TSM102(2004) Просмотр технического описания (PDF) - STMicroelectronics

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TSM102
(Rev.:2004)

VOLTAGE AND CURRENT CONTROLLER

STMicroelectronics 

TSM102/A

Figure 2 : The Application Schematic - Battery Charger Secondary Side

The first comparator ensures the “Low Battery”

signal generation thanks to the comparison of a

part of the charger’s output voltage (resistor

bridge R17, R19) and the reference voltage. Prop-

er hysteresis is given thanks to R20. An improve-

ment to the chargers security and to the battery’s

life time optimization is achieved by lowering the

current control measurement thanks to Q1 that

shunts the resistor R9 when the battery’s voltage

is below the “Low Battery” level.

The second comparator ensures the “End of

Charge” signal generation thanks to the compari-

son of a part of the charger’s output voltage (resis-

tor bridge R1, R2, R3) and the reference voltage.

When either of these two signals is active, the cor-

responding LED is polarized for convenient visual-

ization of the battery status.

3 - CALCULATION OF THE ELEMENTS

All the components values have been chosen for a

two-Lithium-Ion batteries charge application :

s Current Control : 720mA (Low Battery current

control : 250mA)

s Voltage Control : 8.4V (= 2x 4.2V)

s Low Battery : 5.6V (= 2x 2.5V + 0.6V)

s End of Charge : 8.3V (= 2x 4.15V)

Current Control :

The voltage reference is polarized thanks to the

R4 resistor (2.5mA), and the cathode of the refer-

ence gives a fixed 2.500V voltage.

I = U / R = [Vref( R8 + R9 ) / (R7 + R8 + R9) ] / Rs

= [2.5 x (390 + 820) / (10000 + 390 + 820)] / 0.375

= 720mA

7/9

I = 720mA

P = power dissipation through the sense resistor =

R I2 = 0.375 x 0.7202 = 194mW

In case of “Low Battery” conditions, the current

control is lowered thanks to the following

equation :

I = U / R = [ Vref R8 / (R7 + R8) ] / Rs

= [ 2.5 x 390 / (10000 + 390 ) ] / 0.375

= 250mA

I (LoBatt) = 250mA

Voltage Control :

Vout = Vref / [ R2 / (R1 + R2 + R3) ]

= 2.5 / [ 56 / (131.5 + 56 + 0.68 ) ]

= 8.400V

Vout = 8.400V

Low Battery signal :

If R5 = 0Ω and R6 = open :

Vout(LoBatt) = Vref / [ R19 / ( R17 + R19 ) ]

= 2.5 / [ 10 / (12.4 + 10) ]

= 5.6V

Vout(LoBatt) = 5.6V

End of Charge signal :

Vout(EOC) = Vref / [ (R2 + R3 ) / (R1 + R2 + R3) ]

= 2.5 / [(56 + 0.68) / (131.5 + 56 + 0.68)]

= 8.300V

Vout (EOC)= 8.300V

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