ADP2102A-EVALZ Analog Devices Inc, ADP2102A-EVALZ Datasheet - Page 21
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ADP2102A-EVALZ
Manufacturer Part Number
ADP2102A-EVALZ
Description
BOARD EVAL FOR ADP2102 0.8V-1.2V
Manufacturer
Analog Devices Inc
Datasheet
1.ADP2102YCPZ-4-R7.pdf
(24 pages)
Specifications of ADP2102A-EVALZ
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Output Capacitor
For transient applications, assume a droop of 0.1 V. Typically,
it takes two to three cycles for the output to settle from a load
transient because the capacitor alone supplies the load current
until the loop responds.
Under these conditions, a minimum required output
capacitance is calculated as follows:
Choose a 4.7 μF capacitor for this application.
For an instantaneous step decrease in load current, the output
capacitor required to limit the output voltage overshoot (V
during a full load to no load transient must be determined. This
transient requires the excess energy stored in the output inductor
to be absorbed by the output capacitor with a limited overshoot
in the output voltage.
Assuming an overshoot of 50 mV for a full load transient,
Choose a 4.7 μF capacitor for this application.
C
C
P
I
rms
COUT
OUT_MIN
OUT
=
=
= I
2
2
(
1
1
V
= 3 ×
rms
3
3
OUT
2
×
× ESR = (0.045)
×
+
L
2
V
V
×
V
2 .
DROOP
OUT
OS
I
×
L
Δ
OUT
1
)
I
×
2
×
8 .
LOAD
10
−
(
2
f
V
×
SW
−
×
V
6
IN
4 (
f
OUT
×
SW
×
_
2 .
3
MAX
V
2
2
×
× 0.005 = 10.12 μW
−
IN
=
10
=
1
_
−
) 8 .
MAX
0
6
2
V
. 1 (
1 .
2 .
×
OUT
3
×
85
4
μ
×
2 .
3
)
H
)
0
2
×
3 .
−
=
= 45 mA rms
×
10
. 1 (
0 (
6
) 6 .
) 8
= 3 μF
2
2
= 4.33 μF
OS
)
Rev. B | Page 21 of 24
Input Capacitor
Assume an input ripple of 27 mV based on 1% of V
For ceramic capacitors, the typical ESR is from 5 mΩ to 15 mΩ.
Losses
P
LOSS
C
I
P
P
(0.310 × 0.5 + 0.145 × 0.5) × (0.6)
P
(3.6/2) × 0.6 × (5 ns + 5 ns) × 3 × 10
P
(3.6)
P
P
82 mW + 32.4 mW + 7.8 mW + 28.8 mW = 151 mW
T
P
is well below the junction temperature maximum of 125°C.
rms
LOSS
CIN
SW_COND
TRAN
SW
L
LOSS
JMAX
IN
= DCR × I
=
= I
= (C
= I
2
= 85°C + 54°C/W × 151 mW = 93.15°C
= P
= T
= (V
× 3 × 10
(
. 0 (
OUT
Δ
rms
GATE_P
SW_COND
V
027
= (R
A
/2 = 0.3 A rms
2
IN
+ θ
IN
× ESR = (0.3)
/2) × I
0 /
/
OUT
DS (ON)_P
JA
I
6 .
+ C
6
OUT
= 7.8 mW
−
+ P
2
= 0.08 × (0.6)
. 0
GATE_N
OUT
−
005
TRAN
1
× D + R
ESR
1
× (t
)
) × V
+ P
×
2
)
4
R
× 0.005 = 450 μW
×
×
+ t
SW
3
DS (ON)_N
IN
4
×
F
+ P
2
) × f
×
10
× f
2
= 28.8 mW
6
L
f
2
SW
=
SW
SW
= 82 mW
× (1 − D)) × I
= 2.2 μF
6
= (200 pF) ×
=
= 32.4 mW
=
ADP2102
IN_MIN.
OUT
2
=