Texas Instruments LM3404 Evaluation Boards LM3404HVEVAL/NOPB LM3404HVEVAL/NOPB 사용자 설명서
제품 코드
LM3404HVEVAL/NOPB
SNVS465F – OCTOBER 2006 – REVISED MAY 2013
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:
dielectric in an 1210 case size will be used. From the Design Considerations section, input rms current is:
I
IN-RMS
= 0.7 x Sqrt(0.28 x 0.72) = 314 mA
(39)
Ripple current ratings for 1210 size ceramic capacitors are typically higher than 2A, more than enough for this
design.
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 different vendors easier.
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 different 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:
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:
I
D
= 0.706 x 0.72 = 509 mA
(40)
A Schottky with a forward current rating of 1A would be adequate, however reducing the power dissipation is
critical in this example. Higher current diodes have lower forward voltages, hence a 2A-rated diode will be used.
To determine the proper case size, the dissipation and temperature rise in D1 can be calculated as shown in the
Design Considerations section. V
critical in this example. Higher current diodes have lower forward voltages, hence a 2A-rated diode will be used.
To determine the proper case size, the dissipation and temperature rise in D1 can be calculated as shown in the
Design Considerations section. V
D
for a case size such as SMB in a 40V, 2A Schottky diode at 700 mA is
approximately 0.3V and the
θ
JA
is 75°C/W. Power dissipation and temperature rise can be calculated as:
P
D
= 0.509 x 0.3 = 153 mW
(41)
T
RISE
= 0.153 x 75 = 11.5°C
(42)
C
B
AND C
F
The bootstrap capacitor C
B
should always be a 10 nF ceramic capacitor with X7R dielectric. A 25V rating is
appropriate for all application circuits. The linear regulator filter capacitor C
F
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,
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
O
, is calculated as:
P
O
= I
F
x V
O
= 0.706 x 7.1 = 5W
(43)
Conduction loss, P
C
, in the internal MOSFET:
P
C
= (I
F
2
x R
DSON
) x D = (0.706
2
x 0.8) x 0.28 = 112 mW
(44)
Gate charging and VCC loss, P
G
, in the gate drive and linear regulator:
P
G
= (I
IN-OP
+ f
SW
x Q
G
) x V
IN
P
G
= (600 x 10
-6
+ 4 x 10
5
x 6 x 10
-9
) x 24 = 72 mW
(45)
Switching loss, P
S
, in the internal MOSFET:
P
S
= 0.5 x V
IN
x I
F
x (t
R
+ t
F
) x f
SW
P
S
= 0.5 x 24 x 0.706 x 40 x 10
-9
x 4 x 10
5
= 136 mW
(46)
AC rms current loss, P
CIN
, in the input capacitor:
P
CIN
= I
IN(rms)
2
x ESR = 0.317
2
0.003 = 0.3 mW (negligible)
(47)
DCR loss, P
L
, in the inductor
P
L
= I
F
2
x DCR = 0.706
2
x 0.1 = 50 mW
(48)
Recirculating diode loss, P
D
= 153 mW
Current Sense Resistor Loss, P
SNS
= 164 mW
Electrical efficiency,
η
= P
O
/ (P
O
+ Sum of all loss terms) = 5 / (5 + 0.687) = 88%
Temperature Rise in the LM3404 IC is calculated as:
T
LM3404
= (P
C
+ P
G
+ P
S
) x
θ
JA
= (0.112 + 0.072 + 0.136) x 155 = 49.2°C
(49)
26
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