53
11.69
T HD(vk ) [%]
4.22
3.91
4.14
3.76
4.21
3.80
5.47
5.47
4.32
user’s point of view . . . and a conclusion
VOU T [V]
253.32
232.91
216.86
POU T [W]
781.38
1410.35
2041.09
PIN [W]
846.21
1605.31
2305.07
η [%]
92.34
87.86
88.55
really low η; quite unexpected! is this a mistake?
I
the input currents are not so good, but better than without
injection
I
T HD is in the range from 10 % to 20 %
I
absolutely no notches in the input voltages
I
spikes in the output voltage, rich with harmonics
I
the spikes decrease with increases of the output current
I
the spikes increase the output voltage average, VOU T
I
there are some losses in the system, unexpected?
I
the efficiency is much lower than expected!
I
at 9 A the rectifier operates close to the CCM
I
spikes are the answer! what causes them?
I
let’s take a closer look . . .
vA , iA , IOU T ≈ 3 A
vB , iB , IOU T ≈ 3 A
vA , iA , IOU T ≈ 6 A
vB , iB , IOU T ≈ 6 A
vA , iA , IOU T ≈ 9 A
vB , iB , IOU T ≈ 9 A
let’s start from the end . . .
efficiency at the boundary . . .
Q: what the end is?
A: CCM-DCM boundary, close to IOU T = 9 A in our case
at that point:
PIN J
iY = 2 IOU T cos (3ω0 t)
normalize, . . .
jY = JY m cos (3ω0 t) = 2 cos (3ω0 t)
1
JY m MA, 1
2
√
√
1
3 3
3 3
= ×2×
=
2
8π
8π
√
3 3
POU T =
π
PIN J =
η=
POU T
8
= ≈ 88.89 %
POU T + PIN J
9
JY m = 2 instead of 32 , which would be optimal
experimental results make sense now?
j1 at the boundary (and not just there) . . .
at the boundary . . .
to get to the boundary
√
√
MA, 1
3 3 1
3 3
ρ=
=
× =
JY m
8π
2
16π
and at the boundary . . .
wxMaxima will be heavily needed from here . . .
√
7
3
√
9 3
J1m =
4π
√
9 3
J1RM S = √
4π 2
JRM S =
T HD and the P F at the boundary
and what if Q = 0?
r
224π 2
− 1 ≈ 10.43 %
2187
√
27 3
P F = √ ≈ 0.9946
4π 14
T HD =
not that bad . . .
but, the efficiency is bad!
parameters for Q = 0?
η, T HD, and P F at the boundary, Q = 0
this boundary requires a different value of ρ . . .
ρ=
η=
max (mAV )
1 1
1
= × =
max (jY )
4 2
8
√
POU T
MOU T
6 3
√ ≈ 90.53 %
= 3
=
PIN
2π + 3 3
2 J1m
s
at the boundary . . .
T HD =
s
10
2
−√
9
3π
√
3 2
J1m =
+
π
3
√
√
3
2
J1RM S = √ +
3
π 2
JRM S =
partial conclusions . . .
I
I
I
there is some interest in the DCM, due to the relatively
acceptable THD
s
PF =
√
16π 2 − 24π 3 − 27
√
≈ 4.93 %
4π 2 + 12π 3 + 27
√
4π 2 + 12π 3 + 27
√
≈ 0.9988
20π 2 − 12π 3
T HD and P F okay, but η is not . . .
better T HD and P F are achieved in CCM with better η, this
mode does not make any sense in practice . . .
case Q → ∞, R → 0
p
L/C
I
Q = R0 /R, R0 =
I
let’s analyze large R0 and low R case . . . not just huge Q
I
in that case jY ≈ 2 cos (3ω0 t)
and the spikes limit the amplitude of jY . . .
the boundary between the CCM and the DCM suffers from
poor efficiency, not of practical interest
I
I
. . . since there is no other cause
we should analyze the DCM somewhere away from the
boundary . . .
I
let’s model the spikes somehow . . .
I
maybe with Dirac impulses?
I
that’s why I stressed the “flat” spectrum
I
let’s take a look . . .
vA and vB , IOU T ≈ 3 A
vA and vB , IOU T ≈ 3 A
vA and vB , IOU T ≈ 6 A
vA and vB , IOU T ≈ 6 A
vA and vB , IOU T ≈ 9 A
vA and vB , IOU T ≈ 9 A
definitely, the spikes . . .
definitions of mA0 and mB0 , no spikes
definitely, the spikes are the answer for the DCM:
I
the spikes reduce the 3rd harmonic (at 3ω0 ) in mA and mB
I
the spikes depend on the output current
I
the spikes are disappearing as we are getting close to the
CCM
I
the spikes introduce new harmonics, needed to distort iY
I
the spikes have flat-looking spectrum . . .
I
. . . which makes them suitable to model with Dirac δ
impulses!
the same as mA and mB before . . .
I
and I personally like δ impulses and Dirac’s approach . . .
thus, they have the same spectra, . . .
I
P. A. M. Dirac: “The aim of science is to make difficult
things understandable in a simpler way; the aim of poetry
is to state simple things in an incomprehensible way.”
mA0 = max (m1 , m2 , m3 )
mB0 = min (m1 , m2 , m3 )
let’s model the spikes . . .
mA = mA0 + MX
the model of spikes, DSP version . . .
+∞
X
π
2π
δ ω0 t − − n
3
3
n=−∞
mB = mB0 − MX
+∞
X
2π
δ ω0 t − n
3
n=−∞
where VX is yet to be determined . . .
a word about physical dimension of δ impulses . . .
the spectra, mA . . .
a DSP note . . .
I
when it comes to δ impulses, the DSP approach becomes
extremely error prone
I
be VERY careful in forming the spectra, a tiny error could
destroy the results
I
this is not a story, this is an experience . . .
I
much worse than Gibbs phenomenon . . .
I
the higher the level of discontinuity — the worse
I
with δ impulses spectral leakage is a problem, since the is a
lot to leak
mA = max (m1 , m2 , m3 ) + MX
mA = MA, 0 +
π
2π
δ ω0 t − − n
3
3
n=−∞
MA, k cos (3kω0 t)
k=1
MA, 0
MA, k =
the spectra, mB . . .
∞
X
+∞
X
√
3 3
3
=
+
MX
2π
2π
√
3 3 (−1)k+1
3
+ (−1)k MX
π 9k 2 − 1
π
since we are already here, mOU T , the spectrum . . .
mB = min (m1 , m2 , m3 ) − MX
mB = MB, 0 +
∞
X
mOU T = mA − mB
+∞
X
2π
δ ω0 t − n
3
n=−∞
mOU T = MOU T, 0 +
MB, k cos (3kω0 t)
MOU T, k cos (3kω0 t)
k=1
k=1
√
3 3
3
MB, 0 = −
−
MX
2π
2π
√
3 3
1
3
MB, k =
− MX
π 9k 2 − 1 π
let’s use the spectra . . .
∞
X
MOU T, 0 =
MOU T, k =
√
3 3 3
+ MX
π
π


 0
√
6 3
1
6

 −
+ MX
π 9k 2 − 1 π
for k odd
for k even
a double check . . .
to push jY . . .
ρ JY m = MA, 1 = MB, 1 =
√
3 3 3
− MX
8π
π
at ρ = 0:
MOU T =
and we get MX . . .
MX
√
√
3 π
3 2π
=
− ρ JY m =
−
ρ
8
3
8
3
at ρ =
√
3 3
16π :
MOU T =
to get MOU T . . .
MOU T =
√
27 3
− 2ρ
8π
√
27 3
≈ 1.8607
8π
√
√
√
27 3 3 3
3 3
−
×2=
≈ 1.6540
8π
16π
π
MOU T variation within +12.5 % . . .
dependence of VOU T on IOU T
MOU T
a word about efficiency . . .
√
27 3
=
− 2ρ
8π
√
27 3
− 2ρ
8π
√
1
27 3
+ × ρ × 22 =
2
8π
POU T = MOU T =
denormalize . . .
√
VOU T
27 3
R IOU T
=
−2
Vm
8π
Vm
√
27 3
VOU T =
Vm − 2 R IOU T
8π
PIN = POU T
η =1−
16π
√ ρ
27 3
η passes the double check at the CCM-DCM boundary . . .
the output impedance is 2 R, . . .
δ conclusions . . .
simplify the circuit . . .
I
the spikes are modelled by δ impulses
I
prediction of the output voltage obtained, VOU T . . .
I
prediction of the output impedance, 2 R . . .
I
originally, there are 6 diodes
I
prediction of the efficiency obtained . . .
I
deep theory says there are 26 = 64 states
I
however, this is just an approximation . . .
I
fortunately, we do not care about all of them . . .
I
. . . although experimentally verifiable
I
in the CCM only 6 states occur
I
how good the approximation is?
I
in the DCM there are more than 6 . . .
I
is there a better model?
I
first, let’s reduce the problem as much as we can . . .
I
avoid exact solution, it is not available in a closed form . . .
I
but not more than that . . .
I
. . . like it was in the CCM
I
how about “simulation”?
I
but not just in a form of a cheap experiment . . .
an equivalent circuit to study the DCM
vA
DA
+
vA0
vCA
−
+
−
some voltages defined . . .
iY /2
as defined earlier . . . just rename . . .
C/2
vA0 = max (v1 , v2 , v3 )
+
vT
iY
L
vB0 = min (v1 , v2 , v3 )
−
IOU T
+
vT
vB0
−
+
−
DB
vCB
+
−
C/2
and another voltage waveform which would be needed . . .
vAV 0 =
vA0 + vB0
2
iY /2
vB
about DA and DB
I
DA and vA0 represent v1 , v2 , v3 , and D1, D3, D5
I
DB and vB0 represent v1 , v2 , v3 , and D2, D4, D6
I
DA and DB model the DCM
I
from 6 diodes to 2
I
from 26 = 64 states to 22 = 4
I
and out of these four, one is irrelevant . . .
I
an improvement . . . helps us understand . . .
I
only 3 states to take care of!
I
but that’s not that only . . .
how to get the currents?
in the same way as before:
i1 = d1 iA − d2 iB −
1
iY
3
i2 = d3 iA − d4 iB −
1
iY
3
i3 = d5 iA − d6 iB −
1
iY
3
where dn functions are as defined earlier, n ∈ {1, . . . 6}
iY is really important . . .
reduced number of states . . .
iA = IOU T +
1
iY
2
iB = IOU T −
1
iY
2
...
i1 = (d1 − d2 ) IOU T +
3 (d1 + d2 ) − 2
iY
6
I
instead of 26 = 64 we deal with 4 states now
I
states, I mean diode state combinations
I
and it is not 4, but 3 for IOU T > 0
I
which is why the equivalent circuit has been introduced . . .
I
we need iY . . .
I
vA and vB would also be useful . . .
I
let’s find iY , vA , and vB in each of the states . . .
3 (d3 + d4 ) − 2
iY
6
3 (d5 + d6 ) − 2
+
iY
6
i2 = (d3 − d4 ) IOU T +
i3 = (d5 − d6 ) IOU T
back to the circuit . . .
and some circuit theory . . .
vA
DA
+
vA0
vCA
−
+
−
iY /2
C/2
+
vT
L
iY
−
IOU T
+
I
at most, for DA and DB on, the circuit is of the third order
I
when a diode goes off, algebraic degeneration over iL = iY
occurs, since iY = ±2 IOU T in such a case . . .
I
. . . but that’s not a big problem; the capacitors are . . .
vT
vB0
−
+
DB
−
vCB
+
−
C/2
iY /2
vB
and some circuit theory, capacitors . . .
I
the capacitors share the same current, iY /2
I
thus, their voltages differ for a constant . . .
I
the circuit (with DA and DB on) has three poles, one of
them in s = 0
I
the capacitors have the same AC components in their
voltages
I
but the DC components are different!
I
it is not a big deal to find the DC components . . . well . . .
I
especially since VCA = −VCB due to the symmetry . . .
I
I
but to find the AC component is a problem . . .
after some dirty job . . .
let’s introduce
vC =
vCA + vCB
2
which turns out to be the AC component of the voltages across
CA and CB , since the DC components are the opposite . . .
it can be shown that the resistance distribution parameter does
not have any influence . . .
the circuit exposes algebraic degeneration if a diode is off, as
already stated . . .
which we are going to solve!
a lot of effort to solve in a sort of elegant way . . .
state 0, DA is on, DB is on
state −1, DA is off, DB is on
equations:
L
diY
= −R iY − vC + vAV 0
dt
iY = −2 IOU T
C
dvC
= iY
dt
C
conditions:
dvC
= −2 IOU T
dt
condition:
−2 IOU T ≤ iY
if violated switch to state −1
iY ≤ 2 IOU T
if violated switch to state +1
vC > vAV 0 + 2 R IOU T
if violated switch to state 0
state +1, DA is on, DB is off
vA and vB
iY = 2 IOU T
C
dvC
= 2 IOU T
dt
state
0
−1
+1
DA
on
off
on
vA
vA0
vA = vA DCM
vA0
DB
on
on
off
vB
vB0
vB0
vB = vB DCM
vA DCM = −vB0 − 4 R IOU T + 2 vC
condition:
vC < vAV 0 − 2 R IOU T
vB DCM = −vA0 + 4 R IOU T + 2 vC
if violated switch to state 0
new normalization, motivation
new normalization
I
existing normalization of currents, with Ibase = IOU T is
inadequate . . .
I
the problem is in the dependence of ρ on IOU T ,
ρ = R IOU T /Vm
I
R remains constant, while IOU T varies
I
it is inconvenient to consider variations of IOU T as
variations of ρ, but not that big of a deal . . .
I
I
we need a solid foundation for Ibase
p
besides, R0 , L/C plays a significant role now . . .
I
it’s time to renew normalization . . .
the resonance parameter, r
Vbase = Vm
r
Rbase , R0 =
Ibase =
JOU T =
L
C
Vm
R0
IOU T
R0 IOU T
=
Ibase
Vm
ρ=
R
1
=
R0
Q
state 0, DA is on, DB is on, normalized
equations:
ωR , √
r,
djY
= 3 r (−ρ jY − mC + mAV 0 )
dϕ
1
LC
dmC
= 3 r jY
dϕ
ωR
3 ω0
in resonance r = 1, the CIN should be designed to meet this,
this is the resonance constraint
state −1, DA is on, DB is off, normalized
conditions:
−2 JOU T ≤ jY
if violated switch to state −1
jY ≤ 2 JOU T
if violated switch to state +1
state +1, DA is off, DB is on, normalized
jY = −2 JOU T
jY = 2 JOU T
dmC
= −6 r JOU T
dϕ
dmC
= 6 r JOU T
dϕ
condition:
mC > mAV 0 + 2 ρ JOU T
condition:
if violated switch to state 0
mC < mAV 0 − 2 ρ JOU T
if violated switch to state 0
mA and mB
state
0
−1
+1
simulation is an easy task?
DA
on
off
on
DB
on
on
off
mA
mA0
mA = mA DCM
mA0
mB
mB0
mB0
mB = mB DCM
mA DCM = −mB0 − 4 ρ JOU T + 2 mC
mB DCM = −mA0 + 4 ρ JOU T + 2 mC
the steady state acceleration method published in . . .
I
all the equations derived . . .
I
just to solve them . . .
I
equations piecewise-linear, nonhomogeneous . . .
I
trapezoidal rule to integrate . . .
I
simple discretization . . .
I
but the steady state is required!
I
which is a problem of itself!
I
and remains to be a problem . . .
I
a new steady state acceleration method had to be derived
to solve the model in a reasonable time . . .
I
the original intention was to include the method in this
presentation . . .
T HD versus JOU T , the simulation result
Marija Stojsavljević, Predrag Pejović
“An Extrapolation Method for Accelerated Convergence
to Steady State Solution of Power Electronics Circuits”
Power Conversion and Intelligent Motion, PCIM Europe 2005,
pp. 574–578, Nuremberg, Germany, June 2005
MOU T versus JOU T , the simulation result
η versus JOU T , the simulation result
MOU T max versus JOU T , the simulation result
T HD versus r, the simulation result
after the simulation . . .
I
simulation may be used even to draw fairly general
conclusions
I
but this requires analytical preparation and normalization
I
obtained diagrams should be denormalized to apply for a
specific circuit
I
agreement with δ impulse approach?
I
disagreement only at low IOU T
I
which was expected . . . after we got the results
after the DCM . . .
I
finally, there is some understanding of the DCM . . .
I
but is there any use of it?
I
at first, it seems pretty useless . . .
I
but do we study only the things to be applied at the very
moment?
I
actually, we do!
I
but there is resistance emulation . . .
I
where these concepts turned out to be useful
I
although this was not an original idea
I
and this is our next topic . . .
published in . . .
Predrag Božović, Predrag Pejović
“Current Injection Based Low Harmonic Three Phase
Diode Bridge Rectifier Operating in Discontinuous
Conduction Mode”
IEE Proceedings Electric Power Applications,
vol. 152, no. 2, pp. 199–208, March 2005
without any problem!