AN296 Silicon_Laboratories, AN296 Datasheet - Page 13

• Manufacturer Part Number: AN296
• Description: Using THE Si3400 AND Si3401 POE PD Controllers IN Isolated AND Non-isolated Designs
• Manufacturer: Silicon_Laboratories

Solving with the constraint that d
For a given power transformer magnetizing inductance, turns ratio, output voltage, frequency, and input voltage,
this gives the output current at which the current becomes continuous and always flows in either the transformer
primary or secondary.
Lm = 40 µH gives a good compromise between transformer size (larger Lm gives a larger transformer) and peak
current (larger Lm gives smaller peak current at the input and output).
Plugging in V
I
Above this current, the transformer current becomes continuous in that there is always current flow in either the
transformer primary or secondary.
For larger output current, the duty cycle stays fairly constant at
In the continuous mode, the average current while the switcher FET is on is determined by setting the average
input power after an efficiency, ε, to equal the average output power:
In this mode of operation, there is a change in primary current while the FET is on
The peak current that the transformer must handle is
For the same transformer above with I
Increasing the turns ratio decreases peak current, particularly on the primary side. However, the secondary voltage
is reflected back to the primary, and the increased turns ratio also increases the voltage on the switcher FET.
Additionally, transformer leakage inductance causes an additional spike of voltage on the switcher FET, which must
be clamped by a snubber.
The FET maximum drain voltage is 80 V, and the maximum voltage at Vpos is about 55 V; so, the snubber must
clamp to 25 V.
A Zener diode and fast recovery diode are recommended to clamp the output at less than 25 V above V
0
I
= 0.68 A
I
0
peak
I
D
avg
ΔI
=
=
×
=
=
ε
N
V
×
------------------- -
(
I
p
V
×
L
avg
0
---------------------------- -
(
×
p
m
2 f
--------------------------------------------- -
(
+ V
V
×
V
ε
×
×
+
p
0
×
D
f
f
ΔI
---- -
= 4 V, V
+
(
)
2
+
D
×
V
N V
V
L
o
=
(
f
m
+
)
I
V
o
p
O
f
+
×
= 48 V, N = 4, Lm = 40 µH, ε = 0.9, and f = 380 kHz gives:
)
×
V
(
1
------------------------------------------------ -
⎛
⎝
f
V
+d
)
1
)
o
+
2
+
= 1 gives the following:
N
O
V
= 3 A, the peak transformer current is 1.85 A.
f
×
)
N
(
---------------------- -
V
0
V
+
p
V
f
)
Rev. 0.8
⎞
⎠
2
AN296
POS
.
13