IEEE Power Electronics Magazine - September 2017 - 45
voltage-mode buck converter operated in the continuous
conduction mode (CCM). An impedance is a transfer function linking an excitation signal I out to a response signal
Vout . Here, I out is the test generator that we have installed,
while Vout is the resulting voltage produced across its
terminals. To determine the various coefficients from (11),
we can follow the Figure 2 flowchart and start with s = 0:
short the inductor and open the capacitor as shown in the
figure. The circuit is simple, and the resistance, R 0, seen
from the current source is simply the parallel combination
of rL and R load
R 0 = rL < R load .
1
1
1
rC + sC = 0 " s z = - r C .
2
C 2
2
Z out ^ s h =
^ rL < R load h
(14)
Vout(s) = 0?
Rload
Z1
rL
rC
I (sz) = 0
Z2
FIG 8 If impedances Z1 or Z2 become transformed shorts,
the response Vout is nulled.
τ1
τ2
R?
L1
(16)
(18)
Rload
C2
rL
L1
Rload
C2
rL
rC
R = rL + Rload
rC
R = (rL ||Rload) + rC
FIG 9 What resistance do you see between the selected
component terminals when the second is set in its dc state?
(17)
L
C 2 ^ rc + R load h
b 2 = x 1 x 12 = r + R1
L
load
L
b 2 = x 2 x = C 2 6rL < R load + rC@ r + R 1 < r .
load
L
C
Iout(sz)
R?
The second-order coefficient b 2 is determined by using
the notation introduced in (12). Either L 1 is set in its highfrequency state (open circuit) and you look at the resistance
driving C 2 to obtain x 12, or C 2 is put in its high-frequency
state (short circuit) and you look at the resistance driving
L 1 for x 21 . Figure 10 shows the two possible arrangements.
You usually select the one leading to the simplest expression or the one avoiding a product indeterminacy if any
(3 # 0 or 3 3, for instance). The following two definitions for b 2 are identical, and the first one is found to be
the simplest:
2
1
1
sC2
sL1
(15)
The first coefficient, b 1, of the denominator D (s)
is obtained by looking at the resistance offered by L 1 's
terminals while C 2 is in its dc state (open): you have x 1 .
Then, you look at the resistance driving C 2 while L 1 is set
in its dc state (short circuit): you obtain x 2 . As illustrated by
Figure 9, the sketch immediately leads to the definition of b 1
L
+ C 2 6^ rL < R load h + rC@ .
b 1 = x 1 + x 2 = r + R1
L
load
.
r +R
L
+ C 2 6rL < R load + rC@k + s 2 a L 1 C 2 rC + R load k
1 + s a r + R1
L
load
L
load
(19)
The denominator N (s) is, therefore, expressed by
L
N ^ s h = a 1 + s rL1 k^1 + srC C 2h .
1
a1 + s L
rL k^1 + srC C 2h
(13)
Do we have zeros in this circuit? We examine the transformed circuit shown in Figure 8. Let's check what component combinations would bring the response Vout to zero
when the excitation current I out is tuned at a zero angular
frequency s z . We can identify two transformed short circuits involving rL- L 1 and rc-C 2 .
The roots for these two impedances are immediately
determined.
r
rL + sL 1 = 0 " s z = - LL ,
1
We now have all of the ingredients to assemble the final
transfer function that is defined as
L1 in HighFrequency State
C2 in HighFrequency State
τ12 ←R ?
τ12 ←R ?
R?
L1
R?
C2
rL
rC
Rload
L1
rL
C2
Rload
rC
L1
τ21 = C2 (rc + Rload)
τ12 =
b2 = τ1τ21
b2 = τ2τ12
rL + Rload || rC
FIG 10 What resistance do you see between the selected
component terminals when the second is set in its highfrequency state?
September 2017
z IEEE POWER ELECTRONICS MAGAZINE
45
Table of Contents for the Digital Edition of IEEE Power Electronics Magazine - September 2017
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