LM2743MTCX/NOPB National Semiconductor, LM2743MTCX/NOPB Datasheet - Page 20

• Manufacturer Part Number: LM2743MTCX/NOPB
• Description: IC REG CTLR BUCK N-CH 14-TSSOP
• Manufacturer: National Semiconductor
• Series: PowerWise®r
• Type: Step-Down (Buck)r

Specifications of LM2743MTCX/NOPB

• Internal Switch(s): No
• Synchronous Rectifier: Yes
• Number Of Outputs: 1
• Voltage - Output: 0.6 ~ 13.5 V
• Current - Output: 20A
• Frequency - Switching: 50kHz ~ 1MHz
• Voltage - Input: 1 ~ 16 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 14-TSSOP
• For Use With: LM2743EVAL - BOARD EVALUATION LM2743LM2743-19AEVAL - BOARD EVALUATION LM2743-19A
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Power - Output: -
• Other names: LM2743MTCX

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The FDS6898A has a typical turn-on rise time t
fall time t
losses for this type of dual N-Channel MOSFETs are 0.061W.
FET Conduction Loss (P
R
to account for the increasing R
There are few additional losses that are taken into account:
IC Operating Loss (P
where I
FET Gate Charging Loss (P
The value n is the total number of FETs used and Q
typical gate-source charge value, which is 3 nC. For the FD-
S6898A the gate charging loss is 5.94 mW.
Input Capacitor Loss (P
DS(ON)
Q-VCC
= 13 mΩ and the factor is a constant value (k = 1.3)
P
f
P
P
CND
of 15 ns and 16 ns, respectively. The switching
CND2
SW
P
P
P
P
is the typical operating V
CND1
SW
CND2
P
= 98.42 mW + 172 mW = 270.42 mW
GATE
= 0.5 x 3.3V x 4A x 300 kHz x 31 ns
P
CND1
P
= (4A)
IC
= 0.5 x V
GATE
= 1.5 mA x 3.3V = 4.95 mW
= (4A)
= (I
P
= 2 x 3.3V x 3 nC x 300 kHz
= (I
CND
P
P
P
OUT
IC
IC)
2
= n x V
GATE
SW
x 13 mΩ x 1.3 x (1 - 0.364)
OUT
= I
2
= P
CAP
)
IN
CND
2
x 13 mΩ x 1.3 x 0.364
= 61.38 mW
Q_VCC
)
x R
= 5.94 mW
x I
2
CND1
)
GATE
x R
CC
)
DS(ON)
OUT
DS(ON)
x Q
DS(ON)
+ P
x V
)
x (t
GS
of a FET due to heating.
CC
CND2
r
x k x (1-D)
CC
+ t
,
x f
x k x D
current
f
SW
) x f
SW
r
and turn-off
GS
is the
20
where,
Here n is the number of paralleled capacitors, ESR is the
equivalent series resistance of each, and P
tion in each. So for example if we use only one input capacitor
of 24 mΩ.
Output Inductor Loss (P
where DCR is the DC resistance. Therefore, for example
Total System Efficiency
P
TOTAL
= P
P
FET
P
IND
P
IND
P
CAP
+ P
IND
= (4A)
= I
IND
IC
= 88.8 mW
= 176 mW
2
)
OUT
+ P
2
x 11 mΩ
GATE
x DCR
+ P
CAP
CAP
+ P
is the dissipa-
IND