PRECISION_Li－ION_BATTERY_CHARGER

Published:2009/7/13 22:14:00 Author:May | From:SeekIC

An alternative to using expensive 0.25-percent precision resistors is presented in this battery-charger design, which adds two 1-percent resistors and two jumpers to the charger.In constant-voltage-mo de charging, a lithium-ion cell requires 4.1 V ±50 mV. The 1.2-percent requirement represents a tight tolerance. In a regulation loop where a voltage divider is compared against a reference, the accuracy is typically achieved by selecting a 0.7-percent reference and a volt-age divider with 0.25-percent tolerance resistors. Unfortunately, 0.25-percent precision resistors cost three times as much as 1-percent resistors and have very long lead times. One solution for moderate-volume production involves adding two 1-percent resistors and two jumpers to the charger circuit (see the figure). The jumpers are removed as necessary to bring the constant voltage to the required accuracy of 1.2 percent. The charger selected for this example is the LT1510. There are three lithium-ion cells in the. battery. After a value is selected for R4 the values for R1, R2, and R3 can be calculated using the equations given. K is the relative change required for a circuit with all of its tolerances in one direction. For example, in the case of a 0.5-percent reference and two 1-percent resistors, the total tolerance is 2.5 percent. To bring it back to 1.2 percent, the percentage change required is 2.5 percent - 1.2 percent 1.3 percent, and K =0.013. The jumpers (J1 and J2) must be opened based on the following:If VOUT is K/2 below nominal, remove J1.If VOUT is K/2 above nominal, remove J2.The following values were calculated: R1 = 4.99 kΩ, R2 = 324Ω,R 3=80.6 ,Ω and R4= 20 KΩ The voltage below which jumper J1 should be opened is 12.34 V -1.3 percent /2 = 12.22 V. The voltage above which jumper J2 should be opened is 12.34 V + 1.3 percent /2 = 12.42 V.



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