IEEE Power Electronics Magazine - September 2017 - 49

(24)

VT = V^ah - V^ c h,

(25)

V^ah = R load I 1 ,
I T + V^ c h k 2
1 ,
k2 + R
load
V^ c h = k 4 V^ah + k 5 I C + k 6 V^ah - k 6 V^ c h .

I T - I 1 = k 2 ^V^ah - V^ c hh " V^ah =

(27)

(28)

If you rearrange and replace the k coefficients by their
definition from Figure 13, you have the definition for time
constant x 1
x1 =

L1
L1
=
.
R load
R load ^1 - k 4h - k 5
M ^ 1 + M h + 0 .5
k 6 + R load k 2 ^1 - k 4h + 1

(29)

The second time constant implies looking at the resistance seen from C 2 's terminals while L 1 is a short circuit. The new circuit appears in Figure 16. As L 1 shorts
terminals a and c together, simplification occurs updating the circuit to that of the right side of the picture.
Again, a few simple equations will lead you to the result
quickly:
VT - I T R load
,
R load
k5 IC
VT = k 4 VT + k 5 I C " VT = 1 k4 .
VT = ^ I T + I C h R load " I C =

k5
R
R
= 2load " x 2 = 2load C 2 .
k
k4 + R 5 - 1
load

For the second-order coefficient, we will set capacitor C 2 in
its high-frequency state (replace it by a short circuit) while
we determine the resistance driving inductor L 1 . The drawing illustrating this approach is given in Figure 17. Because
the output is shorted by C 2, nodes a and c are at the same
0-V potential. The electric circuit is simplified to that of the
right-side sketch.
We can write a first equation describing the VT voltage.
Observing that 1) I T and I C are identical and 2) VT = - V(c),
we have
VT = - ^ k 5 I C - k 6 V^ c hh = - ^ k 5 I T + k 6 VT h " VT ^1 + k 6h = - k 5 I T .
(34)
Factoring VT /I T , the resistance seen from L 1's terminals is
k5
VT
IT = - 1 - k6 .

1

(35)

The second-order time constant x 21 is defined by
x 21 =

L1
k

a - 1 +5k k

=

6

L1
L1
=
2 .
1
R load Vin2
R
a
k
load
2
1
+
M
^ Vin + Vout h

(36)

(31)
If we consider that Vout = MVin, the b 2 coefficient is
expressed as
b 2 = x 2 x 21 =
(32)

L 1 C 2 ^1 + M h2
.
2

(37)

Assembling the time constants, we have determined
leads to the denominator D (s)

If you try to determine the third time constant involving
C 3, the transformer configuration (perfect coupling) makes
its terminals voltage equal to 0 V: the capacitor plays no role

L1

R
R
L1
+ 2load C 2 . 2load C 2 . (33)
R load
M ^1 + M h + 0.5

The Second-Order Coefficient

(30)

Substitute (30) in (31), then solve for VT , and rearrange.
You should find
VT
IT =

b1 = x1 + x2 =

(26)

You substitute (26) in (27) and then solve for V^ c h . Substitute
V^ c h in (26), and solve for V^ah . Then, write
R load ^1 - k 4h - k 5
V^ah - V^ c h
VT
=
.
IT =
IT
k 6 + R load k 2 ^1 - k 4h + 1

in the dynamic transfer function. The first coefficient b 1 is
thus defined by

D ^ s h = 1 + b 1 s + b 2 s 2 = 1 + s ^x 1 + x 2h + s 2 x 2 x 21 .

(38)

1

a
I1

c
IC
+

k2V(a, c)

IT
VT

a

I3

IT + IC

IC
Rload

IT
+
k4V(a) + k5Ic

VT

Rload

k4V(a) + k5Ic + k6V(a, c)
FIG 16 Shorting the inductor truly simplifies the circuit.

September 2017

z	IEEE POWER ELECTRONICS MAGAZINE

49
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