audioguru said:Connect a low value resistor in series with the negative battery terminal to ground. Then use a comparator that works with its inputs near 0V like an LM393 or LM339 to compare when the current drops to about 3% of the battery cell's rated current.
audioguru said:You don't want to measure the charger's current which includes the relay's current, you want to measure the battery's current.
10 ohms causes a voltage drop of 3V when the current is 0.3A. That is a lot of wasted voltage.
You need a voltage regulator with its output adjusted to 4.2V plus or minus only 1%.
You also need safety features like temperature and pressure sensors and maybe a timer. you don't want the battery to catch on fire during your charging demonstration.
If your teacher knows about the hazzards of a Lithium-Ion battery catching on fire when it is overcharged because the charging circuit has failed then your extremely simple circuit will be shot down (in flames).hardcore misery said:we don't have temperature sensor..
hardcore misery said:can we know the time for fully charging a li-ion battery through computations?
hi Lefty,Leftyretro said:As far as I know most li-ion chargers measure and use a battery end of charge voltage to detemine when to stop charging the battery. The temp sensor is an override shutdown of the charging regardless of state of charge of the battery.
Lefty
ericgibbs said:hi Lefty,
From the graph you will see that the 4.2V state has to be held until the Icharge <= 3% of the 1C rate.
This is where the problem lies with hardcores charger circuit, it now cuts off at 4.2V when it reaches 4.2v!.
audioguru said:Sense and regulate the voltage across the battery to be 4.2V. Then sense the current in the 10 ohm resistor in series with the battery.
audioguru said:Sense and regulate the voltage across the battery to be 4.2V. Then sense the current in the 10 ohm resistor in series with the battery.
hardcore misery said:how can we regulate the voltage across the battery? in charging condition, at this point, there is sudden fluctuation on the voltage at 4.2V. sir if you don't mind, i would like to see some illustrations on what you are trying to explain sir, or just equations or formulas, or examples.. thank you sir
audioguru said:If you want to regulate the charging current like IC chargers do, then you need a higher voltage supply feeding a current regulator in series with a 4.2V voltage regulator. If the current is regulated at 800mA then the battery cell charges with that much until its voltage reaches 4.2V. Then the battery cell automatically reduces its charging current.
Instead of a current regulator you can use a 4 ohm resistor in series with the battery cell to limit the charging current. The charging current will never be 4.2V/4 ohms= 1.05A because a Li-Ion battery cell is not allowed to be charged if its voltage is less than about 3.0V. So then its charging current would start at 300mA and will reduce as the battery charges. The battery cell will take a long time to charge.
You have the comparator's inputs backwards. When the charging current is high then the comparator's inverting input is higher than its other input so its output is low, the transistor is turned off and the relay disconnects the charging. It would then turn on then turn off over and over like a buzzer.
I think sir, you are reffering 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?
The reference voltage made with R1 and R2 is much too high. Now it makes 1.87V but a 3% charging current makes a voltage of only 0.27V. R4 is not needed. sir 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)
by the way sir, 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 values of the input resistors and R7 are much too low. Use values 100 times higher.If you don't mind sir, what is the reason for choosing higher values for input resistors?
R5 overloads the weak output of a low-power LM339. Its minimum output current is only 6mA.
Use 3.3k for R5 then the comparator's output current is about 2.6 mA so it is not overloaded.
R6 is not needed, remove it. The current in the LED is too low for it to be seen. Remove it.
Then the base current in the transistor is 2.5mA which is plenty.
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.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?
I ignored the hysteresis created by R7. The voltage divider if R1 and R2 divide the 8.8V battery to 1.87V.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)
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.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 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.what is the reason for choosing higher values for input resistors?
Yes but the voltage across the 10 ohms or 4 ohms resistor will never reach that high.the Vref(ON)=2.0468V and Vref(OFF)=1.737V with 9V supply.
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