Need a charging circuit for 12V 7.2Ah Amaron Quanta Lead acid Battery

Dear Sir,

I am using 12V 7.2Ah Amaron Quanta lead acid battery for an Emergency Lamp. I am using the following charging circuit which is attached with this post. In this charging circuit, 230 ac voltage is given as input to the 16-0-16 transformer. The output of transformer is filtered through rectifier diodes and then given as input to the LM317 of charging circuit. As of now transformer is needed to step down the ac input voltage. I need a charging circuit without a transformer. So kindly suggest me a circuit which consists of input rectification without a transformer. Please suggest it with a moderate cost. And also I need one more clarification. I am using "Constant Voltage" - trickle charging method to charge the battery. I am using a load of 2 lamps of 110 W lamps which stand for around 50 minutes of full charged battery. The lamps will cut-off at 9.5-9.8 V. Then it takes around 16 hours to make the battery full charge. Could you please confirm me if I can charge the battery soon (min 6 hours). Is so at what rate I should use the load. Please help me in this situation.

One of the advantages of a transformer is that it prevents your lead-acid battery from being in direct electrical contact to mains power.

Mains is high voltage. A transformerless power supply would conceivably make it that much easier for a spark to ignite hydrogen fumes in or around the battery, with a possible explosion.

Furthermore a coil (as used in a transformerless power supply) can generate a high voltage kick when current flow is shut off suddenly. Again the danger is a spark igniting hydrogen fumes. A diode is usually installed across a coil to absorb the high voltage kick.

I'm not sure about commercially-manufactured transformerless chargers for lead-acid batteries. If they're US-made then UL regulations would stipulate safeguards to prevent the above hazards. Do you know how to install such safeguards? I don't.

Hello tvibakar,

I think, there are some faults in your discription.

1. You write, that you are charging the battery with a LM317. This regulator has an output of 1,5A. This is too much for the charghing of a battery of 7,2Ah. The better current for charging will be 720mA.

2. You write that you are using 2 lamps of 110W on this battery. So the battery has 18,33A to deliver. For 50 minutes of using, the battery must have a capacity of app. 16Ah.

3. The cut-off-voltage is also too low. The minimum cutt-off-voltage is 10.5V, else the battery will damage.

4. Your wish to charge the battery into 6 hours will be possible, but it will also shorten the life of the battery.

5. Transformerless circuits are very dangerous, and unsuitable for lead acid batteriey. This batteries can not charge with constant current, and transformerless circuits work with constant current.

The circuit you need must control the charging, the cut-off-voltage and must switch on the lamps, when the powerline is off.

Later on I will post you a compl. circuit, I you wish this.

Best regards

Rainer


PS: Sorry, if my English is not so good.

rfredel said:

Hello tvibakar,

I think, there are some faults in your description.

1. You write, that you are charging the battery with a LM317. This regulator has an output of 1,5A. This is too much for the charging of a battery of 7,2Ah. The better current for charging will be 720mA.

2. You write that you are using 2 lamps of 110W on this battery. So the battery has 18,33A to deliver. For 50 minutes of using, the battery must have a capacity of app. 16Ah.

3. The cut-off-voltage is also too low. The minimum cut-off-voltage is 10.5V, else the battery will damage.

4. Your wish to charge the battery into 6 hours will be possible, but it will also shorten the life of the battery.

5. Transformerless circuits are very dangerous, and unsuitable for lead acid battery. This batteries can not charge with constant current, and transformerless circuits work with constant current.

The circuit you need must control the charging, the cut-off-voltage and must switch on the lamps, when the powerline is off.

Later on I will post you a compl. circuit, I you wish this.

Best regards

Rainer


PS: Sorry, if my English is not so good.

Click to expand...

Thanks for your reply Rainer. Actually we are using 2 lamps of 55W each producing totally around 110W standing for 50 mts. We use halogen lamps and it produces much heat. Could you please specify any lamps which consume less power and more illumination. For a 12V 7.2Ah lead battery, atleast the lamps should glow for 2 to 3 hours. Please specify some lamps which is suitable for the above condition. And also please post the charging circuit without the application of transformer using lead acid battery. If lead acid battery is not suitable, pls specify any other battery which comes under moderate cost.

Hello,

I will try to help you, but I stay in Germany and I don't know, what types of Lamp are aviable in your country. What kind of Lamp are you using now? Can you make a picture?

Could you please specify any lamps which consume less power and more illumination

Click to expand...

I can look for lamps here and give you the specification. But therefore I need more informations, like is it outside or inside.

For example you have a halogen bulb like this 12V / 50W you can change it into a LED bulb like this . The LED needs only 12V / 4W, and it's only a little bit darker.

also please post the charging circuit without the application of transformer using lead acid battery

Click to expand...

You get the circuit diagram from me, but it takes a 2-3 days, because I must draw it. I have calculate the circuit without a transformer, but it will be more expensive then with transformer.

Without a transformer you must rectify the line voltage, and you need a voltage regulator to get 18V for charging.

I think you have a transformer for the 2 lamps. You can use ist also for charging. If you see my circuit, you will see that it is easier with a transformer then without it.

So, I will start to draw the diagram.

Best regards

Rainer

rfredel said:

Hello,

I will try to help you, but I stay in Germany and I don't know, what types of Lamp are aviable in your country. What kind of Lamp are you using now? Can you make a picture?



I can look for lamps here and give you the specification. But therefore I need more informations, like is it outside or inside.

For example you have a halogen bulb like this View attachment 76752 12V / 50W you can change it into a LED bulb like this View attachment 76754. The LED needs only 12V / 4W, and it's only a little bit darker.



You get the circuit diagram from me, but it takes a 2-3 days, because I must draw it. I have calculate the circuit without a transformer, but it will be more expensive then with transformer.

Without a transformer you must rectify the line voltage, and you need a voltage regulator to get 18V for charging.

I think you have a transformer for the 2 lamps. You can use ist also for charging. If you see my circuit, you will see that it is easier with a transformer then without it.

So, I will start to draw the diagram.

Best regards

Rainer

Click to expand...

Dear rainer,
Its almost more than a week we chat. May I know about the circuit you drawn which fullfill my limitations. I am expecting your reply Rainer.

Dear tvibakar,

sorry, that I don't answer in time, but I stay in hospital for 5 days. I ask my son to send you a PN, but I think you don't receive it.

The circuit is finsh draw, I only must write how to adjust it, because my discription is in German. I think it's not so expensive, because their no special devices in it.

I hope I can finish it today afternoon.

Best regards

Rainer

rfredel said:

Dear tvibakar,

sorry, that I don't answer in time, but I stay in hospital for 5 days. I ask my son to send you a PN, but I think you don't receive it.

The circuit is finsh draw, I only must write how to adjust it, because my discription is in German. I think it's not so expensive, because their no special devices in it.

I hope I can finish it today afternoon.

Best regards

Rainer

Click to expand...

Dear Rainer,

Sorry for troubling you again. Pls post the circuit you drawn. I am in a emergency situation. Kindly post your circuit soon

Hello Thomas,

I'll send you first the circuit drawings and a short discription.

Picture No. 1 shows the compl.circiut. With S1 you switch on the emergency lamps with the relays Rel1 and Rel2. It starts also the transformer from Line. So Relay Rel3 can switch up to line. If linevoltage is lost, Rel3 switch down to battery.

Page 2 shows the charging circuit. The current is limited to 720mA for your battery. So it take 10 hrs. for full charging. During charging the LED1 (green) will light.
To adjust the end-of-charge voltage put a resistor of 470 Ohms instead of the battery. Adjust P1 so, that an voltage of 13,8V is across the resistor. When the Battery reach this voltage, the Battery is fully charge and the charge will stop charging. If the voltage drop down, the charging start again. So the battery is hold on full charging.

Page 3 shows the deep discharge protection circuit. It protects the battery for low voltage. If the battery voltage drop to 10.7V the relay REL4 switch off the current to the lamps. To adjust the breaking point put a voltage of 10,7V to point 5 and 6. Turn P2 clockweise to end. Push the button S2 and the turn P2 ccw until the relay switch off and LED3 don't light.
Every time, when the battery was empty, you must push the button S2 after charging to activate it again.

I hope, it will help you. If you have more questions or you want a complete discription, contact me. You can do it here or with PM or mail.

Best regards

Rainer

rfredel said:

Hello Thomas,

I'll send you first the circuit drawings and a short discription.

Picture No. 1 shows the compl.circiut. With S1 you switch on the emergency lamps with the relays Rel1 and Rel2. It starts also the transformer from Line. So Relay Rel3 can switch up to line. If linevoltage is lost, Rel3 switch down to battery.

Page 2 shows the charging circuit. The current is limited to 720mA for your battery. So it take 10 hrs. for full charging. During charging the LED1 (green) will light.
To adjust the end-of-charge voltage put a resistor of 470 Ohms instead of the battery. Adjust P1 so, that an voltage of 13,8V is across the resistor. When the Battery reach this voltage, the Battery is fully charge and the charge will stop charging. If the voltage drop down, the charging start again. So the battery is hold on full charging.

Page 3 shows the deep discharge protection circuit. It protects the battery for low voltage. If the battery voltage drop to 10.7V the relay REL4 switch off the current to the lamps. To adjust the breaking point put a voltage of 10,7V to point 5 and 6. Turn P2 clockweise to end. Push the button S2 and the turn P2 ccw until the relay switch off and LED3 don't light.
Every time, when the battery was empty, you must push the button S2 after charging to activate it again.

I hope, it will help you. If you have more questions or you want a complete discription, contact me. You can do it here or with PM or mail.

Best regards

Rainer

View attachment 77175View attachment 77176View attachment 77177

Click to expand...

Thanks for your circuit Rainer.

Nice Circuit...Could you please explain that how in circuit 2, current is limited to 720mA.

Hello cmsrlabs,

the current is limited by R4 and T2. R4 is calculated that 0,7V drop on it. With this voltage between B and E of T2 this transistor will switch on and reduce the B to E voltage of T1. So the output voltage drop and the current will constantly 0,72A. R4 is not very critical. Here I have take 0,82Ω, but you can also take 1Ω. In my case the current is a little bit higher then 0,72A, in the other case a liitle bit less.

The calculation for R4 is R = U_be / I_load.

T3 stops allso the output voltage in case of wrong polarity.

LED1 shows if loading current is flow. The voltage drop at D6 and R7 switch on T4 if loading current passed D6.

LED2 only lights up, if the polarity of the battery ist wrong.

With P1 you adjust the full loading voltage of the battery (13,8V or 14,1V).

Regards

Rainer

Thanks for explaining. but what is the need of using T1 and R3.

How Output remains at 13.8 Volt after passing through transistors.

- - - Updated - - -

Also please tell me can we add trickle charging in above circuit with or without using Microcontroller

Hello cmsrlabs,
T1 is the regulation transistor to give voltage and power to the output. R3 is the resistor from the output of the regulation IC to the base of T1. R3 limits the base current.

In this moment I see, that I have forgot one connection from rectifier (-) to ground. Look for the red correction in my attached circuit diagram.

How Output remains at 13.8 Volt after passing through transistors.

Click to expand...

That is no problem, because their are 25V at collector of T1.

A special circuit for tickle charging is not necessary, becaus the IC controlls the output (battery) - voltage instandly. As long as the voltage is 13,8V (14,1V) the IC give no voltage to the base of T1. If the voltage dropped, T1 will open and charging starts again. The charging current is dependent frrom the battery voltage. At the lowest battery voltage the max. current, calculate by R4 = U_beT2 / I_load, will flow.

Thanks for your reply. I will try this circuit soon.

What should be the Charging Cut Off Voltage during charging and Low Cut off voltage during discharging to avoid Overcharging and Deep Discharge Mode for 12 Volt 7A Lead Acid Battery.

I have also read that we should follow charging algorithms to charge Lead Acid Battery and also to take care of temperature. Is there any need to follow this.

Hello cmsrlabs,

the charging cut off voltage to avoid overcharging is 13,8V. This voltage is to adjust with P1.

The deep discharge voltage is 10,8V. A circuit to avoid deep discharge is shown in diagram 3. If you need a circuit discription, please contact me.

If you charge lead acid batteries with low current, like this one (1/10 of capacity), you must not take care of the temperatur, only you use quick charging with high current (2 or more times of capacity). Then the battery gets hot (cooking of acid) and can going exploded. The temperature is not depended to the charging condotion.

Temperature control is only necessery to NiCd, NiMH or Li batteries for quick charging.

Regards

Rainer

tvibakar said:

Dear Sir,

I am using 12V 7.2Ah Amaron Quanta lead acid battery for an Emergency Lamp. I am using the following charging circuit which is attached with this post. In this charging circuit, 230 ac voltage is given as input to the 16-0-16 transformer. The output of transformer is filtered through rectifier diodes and then given as input to the LM317 of charging circuit. As of now transformer is needed to step down the ac input voltage. I need a charging circuit without a transformer. So kindly suggest me a circuit which consists of input rectification without a transformer. Please suggest it with a moderate cost. And also I need one more clarification. I am using "Constant Voltage" - trickle charging method to charge the battery. I am using a load of 2 lamps of 110 W lamps which stand for around 50 minutes of full charged battery. The lamps will cut-off at 9.5-9.8 V. Then it takes around 16 hours to make the battery full charge. Could you please confirm me if I can charge the battery soon (min 6 hours). Is so at what rate I should use the load. Please help me in this situation.

Click to expand...

At first you should take adequate battery for this job, for sure 7,2Ah is not good for 110W. Full capacity of 7,2Ah of lead acid is declared by at least C/10 or C/20 (your case) discharge time (7,2Ah discharging in 10h by 0,72A). Additional you put battery on torture discharging it below 10,5V maybe you kill bat, but for sure you drastically decrease her life. Maximum allowed discharge current and charging current is given in manufacturers datasheets, I suggest you to use this papers.

Your battery capacity of 7,2Ah is declared for C/20 !!!!.
That means guaranted 7,2Ah at 360mA load during discharge in 20h of time!
According to manufacturer data, your battery can give you max 4,5Ah capacity to 10,5V level, if you want to discharge it in 1h time.
All of this at 27C with fresh new battery.

I dont know which battery you discharge with 110W (9A current) for 50min but 7,2Ah 12V lead acid is not for sure, or maybe you mix with some other battery. :wink:

Your battery datasheet is in file named "Quanta 12V 7,2Ah.pdf" it say clearly that you can get for 10,5V in 1h discharge time up to 4,5Ah (we speak for new battery). And max continious current for 60min is around 4,5A or around 50W.

How you get 50min with load 110W on 7,2Ah?

In best case you will get around 17min with new fresh battery! And after that first shot your battery can not be considered as healthy because you didnt comply with the manufacturers recommendations about initally and continous discharging current!. :roll:

When I read first post I have feeling and realize that something is wrong and we have fishing story, Lead Acid dont have that amount of energy density for that small battery 7,2Ah.



Charge this battery with controlled current 0,7A initially current can go up to 1,8A not more. Life time for this small batteries are in range from 3 to 5 years, in manufacturer declared charging/discharging/temperature recommendations. Take in mind like manufacturer datasheet says, this battery give best results in capacity at 27C temperature, keep this battery away from some heat source, specially from transformers cores.


My advice like to friend is to keep away from fast chargers, its smart but only for battery sellers rest is fishing/hunting stories.

Did you know that mobile Li-Ion 3,7V can be charged fast with direct on 12V car lead acid ?
I dont joking this is for real, phone works, bat works! But what happen later with this battery, what is good bat operation life,....

Use LED lights tech like previous thread posters advise. For working and living environment choose warm white leds light.

Keep manufacturers recommendations for design.



Why I always talk that lead acid should be used as float battery, see this diagram (diagram is from your battery datasheet), look what is life if you discharge it max up to 30% :




Constant Voltage Charging diagram :




:wink:



Other Manufacturer Battery Datasheet as Examples:

Hello

I have designed above circuit and control deep discharge 10.8 Volt with PIC Microcontroller.

The Battery Charges starts from current 750ma and goes on decreasing as voltage increasing on Battery and becomes zero after battery fully charged. From Voltage 13.2 to 13.8 the current is very low from 0.10A to 0.03A. It also takes long time to reach current to 0.03A.

Is the above value are correct as per requirement of Lead Acid Battery.

Also how to predict the life of battery in years.

- - - Updated - - -

Also should we require any special pure DC power Supply to charge battery or our simple bridge with above circuit maintains the life of battery. Is there any need to design SMPS to design Power Supply for battery charging.

Hello cmsrlabs,

no, you don't need a SMPS only transformer and bridge rectifier, like I draw it.

It's right, that the charging current dropped down, after reaching 13,2V. It's a protection, that the battery will not be overcharge. So you must not take care for overcharging and the charger can be connect everytime. With this low current the self discharging of the battery will be eliminate.

I can not predict the lifetime of the battery. I am not a prognosticator and my cristal ball to look for the future is for repair. ;-) :grin:

Regards

Rainer

:smile:
actually i want to define the lifetime of product so that i can decide warranty etc.

For this i planned to decide life of components present in project and then find out the average life of all components to decide life of whole project.

Any idea to decide the warranty etc.