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Lead acid battery

Bluejets

Oct 5, 2014
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Actually I was having this module but didn't know that it can limit the current after the battery has reached the full charge.Thanks for the help.

Think you will find there are a couple of different versions.
You will have to check whatever module you have to be absolutely certain.

This three pot version does for sure.

 

Karthik rajagopal

May 9, 2016
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Think you will find there are a couple of different versions.
You will have to check whatever module you have to be absolutely certain.

This three pot version does for sure.

Thank you so much for the help. Buying one saves time rather than building one. Made my work easier. :)
 

Harald Kapp

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Could never start to build one for that, besides I always have my time better spent on other things.
DIY is not always about saving money or time. It is also a matter of self satisfaction having achieved a goal by yourself.
 

Bluejets

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Yes, I've seen that response previously but quite obviously I simply do not have the time to waste. Too many other things on the go.
 

WHONOES

May 20, 2017
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DIY is not always about saving money or time. It is also a matter of self satisfaction having achieved a goal by yourself.
Couldn't agree more. You learn so much tackling even simple projects.
 

gorgon

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You should be aware of the fact that the charge voltage is dependent on the ambient temperature.
This means that the higher the ambient temperature is, the lower the charge voltage. I don't know your environment and temperature, but if it much higher than 25 C you should take this into the calculation.
 

WHONOES

May 20, 2017
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Could probably provide a circuit diagram for a 6V charger if you are interested.
 

WHONOES

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OK. It will take a few days, but I will see what I can rustle up.
 

Karthik rajagopal

May 9, 2016
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OK. It will take a few days, but I will see what I can rustle up.
Hello sir,
I tried out my own circuit but not sure whether it would work.Ihave attached the image below . Plz tell me if there any modifications or errors in my circuit .IMG_20190520_193916.JPG

Here I have connected the first lm317 to act as constant voltage source (6.8v)and the second one for the constant current source (0.5A). The comparator has its non inverting terminal connected to the battery and it's inverting terminal to the variable Zener diode. I have set the Zener diode to have a reverse breakdown voltage of 6.6v. So when my battery reaches a voltage greater than 6.6v , the opamp turns the relay on there by connecting the battery to the constant voltage source . Here I have the voltage fixed at 6.8v, so that I can use it as a float charger connected indefinitely to the battery.
 

Audioguru

Sep 24, 2016
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The LM317 current regulator has a voltage loss of about 3V (the dropout voltage of 1.9V plus the 1.1V across 2.2 ohms) so it should come before the voltage regulator. The voltage regulator also has a dropout voltage of 1.9V so the adapter voltage should be about 12.8VDC.

You are charging with unlimited current (the maximum current from the LM317 voltage regulator but you should read about charging a lead-acid battery mainly with a current regulator at www.batteryuniversity.com .

When the relay turns off then it produces a high voltage that will zap the BC547 transistor. Add a diode across the coil. Why use a relay anyway??

Another problem you have is that you are sensing the voltage of the charging battery. When it reaches 6.6V then it is only about half-charged. The current must be sensed and when it drops to a steady low amount then the battery is fully charged.
 
Last edited:

Karthik rajagopal

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The LM317 current regulator has a voltage loss of about 3V (the dropout voltage of 1.9V plus the 1.1V across 2.2 ohms) so it should come before the voltage regulator. The voltage regulator also has a dropout voltage of 1.9V so the adapter voltage should be about 12.8VDC.

You are charging with unlimited current (the maximum current from the LM317 voltage regulator but you should read about charging a lead-acid battery mainly with a current regulator at www.batteryuniversity.com .

When the relay turns off then it produces a high voltage that will zap the BC547 transistor. Add a diode across the coil. Why use a relay anyway??
Thanks for the reply .
If I connect the current regulator before the voltage regulator ,then how will I switch from CC to CV ? I used the relay to switch from CC to CV mode . I forgot to add a flyback diode in the diagram. I had already seen these charging information on battery university but I am not able to get it. That's a lot of information . It would be great if you guide me in that. Where does that 1.9v loss occur in the circuit ?
 

Karthik rajagopal

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Another problem you have is that you are sensing the voltage of the charging battery. When it reaches 6.6V then it is only about half-charged. The current must be sensed and when it drops to a steady low amount then the battery is fully charged.
Let me assume that the low amount of current that the battery gets to when it is fully charged is 10mA. Then I can use a 2.2 ohm 5w resistor in series to generate a voltage drop of 0.022v. This can be fed to an op amp in differential amplifier configuration with a gain of 100( I am not sure whether it can provide a gain of 100). Then I will get an output voltage of 2.2v. This can be fed to the second op amp which acts as comparator with a reference voltage of 2.1v. This will activate the relay when fully charged. Hope my theory is right .
 

Audioguru

Sep 24, 2016
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The datasheet of an LM317 has a graph showing its "dropout voltage" which is about 1.9V. The dropout is when the input voltage is only 1.9V higher than its output voltage and it is already doing a poor voltage regulation. The datasheet shows at least 3V from input to output for its excellent voltage regulation so that it is far from dropout. The current regulator also has this 1.9V dropout voltage plus the voltage lost across its resistor.

You do not need an amplifier plus a comparator since a comparator can have a lot more gain than 100. But a comparator (and most opamps) have an input offset voltage of up to 10mV so your selected 22mV detection might be 12mV or it might be 32mV.
Why such a low 10mA detection and why such a low 0.22 ohms resistor? Use 50mV and 1 ohm instead.

The type of battery, how much acid is in it, the temperature and other things determine the voltage of a charged battery.
The Battery University says the voltage range per cell (your 6V battery has 3 cells).

A lead-acid battery is antique. Does anybody use them anymore so that a company can make a battery charger IC for them?
 

Karthik rajagopal

May 9, 2016
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The datasheet of an LM317 has a graph showing its "dropout voltage" which is about 1.9V. The dropout is when the input voltage is only 1.9V higher than its output voltage and it is already doing a poor voltage regulation. The datasheet shows at least 3V from input to output for its excellent voltage regulation so that it is far from dropout. The current regulator also has this 1.9V dropout voltage plus the voltage lost across its resistor.

You do not need an amplifier plus a comparator since a comparator can have a lot more gain than 100. But a comparator (and most opamps) have an input offset voltage of up to 10mV so your selected 22mV detection might be 12mV or it might be 32mV.
Why such a low 10mA detection and why such a low 0.22 ohms resistor? Use 50mV and 1 ohm instead.

The type of battery, how much acid is in it, the temperature and other things determine the voltage of a charged battery.
The Battery University says the voltage range per cell (your 6V battery has 3 cells).

A lead-acid battery is antique. Does anybody use them anymore so that a company can make a battery charger IC for them?
Thank you so much for this crystal clear explanation . That was truly helpful .
I assumed 10mA as I don't know the current drawn by a Battery when it has attained it's full charge state .
I used a comparator with the amplifier because setting a reference voltage for the comparator will be difficult . Let's say I have a voltage drop of 0.11v across a 2.2ohm resistor for 50mA ,then setting a reference like 0.12v is not possible . So multiplying by a gain of 50 we get 5.5v. Now I can set my reference using a multi turn pot to 5.4v or use an adjustable Zener diode to set the reference .IMG_20190521_010604.JPG
So, if this circuit is correct ,how will I switch to CV mode for float charging?
 

Audioguru

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You did not read what The Battery University said. Constant current until the upper voltage limit is reached (you must detect it) then keep that voltage with a voltage regulator until you detect that the charging current has dropped low, then switch to a trickle charge.
 

Karthik rajagopal

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You did not read what The Battery University said. Constant current until the upper voltage limit is reached (you must detect it) then keep that voltage with a voltage regulator until you detect that the charging current has dropped low, then switch to a trickle charge.
When should I detect the drop in current ?
As far as I have understood ,in constant current mode I will detect the upper voltage limit ,here for me my charging voltage is 6.8v. So 6.8v will be my upper voltage limit . Then I will switch to constant voltage mode where my charging voltage will be 6.8v with a Max current of 0.5A , which is the trickle charge mode . I don't have the 3rd stage of battery charging where the voltage drops to a much lower value than the charging voltage as my charging 6.8v I think can directly be used to trickle charge.
Am I right ?
 
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