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BQ24704 Datasheet, PDF (21/31 Pages) Texas Instruments – Low Cost, Host-Controlled Li-Ion and Li-Polymer Battery Charger With Low Iq
bq24704
www.ti.com........................................................................................................................................................................................................ SLUS838 – MAY 2009
Table 2. Component List for Typical System Circuit of Figure 1
PART DESIGNATOR
Q1, Q2, Q3
Q4, Q5
D1, D2
RAC, RSR
L1
C6, C7, C11, C12
C4, C10
C2, C3, C8, C9, C13, C14, C15
C5
C1
R3, R4, R15
R1
R2
R10
R6
QTY
3
2
2
2
1
4
2
7
1
1
3
1
1
1
1
DESCRIPTION
P-channel MOSFET, –30V,-6A, SO-8, Vishay-Siliconix, Si4435
N-channel MOSFET, 30V, 12.5A, SO-8, Fairchild, FDS6680A
Diode, Dual Schottky, 30V, 200mA, SOT23, Fairchild, BAT54C
Sense Resistor, 10 mΩ, 1%, 1W, 2010, Vishay-Dale, WSL2010R0100F
Inductor, 8.2µH, Vishay-Dale, IHLP5050CE-01
Capacitor, Ceramic, 10µF, 25V, 20%, X5R, 1206, Panasonic, ECJ-3YB1E106M
Capacitor, Ceramic, 1µF, 25V, 10%, X7R, 2012, TDK, C2012X7R1E105K
Capacitor, Ceramic, 0.1µF, 50V, 10%, X7R, 0805, Kemet, C0805C104K5RACTU
Capacitor, Ceramic, 100pF, 25V, 10%, X7R, 0805, Kemet, C0805C101K5RACTU
Capacitor, Ceramic, 2.2µF, 25V, 10%, X7R, 2012, TDK, C2012X7R1E225K
Resistor, Chip, 10 kΩ, 1/16W, 5%, 0402
Resistor, Chip, 432 kΩ, 1/16W, 1%, 0402
Resistor, Chip, 66.5 kΩ, 1/16W, 1%, 0402
Resistor, Chip, 2 Ω, 1W, 1%, 1210
Resistor, Chip, 20 kΩ, 1/16W, 1%, 0402
APPLICATION INFORMATION
Input Capacitance Calculation
During the adapter hot plug-in, the ACFET has not been turned on. The AC switch is off and the simplified
equivalent circuit of the input is shown in Figure 29.
IIN
VIN
Charger
Ri
Vi
Li
Rc
Ci
A. Ri: Equivalent resistance of cable
B. Li: Equivalent inductance of cable
C. RC ESR of Ci
D. Ci: Decoupling capacitor
Figure 29. Simplified Equivalent Circuit During Adapter Insertion
The voltage on the input capacitor(s) is given by:
VIN(t) =
IIN (t)
x
RC
+
VCi (t)
=
Vi
-
Rt t
VIe 2Li
êëéê
Ri - RC
wLi
sinwt + coswtùúúû
VCi(t) =
Vi
-
Rt t
VIe 2Li
éê Rt
êë 2wLi
sinwt + coswtùúúû
Rt = Ri + RC
w=
I
LiCi
-çççèæç
Rt
2Li
÷÷ø÷÷ö2
IIn(t) =
Vi
wLi
Rt t
e 2Li
sinwt
(6)
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