LM3404FSTDIMEV National Semiconductor, LM3404FSTDIMEV Datasheet - Page 21

• Manufacturer Part Number: LM3404FSTDIMEV
• Description: BOARD, EVAL, FAST DIMMING, LM3404
• Manufacturer: National Semiconductor

Specifications of LM3404FSTDIMEV

• Kit Contents: LM3404FSTDIM Assembled Evaluation Board, LM3404 Datasheet, Application Note
• Svhc: No SVHC (15-Dec-2010)
• Kit Features: Designed To Drive A Series String Of High Power, High Brightness

R
A preliminary value for R
Δi
tions for Δi
Sub-1Ω resistors are available in both 1% and 5% tolerance.
A 1%, 0.33Ω device is the closest value, and a 0.33W, 1206
size device will handle the power dissipation of 162 mW. With
the resistance selected, the average value of LED current is
re-calculated to ensure that current is within the ±5% toler-
ance requirement. From the expression for average LED
current:
INPUT CAPACITOR
Following the calculations from the Input Capacitor section,
Δv
quired capacitance is:
To provide additional safety margin the a higher value of 3.3
µF ceramic capacitor rated to 50V with X7R dielectric in an
1210 case size will be used. From the Design Considerations
section, input rms current is:
Ripple current ratings for 1210 size ceramic capacitors are
typically higher than 2A, more than enough for this design.
RECIRCULATING DIODE
The input voltage of 24V ±5% requires Schottky diodes with
a reverse voltage rating greater than 30V. The next highest
standard voltage rating is 40V. Selecting a 40V rated diode
provides a large safety margin for the ringing of the switch
node and also makes cross-referencing of diodes from differ-
ent vendors easier.
The next parameters to be determined are the forward current
rating and case size. In this example the low duty cycle (D =
7.1 / 24 = 28%) places a greater thermal stress on D1 than
on the internal power MOSFET of the LM3404. The estimated
average diode current is:
A Schottky with a forward current rating of 1A would be ade-
quate, however reducing the power dissipation is critical in
this example. Higher current diodes have lower forward volt-
ages, hence a 2A-rated diode will be used. To determine the
proper case size, the dissipation and temperature rise in D1
SNS
L
IN(MAX)
. This value should be re-evaluated based on the calcula-
I
F
= 0.2 / 0.33 - (7.1 x 2.2 x 10
C
I
will be 24V x 2%
IN-RMS
F
IN(MIN)
:
= 706 mA, 1% above 700 mA
t
= 0.7 x Sqrt(0.28 x 0.72) = 314 mA
SNS
I
= (0.7 x 7.4 x 10
D
= 0.706 x 0.72 = 509 mA
= 220 ns, R
SNS
P-P
was determined in selecting
= 480 mV. The minimum re-
SNS
-7
-7
) / 47 x 10
) / 0.48 = 1.1 µF
= 0.33Ω
-6
+ 0.266 / 2
21
can be calculated as shown in the Design Considerations
section. V
diode at 700 mA is approximately 0.3V and the θ
W. Power dissipation and temperature rise can be calculated
as:
C
The bootstrap capacitor C
capacitor with X7R dielectric. A 25V rating is appropriate for
all application circuits. The linear regulator filter capacitor C
should always be a 100 nF ceramic capacitor, also with X7R
dielectric and a 25V rating.
EFFICIENCY
To estimate the electrical efficiency of this example the power
dissipation in each current carrying element can be calculated
and summed. Electrical efficiency, η, should not be confused
with the optical efficacy of the circuit, which depends upon the
LEDs themselves.
Total output power, P
Conduction loss, P
Gate charging and VCC loss, P
regulator:
Switching loss, P
AC rms current loss, P
DCR loss, P
Recirculating diode loss, P
Current Sense Resistor Loss, P
Electrical efficiency, η = P
5 / (5 + 0.687) = 88%
Temperature Rise in the LM3404 IC is calculated as:
P
T
B
P
CIN
LM3404
AND C
C
P
P
S
= (I
= I
G
= 0.5 x 24 x 0.706 x 40 x 10
= (600 x 10
IN(rms)
F
= (P
D
2
F
P
x R
for a case size such as SMB in a 40V, 2A Schottky
L
L
, in the inductor
C
P
= I
2
P
DSON
S
+ P
x ESR = 0.317
O
P
T
F
P
= 0.5 x V
S
2
G
RISE
= I
D
, in the internal MOSFET:
G
x DCR = 0.706
C
= (I
-6
) x D = (0.706
= 0.509 x 0.3 = 153 mW
, in the internal MOSFET:
F
+ P
+ 4 x 10
O
x V
= 0.153 x 75 = 11.5°C
IN-OP
, is calculated as:
CIN
155 = 49.2°C
S
) x θ
O
B
, in the input capacitor:
IN
O
D
= 0.706 x 7.1 = 5W
should always be a 10 nF ceramic
+ f
x I
/ (P
= 153 mW
JA
5
2
SW
F
x 6 x 10
G
0.003 = 0.3 mW (negligible)
SNS
x (t
O
= (0.112 + 0.072 + 0.136) x
, in the gate drive and linear
2
x Q
2
+ Sum of all loss terms) =
x 0.8) x 0.28 = 112 mW
-9
R
x 0.1 = 50 mW
= 164 mW
x 4 x 10
+ t
G
) x V
-9
F
) x 24 = 72 mW
) x f
IN
SW
5
= 136 mW
www.national.com
JA
is 75°C/
F