Monitoring the status of a battery (or batteries in a series)

[Face]

I was thinking about some design ideas, and I was wondering,
How can I go about making a device (probably some kind of circuit)
that I can integrate into something else, to give it a feed of how
much battery is left? That sounds a bit unclear, so let me give an
example: I have four AA cell batteries in a series(=6V), and I'm
using them to power an LED (We'll say 3V). Obviously that would be
too much voltage, so I'd want to add a resistor. But what I'd like to
do, is use a variable resistor, such that as the batteries run out of
juice, the resistor will resist less and less, so the LED will always
have a constant source of 3V, until the batteries reach 50%, in which
case they will supply less than 3V, and the resistor will do nothing.

I don't know much about batteries, or even which part of them declines
over usage--wattage? voltage? both?

So what I would like to know, is how I can do something similar to the
scenario I explained Above.

Thanks...Any feedback on anything I asked would be greatly appreciated .

 

[MikeM]

I was thinking about some design ideas, and I was wondering,
How can I go about making a device (probably some kind of circuit)
that I can integrate into something else, to give it a feed of how
much battery is left? That sounds a bit unclear, so let me give an
example: I have four AA cell batteries in a series(=6V), and I'm
using them to power an LED (We'll say 3V). Obviously that would be
too much voltage, so I'd want to add a resistor. But what I'd like to
do, is use a variable resistor, such that as the batteries run out of
juice, the resistor will resist less and less, so the LED will always
have a constant source of 3V, until the batteries reach 50%, in which
case they will supply less than 3V, and the resistor will do nothing.

I don't know much about batteries, or even which part of them declines
over usage--wattage? voltage? both?

So what I would like to know, is how I can do something similar to the
scenario I explained Above.

Thanks...Any feedback on anything I asked would be greatly appreciated .

Read this, understand what it does.

http://www.national.com/ds/LM/LM317L.pdf

 

[Ban]

I was thinking about some design ideas, and I was wondering,
How can I go about making a device (probably some kind of circuit)
that I can integrate into something else, to give it a feed of how
much battery is left? That sounds a bit unclear, so let me give an
example: I have four AA cell batteries in a series(=6V), and I'm
using them to power an LED (We'll say 3V). Obviously that would be
too much voltage, so I'd want to add a resistor. But what I'd like to
do, is use a variable resistor, such that as the batteries run out of
juice, the resistor will resist less and less, so the LED will always
have a constant source of 3V, until the batteries reach 50%, in which
case they will supply less than 3V, and the resistor will do nothing.

What you describe is called constant current source. It is a widly used
circuit topology and can be made with a few inexpensive parts. Most simple
would be 2 resistors, 2 diodes and a transistor:

+Bat
o--------------+
| |
| D3 V
.-. - LED
R1| | |
3k9| | o-----+
'-' |
| |/
+------| T1
| |>
D1 V |
- |
| .-.
| R2| |
| 27R| |
D2 V '-'
- |
| |
| |
=== ===
GND GND

1N4148 2N2904

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
The voltage on the base of the transistor is constant regardless of battery
voltage and so is its emitter current. Of course there will be some
fluctuations with temperature, but using an opamp and a voltage reference
can increase accuracy and stability.

 

[dB]

example: I have four AA cell batteries in a series(=6V),

You have a battery consisting of four AA cells.

, in which
case they will supply less than 3V, and the resistor will do nothing.

Wrong. The resistor will still limit the current through the l.e.d.

I don't know much about batteries

Nor about l.e.ds. They are current, not voltage, operated devices.
Read this series of articles:
http://p199.ezboard.com/fbasicelectronicsfrm5.showMessage?topicID=16.topic

I don't know much about batteries, or even which part of them declines
over usage--wattage? voltage? both?

Then read about them.

 

[Face]

Read this, understand what it does.

http://www.national.com/ds/LM/LM317L.pdf

Using the design on the left middle of page 9 in that PDF, I came up
with the following (Thanks, Ban):


---------LED------
| |
| |
| |
--- |
4.5V - |
| |
| |
| |
.-----------. |
| VIN | |
| | |
|LM317L | |
| ADJ | ---------+
| | |
| | |
| VOUT | |
'-----------' |
| |
| ___ |
-----------|___|--|

????

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

What I don't know, is how do I figure out what value of resistor I
need there, if I want to give the LED 3.5V at ~20mA. Maybe I used the
wrong schematic alltogether, but I don't think that I did.

Also, that design that Ban posted:

+Bat
o--------------+
| |
| D3 V
.-. - LED
R1| | |
3k9| | o-----+
'-' |
| |/ +------| T1
| |>
D1 V |
- |
| .-.
| R2| |
| 27R| |
D2 V '-'
- |
| |
| |
=== ===
GND GND

1N4148 2N2904

If this is more useful for what I want, would you mind explaining it a
little better, I don't understand it...also, I have no idea what parts
I should get for it, like what kind of diodes, etc.

Sorry if I'm being a bit dense, I'm relatively new to this sort of
thing. And also, I apologize for any wrong terminology or phrasing
(which Db so kindly pointed out to me), for the same reason.


Thanks very much,
Face.

 

[Kevin R]

Normal LEDs have a forward voltage drop of around 1.2V.
maybe you have a couple of LEDs in series ?

The trouble with linear regulators such as the LM317 is that
if you drop the voltage by 50%, you lose at least 50% of the
power as heat. That's up to half the battery power just being
wasted. If you are using a single LED, then you will lose 80%
of the power while the supply is at 6V.

Linear Technologies make a range of switched mode regulators
specially designed for low power applications such as this.
If you could put in a little more effort and put together something like
what's on this web site, then you could probably double the battery life.
http://www.linear.com/prod/prod_home.html?product_family=power

I have used LT's regulators quite a few times and they are well documented,
widely available and reasonably inexpensive. They even have a free software
package Switcher Cad II which allows you to simulate regulator circuit designs.

Only trouble is, you'd probably have to make a PCB, I don't think they'd take
too well to vero board.

Just a thought.
Kev.

 

[Michael]

Can't locate the original post so I'm tacking this onto Kev's post.

MikeM, what you want is a pulse-width-modulated current source. Cavers
(spelunkers) use a variety of such devices in order to waste little
battery energy. Google "caving", "cavers", "spelunk" and eventually you
will hit what you're looking for.

 

[John Beardmore]

I have used LT's regulators quite a few times and they are well documented,
widely available and reasonably inexpensive. They even have a free software
package Switcher Cad II which allows you to simulate regulator circuit designs.

Only trouble is, you'd probably have to make a PCB, I don't think they'd take
too well to vero board.

Only switchers I have used have been on veroboard. Never seen any
problems.


Cheers, J/.