IEEE Power Electronics Magazine - September 2017 - 50
IT
VT
c
Simplify IC
a
c
+
k5Ic-k6V(c)
0V
0V
IC
k2V(a, c)
+
VT
IT
Rload
C2
k4V(a) + k5Ic + k6V(a, c)
FIG 17 The second-order coefficient sets one of the energy-storing elements in its high-frequency state, C2, while you determine
the resistance seen from inductor terminals.
If we consider a low-Q approximation, this second-order
denominator can be approximated by two cascaded poles
defined as
1
1
1
2
~ p = b = x1 + x2 . x2 " ~ p = R C
1
load 2
(39)
2
R
x +x
b
1
1
~ p = b 1 = 1 2 2 . 2 " ~ p = Lload a 1 + M k
2
1
x2 x1
x1
(40)
1
1
2
2
and combined as
s
s
D ^ s h . ` 1 + ~ p j` 1 + ~ p j.
1
(41)
2
Determination of the Zero
as shown in the figure. If the output is nulled, then current I 1 is also null, implying that I c = I 3 .
The voltage of node c is defined by
V^ c h ^ s h =
D^ s hk 3 + I c ^ s hk 5
.
1 + k6
The current I c is therefore equal to the voltage at node c
divided by the impedance of L 1
D^ s hk 3 + I c ^ s hk 5
D^ s hk 3
1 + k6
Ic ^sh =
" Ic ^sh =
, (43)
sL 1
sL 1 ^1 + k 6 h - k 5
while current I 3 is equal to
As explained at the beginning of this article, when the
excitation is tuned to the zero angular frequency s z, the
response of the transformed network is nulled (see Figure 1). The exercise will now consist of bringing the excitation back and determining the condition in the transformed circuit that creates an output null. Figure 18
shows the updated circuit that we need to study. The
interesting thing with the output null is its propagation to
other nodes. For instance, if Vout = 0 V, then, because of
the transformer upper connection, node a is also at 0 V,
and all expressions involving this node can be simplified
I3
sL1 1 1
a
0V
I3-I1 I1
Ic (s)
1/sCc Iout (sz) = 0
0V
c
I=0
D(s)
k1D(s)k2V(a, c) 1
+
+
sC2
I 3 ^ s h = k 1 D ^ s h - k 2 V^ c h = k 1 D ^ s h - k 2 sL 1 I c ^ s h .
Now, substitute (43) in (44), and then equate I c and I 3:
k 1 D ^ s h - k 2 sL 1
D^ s hk 3
D^ s hk 3
=
. (45)
sL 1 ^1 + k 6h - k 5 sL 1 ^1 + k 6h - k 5
Solve for s, replace the k-coefficients by their values
from Figure 13, rearrange, and you find
sz =
R load
.
L 1 M^1 + M h
(46)
s
1 - ~z
H^ s h = H 0
,
` 1 + ~s j` 1 + ~s j
p
p
Vout(sz) = 0
Rload
1
k3D(s) + k4V(a) + k5Ic(s) + k6V(a, c)
2
~p = R C
load 2
(48)
2
R
1
~ p = Lload a 1 + M k
1
(49)
2
FIG 18 A specific condition in the transformed network
observed at s = sz nulls the response.
z September 2017
(47)
2
with
1
IEEE POWER ELECTRONICS MAGAZINE
(44)
This is a positive root and, therefore, a right half-plane
zero. The complete transfer function is obtained by gathering all the pieces and recognizing that poles and zeros
are actually those of a DCM buck-boost converter:
I1
50
(42)
~z =
R load
,
L 1 M^1 + M h
(50)
Table of Contents for the Digital Edition of IEEE Power Electronics Magazine - September 2017
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