Maker Pro
Maker Pro

Battery level tester.

J

James Meyer

Jan 1, 1970
0
Or maybe you want something to let you know how much charge is still
in the battery or how much time you've got left until it goes flat? A
battery "gas gauge" kind of thingy?

Well, boss, if that's what yuh wants, just ax fo' it an' ahm sho' some
of us ol' niggers'll jump at the chance to serve yuh.

Sho 'nuf...

Measure the current draw with a current sensor. Convert the sensed
current to an appropriate voltage and use the voltage to drive a small, toy
sized, motor with a gear train on its output. Attach pointers to a couple of
the gears and you will have an amp/hour readout similar to the gas or electric
meter on the side of your house. Reset your meter to zero before use, always
start your observing sessions with a fully charged battery, and Bob's yer uncle!

Jim
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that James Meyer <[email protected]>
wrote (in said:
An integrating amp-hour meter **is too** trivial....

What is trivial for you may well not be for an amateur astronomer. (;-)
Measure the current draw with a current sensor. Convert the sensed
current to an appropriate voltage and use the voltage to drive a small, toy
sized, motor with a gear train on its output. Attach pointers to a couple of
the gears and you will have an amp/hour readout similar to the gas or electric
meter on the side of your house. Reset your meter to zero before use,

Yeah, right! (Hint: that ain't as easy to do as to say.)
 
C

Colin Dawson

Jan 1, 1970
0
John Fields said:
---
I don't think so. What's happened is that you have fixed in your mind
what you want to hear and you don't understand what you've being told
so you've concluded that since it doesn't sound like what you want to
hear it must be bogus.
---

Maybe your right. I have locked my mind on a possible solution and refuse
to accept "No" for an answer. I'm a computer programmer by trade and find
that other developers (and me sometimes) say that something cannot be done,
only to find that with a little more thought and creativity, that there is a
way (all-be-it complicated) to accomplish whatever is was.

I do see electronics as being any different to a computer program, after all
they're not that unrelated. A program is, to put it simplistically is
basically a sequence of flipping switches. (ok ok, that's probably too
simple)
---
If you increase the number of devices demanding current from the
battery it won't "overrate" the rest of the stuff, all it'll do is
reduce the amount of time until the battery voltage drops to some
arbitrary point.
---

oops, sorry. I should have been talking Amperage, the leads that run from
the battery can carry about 40Amps (Not sure the exact thickness, but it's
the stuff that's used in model power boats to connect from the battery to
speed controller. I think it's about 4 or 5 mm thick)
---
Yes, it will. There is a voltage below which current shouldn't be
taken from the battery, and once the battery voltage decays to that
point it should either be disconnected or recharged.
---

OK, I need to get more litteral about this. At the moment, the circuit
that I'm using measures a voltage range between 11v and 14v. I know from
previous experiments that I want to recharge my battery when the voltage
drops to about 11v with no load (other than the monitor circuit). Also I
know that as I increase the load the voltage that the monitor reads is
forced lower by .x of a volt depending on the load. The higher the load,
the higher that this .x is. At the moment my circuit, when under high load,
shows the 11V reading pretty quickly when the battery is under a high load.
When I shut off the load, the voltage returns to a higher value and gives
it's true reading, which match the chemical thingy in the battery (which I
can't see in normal use as it's not illuminated, and shut in the boot of a
car)
---
Yes, it does. If you look at the discharge curves for _any_ battery
you'll find that as the rate of discharge increases (as more current
is drawn from it) the smaller its capacity becomes. The capacity of
most batteries (C) is rated in Ampere Hours, but full capacity can
only be achieved if some fraction of the one hour rate of current is
drawn, ususally C/10 or C/20 for lead-acid batteries. That means that
if you have a fully charged 100AH battery rated for C/10 and you draw
10 amperes from it, its voltage will decay to the cutoff point (say
10V for a 12V lead-acid battery) in 10 hours. However, if you take
100A from it its voltage will decay to 10V in substantially less than
1 hour. Also, since the battery's internal impedance will cause its
voltage to fall more and more as more and more current is drawn from
it, that will futher shorten the time until it reaches cutoff.

Let me put this another way. Say I've been using my battery for a while. I
will consider it completely flat when is reaches 10v either under load or
not. At the moment, my battery is reading 11.3v and is under load. My
circurit, is showing 11.3v. When I turn off the load, the battery voltage
immediatly jumps up to 11.7v. When I turn on the load it slowly returns to
11.3v. One of the devices that I'm using is a laptop. When the hard drive
is working, the current rapidly changes as the drive heads move across the
disk and data is read/written. This turns the LEDS on the circuit into
quite a good light show, which is really annoying and I want to stop it
doing that.
---
Well, you should, and that's precisely why I said that you want to
hear what you want to hear, not what's at variance with what you
believe.
---

hmm, I shouldn't have said that. It's why I built the circuit in the first
place. What I want is a steady reading, not one that makes the LED's look
like a reject from a bad SCI FI Film.


I asked for that.
---
Won't anyone give me a straight answer on how the hell to build an Ammeter
circuit, so that I can get the "BATTERY LEVEL MONITOR" to give a correct
reading?

---
Yowzah boss!!!


+-------------------+
/ |
A--+ [FUSE]
| |
[AMMETER] +----------+
| | |
+-----------+ [DRIVE] [VOLTMETER]
| | | |
[BATTERY] [VOLTMETER] +----------+
| | |
B--+-----------+ |
\ |
+-------------------+

Everything else connects (just like the drive with its own set of
wires and its own fuse) to points A and B. That is, directly to the
ammeter and the battery.

But... That's still only going to give you voltage readings and
current readings, so you'll still have to disconnect loads depending
on battery voltage.

Or maybe you want something to let you know how much charge is still
in the battery or how much time you've got left until it goes flat? A
battery "gas gauge" kind of thingy?

Yes, that's exactly what I'm trying to achieve.
Well, boss, if that's what yuh wants, just ax fo' it an' ahm sho' some
of us ol' niggers'll jump at the chance to serve yuh.

Colin Dawson
www.cjdawson.com
 
C

Colin Dawson

Jan 1, 1970
0
James Meyer said:
Measure the current draw with a current sensor. Convert the sensed
current to an appropriate voltage and use the voltage to drive a small,
toy
sized, motor with a gear train on its output. Attach pointers to a couple
of
the gears and you will have an amp/hour readout similar to the gas or
electric
meter on the side of your house. Reset your meter to zero before use,
always
start your observing sessions with a fully charged battery, and Bob's yer
uncle!

Jim

As other's have said on this thread, that idea won't really tell me when to
re-charge, as different current will change the characteristics of the
battery, so I won't actually know when the Ah usage is too much. Hence
basing the reading primarily on voltage. I do see how that would work
though.

Regards

Colin Dawson
www.cjdawson.com
 
F

Fred Bloggs

Jan 1, 1970
0
All you need to know is that you can't make a simple determination of
battery capacity while it is under heavy load. The decrease in voltage
you are seeing results from the battery internal resistance. This
internal resistance is a complicated and usually unknown function of
many parameters: state of discharge, current level, temperature, battery
age, history and method of battery recharging, manufacturing control
over component materials, day of the week...and it is transiently time
dependent too. You will be better off "wishing upon a star" than
building anything.
 
R

Rich Grise

Jan 1, 1970
0
I think everyone here has managed to completly miss the point of what I'm
asking.

I think you're missing the point of the answers.
There is enough voltage in my setup to power everything that I want to
power. The cables are thick enough to power everything that I want to
power, and then some. (I can easily double the number of devices without
ricking overrating the stuff that I've used. I've done it)

At the moment my "Battery Monitor" is actually a "VoltMeter". I don't
want a VoltMeter connected to the battery, as is doesn't tell me when it's
time to start thinking about recharging the battery.

Um, that's exactly what a "Battery Monitor" is - a voltmeter, which, by
the indicated voltage, gives a rough idea of the state of charge of the
battery.
What I want is a "Battery Level Meter".

That's a voltmeter, connected across the battery terminals.
Just because I start pulling 10A
from my battery doesn't mean that it's capacity suddenly drops, as a
VoltMeter shows.

Yes, as a matter of fact, it does. That's what the voltmeter is showing
you. When you're drawing 10 amps, and the voltage droops to 10 Volts,
_then you have reached the capacity of that battery to supply a 10 amp
load_!
I don't care what the Voltage of the battery is.

Maybe not, but you'd better learn to.
I
WANT TO KNOW WHEN I NEET TO START TURNING OFF DEVICES BECAUSE THE DAMMED
BATTERY
IS ALMOST FLAT AND MY TELESCOPE IS ABOUT TO LOOSE IT'S ALIGNMENT.

When the meter reads "Low Battery," It's pretty obvious that it's time
to start turning off equipment. The battery voltage pops up, it means
that the battery _now_ has the capacity to drive the _remaining_ loads,
until it droops to your "Low Battery" level again.
(at
this point Colin has thrown his teddy out of the cot)

This explains a _HELL_ of a lot.
Get the point now?

I got the point two rounds ago. You're not listening.
Won't anyone give me a straight answer on how the hell to build an Ammeter
circuit,

You never asked anybody for an ammeter circuit. And an ammeter doesn't
measure the battery's level anyway - it measures how much current you're
drawing from it.
so that I can get the "BATTERY LEVEL MONITOR" to give a correct
reading?

They're not directly related. If you insist on using an ammeter, then record
its output over time, to get actual amp hours. Monitoring the voltage during
this process can be instructive.

Good Luck!
Rich
 
C

Charles W. Johson Jr.

Jan 1, 1970
0
Colin Dawson said:
John Fields said:
---
I don't think so. What's happened is that you have fixed in your mind
what you want to hear and you don't understand what you've being told
so you've concluded that since it doesn't sound like what you want to
hear it must be bogus.
---

Maybe your right. I have locked my mind on a possible solution and refuse
to accept "No" for an answer. I'm a computer programmer by trade and
find that other developers (and me sometimes) say that something cannot be
done, only to find that with a little more thought and creativity, that
there is a way (all-be-it complicated) to accomplish whatever is was.

I do see electronics as being any different to a computer program, after
all they're not that unrelated. A program is, to put it simplistically is
basically a sequence of flipping switches. (ok ok, that's probably too
simple)
---
If you increase the number of devices demanding current from the
battery it won't "overrate" the rest of the stuff, all it'll do is
reduce the amount of time until the battery voltage drops to some
arbitrary point.
---

oops, sorry. I should have been talking Amperage, the leads that run from
the battery can carry about 40Amps (Not sure the exact thickness, but it's
the stuff that's used in model power boats to connect from the battery to
speed controller. I think it's about 4 or 5 mm thick)
---
Yes, it will. There is a voltage below which current shouldn't be
taken from the battery, and once the battery voltage decays to that
point it should either be disconnected or recharged.
---

OK, I need to get more litteral about this. At the moment, the circuit
that I'm using measures a voltage range between 11v and 14v. I know from
previous experiments that I want to recharge my battery when the voltage
drops to about 11v with no load (other than the monitor circuit). Also I
know that as I increase the load the voltage that the monitor reads is
forced lower by .x of a volt depending on the load. The higher the load,
the higher that this .x is. At the moment my circuit, when under high
load, shows the 11V reading pretty quickly when the battery is under a
high load. When I shut off the load, the voltage returns to a higher value
and gives it's true reading, which match the chemical thingy in the
battery (which I can't see in normal use as it's not illuminated, and shut
in the boot of a car)
---
Yes, it does. If you look at the discharge curves for _any_ battery
you'll find that as the rate of discharge increases (as more current
is drawn from it) the smaller its capacity becomes. The capacity of
most batteries (C) is rated in Ampere Hours, but full capacity can
only be achieved if some fraction of the one hour rate of current is
drawn, ususally C/10 or C/20 for lead-acid batteries. That means that
if you have a fully charged 100AH battery rated for C/10 and you draw
10 amperes from it, its voltage will decay to the cutoff point (say
10V for a 12V lead-acid battery) in 10 hours. However, if you take
100A from it its voltage will decay to 10V in substantially less than
1 hour. Also, since the battery's internal impedance will cause its
voltage to fall more and more as more and more current is drawn from
it, that will futher shorten the time until it reaches cutoff.

Let me put this another way. Say I've been using my battery for a while.
I will consider it completely flat when is reaches 10v either under load
or not. At the moment, my battery is reading 11.3v and is under load. My
circurit, is showing 11.3v. When I turn off the load, the battery voltage
immediatly jumps up to 11.7v. When I turn on the load it slowly returns
to 11.3v. One of the devices that I'm using is a laptop. When the hard
drive is working, the current rapidly changes as the drive heads move
across the disk and data is read/written. This turns the LEDS on the
circuit into quite a good light show, which is really annoying and I want
to stop it doing that.
---
Well, you should, and that's precisely why I said that you want to
hear what you want to hear, not what's at variance with what you
believe.
---

hmm, I shouldn't have said that. It's why I built the circuit in the
first place. What I want is a steady reading, not one that makes the
LED's look like a reject from a bad SCI FI Film.


I asked for that.
---
Won't anyone give me a straight answer on how the hell to build an
Ammeter
circuit, so that I can get the "BATTERY LEVEL MONITOR" to give a correct
reading?

---
Yowzah boss!!!


+-------------------+
/ |
A--+ [FUSE]
| |
[AMMETER] +----------+
| | |
+-----------+ [DRIVE] [VOLTMETER]
| | | |
[BATTERY] [VOLTMETER] +----------+
| | |
B--+-----------+ |
\ |
+-------------------+

Everything else connects (just like the drive with its own set of
wires and its own fuse) to points A and B. That is, directly to the
ammeter and the battery.

But... That's still only going to give you voltage readings and
current readings, so you'll still have to disconnect loads depending
on battery voltage.

Or maybe you want something to let you know how much charge is still
in the battery or how much time you've got left until it goes flat? A
battery "gas gauge" kind of thingy?

Yes, that's exactly what I'm trying to achieve.
Well, boss, if that's what yuh wants, just ax fo' it an' ahm sho' some
of us ol' niggers'll jump at the chance to serve yuh.

Colin Dawson
www.cjdawson.com

Colin
What you want is called an Integrator, basicly it'll be a lowpass
filter, to filter the spikes caused by powering on and off.

Charles
 
C

Colin Dawson

Jan 1, 1970
0
You never asked anybody for an ammeter circuit. And an ammeter doesn't
measure the battery's level anyway - it measures how much current you're
drawing from it.

Didn't I? I take it that you read the original message.....

"Is there something that I can do to counter balance this high current
voltage drop?

My thought on this is to place a kind of Ammeter into the circuit, that will
adjust the value of VR1, in accordance to the amount of current drawn
through the circuit. Changes in current would would effect this part of
the circuit, and continuously trim the battery meter, so that the readout
remains stable. (and hopefully correct)"


Regards

Colin Dawson
www.cjdawson.com
 
C

ChrisGibboGibson

Jan 1, 1970
0
:

[snip]
I've got a problem with the battery meter, in that as I turn stuff on, and
the amount of current drawn increases, the meter shows a voltage drop, and
since the difference between a full and empty battery reading is about 2V
(12V= full 10V = empty) this shows a significant drop on the readout.

What you've done is prove to yourself that battery voltage is not much use as a
state of charge indicator when the battery is actually being used.

That's precisely why there's such a huge market in very expensive amp hours
counters. It's the only reliable way of doing what you want to do. A volt meter
simply won't do it.

As another poster pointed out, a *very* basic, discharge only, amphours counter
can be built relatively easily.

Gibbo
 
R

R Adsett

Jan 1, 1970
0
Why not just use a Curtis gauge? Short of an amp-hour meter thay are
AFAIK as good as they get.

Robert
 
J

John Fields

Jan 1, 1970
0
Let me put this another way. Say I've been using my battery for a while. I
will consider it completely flat when is reaches 10v either under load or
not. At the moment, my battery is reading 11.3v and is under load. My
circurit, is showing 11.3v. When I turn off the load, the battery voltage
immediatly jumps up to 11.7v. When I turn on the load it slowly returns to
11.3v. One of the devices that I'm using is a laptop. When the hard drive
is working, the current rapidly changes as the drive heads move across the
disk and data is read/written. This turns the LEDS on the circuit into
quite a good light show, which is really annoying and I want to stop it
doing that.

---
A really simple way to do that would be to partition your voltmeter so
that it only displays mutually exclusive LEDs which correspond to
three voltages; say, >11V, which would light a green LED, 11V to >10V,
which would light a yellow LED, and <=10V, which would light a red
LED. Very easy to do with a single LM339, a voltage reference, and a
handful of resistors. Interested?
---
hmm, I shouldn't have said that. It's why I built the circuit in the first
place. What I want is a steady reading, not one that makes the LED's look
like a reject from a bad SCI FI Film.

---
Well, that's a little tougher than three LEDs, for all the reasons
Fred Bloggs enumerated, but if you wanted to try it what you'd need to
do would be to monitor the current coming out of the battery an the
battery voltage while keeping track of time in order to determine the
total energy delivered into the load. Knowing how much energy was in
the battery to start with, all you'd have to do would be to
continuously subtract what was used from what was left after the last
subtraction and sound an alarm when you emptied the battery. Of
course, to do it properly you'd also have to keep track of everything
else Fred mentioned, which may really be overkill for your
application, since all you really care about is when it's getting
close to shutdown, and the yellow LED will tell you that.
 
R

Rich Grise

Jan 1, 1970
0
Didn't I? I take it that you read the original message.....

"Is there something that I can do to counter balance this high current
voltage drop?

My thought on this is to place a kind of Ammeter into the circuit, that
will adjust the value of VR1, in accordance to the amount of current drawn
through the circuit. Changes in current would would effect this part of
the circuit, and continuously trim the battery meter, so that the readout
remains stable. (and hopefully correct)"
You've thrown yourself a red herring, and have been chasing your tail ever
since.

And you seem determined to continue to do so.

I gave it my best shot.

Good Luck.
Rich
 
J

John Crighton

Jan 1, 1970
0
I read in sci.electronics.design that John Crighton <[email protected]>
wrote (in said:
I think everyone here has managed to completly miss the point of what I'm
asking. [snip]
Hello Colin,
I think I know what you are on about. You want some warning
that your battery is about to fail/run out of capacity. Is that it?

I saw a little circuit that was used to test lead acid back up
batteries, that were on constant trickle charge. Every hour
a heavy load, lasting about a millisecond was connected
across the battery bank. An op amp was used as a comparator.
One input was connected to the battery using a suitable divider
and the other input was connected to a referece voltage.

If the battery voltage drooped below a preset level when
the heavy current flowed in the load, the comparator's output
triggered a buzzer. The workers then had a warning that
one or more of the batteries in the bank was not performing
well and would be no good for back up power when required.

All this is, is a load, which he already has, and a 'voltmeter' that has
just a 'go and a 'no-go' mark. He's already got a better voltmeter than
that.

Hello John W you have missed the point of my suggestion.
Colin doesn't have a high current load tester which in the blink
of an eye can check the fitness of the battery to continue its job.
It is a warning device not a voltmeter. Read what I wrote again.

There is absolutely no point in all this complication.

Yes there is. John W. you just haven't seen the point yet.
By zapping the battery with a quick 1 millisecond heavy
load, the battery's fitness to continue is checked.
There is nothing complicated about a resistor and a
high power mosfet, switched by a couple of 555s
supplying a 1msec pulse every 10 minutes?
If the triggerd buzzer doesn't sound, Colin will know
that the battery is still fit to continue.


He can already do this with the loads he's got. He doesn't need another
contrived test.

John W, How can he perform a high current test with
the loads that he has? You are saying this because you have
missed the point of the rapid discharge test. The load he has
is a light load. You can not do a quick high current test with a light
load, You are not understanding what I said. Read what I said
again. I am suggesting using a very heavy load directly across the
battery terminals for a very short period of time (1mSec) which will
check the fitness of the battery to continue.
You haven't read what I said, just pooh poohed without reading
carefully.

What the original poster "needs" is up to him. You sound a bit
like a school master dismissing peoples suggestions.
Let's hear your suggestion for testing the fitness of Colin's
battery to continue its job.
Yes. The glitching would create havoc.
It's all totally unnecessary.

Blooming Heck! That sounds a bit arrogant!
I guess most hobbies and pastimes are unnecessary :)
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

Regards,
John Crighton
Sydney
 
J

John Crighton

Jan 1, 1970
0
Didn't I? I take it that you read the original message.....

"Is there something that I can do to counter balance this high current
voltage drop?

My thought on this is to place a kind of Ammeter into the circuit, that will
adjust the value of VR1, in accordance to the amount of current drawn
through the circuit. Changes in current would would effect this part of
the circuit, and continuously trim the battery meter, so that the readout
remains stable. (and hopefully correct)"


Regards

Colin Dawson
www.cjdawson.com
Hello Colin,
I don't fully understand what you are saying here but it
sounds like that you are talking about a voltage regulator
circuit.

If you are talking about a voltage regulator to give you
a constant Voltage to run your gear, here is a brute
force simple method.

(Go to the motor car wreckers and pick up a second hand
car battery for nothing or next to nothing. Drill small holes into
the battery top to find the links that connect each cell. Screw in
some self tapping screws. You now have a battery with various
voltage taps. Test the cells and short the dud cells out with
heavy wire links because they are dud anyway.
So now your free 12 volt battey is a 10V or 8V or 6V battery,
depending on how many dud cells that you find.)

Now your main power source is let's say 18Volts.
(your good 12 V battery and some cells from old battery in series)
The outpur of your regulator is say 12 volts.
You will need to allow a couple of volts for your
regulator to work.

Use a voltmeter to monitor the input voltage to
your regulator. When it starts to drop close to 14V
you know that is is time to pack up and stop playing
with your telescope gear.

Google up regulator shematics and find something
that you like.
Here is one to give you an idea.
http://www.zen22142.zen.co.uk/Circuits/Power/1230psu.htm
You don't need the transformer and the big capacitor.
You don't need to use all six outboard transistors as shown.

Regards,
John Crighton
Sydney
 
L

legg

Jan 1, 1970
0
Didn't I? I take it that you read the original message.....

"Is there something that I can do to counter balance this high current
voltage drop?

My thought on this is to place a kind of Ammeter into the circuit, that will
adjust the value of VR1, in accordance to the amount of current drawn
through the circuit. Changes in current would would effect this part of
the circuit, and continuously trim the battery meter, so that the readout
remains stable. (and hopefully correct)"

From previous;

"Run two .... blah blah.............Kelvin connection.

"Alternately, one lighter-guage wire from the negative battery
terminal could be used to measure cable drop. Inverted (using
a single supply op amp) and resistively summed into your IC's
input pin (SIG), it could compensate for cable drops in the
measurement.

"In any event.......blah blah....... need."

The cable drop is the current sensing shunt betwen the battery and the
meter's ground reference at the load side. This sensing voltage is
easily inverted using a 358 or similar single supply op-amp, with or
without gain and resistively weighted to modify your meter's input, as
you request.

RL
 
J

James Meyer

Jan 1, 1970
0
As other's have said on this thread, that idea won't really tell me when to
re-charge, as different current will change the characteristics of the
battery, so I won't actually know when the Ah usage is too much. Hence
basing the reading primarily on voltage. I do see how that would work
though.

Regards

Colin Dawson
www.cjdawson.com

I do this sort of thing for a living, and I can tell you that measuring
the current out of the battery and accumulating a running total of amp-hours
will come *much* closer to telling you how much charge is still left in it than
any voltmeter reading no matter how you massage the voltage.

One op-amp, a transistor, a motor with some gears, a handful of passive
parts, and a day to put the thing together.

Either that, or a PIC microprocessor.........

Jim
 
H

Hal Murray

Jan 1, 1970
0
I do this sort of thing for a living, and I can tell you that measuring
the current out of the battery and accumulating a running total of amp-hours
will come *much* closer to telling you how much charge is still left in it than
any voltmeter reading no matter how you massage the voltage.

How important is the temperature?

Is 1A at room temp the same amount of "life" out of the battery
as 1A at freezing?
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that John Crighton <[email protected]>
I am suggesting using a very heavy load
directly across the battery terminals for a very short period of time
(1mSec) which will check the fitness of the battery to continue.

I know that you are. I do read carefully and I do not criticize lightly.
A heavy-current test is simply not necessary for a deep discharge
battery. It could be useful for a float battery that is not intended to
be deeply discharged, or for a capacity test on a stand-by battery that
is rarely used.

The OP only needs to know whether the battery will run *HIS* loads, not
one five or more times larger. When the on-load voltage reaches 10 V or
so, it's time to re-charge. What calls his attention to something odd in
his set-up is that the voltage fluctuates rapidly when he uses his
laptop. That MUST be due to cable and/or termination resistance; nothing
else can cause it. But he doesn't seem willing to accept that.

Your high-current tester may be simple for you, but it seems to me that
the OP has quite limited knowledge of electronics and it wouldn't be
simple for him to design.
 
J

James Meyer

Jan 1, 1970
0
How important is the temperature?

Is 1A at room temp the same amount of "life" out of the battery
as 1A at freezing?

Current is always more accurate than voltage in predicting battery
capacity. And that takes into account temperature and how fast you take the
current out. You will never get 100% accuracy with any scheme, but measuring
the current out and keeping a running total of amp-hours will come closer than
any voltage reading.

After all, that's how the battery's maker rates size and capacity of the
battery in the first place. In amp-hours.

Jim
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that James Meyer <[email protected]>
Current is always more accurate than voltage in predicting
battery
capacity. And that takes into account temperature and how fast you take
the current out. You will never get 100% accuracy with any scheme, but
measuring the current out and keeping a running total of amp-hours will
come closer than any voltage reading.

I've never come across a thread with so many misleading diversions! The
OP doesn't need to know the capacity. He needs to know when the
*voltage* is too low to work his equipment reliably. He has a suitable
voltmeter, but it gives readings he doesn't understand, almost certainly
because of cable resistance.

What you say above is correct but not relevant to the issue.
After all, that's how the battery's maker rates size and
capacity of the
battery in the first place. In amp-hours.

Indeed, but, as you imply above, the specified capacity is given for one
particular discharge current and temperature, and the capacity is
different at other currents (and temperatures, which may or may not be
an issue here).
 
Top