Datasheet

( )
D1
4
I
IVP
RIPPLE
OUTfCOND
-´
÷
÷
ø
ö
ç
ç
è
æ
+´= =653mW
( )[ ]
fVVC
2
1
P
2
fINSW
= 6.8mW´+´=
Thisadditionallossraisesthetotallossto: 660mW
mW491
0.725
54.0
660 =
÷
ø
ö
ç
è
æ
TPS40200
SLUS659F –FEBRUARY 2006–REVISED MARCH 2012 www.ti.com
Rectifier Selection Criteria
1. Rectifier Breakdown Voltage
The rectifier has to withstand the maximum input voltage which in this case is 16 V. To allow for switching
transients which can approach the switching voltage a 30 V rectifier was selected.
2. 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 paramount importance, choose a diode with a minimum of forward drop. 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 will provide 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:
where
– D = the duty cycle
– I
RIPPLE
is the ripple current
– I
OUT
is the output current
– V
F
is the forward voltage
– P
COND
is the conduction power loss
(12)
The switching capacitance of this diode adds an ac loss, given by:
(13)
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
for a savings in power of 660 - 491 = 169 mW.
To illustrate the relevance of this power savings we measured the full load module Efficiency for this
application at 3.3 and 5 V. The 5 volt output efficiency is 92% vs. 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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