LM3404EVAL/NOPB National Semiconductor, LM3404EVAL/NOPB Datasheet - Page 24

• Manufacturer Part Number: LM3404EVAL/NOPB
• Description: BOARD EVALUATION LM3404
• Manufacturer: National Semiconductor
• Series: PowerWise®r

Specifications of LM3404EVAL/NOPB

• Current - Output / Channel: 1A
• Outputs And Type: 1, Non-Isolated
• Voltage - Output: 4 V
• Features: Dimmable
• Voltage - Input: 6 ~ 42 V
• Utilized Ic / Part: LM3404
• Silicon Manufacturer: National
• Silicon Core Number: LM3404
• Kit Application Type: Power Management - Voltage Regulator
• Application Sub Type: Buck Regulator
• Kit Contents: Board, User Guide, LM3404 Datasheet
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: *LM3404EVAL; LM3404EVAL

www.national.com
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
The dark grey, inner loop represents the high current path
during the MOSFET on-time. The light grey, outer loop rep-
resents the high current path during the off-time.
GROUND PLANE AND SHAPE ROUTING
The diagram of
of continuous current vs. the flow of pulsating currents. The
circuit paths with current flow during both the on-time and off-
time are considered to be continuous current, while those that
carry current during the on-time or off-time only are pulsating
currents. Preference in routing should be given to the pulsat-
B
P
P
AND C
P
G
C
S
= (600 x 10
= 0.5 x 48 x 0.5 x 40 x 10
= (I
F
F
2
x R
P
P
O
S
Figure 14
P
DSON
= I
= 0.5 x V
S
G
-6
, in the internal MOSFET:
C
F
= (I
+ 2.23 x 10
, in the internal MOSFET:
) x D = (0.5
x V
O
IN-OP
CIN
, is calculated as:
O
B
, in the input capacitor:
IN
is also useful for analyzing the flow
= 0.5 x 35.2 = 17.6W
should always be a 10 nF ceramic
+ f
x I
SW
F
5
G
2
x (t
-9
, in the gate drive and linear
x 6 x 10
x 0.8) x 0.73 = 146 mW
x Q
x 2.23 x 10
R
+ t
G
) x V
F
-9
) x f
FIGURE 14. Buck Converter Current Loops
) x 48 = 94 mW
IN
SW
5
= 107 mW
F
24
DCR loss, P
Recirculating diode loss, P
Current Sense Resistor Loss, P
Electrical efficiency, η = P
17.6 / (17.6 + 0.644) = 96%
Temperature Rise in the LM3404HV IC is calculated as:
Layout Considerations
The performance of any switching converter depends as
much upon the layout of the PCB as the component selection.
The following guidelines will help the user design a circuit with
maximum rejection of outside EMI and minimum generation
of unwanted EMI.
COMPACT LAYOUT
Parasitic inductance can be reduced by keeping the power
path components close together and keeping the area of the
loops that high currents travel small. Short, thick traces or
copper pours (shapes) are best. In particular, the switch node
(where L1, D1, and the SW pin connect) should be just large
enough to connect all three components without excessive
heating from the current it carries. The LM3404/04HV oper-
ates in two distinct cycles whose high current paths are shown
in
ing current paths, as these are the portions of the circuit most
likely to emit EMI. The ground plane of a PCB is a conductor
and return path, and it is susceptible to noise injection just as
any other circuit path. The continuous current paths on the
ground net can be routed on the system ground plane with
less risk of injecting noise into other circuits. The path be-
tween the input source and the input capacitor and the path
between the recirculating diode and the LEDs/current sense
resistor are examples of continuous current paths. In contrast,
the path between the recirculating diode and the input capac-
itor carries a large pulsating current. This path should be
P
T
CIN
Figure
LM3404
= I
IN(rms)
= (P
14:
P
L
L
C
, in the inductor
= I
2
+ P
x ESR = 0.222
F
2
G
x DCR = 0.5
+ P
S
155 = 54°C
) x θ
O
D
/ (P
= 47 mW
JA
2
2
0.003 = 0.1 mW (negligible)
SNS
20205428
O
= (0.146 + 0.094 + 0.107) x
x 0.56 = 140 mW
+ Sum of all loss terms) =
= 110 mW