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LM2743MTC/NOPB датащи(PDF) 27 Page - Texas Instruments

номер детали LM2743MTC/NOPB
подробное описание детали  LM2743 2.2-V to 16-V Input, voltage mode, synchronous buck controller with tracking
PDF  44 Pages
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производитель  TI2 [Texas Instruments]
домашняя страница  https://www.ti.com
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LM2743MTC/NOPB датащи(HTML) 27 Page - Texas Instruments

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K =
POUT
POUT + PTOTAL
x 100%
-180
-156
-132
-108
-84
-60
100
1k
10k
100k
1M
FREQUENCY (Hz)
-40
-20
0
20
40
60
100
1k
10k
100k
1M
FREQUENCY (Hz)
27
LM2743
www.ti.com
SNVS276I – APRIL 2004 – REVISED FEBRUARY 2019
Product Folder Links: LM2743
Submit Documentation Feedback
Copyright © 2004–2019, Texas Instruments Incorporated
Typical Applications (continued)
The bandwidth and phase margin can be read graphically from Bode plots of HEA are shown in Figure 38.
Figure 37. Gain vs Frequency
Figure 38. Overall Loop Gain and Phase
The bandwidth of this example circuit is 59 kHz, with a phase margin of 60°.
8.2.1.2.9
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 (POUT) loss and the Total Power (PTOTAL) loss:
(47)
The Output Power (POUT) for the Figure 30 design is (1.2 V x 4 A) = 4.8 W. The Total Power (PTOTAL), with 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 (PSW) and conduction losses(PCND).
PFET = PSW + PCND = 61.38 mW + 270.42 mW
(48)
PFET = 331.8 mW
(49)
The following equations show FET Switching Loss (PSW).
PSW = PSW(ON) + PSW(OFF)
(50)
PSW = 0.5 x VIN x IOUT x (tr + tf) x fSW
(51)
PSW = 0.5 x 3.3 V x 4 A x 300 kHz x 31 ns
(52)
PSW = 61.38 mW
(53)
The FDS6898A has a typical turn-on rise time tr and turn-off fall time tf of 15 ns and 16 ns, respectively. The
switching losses for this type of dual N-Channel MOSFETs are 0.061 W.
The following equations show FET Conduction Loss (PCND).
PCND = PCND1 + PCND2
(54)
PCND1 = (IOUT)
2 x R
DS(ON) x k x D
(55)
PCND2 = (IOUT)
2 x R
DS(ON) x k x (1-D)
(56)
RDS(ON) = 13 mΩ and the factor is a constant value (k = 1.3) to account for the increasing RDS(ON) of a FET due to
heating.
PCND1 = (4A)
2 x 13 mΩ x 1.3 x 0.364
(57)



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