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| 型号： | ADP1879-0.3-EVALZ |
| 厂商： | Analog Devices Inc |
| 文件页数： | 29/40页 |
| 文件大小： | 0K |
| 描述： | BOARD EVAL FOR ADP1879-0.3 |
| 标准包装： | 1 |
| 系列： | * |
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?? ???????? , ?????? 15 A
?? ???? , ?????? =
=
?? ????????
??? ?? ≈
=5 A
??? ???? , ?????? ? ?? ?????? ?
?? ??????
1.8 V
13.2 V – 1.8 V
×
=
?? ?????? ( ???????? ) = ?????? × ?? ??2
??? ????????
?? ?????? = 2 ×
?? ???? × ( ??? ?????????? )
300 × 10 3 × (90 mV )
( ?? × ?? 2 ???????? )
?? ?????? =
(( ?? ?????? ? ??? ?????????? ) 2 ? ( ?? ?????? ) 2 )
= 1 × 10 ?6 × (15 A ) 2
1 1 ??? ???? , ?????? ? ?? ?????? ? ?? ??????
?? ?????? =
×
×
1 (13.2 V – 1.8 V)
1
1.8 V
=
×
×
=1.49 A
2
3 1 μ F×300×10 3
13.2 V
√
Data Sheet
The rise in package temperature (for a 14-lead LFCSP_WD) is
T R = θ JA × P DR(LOSS)
= 30°C × 132.05 mW
= 4.0°C
Assuming a maximum ambient temperature environment of 85°C,
T J = T R × T A = 4.0°C + 85°C = 89.0°C,
which is below the maximum junction temperature of 125°C.
DESIGN EXAMPLE
The ADP1878 / ADP1879 are easy to use, requiring only a few
design criteria. For example, the example outlined in this section
uses only four design criteria: V OUT = 1.8 V, I LOAD = 15 A (pulsing),
V IN = 12 V (typical), and f SW = 300 kHz.
Input Capacitor
The maximum input voltage ripple is usually 1% of the
minimum input voltage (11.8 V × 0.01 = 120 mV).
V RIPP = 120 mV
V MAX,RIPPLE = V RIPP ? ( I LOAD,MAX × ESR )
= 120 mV ? (15 A × 0.001) = 45 mV
= 120 μF 4 ?? ???? ?? ?????? , ???????????? 4 × 300 × 10 3 × 105 mV
Choose five 22 μF ceramic capacitors. The overall ESR of five
22 μF ceramic capacitors is less than 1 mΩ.
I RMS = I LOAD /2 = 7.5 A
P CIN = ( I RMS ) 2 × ESR = (7.5 A) 2 × 1 mΩ = 56.25 mW
Inductor
Determining inductor ripple current amplitude:
3
Then, calculating for the inductor value
?? = ??? ?? × ?? ???? , ??????
5 V ×
= 1.03 μH 300 × 10 3 13.2 V
The inductor peak current is approximately
15 A + (5 A × 0.5) = 17.5 A
Therefore, an appropriate inductor selection is 1.0 μH with
DCR = 3.3 mΩ (Würth Elektronik 7443552100) with a peak
current handling of 20 A.
= 0.003 × (15 A) 2 = 675 mW
ADP1878/ADP1879
Current-Limit Programming
The valley current is approximately
15 A ? (5 A × 0.5) = 12.5 A
Assuming a low-side MOSFET R ON of 4.5 mΩ and 13 A, as the
valley current limit from Table 7 and Figure 71 indicate, a pro-
gramming resistor (RES) of 100 kΩ corresponds to an A CS
of 24 V/V.
Choose a programmable resistor of R RES = 100 k? for a current
sense gain of 24 V/V.
Output Capacitor
Assume that a load step of 15 A occurs at the output and no more
than 5% output deviation is allowed from the steady state operating
point. In this case, the advantage of the ADP1878 is that because
the frequency is pseudo fixed, the converter is able to respond
quickly because of the immediate, though temporary, increase
in switching frequency.
Δ V DROOP = 0.05 × 1.8 V = 90 mV
Assuming the overall ESR of the output capacitor ranges from
5 mΩ to 10 mΩ,
=2× 15 A
= 1.11 mF
Therefore, an appropriate inductor selection is five 270 μF
polymer capacitors with a combined ESR of 3.5 mΩ.
Assuming an overshoot of 45 mV, determine if the output
capacitor that was calculated previously is adequate
1.4 ?
= (1.8 mF 45 mV ) 2 ? (1.8) 2
Choose five 270 μF polymer capacitors.
The rms current through the output capacitor is
2 √3 ?? × ?? ???? ?? ???? , ??????
The power loss dissipated through the ESR of the output
capacitor is
P COUT = ( I RMS ) 2 × ESR = (1.5 A) 2 × 1.4 mΩ = 3.15 mW
Rev. B | Page 29 of 40
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| ADP1880ACPZ-R7 | 制造商:Analog Devices 功能描述: |
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