voltage regulation

[majorkuso]

I am on a budget let's assume I don't have any money to spend and must use the following parts huge supply of 1/8W to 1/4 resistors in all values, 3 lm317t voltage regulators, fans, and heat sinks. i am trying to get a power supply that outputs 14.6 volts at 2.5 amps down to 7 volts at .1 to .5 amps to charge a battery.

One solution I am thinking is putting two regulators in series and have each one output down to .5 amps out, operating as current sources. then the third adjusts the voltage to 6.9 to 7.

An alternative would be to take several resistors and connect them in parallel to divide current up but how would i get the number of resistors needed.

[Balaur]

One solution I am thinking is putting two regulators in series and have each one output down to .5 amps out, operating as current sources. then the third adjusts the voltage to 6.9 to 7.

Que?

It doesn't work like that. You achieve absolutely nothing by connecting two current sources in series. Why you would connect a third one (voltage regulator? he, he) in that configuration?

You cannot control both the current and voltage at the same time. You can control (set) one and limit the other.

In any case, I assume that you are trying to recharge some kind of lead-acid (maybe sealed) battery? If you want to keep the battery in good health, please familiarize yourself with the recharging algorithm.

Basically, you have two main approaches for the battery recharging, based on their intended usage: cycle charging for applications where the battery will get quite an exercise and top-up charging where the battery is an auxiliary power source.

For cycle charging, you can either do constant voltage or constant voltage/constant current. Let's  see:

I) Constant voltage
-  apply a nominal voltage (let's say 2.45 V) per cell until the measured charge current is stable for a few hours.
In your case, you can build a parallel voltage regulator with your LM317s (1-3 depending on the maximal current requirements) BUT you need to monitor in some ways the charge current to cut the charging process when the battery is full. If your source doesn't have enough current capability, it's not that nice: the supplied voltage will be lower at first (and the poor LMs, transformer will be very hot) and rising until the regulating voltage is met. Recharge time will be longer.

A saner/safer implementation of this scheme is the following:

II) Constant current / Constant voltage
- you start with a constant current (let's say 0.4C) and monitor the battery voltage until it reaches the 2.45V per cell (it takes ~2hours). Then continue for 3-4 hours with a constant voltage of 2.45V/cell
In your case, that means that you use one of the LMs as a constant current source, recharge the battery with that and then switch to another circuit configured as a constant voltage.

This scheme is safer in the sense that the maximal  required current is always limited to a given value.

III) Two-step constant voltage and trickle charge are also possible.

All this to say that if you care about the health of your batteries, please charge them safely. Most methods require some additional control or monitoring. The constant current / constant voltage method is the most reliable but your charging circuit needs to monitor the voltage to detect when to change the charging method.

You can also find good and relatively affordable chargers.

[Rerouter]

well you never actually said what your supply voltage was, but lets just start off in the ballpark, 14.6 and 7V seems to be lead acid battery voltages, though i may be wrong, if so correct me (i wont be able to talk you through a 2 part lithium charger nor would i reccomend it with those parts)

if so you first make use of some of your resistors to drop the voltage at the input of the regulator to 16V at your maximum current, this then gets split between 2 of your regulators minimum (each can handle 1.5A max) and will feed through current to try and match the feedback voltage, note if you want both constant current and voltage, you need either a transistor, an op amp or a 4th regualtor, there is no escaping it,

your buffer resistance on the output of each of your regulators has to be 0.2 ohm minimum,

and if you looked over the datasheet that not only describes half of this but also how to implement using the same 2 feedback resistors across 3 regulators,

http://www.ee.buffalo.edu/courses/elab/LM117.pdf

[majorkuso]

The voltage from the dc source is 14.6 at 2.5 amps the batteries are lithium polymer  they are used for a deer feeder and they are rated at 6.9 volts  4.6AH I  have some more lm317 10 total so adding a fourth won't hurt I know i could just buy one, but that would defeat the purpose of the learning experience. I am thinking of limiting the current and allowing the voltage to stay at 7.0. I appreciate the help you guys are giving me.

[PSR B1257]

adding a fourth won't hurt

Now it is getting absolutely absurd.

Two regulators in one circuit is the reasonable maximum. One for current and one for voltage limiting.

[majorkuso]

Maybe i am over thinking the whole thing I have managed to find a better dc power supply operating at 12.4 volts I will probably use it instead at least it is closer to the targeted 7 volts.

adding a fourth won't hurt

Now it is getting absolutely absurd.

Two regulators in one circuit is the reasonable maximum. One for current and one for voltage limiting.

so what your saying is that i should just use one as a current source and the other to limit the voltage.

[PSR B1257]

use one as a current source and the other to limit the voltage.

That's a typical application for linear power supplies for instance, but not essentially for charging LiPo's. In theory it should work, but it is easier to use a IC which is designed to do this.

They need a constant current at the beginning, till they reach there nominal voltage. Then they are charged with these constant voltage.

[majorkuso]

Ok what one would you recommend in terms of a charger ic  I set the current with one and used the second to set the voltage to 7 the output is 7v at .145A  oh and I forgot to update the batteries are sealed lead acid

[majorkuso]

Ok when I try to charge the battery the voltage drops to about 5.54 volts I included a pic of the schem as a voltage source  how can i modify it to limit the current to .7A

[ptricks]

Save yourself the headache and pick up a cheap transistor you can use as a pass transistor on the lm317. You can probably get one for under $1 USD, or in an old stereo, tv, or power supply. LM317, resistors , transistor and capacitors is all you need for a high current LM317.
[To limit voltage and current you need an LM317 configured for voltage mode, then a second LM317 configured for current mode connected to the output of the first.
http://diyaudioprojects.com/Technical/Voltage-Regulator/

[majorkuso]

is this what you mean when you set the second to current limiting

[ptricks]

is this what you mean when you set the second to current limiting

yep that is voltage regulation on the first and current regulation on the second.

[majorkuso]

ok so now i need to find a resistor or potentiometer that can handle the wattage I need to see if i have any for the current regulation

[Bored@Work]

If you go that way (I wouldn't, but if):

1. First the current limit, then the voltage regulation. Otherwise you won't get proper voltage regulation due to the drop the current limiting LM317 causes.

2. Using a pot to adjust the current limit in that particular circuit is a bad idea. You need a rather beefy one. If you want to stick with the LM317 for current limiting, there are some circuits around which work around that requirement (an adjustable voltage reference is placed in the feedback loop. The reference is adjusted with a normal, low wattage pot then). IIRC some LM317 datasheets contain this circuit as an example.

[majorkuso]

I remembered that I had a couple of 6.8 ohm 1 watt flameproof resistors I put two in parallel not only does that give me the headroom for dissipation, but it will work for my values too.

[ptricks]

2. Using a pot to adjust the current limit in that particular circuit is a bad idea. You need a rather beefy one. If you want to stick with the LM317 for current limiting, there are some circuits around which work around that requirement (an adjustable voltage reference is placed in the feedback loop. The reference is adjusted with a normal, low wattage pot then). IIRC some LM317 datasheets contain this circuit as an example.

A pot placed between the adj pin and the resistor will work. The amount of current used by the adj pin isn't even 1ma

[Mechatrommer]

I remembered that I had a couple of 6.8 ohm 1 watt flameproof resistors I put two in parallel not only does that give me the headroom for dissipation, but it will work for my values too.

say paralleling two you get 3.4ohm 2watt. your spec is 2.5A, thats 7V across resistor, ie more than 14W, enough to toss your bulletproof resistors from inside in no time. we are not yet talking how bulky the potentiometer is if you decide to put it there.

[IanB]

The voltage from the dc source is 14.6 at 2.5 amps the batteries are lithium polymer  they are used for a deer feeder and they are rated at 6.9 volts  4.6AH I  have some more lm317 10 total so adding a fourth won't hurt I know i could just buy one, but that would defeat the purpose of the learning experience. I am thinking of limiting the current and allowing the voltage to stay at 7.0. I appreciate the help you guys are giving me.

If you have lithium polymer batteries to charge you should purchase a purpose designed lithium ion charger or if you really want to make a charger then use a proper lithium ion charger IC.

If you try to construct a charging circuit for big lithium polymer batteries and you are not an electronics expert you may cause the batteries to explode. You will be out of pocket for a new battery, and if you happened to have the batteries indoors you might have fire damage as well. People have burned their house down doing such things.

Watch this and be careful:

[T4P]

Yeah, balance chargers are the right tool for the job