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参数资料
| 型号： | ADP1872-BL2-EVZ |
| 厂商： | Analog Devices Inc |
| 文件页数： | 28/40页 |
| 文件大小： | 0K |
| 描述： | EVAL BOARD FOR ADP1872 |
| 标准包装： | 1 |
| 主要目的： | DC/DC，步降 |
| 稳压器拓扑结构： | 降压 |
| 板类型： | 裸（未填充） |
| 已供物品： | 裸板 |
| 已用 IC / 零件： | ADP1872 |
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ADP1872/ADP1873
The inductor peak current is approximately
15 A + (5 A × 0.5) = 17.5 A
Data Sheet
Choose five 270 μF polymer capacitors.
The rms current through the output capacitor is
P DCR ( LOSS ) =
DCR × I LOAD
1
1 1 ( V IN , MAX ? V OUT )
1 ( 13 . 2 V ? 1 . 8 V ) 1 . 8 V
×
3 1 μF × 300 × 10
13 . 2 V
Therefore, an appropriate inductor selection is 1.0 μH with
DCR = 3.3 mΩ (7443552100) from Table 7 with peak current
handling of 20 A.
2
I RMS =
= ×
2
× ×
2 3 L × f SW
3
V OUT
V IN , MAX
= 1 . 49 A
R T = 15 kΩ ×
= 30 kΩ
G CS =
1 1
= 0.003 × (15 A) 2 = 675 mW
Current Limit Programming
The valley current is approximately
15 A ? (5 A × 0.5) = 12.5 A
Assuming a lower side MOSFET R ON of 4.5 mΩ, choosing 13 A
as the valley current limit from Table 6 and Figure 70 indicates
that a programming 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 a load step of 15 A occurs at the output and no more
than 5% is allowed for the output to deviate from the steady
state operating point. The ADP1872’s advantage is, 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
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
Feedback Resistor Network Setup
It is recommended to use R B = 15 kΩ. Calculate R T as
(1 . 8 V ? 0 . 6 V)
0 . 6 V
Compensation Network
To calculate R COMP , C COMP , and C PAR , the transconductance
parameter and the current-sense gain variable are required. The
transconductance parameter (G M ) is 500 μA/V, and the current-
sense loop gain is
= = 8 . 33 A/V
A CS R ON 24 × 0 . 005
where A CS and R ON are taken from setting up the current limit
(see the Programming Resistor (RES) Detect Circuit and Valley
Current-Limit Setting sections).
The crossover frequency is 1/12 th of the switching frequency:
C OUT = 2 ×
= 2 ×
C OUT =
1 × 10 ? 6 × ( 15 A ) 2
f CROSS 2 π f CROSS C OUT V
f CROSS + f ZERO
R COMP = × × OUT
25 × 10 3 2 × 3 . 141 × 25 × 10 3 × 1 . 11 × 10 ? 3
= ×
25 × 10 + 6 . 25 × 10
500 × 10 × 8 . 3
C COMP =
1
2 × 3 . 14 × 100 × 10 3 × 6 . 25 × 10 3
5 mΩ to 10 mΩ,
? I LOAD
f SW × ( ? V DROOP )
15 A
300 × 10 3 × ( 90 mV )
= 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:
( L × I 2 LOAD )
( ( V OUT ? ? V OVSHT ) 2 ? ( V OUT ) 2 )
=
( 1 . 8 ? 45 mV ) 2 ? ( 1 . 8 ) 2
300 kHz/12 = 25 kHz
The zero frequency is 1/4 th of the crossover frequency:
25 kHz/4 = 6.25 kHz
G M G CS V REF
3 3 ? 6
= 100 kΩ
1
2 π R COMP f ZERO
=
= 250 pF
×
1 . 8
0 . 6
= 1.4 mF
Rev. B | Page 28 of 40
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