Datasheet

LT3478/LT3478-1
14
34781f
APPLICATIO S I FOR ATIO
WUU
U
allowed level at higher temperatures, the CTRL2 voltage
curve may require a greater downward slope between
25°C and 50°C to compensate for that loss of slope at
higher temperatures.
Example: Calculate the resistor values required for generat-
ing CTRL1 and CTRL2 from V
REF
based on the following
requirements:
(a) I
LED
= 700mA at 25°C
(b) I
LED
derating curve breakpoint occurs at 25°C
(c) I
LED
derating curve has a slope of –200mA/25°C be-
tween 25°C and 50°C ambient temperature
Step1: Choose CTRL1 = 700mV for I
LED
= 700mA
CTRL1 = V
REF
/(1 + R2/R1)
R2 = R1 • [(V
REF
/CTRL1) – 1]
For V
REF
= 1.24V and choosing R1 = 22.1k,
R2 = 22.1k [(1.24/0.7) – 1]
R2 = 17k (choose 16.9k)
CTRL1 = 1.24/(1 + (16.9/22.1))
CTRL1 = 703mV (I
LED
= 703mA)
Step 2: Choose resistor network option A (Figure 7) and
CTRL2 = CTRL1 for 25°C breakpoint
start with R4 = R2 = 16.9k, R
NTC
= 22k (closest value
available)
CTRL2 = 701mV (I
LED
= Min(CTRL1, CTRL2) • 1A =
701mA)
Step 3: Calculate CTRL2 slope between 25°C and 50°C
CTRL2 (T) = 1.24/(1 + R4/R
NTC
(T))
at T = T
O
= 25°C, CTRL2 = 701mV
at T = 50°C, R
NTC
(T) = R
NTC
(T
O
).e
x
, x = B [(1/(T + 273)
– 1/298)]
(B = B-constant; linear over the 25°C to 50°C temperature
range)
For R
NTC
B-constant = 3950 and T = 50°C
x = 3950 [(1/323) – 1/298] = –1.026
R
NTC
(50°C) = R
NTC
(25°C).e
–1.026
R
NTC
(50°C) = 22k • 0.358
R
NTC
(50°C) = 7.9k
CTRL2(50°C) = 1.24/(1 + 16.9/7.9) = 395mV
CTRL2 slope (25°C to 50°C) = [CTRL2(50°C)
– CTRL2(25°C)]/25°C
= (395 – 701)/25
= –306mV/25°C
I
LED
slope = –306mA/25°C
The required I
LED
slope is –200mA/25°C. To reduce the
slope of CTRL2 versus temperature it is easier to keep
the exact same NTC resistor value and B-constant (there
are limited choices) and simply adjust R4 and the type
of resistor network used for the CTRL2 pin. By changing
the resistor network to option C it is possible to place a
temperature independent resistor in series with R
NTC
to
reduce the effects of R
NTC
on the CTRL2 pin voltage over
temperature.
Step 4: Calculate the resistor value required for R
Y
in
resistor network option (c) (Figure 7) to provide an I
LED
slope of –200mA/25°C between 25°C and 50°C ambient
temperature.
CTRL2 (25°C) = 0.7V = 1.24/(1 + (R4/(R
NTC
(25°C)+
R
Y
))
R4 = 0.77 (R
NTC
(25°C) + R
Y
) (a)
for –200mA/25°C slope ≥ CTRL2(50°C) = 0.7 – 0.2 =
0.5
CTRL2(50°C) = 0.5V = 1.24/(1 + (R4/(R
NTC
+ R
Y
))
R4 = 1.48 (R
NTC
(50°C) + R
Y
) (b)
Equating (a) = (b) and knowing R
NTC
(25°C) = 22k and
R
NTC
(50°C) = 7.9k gives,
0.77 (22k + R
Y
) = 1.48 (7.9k + R
Y
)
17k + 0.77 R
Y
= 11.7 k + 1.48 R
Y
R
Y
= (17k – 11.7k)/(1.48 – 0.77)
R
Y
= 7.5k