AAT2552IRN-CAE-T1 Analogic Tech, AAT2552IRN-CAE-T1 Datasheet - Page 20

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AAT2552IRN-CAE-T1

Manufacturer Part Number
AAT2552IRN-CAE-T1
Description
Charge Pumps Total Pwr Solution for Portable App
Manufacturer
Analogic Tech
Datasheet

Specifications of AAT2552IRN-CAE-T1

Package / Case
TDFN-16 EP
Mounting Style
SMD/SMT
Lead Free Status / Rohs Status
Lead free / RoHS Compliant

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
AAT2552IRN-CAE-T1
Manufacturer:
ON
Quantity:
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Part Number:
AAT2552IRN-CAE-T1
Manufacturer:
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Quantity:
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There are three types of losses associated with the step-
down converter: switching losses, conduction losses, and
quiescent current losses. Conduction losses are associ-
ated with the R
switching devices. Switching losses are dominated by the
gate charge of the power output switching devices. At full
load, assuming continuous conduction mode (CCM), a
simplified form of the losses is given by:
I
term t
verter switching losses.
For the condition where the step-down converter is in
dropout at 100% duty cycle, the total device dissipation
reduces to:
Since R
vary with input voltage, the total losses should be inves-
tigated over the complete input voltage range.
Given the total losses, the maximum junction tempera-
ture can be derived from the θ
package which is 50°C/W.
20
SystemPower
Q
is the step-down converter quiescent current. The
Figure 4: Maximum Charging Current Before
P
sw
TOTAL
500
450
400
350
300
250
200
150
100
DS(ON)
50
4.25
is used to estimate the full load step-down con-
0
Thermal Cycling Becomes Active.
+ (t
=
, quiescent current, and switching losses all
4.5
I
P
O
sw
2
TOTAL
T
DS(ON)
4.75
· (R
J(MAX)
· F
TM
T
A
S
DSON(H)
= I
= 85°C
5
· I
characteristics of the power output
= P
O
O
5.25
2
T
+ I
TOTAL
A
· R
= 60°C
· V
Q
5.5
) · V
DSON(H)
O
V
· Θ
IN
+ R
5.75
IN
(V)
V
JA
JA
DSON(L)
IN
+ I
+ T
6
for the TDFN34-16
Q
T
AMB
· V
A
6.25
T
= 25°C
A
· [V
IN
= 45°C
6.5
IN
- V
w w w . a n a l o g i c t e c h . c o m
6.75
O
])
7
Total Power Solution for Portable Applications
Capacitor Selection
Linear Regulator Input Capacitor (C6)
An input capacitor greater than 1μF will offer superior
input line transient response and maximize power sup-
ply ripple rejection.
electrolytic capacitors may be selected for C
no specific capacitor ESR requirement for C
for 300mA LDO regulator output operation, ceramic
capacitors are recommended for C
ent capability over tantalum capacitors to withstand
input current surges from low impedance sources such
as batteries in portable devices.
Battery Charger Input Capacitor (C1)
In general, it is good design practice to place a decou-
pling capacitor between the ADP pin and GND. An input
capacitor in the range of 1μF to 22μF is recommended.
If the source supply is unregulated, it may be necessary
to increase the capacitance to keep the input voltage
above the under-voltage lockout threshold during device
enable and when battery charging is initiated. If the
adapter input is to be used in a system with an external
power supply source, such as a typical AC-to-DC wall
adapter, then a C
be used. A larger input capacitor in this application will
minimize switching or power transient effects when the
power supply is “hot plugged” in.
Step-Down Converter Input Capacitor (C6)
Select a 4.7μF to 10μF X7R or X5R ceramic capacitor for
the input. To estimate the required input capacitor size,
determine the acceptable input ripple level (V
for C
is a maximum when V
Always examine the ceramic capacitor DC voltage coeffi-
cient characteristics when selecting the proper value. For
example, the capacitance of a 10μF, 6.3V, X5R ceramic
capacitor with 5.0V DC applied is actually about 6μF.
IN
. The calculated value varies with input voltage and
V
V
IN
O
C
C
· 1 -
IN
IN(MIN)
IN
=
capacitor in the range of 10μF should
=
V
V
PRODUCT DATASHEET
V
V
V
IN
I
Ceramic, tantalum, or aluminum
IN
O
O
PP
IN
O
is double the output voltage.
V
· 1 -
I
O
PP
- ESR · F
=
1
4
- ESR · 4 · F
AAT2552178
for V
V
V
1
IN
O
IN
S
IN
= 2 · V
due to their inher-
2552.2009.01.1.3
S
O
IN
PP
IN
. However,
) and solve
. There is

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