Datasheet
Io 25 mA
Iavg = = = 227 mA
1 D 1 0.89- -
L (Vin Vsw) dt (2.8V 0.14V) 792 nsV dt
di = = = = 117 mA
L L 18 H
´ - ´ - ´
m
(on) AVG (on)
Io 25 mA
Vsw = Rds I = Rds = 0.6 140 mV
1 D 1 0.89
´ ´ W ´ »
- -
2.8 V
D 1 0.8 89%
19.2 V + 0.5 V
æ ö
» - »
ç ÷
è ø
Vi
D 1
Vo + V
æ ö
» - h ´
ç ÷
¦
è ø
(on) AVG
Io
Vsw = Rds I Iavg =
1 D
´
-
Vo + V Vin
D =
Vo + V Vsw
¦ -
¦ -
TPS65070, TPS65072, TPS65073
TPS650731, TPS650732, TPS650701, TPS650702, TPS650721
SLVS950G –JULY 2009–REVISED MAY 2013
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Connecting both strings in parallel is required because the wLED converter generates its output voltage
dependant on the current in ISINK1 and ISINK2. If the current falls below the target, the output voltage is
increased. If one string is open, the wLED driver will boost the output voltage to its maximum because it
assumes the voltage is not high enough to drive current into this string (there could be different numbers of
wLEDs in the two strings).
Inductor Design
The inductor in a boost converter serves as an energy storage element. The energy stored equals ½ L × I
2
.
Therefore, the inductor must not be saturated as the inductance will drop and the energy stored will be reduced
causing bad efficiency. The converter operates with typically 15μH to 22μH inductors. A design example for an
application powering 6LEDs in one string given below:
Vin = 2.8 V — minimum input voltage for the boost converter
Vo = 6 × 3.2 V = 19.2 V — assuming a forward voltage of 3.2V per LED
Vf = 0.5 V — forward voltage of the Schottky diode
Io = 25 mA maximum LED current
Fsw = 1.125 MHz — switching frequency — T=890ns
Rds(on) = 0.6R — drain-source resistance of the internal NMOS switch
Vsw — voltage drop at the internal NMOS switch
I
AVG
— average current in NMOS when turned on
The duty cycle for a boost converter is:
(10)
With:
(11)
A different approach to calculate the duty cycle is based on the efficiency of the converter. The typical number
can be found in the graphs, or as a first approach, we can assume to get an efficiency of about 80% as a typical
value.
(12)
With the values given above
(13)
ton = T × D = 890 ns × 0.89 = 792 ns
toff = 890 ns – 792 ns = 98ns
(14)
When the NMOS switch is turned on, the input voltage is forcing a current into the inductor. The current slope
can be calculated with:
(15)
(16)
The minimum and maximum inductor current can be found by adding half of the inductor current ripple (di) to the
average value, which gives:
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