Datasheet
0 50 100 150 200 250 300
Load-mA
40
45
50
60
70
85
90
100
Efficiency-%
95
80
75
65
55
Withoutisolation
Withisolation
V =4.2V,
Output=15V
IN
SS
0.5 V C5
t =
5 A
´
m
TPS61093
www.ti.com
SLVS992A –SEPTEMBER 2009–REVISED NOVEMBER 2009
WITHOUT ISOLATION FET
The efficiency of the TPS61093 can be improved by connecting the load to the VO pin instead of the OUT pin.
The power loss in the isolation FET is then negligible, as shown in Figure 13. The tradeoffs when bypassing the
isolation FET are:
• Leakage path between input and output causes the output to be a diode drop below the input voltage when
the IC is in shutdown
• No overload circuit protection
When the load is connected to the VO pin, the output capacitor on the VO pin should be above 1-μF.
START UP
The TPS61093 turns on the isolation FET and PWM switch when the EN pin is pulled high. During the soft start
period, the R and C network on the SS pin is charged by an internal bias current of 5-μA (typ). The RC network
sets the reference voltage ramp up slope. Since the output voltage follows the reference voltage via the FB pin,
the output voltage rise time follows the SS pin voltage until the SS pin voltage reaches 0.5-V. The soft start time
is given by Equation 4.
(4)
Where C5 is the capacitor connected to the SS pin.
When the EN pin is pulled low to switch the IC off, the SS pin voltage is discharged to zero by the resistor R3.
The discharge period depends on the RC time constant. Note that if the SS pin voltage is not discharged to zero
before the IC is enabled again, the soft start circuit may not slow the output voltage startup and may not reduce
the startup inrush current.
INDUCTOR SELECTION
Because the selection of the inductor affects steady state operation, transient behavior, and loop stability, the
inductor is the most important component in power regulator design. There are three important inductor
specifications, inductor value, saturation current, and dc resistance. Considering inductor value alone is not
enough.
The saturation current of the inductor should be higher than the peak switch current as calculated in Equation 5.
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