Battery Charger with auto shut-off feature

[ericgibbs]

sir i think theres no problem on the ratio of resistors for vb/2, (1k/1k) the problem is on setting the Vref of ON/OFF states with the correct values of resistors

hi, hcm,
Tell me the type of opa/comparator you are using?

The quickest solution is going to be if I make a working version of the project,
so that I can tell you the resistor values.

Is the power supply still at +10Vdc?

Is the charger voltage still set for +4.2V?

EDIT: also give me details of the battery type, voltage, AHr,,,etc.

 

[hardcore misery]

hi, hcm,
Tell me the type of opa/comparator you are using?

The quickest solution is going to be if I make a working version of the project,
so that I can tell you the resistor values.

Is the power supply still at +10Vdc?

Is the charger voltage still set for +4.2V?

thats great sir!!! we are 6 members on a group, and i'm the only one who works for our prototype..

we are using LM339 as a comparator, our supply is variable, but on my calculations we use 9.7V as a supply, and 4.2V as the charger voltage.

i think i got the correct resistors to be used.

R1=1k
R2=270
Rh=2200
RL=100k (r5 on the given schematic on previous post, is it ok to use this? previously 1k)

i got a vref(on state)=2.078V and a vref(off state)=1.8907V ( so we can charge batteries with voltage lower than 3.7V or 3.8V )

btw sir, how can we safely discharge li-ion batteries? on our sample batteries we can only got 3.7V.

quick question, our charging voltage is 4.2V and the limitting resistor is 10ohms, and we charge at least 3.7V, the charging current is only 50mAh, is this ok? or is it advisable to charge batteries with constant current? at least 300mah?

EDIT: Original Nokia Li-ion Battery BLC-2

950 mAh LI-ION / 3.6V
Weight: 108g
Talk time: 2h 40 min - 4 h 45 min
Standby time: 55 - 260 h
Variation in operation times will occur depending on SIM card, network settings and usage. Talk times is increased by up to 30% if Half Rate is active, and reduced by 5% if Enhanced Full Rate is active
Picture for reference only

Compatability:
3310/3330, 3410, 3510/3510i, 5510, 6650, 6800, 6810

 

[ericgibbs]

i think i got the correct resistors to be used.

R1=1k
R2=270
Rh=2200;;; This is lower than I would expect


i got a vref(on state)=2.078V and a vref(off state)=1.8907V ( so we can charge batteries with voltage lower than 3.7V or 3.8V )

btw sir, how can we safely discharge li-ion batteries? on our sample batteries we can only got 3.7V.
Never try to discharge this type of battery to less than 2.8V, you could damage the battery.
Use a 10R 3Watt resistor as a discharge load.. DONT go less than 2.8V

quick question, our charging voltage is 4.2V and the limitting resistor is 10ohms, and we charge at least 3.7V, the charging current is only 50mAh, is this ok? or is it advisable to charge batteries with constant current? at least 300mah?
The manufacturers claim that this type of battery, for a phone, can be charged a the 1C rate, but 0.9C is recommended.
EDIT: a 300mA charge current is about the 0.3C rate for this battery.

Hi,
I'll build a prototype of your circuit, give me 24hrs.

 

[hardcore misery]

ok sir, that would be nice... so i can verify the status of our prototype (mainly the status at NO LOAD condition, which the the charging voltage is 4.2V and the Vref is lower than 4.2V (2.1V or 2.05V)

 

[ericgibbs]

ok sir, that would be nice... so i can verify the status of our prototype
(mainly the status at NO LOAD condition, which the the charging voltage is 4.2V and the Vref is lower than 4.2V (2.1V or 2.05V)[/QUOTE]

hi,
Managed to get a LM339 for testing!
Will build today.

The charge voltage is 4.2v, when the charger is disconnected by the relay the battery voltage will be about 3.2v.

In your circuit you have chosen your resistor divider to give a Vbattery/2, that is,, 2.1V [when charging]
and 1.6V [when charged and relay just switched off].

From your original OP you wanted a circuit that stopped charging when the battery was fully charged?
Will the battery be removed from the charger when its fully charged?
If yes, why are you concerned about measuring and detecting with the comparator the lowest discharged voltage of a battery?

In my simulation for the battery on charge is:
from +5v to 0v I use a 2K, 20 turn variable resistor, the wiper of the variable is the simulated battery voltage.
I can adjust the variable to simulate the different battery voltage levels, this is much easier than trying to use an actual battery during testing.

 

[ericgibbs]

hi,
Added a 10K [P2] variable pot from the output of the comparator to 0V,
the wiper is connected back to the non inverting input of the comp via a 1K5 resistor.
Added a 1K [P1] variable pot to the junction of R1/R2,, Added a 0.1uF

Both pots should be multiturn to ensure setting accuracy.

Also moved the LED from the output of the comp to the relay drive collector.

As a temporary simulated battery voltage source connect a 2K [P3] variable pot from +5V to 0v
use the wiper voltage as the battery voltage.[ no actual battery connected at this time]

LOOP:
Set P3 , the test voltage to 1.4v
Set P1. junction of R1/2 to 2.10v
Set P2, the hysteresis pot to half its value.
[at this stage the relay should be energised,, charging]

Increase the test voltage [slowly] pot P3 output to 2.1v, close to this value the relay should be de-energised.
Slowly decrease the test voltage down to about 1.4v, the relay should be energised about this voltage.

If the difference between the On/Off voltages is not [ 1.4 and 2.1] adjust P2 the hysteresis pot.

Repeat this LOOP until the relay switches at 1.4v and 2.1v

It does set up OK and it works with the circuit I have posted.
Also a graph of measurements.

I would recommend that you use a Vref voltage source to drive the R1/2 P1 pots, this voltage MUST be stable.

Try it let me know.

 

[hardcore misery]

ok sir, that would be nice... so i can verify the status of our prototype
(mainly the status at NO LOAD condition, which the the charging voltage is 4.2V and the Vref is lower than 4.2V (2.1V or 2.05V)[/QUOTE]

hi,
Managed to get a LM339 for testing!
Will build today.

The charge voltage is 4.2v, when the charger is disconnected by the relay the battery voltage will be about 3.2v.

In your circuit you have chosen your resistor divider to give a Vbattery/2, that is,, 2.1V [when charging]
and 1.6V [when charged and relay just switched off].

From your original OP you wanted a circuit that stopped charging when the battery was fully charged?
Will the battery be removed from the charger when its fully charged?
If yes, why are you concerned about measuring and detecting with the comparator the lowest discharged voltage of a battery?

In my simulation for the battery on charge is:
from +5v to 0v I use a 2K, 20 turn variable resistor, the wiper of the variable is the simulated battery voltage.
I can adjust the variable to simulate the different battery voltage levels, this is much easier than trying to use an actual battery during testing.

we did an experiment a while ago, and arrive with the expected outputs, but there is a problem on the ACTUAL battery voltage, again, the charging voltage is 4.2V while the battery voltage of a discharge battery is 3.4V, when the Relay is ON, the voltage measured by the comparator is at approx 3.8V to 3.9V then running (which i don't know the computation why is that the voltage is 3.9V where the battery voltage is 3.8V and the charging voltage is 4.2V) when the relay is OFF, we checked the battery voltage and its not fully charged (still at 3.9V) maybe we did a mistake? is it expected? we thought that as soon as the voltage measured by the comparator exceeds 4.2V(which makes the relay OFF) the actual battery voltage is also 4.2V, is this really expected? i will try again this method..
anyway thanks sir for the effort, i will make some revision on our prototype based on your given schematic

question: what formula is used to determine the current or voltage across the base of the transistor and the RL, when i tried to use a 100k as the RL, the transistor is not driven...

so sir, would you suggest a current regulated charger? or we will stick to our charger with a current limiter?(10ohms/5W)

 

[ericgibbs]

hi,

we did an experiment a while ago, and arrive with the expected outputs, but there is a problem on the ACTUAL battery voltage, again, the charging voltage is 4.2V while the battery voltage of a discharge battery is 3.4V, when the Relay is ON, the voltage measured by the comparator is at approx 3.8V to 3.9V then running (which i don't know the computation why is that the voltage is 3.9V where the battery voltage is 3.8V and the charging voltage is 4.2V) when the relay is OFF, we checked the battery voltage and its not fully charged (still at 3.9V) maybe we did a mistake? is it expected? we thought that as soon as the voltage measured by the comparator exceeds 4.2V(which makes the relay OFF) the actual battery voltage is also 4.2V, is this really expected? i will try again this method..
anyway thanks sir for the effort, i will make some revision on our prototype based on your given schematic

Using the circuit I posted this morning, I have the project running on my bench, it works perfectly and is fairly easy to set up.

question: what formula is used to determine the current or voltage across the base of the transistor and the RL, when i tried to use a 100k as the RL, the transistor is not driven...

Do you mean a transistor base drive resistor?
If it a general purpose low.medium power transistor you would expect a forward drop from the base of the transistor to the emitter of the transistor of about 0.7V.

If the transistor has a gain of say, 50 and you required a collector current of 50mA, the Rb resistor would have to be calculated to supply 1mA from the circuit voltage driving it.
Example: if the driving voltage was +10v then 10v - 0.7Vbe = 9.3v/0.001 = 9300hm: , I would choose nearest pref value of 8K2.

so sir, would you suggest a current regulated charger? or we will stick to our charger with a current limiter?(10ohms/5W)

How many batteries do you plan to charge over what period?

 

[hardcore misery]

i forgot to tell you sir that our prototype is designed to charge 3 cellphone batteries

 

[hardcore misery]

so sir is it expected that when the voltage measured by the + feedback of lm339 reaches 4.2V(which turns off the relay) the actual voltage of the battery is also 4.2V?

 

[ericgibbs]

so sir is it expected that when the voltage measured by the + feedback of lm339 reaches 4.2V(which turns off the relay) the actual voltage of the battery is also 4.2V?

At the instant just before the charger turns off the battery will be 4.2V, the instant after the charger turns off, the battery voltage will fall to about 3.2V.

i forgot to tell you sir that our prototype is designed to charge 3 cellphone batteries

Are you trying to charge 3 cells in parallel?????

 

[audioguru]

Instead of just guessing, why don't you read a tutorial about charging Li-Ion cells?
Maxim-IC and Texas Instruments make charger ICs that have a full explanation about charging in their datasheets.

The battery cell is not fully charged when its voltage reaches 4.2V. It must remain charging at 4.2V for some time then when its current drops to a certain low amount it is fully charged.

 

[ericgibbs]

hi hardcore,
As we have been trying to point out to you, this is not a 'smart charger', it simply detects when the battery under charge, reaches 4.2V, then it stops charging.

If you need a 'smart/intelligent' charger, I would suggest a low cost PIC, with an ADC input.

 

[hardcore misery]

At the instant just before the charger turns off the battery will be 4.2V, the instant after the charger turns off, the battery voltage will fall to about 3.2V.



Are you trying to charge 3 cells in parallel?????

so therefore, the battery is not fully charged when the comparator reads a voltage about 4.2V, do we need another circuit?

nope sir, we have separate charger circuit for each cellphone battery (composed of a voltage regulator)

 

[ericgibbs]

so therefore, the battery is not fully charged when the comparator reads a voltage about 4.2V, do we need another circuit?

nope sir, we have separate charger circuit for each cellphone battery (composed of a voltage regulator)

hi,
Please read this web doucment.

C:\Documents and Settings\Administrator\Desktop\Charging lithium-ion batteries.htm

 

[hardcore misery]

Instead of just guessing, why don't you read a tutorial about charging Li-Ion cells?
Maxim-IC and Texas Instruments make charger ICs that have a full explanation about charging in their datasheets.

The battery cell is not fully charged when its voltage reaches 4.2V. It must remain charging at 4.2V for some time then when its current drops to a certain low amount it is fully charged.

therefore, a voltage comparator is not the solution? were running out of time, and our final defense is set on saturday... so how can we measure the current drops? or can we use this to compare on a certain reference voltage?

 

[hardcore misery]

hi hardcore,
As we have been trying to point out to you, this is not a 'smart charger', it simply detects when the battery under charge, reaches 4.2V, then it stops charging.

If you need a 'smart/intelligent' charger, I would suggest a low cost PIC, with an ADC input.

so if we will use PIC's we will do another circuit, and unfortunately, we don't have any background on microcontrollers...

 

[ericgibbs]

so if we will use PIC's we will do another circuit, and unfortunately, we don't have any background on microcontrollers...

hi,
Please read:
**broken link removed**

Also Google for 'charging LI ION batteries'

 

[hardcore misery]

do we need a differential amplifier?

EDITED: on what area of our circuit can we measure the current drops until the battery is fully charged?

 

[audioguru]

do we need a differential amplifier?

EDITED: on what area of our circuit can we measure the current drops until the battery is fully charged?

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.