Datasheet

mW491
0.725
54.0
660 =
÷
ø
ö
ç
è
æ
P
SW
+
1
2
[C (V
IN
) V
f
)
2
f] + 6.8 mW
( )
D1
4
I
IVP
RIPPLE
OUTfCOND
-´
÷
÷
ø
ö
ç
ç
è
æ
+´= =653mW
TPS40200-EP
SGLS371A –JANUARY 2007–REVISED JANUARY 2013 www.ti.com
Rectifier Selection Criteria
• Rectifier breakdown voltage
The rectifier has to withstand the maximum input voltage which, in this case, is 16 V. To allow for switching
transients that can approach the switching voltage, a 30-V rectifier was selected.
• Diode size
The importance of power losses from the Schottky rectifier (D2) is determined by the duty cycle. For a low
duty-cycle application, the rectifier is conducting most of the time, and the current that flows through it times
its forward drop can be the largest component of loss in the entire controller. In this application, the duty cycle
ranges from 20% to 40%, which in the worst case means that the diode is conducting 80% of the time. Where
efficiency is of major importance, choose a diode with as low a forward drop as possible. In more cost-
sensitive applications, size may be reduced to the point of the thermal limitations of the diode package.
The device in this application is large, relative to the current required by the application. In a more cost-
sensitive application, a smaller diode in a less-expensive package provides a less-efficient, but appropriate,
solution.
The device used has the following characteristics:
• V
f
= 0.3 V at 3 A
• C
t
= 300 pF (C
t
= the effective reverse-voltage capacitance of the synchronous rectifier, D2)
The two components of the losses from the diode D2 are:
(11)
Where:
D = Duty cycle
I
RIPPLE
= Ripple current
I
OUT
= Output current
V
F
= Forward voltage
P
COND
= Conduction power loss
The switching capacitance of this diode adds an AC loss, given by:
(12)
This additional loss raises the total loss to: 660 mW.
At an output voltage of 3.3 V, the application runs at a nominal duty cycle of 27%, and the diode is conducting
72.5% of the time. As the output voltage is moved up to 5 V, the on time increases to 46%, and the diode is
conducting only 54% of the time during each clock cycle. This change in duty cycle proportionately reduces the
conduction power losses in the diode. This reduction may be expressed as:
(13)
for a savings in power of 660 – 491 = 169 mW.
To illustrate the relevance of this power savings, the full-load module efficiency was measured for this application
at 3.3 V and 5 V. The 5-V output efficiency is 92% versus 89% for the 3.3-V design. This difference in efficiency
represents a 456-mW reduction in losses between the two conditions. This 169-mW power-loss reduction in the
rectifier represents 37% of the difference.
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