# battery charger

Discussion in 'The Projects Forum' started by Ouch, Jul 20, 2008.

1. ### Ouch Thread Starter New Member

Jul 20, 2008
4
0
Hi,
I made a 12v battery charger. It is working fine. But I need to design a current control section for 1A. It can be done with a 10 ohm resistor but wouldnt that require a 10watt resistor? So else could be done?

Apr 5, 2008
15,799
2,386
Hello,

What is the output voltage of the charger?
What is the voltage of the battery to be charged?

You could try a currentsource with an LM338. (see attachment).

Greetings,
Bertus

• ###### currentsource.jpg
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Last edited: Jul 20, 2008
3. ### Ouch Thread Starter New Member

Jul 20, 2008
4
0
The battery is a 12volt battery, to be charged at 13v. I need current limiting with easily available parts like resistors, caps, and transistor.

Thanks.

Apr 5, 2008
15,799
2,386
Hello,

With the 10 Ohm resistor you will take 10 Volts away at 1 Ampere.
This can only be done when the charger gives 22 Volts and the battery is 12 Volts.
(this puzzles me when you say the charger is 13 Volts)

So I ask you again what is the output voltage of the charger?

Greetings,
Bertus

5. ### Ouch Thread Starter New Member

Jul 20, 2008
4
0
Hello,

The battery is 12v 7A lead acid.
The charger outputs 13v and is capable of 3A, but I want to limit that to say, 1A or 700mA.

Thanks.

6. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
It's more likely that your charger is putting out somewhere between 13.2-13.6v.

If your charger has any kind of internal regulation, a 1 Ohm 2W resistor in series with the charger-battery connection would likely limit the charging current to be within the range you're looking for.

7. ### Tahmid Active Member

Jul 2, 2008
344
27
Hi,

I have also seen this 1 ohm 2watt resistor at the output of a battery charger to regulate current, in a magazine. Please could you elaborate on how you have calculated this.

When you say internal regulation, do you mean voltage regulation or current regulation?

Thank you.
Tahmid.

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8. ### Ouch Thread Starter New Member

Jul 20, 2008
4
0
Hi,

I would also like to know.

Plus, I would also like to know if, for battery overcharging, I can short the output at the charger.
A diode will be placed between the charger output and battery. A zener of 13v will detect overvoltage and will be connected to an npn transistor. When the battery voltage reaches 13.6v, the npn transistor will short the output of the charger before the diode.
Will this technique work?
Thanks for any help I can get.

9. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Actually, I meant internal voltage regulation. If it were only current regulated, then a simple resistor wouldn't work, as the voltage across the battery & resistor would be increased until the current was again up to perhaps 3A.

But frequently, chargers are built so that when battery voltage is low, they charge in a regulated current or unregulated current mode until the battery is nearly charged, then provide a "float" charge. Adding a resistor in the charge path increases the voltage that the charger "sees".

10. ### Tahmid Active Member

Jul 2, 2008
344
27
SgtWookie,

Hello,
I haven't understood how the 1ohm 2watt resistor will limit the current to 1A.

Thank you.
Tahmid.

11. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
If the battery is highly discharged (say, down to 9v) and the charger is capable of putting out 3A and regulated by current, then my initial recommendation would not apply (it was late, and I was tired).

3A across 1 Ohm results in 3V drop, and 9 Watts of power dissipation. In that case, a resistor rated for 20W of power dissipation should be selected, as that is the nearest standard size that is 2x the actual power dissipation. A 10W resistor would work, but it would get quite hot and not last very long.

As the battery reached "float charge" condition, with the charger running in voltage regulated mode, the battery would be at 12.6v and the charger around 13.4v to 13.6v. That leaves a 0.8v to 1v drop across the resistor, which results in a current of 800mA to 1A.

However, if the charger is still in current regulation mode, the resistor would not have an effect. A shunt clamping circuit or the like would be needed. A shunt clamp would be wasteful, and should be avoided.

Last edited: Jul 22, 2008
12. ### Tahmid Active Member

Jul 2, 2008
344
27
Hi,

Consider a 12volt 7ampere-hour lead-acid battery. It stands at 12V X 7A = 84VA. If load is 84VA, can the battery power the load for an hour and would the battery still remain in float charge condition?
What is the difference between a normal lead-acid battery and a deep discharge type? Performance of which is better?

Thanks.

13. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
Yes.

No.

Basically, the construction of the plates.
Normal = more plates, thinner.
Deep-cycle = fewer plates, thicker.

It depends upon the application.
An automotive-type battery will not last long if used in a deep cycle application. It's designed to power the starter motor for perhaps 30 seconds, and then be fully recharged.

A deep-cycle battery is designed to be discharged over a much longer period of time. It won't have peak output capability anywhere near the automotive type.

They're very different.

14. ### Tahmid Active Member

Jul 2, 2008
344
27
Sgt Wookie,

If the battery is still not in float condition, how would it be charged again? What could be its probable voltage level? Would it cause any harm to the battery?
In my circuit, low cut level is 11.6volt. If it is not in float condition, then, at 11.6v circuit will be stopped. So, at the low-cut voltage level, what could be the probable VA rating of the battery to be given to the load instead of 84VA?
Why is deep-cycle type battery used in solar system?
Thanks.

15. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
If the charger is supplying current at a 3A rate, but it is being discharged at a 7A rate, that is a net loss of 4A.

In order to keep the battery at float, you would have to start off with the battery in float, and supply at least enough current to power the external load.

A lead-acid 12.6v battery is considered completely discharged at 11.4v.

In lead-acid batteries, plate sulphation begins at about 12.4v. Sulphation will eventually result in the battery not being able to accept or release a charge.

If a 12.6v rated battery is at 11.6v, the battery is nearly completely discharged, and will have a short service life if not soon recharged.

I explained deep-cycle vs automotive in my last post.

Why do you think deep cycle batteries would be used in a solar system?

16. ### Tahmid Active Member

Jul 2, 2008
344
27
Hi,
Thank you for the replies.
Then, in a circuit, what should be the ideal hi-cut and lo-cut voltages for a 12v lead-acid battery?

Thanks.

17. ### Tahmid Active Member

Jul 2, 2008
344
27
Hi,
I did not understand. Please make it clear.
Thank you.

18. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
You had a 7AH battery, and subjected it to a load of 7 Amperes for an hour.

Yes, it should be able to do that IF it was in good condition AND fully charged prior to applying the 7A load.

At the end of the hour, the battery would NOT have a "float charge" - it would be COMPLETELY discharged. The voltage at the terminals would be 11.4 or less.

The battery would need to be completely recharged. At a 3 Ampere rate, this would take an absolute minimum of 2 1/3 hours.

If the battery voltage is less than 12.4v, use a high rate of charge. Since 11.4v is considered fully discharged, 12.6 is fully charged, you can extrapolate the charging time if the AH rate of the battery is known. Alternatively, the AH capacity can be determined by measuring the voltage, charging with a constant current for a period of time, then measuring the battery voltage again, and extraplating how long it will take to reach 12.6v.

At 12.6v, cut back to a 2A charge rate.

At 13.4v, cut back to a 600mA "float" charge rate.

If the battery voltage is 10.3v to 10.5v, it is highly likely that a cell is shorted or open. The battery should be re-cycled.

Last edited: Jul 24, 2008