LX2203

Microsemi

Integrated Products Division

11861 Western Avenue, Garden Grove, CA. 92841, 714-898-8121, Fax: 714-893-2570

Page 9

Copyright

© 2004

Rev. 1.0a, 2005-08-10

Li-Ion Battery Charger

TM

®

APPLICATION NOTE (CONTINU E D)

The table below lists some popular Constant Current

Settings along with the associated CCP pin current and

programming resistor:

Charge Current

Iccp Current

Rccp Value

1.0A

13.75µA

90.9K

500mA

6.22µA

200K

100mA

1.00µA

1200K

It is possible to change the constant current setting by

changing the Rccp resistor while in charge mode. Since the

termination current is independent of the charge current,

lowering the constant charge current will increase the charge

time, but will not reduce the stored charge in the battery at

the charge termination point.

The circuit in Figure 4 is an example of a battery charger

configured to charge at 1A, 500mA or 100mA. The switches

are logically controlled and reduce the resistance at the CCP

pin when switched in. It is possible to eliminate the

MOSFET devices if open drain logic is available.

The logic for the CP1 and CP2 would normally come

from the appliance processor which would need to have the

capability to communicate over the USB interface. If there

is only one power connector and the USB interface is active,

the logic could assume the power was coming from the USB

bus and not the wall adapter. If the USB interface is active,

the USB application will know if the appliance has been

enumerated as a high or low load and would set CP1 and

CP2 appropriately.

It is possible to change the current programming circuit to

drive it directly with CMOS logic, but this requires that the

CMOS logic power supply be well regulated (+/-2%). Each

switch resistor leg in Figure 4 can be replaced with a two

resistor network tied to the output of a CMOS gate.

1.21M

R1

CCP

R2

Vcc

500mA/100mA

1.21M

243k

CCP

500mA/100mA

Rsw

Figure 6 – Circuits to provide 100mA and 500mA

constant charge currents.

The values of R1 and R2 are selected such that:

25

.

1

R

R

R

V

2

1

2

CC

=

+

×

&

SW

2

1

2

1

R

R

R

R

R

=

+

×

Solving these equations:

25

.

1

R

V

R

SW

CC

1

×

=

&

SW

1

SW

1

2

R

R

R

R

R

−

×

=

For Vcc = 3.3V and Rsw = 243k; R1 = 649k; R2 = 392k.

These values should provide charge currents of 100mA and

500mA

.

A compensation capacitor of value 0.01uF is required