LM2744EVAL National Semiconductor, LM2744EVAL Datasheet - Page 18

• Manufacturer Part Number: LM2744EVAL
• Description: BOARD EVALUATION LM2744
• Manufacturer: National Semiconductor
• Series: PowerWise®r

Specifications of LM2744EVAL

• Main Purpose: DC/DC, Step Down
• Outputs And Type: 1, Non-Isolated
• Voltage - Output: 1.2V
• Current - Output: 3.5A
• Voltage - Input: 1.8 ~ 5.5V
• Regulator Topology: Buck
• Frequency - Switching: 1MHz
• Board Type: Fully Populated
• Utilized Ic / Part: LM2744
• Lead Free Status / RoHS Status: Not applicable / Not applicable
• Power - Output: -

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As with the generic equation, G
take into account the limited bandwidth of the error amplifier.
The result is:
The total control loop transfer function H is equal to the power
stage transfer function multiplied by the error amplifier trans-
fer function.
The bandwidth and phase margin can be read graphically
from Bode plots of H
EA
H = G
are shown in Figure 14.
PS
EA-ACTUAL
x H
EA
must be modified to
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The bandwidth of this example circuit is 59kHz, with a phase
margin of 60°.
EFFICIENCY CALCULATIONS
The following is a sample calculation.
A reasonable estimation of the efficiency of a switching buck
controller can be obtained by adding together the Output
Power (P
The Output Power (P
design is (1.2V * 4A) = 4.8W. The Total Power (P
an efficiency calculation to complement the design, is shown
below.
The majority of the power losses are due to low and high-side
of MOSFET’s losses. The losses in any MOSFET are group
of switching (P
FET Switching Loss (P
The FDS6898A has a typical turn-on rise time t
fall time t
es for this type of dual N-Channel MOSFETs are 0.061W.
FET Conduction Loss (P
R
account for the increasing R
DS(ON)
P
FIGURE 14. Overall Loop Gain and Phase
= 13mΩ and the factor is a constant value (k = 1.3) to
f
OUT
FET
P
of 15ns and 16ns, respectively. The switching loss-
P
P
CND
CND2
SW
P
P
) loss and the Total Power (P
= P
P
SW
CND1
CND2
SW
P
= 0.5 x 3.3V x 4A x 300kHz x 31ns
= 98.42mW + 172mW = 270.42mW
CND1
SW
= (4A)
= 0.5 * V
) and conduction losses(P
P
= (4A)
SW
P
= I
+ P
CND
= I
P
P
OUT
2
= P
2
OUT
CND
FET
SW
2
x 13mΩ x 1.3 x (1 - 0.364)
SW
= P
OUT
2
) for theTypical Application Circuit
IN
SW(ON)
CND
x 13mΩ x 1.3 x 0.364
= 61.38mW
= 331.8mW
)
x R
= 61.38mW + 270.42mW
* I
DS(ON)
CND1
x R
)
OUT
DS(ON)
DS(ON)
+ P
+ P
* (t
of a FET due to heating.
SW(OFF)
CND2
r
x k x (1-D)
+ t
x k x D
f
)* F
TOTAL
CND
OSC
).
) loss:
r
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and turn-off
TOTAL
), with