Current Regulator .... Help!!

[alesteryew]

I am doing a school project. There are a battery package with 12v and 2 AH (amphere hour - capacity) supply. Now, i need to design a circuit that will output 12v and 5A. I just came out of an idea of using LM338. Is it ok for this project? and is there any other ways to do it? Thanks

 

[audioguru]

Hi Alester,
Linear voltage regulators have a minimum voltage across them (input to output) in order to function, and because of losses. The minimum voltage is called the "dropout" voltage. the LM338 has a dropout voltage of about 2.7V with 5A of output current. So it needs at least 14.7V input to output 12 at 5A.
There are "low dropout" regulators available, but their dropout voltage may not be low enough for you.

 

[panic mode]

What are you trying to power? Maybe it is not nececary to regulate voltage.
How about using simple 5A fuse since your batteries are already 12V?

 

[Nigel Goodwin]

What are you trying to power? Maybe it is not nececary to regulate voltage.
How about using simple 5A fuse since your batteries are already 12V?

I agree, simply use the battery 'as is', it's rarely necessary to regulate a lead acid battery - unless there's some specific reason to do so, and then you regulate lower (or get far more complicated).

 

[audioguru]

Well, the title does specify "Current Regulator", so I thought that he wanted to use the LM338 as a "current-limiter", since it is rated at 5A.
But now I realize that its current-limiting has a tolerance, so a few ICs will current-limit at 5A, many at 8A, and maybe even some at 10A, and the value probably changes with temperature.
Perhaps the LM338 should be wired as a true current regulator, but that would need an additional 1.25V (across the current-setting resistor) in addition to its dropout voltage.
Or how about a self-resetting "Polyfuse"? Many old circuits even used a lightbulb as a current-limiter.

 

[alesteryew]

Thanks to all of your kindly help. I am not that strong in analog electronic. the sentence "but that would need an additional 1.25V (across the current-setting resistor) in addition to its dropout voltage.", how could i find out the dropout voltage of that LM338? is it stated on documentation? thanks. I am considering to use polyfuse..thanks a lot

 

[Roff]

You could potentially save yourself a lot of time, and possibly grief, by telling us what you are going to be powering with the 12v, 5 amp supply. You might not need any regulator.

 

[alesteryew]

Ok, the main purpose is to output 12v 5A to certain system. I am not sure what would be the system (not mentioned).

 

[checkmate]

In most applications, 12V 5A would normally mean 12V voltage supply, with a MAX current output of 5A. Note that it's not CONSTANT current output of 5A. It's no mean task to regulate both current and voltage, as these parameters are highly related to the circuit the supply is powering up. So, it's best you go and figure out your "certain system" first.

 

[BartSimpson]

Okay, it sounds as though the main thing behind this "school project" (from the viewpoint of you your teacher/lecturer) is for you to think about it,use your brain and learn something. If we do your assignment for you, then this doesn't happen.

From what you have told us.

1) You have a 12 volt battery
2) You need to provide a regulated 12v output with a current capacity of 5A

You need to consider the following.

1) The terminal voltage of a battery may be higher than 12volts when fresh.
2) The terminal voltage of a battery will fall when the battery becomes depleted.
3) A linear regulator such as the LM338 or LM7812 will require a higher input voltage to produce a 12v output. This will usually be somewhere in the order of 2-3 volts higher than the output voltage. So, the easy way isn't really an option. Which is probably why you were given this project

This Link might give you a start on batteries. Remember ... Google is your friend.

We can't do your assignment for you (that would just be wrong) but we might be able to point you in the right direction. I would suggest that you start looking for a switched-mode boost controller. It should be easy enough to design a small and simple Power Supply which could provide a regulated 12v output with 5A capacity from your battery. I would look to design it with an input range of between 8 and 14 volts. Using currently available components, you are probably going to need a simple switching controller, a coil, flyback catch diode and possible an output FET along with the odd current sense / limit resistor. You should be able to find all the info you need at the **broken link removed**

 

[ante]

Alesteryew,

To have both 12V regulated and at the same time provide 5A form a 12V 2Ah lead acid can only be made with a switched converter. The battery will drop below 12V after a very short time at 5A, and even with a switcher the battery will soon be drained.

Ante :roll:

 

[alesteryew]

Thanks to all of you. I really appreciate all ideas that given by your guys/gals. Now, i had figure out the direction for this project. Thanks