hardcore misery said:
I think sir, you are refering at the NO LOAD condition, which the charging voltage is 4.2(vref=2.1V) while the Vref of +feedback is 1.8V, the Relay will turn ON when a battery with voltage less than 3.6V is placed on the charger, am i on the right track sir?
When the battery is 3V and is charging then the current in the 10 ohms resistor is 120mA and its voltage is only 1.2V. Then R3 and R4 reduce the (-) input of the comparator to 0.6V. But since the (+) input is 1.87V then the output of the comparator is high and the transistor and relay are on.
If the resistor is 4 ohms then its current is 300mA and its voltage is also 1.2V.
So the charger will never stop charging.
When the battery is fully charged then its current in the 10 ohms or 4 ohms resistor will be low.
If the inputs are reversed and the 1.87V reference is much less then the battery will charge until its current drops to a certain low current.
how did you get the value of 1.87V? the V+ is 8.8V which makes 1.8V (in reference to the formulas and program we made for computing Vref's)
I ignored the hysteresis created by R7. The voltage divider if R1 and R2 divide the 8.8V battery to 1.87V.
does the - feedback senses the battery voltage even if the - feedback is tapped between the negative terminal of the battery and the 4ohms? what is the voltage input at -feedback if R4 is removed?
The (-) input of the comparator measures half the voltage across the 10 ohms resistor R8 which is caused by the battery's charging current. It does not measure the battery's voltage. The current in R8 is used to turn off the charging when the current drops to 3% of the battery's current rating.
You don't need R3 and R4 to make a divider and cut the voltage across R8 in half. Remove R3 and R4 then the (-) input of the comparator will have the full voltage across R8.
what is the reason for choosing higher values for input resistors?
The comparator has a low input current and works fine with resistors that are hundreds of thousands of ohms. The comparator is low-power so its output current is weak. The small amount of output current should be used to drive the transistor, not the very low resistance of R7.
the Vref(ON)=2.0468V and Vref(OFF)=1.737V with 9V supply.
Yes but the voltage across the 10 ohms or 4 ohms resistor will never reach that high.
Read the tutorial about charging Li-Ion cells. You need to detect when the charging current drops to 3% of the battery's current rating which is a voltage across a series 4 ohm resistor of only 108mV without a voltage divider. Then the reference voltage should be 0.108V.