LT3754
13
3754fc
In addition, the current drive required for GATE switching
should also be kept low in the case of high V
IN
voltages
(see “Thermal Considerations” in the Applications Informa-
tion section). The R
DS(ON)
of the MOSFET will determine
d.c. power losses but will usually be less significant
compared to switching losses. Be aware of the power
dissipation within the MOSFET by calculating d.c. and
switching losses and deciding if the thermal resistance
of the MOSFET package causes the junction temperature
to exceed maximum ratings.
Table 4. MOSFET Manufacturers
MANUFACTURER PHONE NUMBER WEB
Vishay Siliconix 402-563-6866 www.vishay.com
International Rectifier 310-252-7105 www.irf.com
Fairchild 972-910-8000 www.fairchildsemi.com
Power MOSFET: Current Sense Resistor
The LT3754 current mode boost converter controls peak
current in the inductor by controlling peak MOSFET current
in each switching cycle. The LT3754 monitors current in the
external N-channel power MOSFET by sensing the voltage
across a sense resistor (RS) connected between the source
of the FET and the power ground in the application. The
length of these tracks should be minimized and a Kelvin
sense should be taken from the top of RS to the sense
pin. A 52mV sense pin threshold combined with the value
of RS sets the maximum cycle-by-cycle peak MOSFET
current. The low 52mV threshold improves efficiency and
determines the value for RS given by:
applicaTions inForMaTion
RS ≤
52mV • 0.7
I
L(PEAK)
where
I
L(PEAK)
=
1
1− D
• 16 • I
LEDx
• 1+
0.5
2
D = MOSFET duty cycle = 1–
V
IN(MIN)
V
OUT(MAX)
V
OUT(MAX)
= N • V
F(MAX)
( )
+ 1V
N = number of LEDs in each string,
V
F(MAX)
= max imum LED forward voltage drop,
V
IN(MIN)
= minimum input voltage to the inductor,
and the 0.5 term represents an inductor peak-to-peak ripple
current of 50% of average inductor current.
The scale factor of • 0.7 ensures the boost converter
can meet the peak inductor requirements of the loop by
accounting for the combined errors of the 52mV sense
threshold, I
LEDx
, RS and circuit efficiency.
Example: For a 12W LED driver application requiring 16
strings of 10 LEDs each driven with 20mA, and choosing
V
IN(MIN)
= 8V, V
OUT(MAX)
= (4V • 10)+1V = 41V and I
LEDx
= 20mA, the value for RS is chosen as:
RS ≤
52mV • 0.7
I
L(PEAK)
≤
52mV • 0.7
41
8
• 16 • 0.02
• 1+ 0.25
( )
≤
52mV • 0.7
2.05
≤ 17.7 mΩ