Battery Charger with auto shut-off feature

[hardcore misery]

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.

this was suggested by a friend of mine, but during that time i can't understand the principle of it, referring to this schematic, a 10 ohms/5W is added on the negative side of the battery charger(4.2V, on the actual circuit) so can we use this resistor for the comaparator?(btw, are we going to use another comparator?) so therefore we will disconnect the transistor and relay circuit on the given schematic, and can we use logic gates to compare two conditions? a condition for the OFF state when the comparator reaches 4.2V(but battery voltage is not fully charge) and another condition for the measurement of current drop until the battery is fully charged?

what can you advice sir? hoping that we can finish these all before thursday for to be ready with mock defense and finalize our documentations. thank you very much.

**broken link removed**

 

[audioguru]

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.

 

[hardcore misery]

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.

we are using a voltage regulator for the charger of battery (set to 4.2V)
we don't have temperature sensor.. so what if the voltage across battery in charging condition reaches 4.2V( as we all know, if we remove the battery, its voltage is not equal to 4.2V) what is the voltage drop across 10ohms?, where can i measure the current drop?

 

[audioguru]

we don't have temperature sensor..

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).

Most Lithium-ion battery cells have a temperature sensor inside. Use it as a shutdown backup in case your circuit fails.

 

[hardcore misery]

so there will be another circuit for the temperature sensor? is it a voltage comparator again?

 

[hardcore misery]

can we know the time for fully charging a li-ion battery through computations?

 

[Leftyretro]

can we know the time for fully charging a li-ion battery through computations?

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]

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

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]

It is too bad that the school kids can't use a lithium battery charger IC that does everything that should be done to charge a dangerous lithium battery.

 

[hardcore misery]

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!.

is there a way to measure the Icharge? this is very confusing to me because if you connect a 10ohms resistor between the negative terminal of the charger and battery,

the charge voltage is 4.2V, while for example the battery voltage is 3.7V

at noload, the load current is 420maH, when the 3.7V is loaded to the charger, there will be a voltage drop on the 10ohms resistor (4.2V - 3.7V)
which is equal to 0.5V and has a current of 50maH, so if the battery voltage reaches 4.2V, theres no voltage drop on the 10 ohms resistor and the current drop on the 10 ohms is ZERO?

where can i measure the current drop illustrated on the graph? is it possible to have two conditions on the charger?

condition1: when the battery voltage reaches 4.2V while charging the transistor will turn OFF.
condition2: when current drops at < 3% (which means that the battery is said to be fully charged) another transistor turns OFF

so a RELAY will turn OFF when these to conditions are met.

is this possible? if yes, what way can we do this? through logic gates?

 

[audioguru]

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]

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.

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..

 

[hardcore misery]

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.

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

 

[ericgibbs]

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

hi,
A block diagram of the circuit you require.

 

[audioguru]

I was thinking about having the 4.2V regulator sense and regulate the 4.2V across the battery cell like this:

 

[hardcore misery]

reffering to the attached image,

i followed the block diagram, so the minor change is to put the Vsense of the negative feedback between the negative terminal of the battery and the Rsense.

are there any wrong connections on the schematic? ( i wonder that theres a voltage can be measured between the battery and the Rsense, will see on the actual experiment)

do we have to regulate the current of the charger?(reffering to the figure7)

is it correct to use 10ohms? or we will use at least 4ohms to limit the current?(4.2V/4 = 1.05A at no load condition)

our project is designed to charge 3 batteries(with separate voltage regulators) are there any effects on the charging voltage or the supply voltage?(caused by 3 charging batteries at the same time)

 

[audioguru]

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.

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.

The values of the input resistors and R7 are much too low. Use values 100 times higher.

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.

 

[hardcore misery]

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.

________________________________________________________

 

[hardcore misery]

In reference to the attached schematic, the Vref(ON)=2.0468V and Vref(OFF)=1.737V with 9V supply.

are there any wrong values there? (except for the R4 which should be removed, i can't understand why it should be removed and i don't know the voltage input at the -feedback, perhaps you can elaborate it sir?) thank you sir..

 

[audioguru]

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.