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ADP1874ARQZ-1.0-R7 датащи(PDF) 32 Page - Analog Devices

номер детали ADP1874ARQZ-1.0-R7
подробное описание детали  Synchronous Buck Controller with Constant On-Time and Valley Current Mode
PDF  44 Pages
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производитель  AD [Analog Devices]
домашняя страница  http://www.analog.com
Logo AD - Analog Devices

ADP1874ARQZ-1.0-R7 датащи(HTML) 32 Page - Analog Devices

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ADP1874/ADP1875
Rev. 0 | Page 32 of 44
For example, if the external MOSFET characteristics are θJA
(16-lead QSOP) = 104°C/W, fSW = 300 kHz, IBIAS = 2 mA, CupperFET =
3.3 nF, ClowerFET = 3.3 nF, VDR = 4.62 V, and VREG = 5.0 V, then the
power loss is
(
)
[
]
()
[
))
002
.
0
0
.
5
10
3
.
3
10
300
(
0
.
5
(
))
002
.
0
62
.
4
10
3
.
3
10
300
(
62
.
4
(
9
3
9
3
)
(
+
×
×
×
×
×
+
+
×
×
×
×
×
=
+
×
+
+
×
=
−
−
BIAS
lowerFET
SW
BIAS
DR
upperFET
SW
DR
LOSS
DR
I
VREG
C
f
VREG
I
V
C
f
V
P
]
= 57.12 mW
)
002
.
0
5
10
3
.
3
10
300
(
)
V
5
V
13
(
)
(
)
(
9
3
)
(
+
×
×
×
×
×
−
=
+
×
×
×
−
=
−
BIAS
total
SW
IN
LDO
DISS
I
VREG
C
f
VREG
V
P
= 55.6 mW
mW
6
.
55
mW
13
.
77
)
(
)
(
)
(
+
=
+
=
LOSS
DR
LDO
DISS
TOTAL
DISS
P
P
P
= 132.73 mW
The rise in package temperature (for 16-lead QSOP) is
mW
05
.
132
°C
104
)
(
×
=
×
θ
=
LOSS
DR
JA
R
P
T
= 13.7°C
Assuming a maximum ambient temperature environment of 85°C,
TJ = TR × TA = 13.7°C + 85°C = 98.7°C
which is below the maximum junction temperature of 125°C.
DESIGN EXAMPLE
The ADP1874/ADP1875 are easy to use, requiring only a few
design criteria. For example, the example outlined in this section
uses only four design criteria: VOUT = 1.8 V, ILOAD = 15 A (pulsing),
VIN = 12 V (typical), and fSW = 300 kHz.
Input Capacitor
The maximum input voltage ripple is usually 1% of the
minimum input voltage (11.8 V × 0.01 = 120 mV).
VRIPP = 120 mV
VMAX,RIPPLE = VRIPP − (ILOAD,MAX × ESR)
= 120 mV − (15 A × 0.001) = 45 mV
mV
105
10
300
4
A
15
4
3
,
,
×
×
×
=
=
RIPPLE
MAX
SW
MAX
LOAD
IN,min
V
f
I
C
= 120 μF
Choose five 22 μF ceramic capacitors. The overall ESR of five
22 μF ceramic capacitors is less than 1 mΩ.
IRMS = ILOAD/2 = 7.5 A
PCIN = (IRMS)2 × ESR = (7.5 A)2 × 1 mΩ = 56.25 mW
Inductor
Determine inductor ripple current amplitude as follows:
3
LOAD
L
I
I ≈
Δ
= 5 A
Therefore, calculating for the inductor value
V
2
.
13
V
8
.
1
10
300
V
5
)
V
8
.
1
V
2
.
13
(
)
(
3 ×
×
×
−
=
×
×
Δ
−
=
IN,MAX
OUT
SW
L
OUT
IN,MAX
V
V
f
I
V
V
L
= 1.03 μH
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) from Table 10
with peak current handling of 20 A.
2
)
(
L
LOSS
DCR
I
DCR
P
×
=
= 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 RON of 4.5 mΩ and 13 A as
the valley current limit from Table 7 and Figure 71 indicates, a
programming resistor (RES) of 100 kΩ corresponds to an ACS
of 24 V/V.
Choose a programmable resistor of RRES = 100 kΩ 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 ADP1874 advantage 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.
ΔVDROOP = 0.05 × 1.8 V = 90 mV
Assuming that the overall ESR of the output capacitor ranges
from 5 mΩ to 10 mΩ,
)
mV
90
(
10
300
A
15
2
)
(
2
3
×
×
×
=
Δ
×
Δ
×
=
DROOP
SW
LOAD
OUT
V
f
I
C
= 1.11 mF
Therefore, an appropriate inductor selection is five 270 μF
polymer capacitors with a combined ESR of 3.5 mΩ.



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