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'[EE] How long does it take to drain a car battery?'
2009\06\14@121202 by

Let's say you have a device that you leave permanently plugged in to the
vehicle, and that draws about 100 mA. How do you calculate (or estimate) the
time it would take for the device to drain the battery to the point where it
won't have enough power left to start the engine?

To clarify, I am more interested in determining the number of hours it is
safe to leave the device plugged in, rather than the number of hours it
takes to kill the battery for sure (although both values are useful). :)

Vitaliy

It's all relative, but my rule of thumb is that car batteries are sized
such that there is a 50% reserve when the battery is new. A medium-sized
car has a battery that is about 50AH, so you can take 25AH out of it and
still start the car. So your 100mA load can go 250 hours and still leave
enough to start the car.

But after the car battery is two years old in a rough environment, the
battery capacity has dropped to perhaps 30AH so 50 hours is all you can
expect to run your device and still leave 25AH to start the car.

These are just my rules of thumb because cars and batteries are real
systems and not calculable. The reality is that it's not the AH that
start the car. But this simplification is as good as any other.

Vitaliy wrote:
{Quote hidden}

It takes much less than half the charge of the battery to start the engine.
Usually you can start the engine lots of times in a short period without
time for recharging.

I don't know for small cars, but tractors and light trucks with 4 or 6
cylinders diesel engines need around 800A of current at 12V to start.

I think the capability to properly start an engine is more related to
the internal resistance. While the battery is able to source the needed
current (hundreds of Amperes) with an acceptable voltage drop (anybody
wants to investigate?) it will be able to start the engine.

Suppose a 5s start-up: 5s * 800A / 3600s/h = 1.11Ah

Regards,

Isaac

Bob Blick escreveu:
{Quote hidden}

__________________________________________________
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Isaac Marino Bavaresco wrote:
> It takes much less than half the charge of the battery to start the engine.
> Usually you can start the engine lots of times in a short period without
> time for recharging.

Yes, but if you have a car battery that has been measured at 52AH and
draw 45AH from it, it will probably not start the car. Back when I
designed battery chargers I did lots of measurements of battery
capacity. And as I said, rules of thumb. If you take 26AH out of a car
battery, don't expect it to start the car unless it's under a year or
two old.

> I think the capability to properly start an engine is more related to
> the internal resistance. While the battery is able to source the needed
> current (hundreds of Amperes) with an acceptable voltage drop (anybody
> wants to investigate?) it will be able to start the engine.
>
> Suppose a 5s start-up: 5s * 800A / 3600s/h = 1.11Ah

Very good. I think there is no conflict in our theories. With your
numbers, a fresh battery and my rule of thumb, you could start the car
25 times before recharging. I believe that. But not 50 times. Those last
few AH in the battery will not start the car.

The internal resistance changes in a lead-acid battery depending on the
state of charge, temperature, and general condition of the battery.

Cheerful regards,

Bob
A scheme I've employed is to monitor OCV when the car is
not running - State-of-charge in the most common lead-acid
chemistry is approximately

( OpenCircuitVoltage - 11 ) / 2.8

Can be temperature corrected with a modest look-up table.

Jack
John Gardner wrote:
> A scheme I've employed is to monitor OCV when the car is
> not running - State-of-charge in the most common lead-acid
> chemistry is approximately
>
> ( OpenCircuitVoltage - 11 ) / 2.8

This is clearly wrong.

First, subtracting a dimensionless number from volts makes no sense.  Then
even assuming you meant to subtract 11V you are left with a value in volts
that is somehow supposed to indicate state of charge, still making no sense.

********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.
Bob Blick wrote:
{Quote hidden}

I once was doing some system-level design of a car entertainment system,
part of which needed to be running even when the engine was off. On the
assumption that most cars are started at least once a day, I decided that
I could consume 10% of the nominal battery capacity over the course of 24
hours, after which, the equipment would put itself into a very low power
standby state. This should leave enough reserve for even an older battery
to start the car after sitting for potentially weeks. Therefore, if a car
battery is, say 48 A-H, I could consume 4.8 A-H over 24 hours, or 200 mA
(2.4 W @ 12 V).

I also toyed around with ideas like having the system "learn" the usage
patterns of the car over the course of multiple weeks, and have it power
itself up in time for the next anticipated drive.

-- Dave Tweed
> This is clearly wrong.

If you can't figure it out don't use it  :)
Bob Blick wrote:

>> I think the capability to properly start an engine is more related to
>> the internal resistance. While the battery is able to source the
>> needed current (hundreds of Amperes) with an acceptable voltage drop
>> (anybody wants to investigate?) it will be able to start the engine.
>>
>> Suppose a 5s start-up: 5s * 800A / 3600s/h = 1.11Ah
>
> Very good. I think there is no conflict in our theories. With your
> numbers, a fresh battery and my rule of thumb, you could start the
> car 25 times before recharging. I believe that. But not 50 times.
> Those last few AH in the battery will not start the car.

I think that's a realistic best-case scenario. As a battery runs down,
it seems to have more and more trouble providing those 800A. You can
still listen to the radio, but the starter motor won't crank :)

There's also the state of the engine (how often and how long do you have
to crank for it to start?), the temperature (battery capacity goes down
quickly as it gets cold -- and I don't mean Brazil-cold, I mean
Canada-cold :), economy-sized batteries and other factors.

Gerhard
Dave Tweed wrote:

> I once was doing some system-level design of a car entertainment system,
> part of which needed to be running even when the engine was off. On the
> assumption that most cars are started at least once a day, I decided that
> I could consume 10% of the nominal battery capacity over the course of 24
> hours, after which, the equipment would put itself into a very low power
> standby state. This should leave enough reserve for even an older battery
> to start the car after sitting for potentially weeks. Therefore, if a car
> battery is, say 48 A-H, I could consume 4.8 A-H over 24 hours, or 200 mA
> (2.4 W @ 12 V).
>
> I also toyed around with ideas like having the system "learn" the usage
> patterns of the car over the course of multiple weeks, and have it power
> itself up in time for the next anticipated drive.

A long long way back(1999) I built a computer for my car to play mp3's
and I made it shut down after the car had been off for a half hour, so
it wouldn't need to reboot if I was just running a short errand. I
figured any longer than that was risking the interior temperature of the
car getting too hot for the computer in the summer. I also toyed with
the idea of the system booting when a doorlock opened, so it would have
a head start(the power supply was robust enough to boot/run even when
the engine was starting), but never did that.

But before I finished the software I'd have to manually shut down, and I
would frequently forget after coming home from work. So from about 6pm
to 7:30am there would be a 2.5 amp load on the battery. The car would
start, but you could tell it was low. The car was only a year old so it
was in good tune and the battery fairly fresh.

It wouldn't survive a Friday-to-Monday drain, but I only did two or
three of those.

Cheers,

Bob
Car batteries are rated by "Reserve Capacity". To convert from
reserve capacity to amp hours one must first understand the
definition of "Reserve Capacity" which is the time in minutes that a
new fully charged battery will deliver 25A at 80 degrees Fahrenheit
and until it discharges to 1.75V per cell. So you take the reserve
capacity and divide it by 60 to get the hours. Then take the hours
and multiply it by 25 to get the amp hour capacity.

For example a Sears Marine Deep Cycle Battery (model 27494) is rated
at 135 minutes of reserve capacity. 135 minutes divided by 60 minutes
= 2.25 hours, 2.25 hours times 25 amps = 56.25 amp hours.

I have found in real life the calculation comes out higher than an
actual battery can deliver due to temperature, physical condition of
the plates and electrolyte, state of existing charge, etc. It does
get you in the ball park though.

Hope that helps.

David

At 11:10 AM 6/14/2009, you wrote:
{Quote hidden}

>
John Gardner wrote:
>> This is clearly wrong.
>
> If you can't figure it out don't use it  :)

There is nothing to figure out.  Your equation is just plain wrong on the
grounds of basic dimensional analisys, and couldn't be left in the archives
without a comment to that effect.

I can guess what you probably meant, but that requires assuming you made two
mistakes in the equation.  That leaves open the very real possibility of
other mistakes, or not guessing the right ones.

Hint: If you get caught saying something incorrect, it will blow over a lot
faster if you say "oops" and correct it instead of drawing more attention to
it by trying to blame it on the messenger.

********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.
Bob Blick wrote:
> It's all relative, but my rule of thumb is that car batteries are sized
> such that there is a 50% reserve when the battery is new. A medium-sized
> car has a battery that is about 50AH, so you can take 25AH out of it and
> still start the car. So your 100mA load can go 250 hours and still leave
> enough to start the car.
>
> But after the car battery is two years old in a rough environment, the
> battery capacity has dropped to perhaps 30AH so 50 hours is all you can
> expect to run your device and still leave 25AH to start the car.

This agrees with the limited data I have.

The device I'm talking about is an OBD interface, that is connected to a car
PC. We've had several reports that when used in fleets, the interface
sometimes drains the battery if the car is parked for extended periods of
time (one week or longer).

In some cases, the battery was drained overnight, because the interface kept
sending keep-alive messages to the ECU, which in turn stayed awake and drew
several amps to power the actuators. This particular problem can be remedied
by simply issuing a "device reset" command to the interface.

What about the impact of this small drain on the battery life? Say, if you
drain the battery to 50% capacity on a regular basis?

Vitaliy

Dave Tweed wrote:
> I once was doing some system-level design of a car entertainment system,
> part of which needed to be running even when the engine was off. On the
> assumption that most cars are started at least once a day, I decided that
> I could consume 10% of the nominal battery capacity over the course of 24
> hours, after which, the equipment would put itself into a very low power
> standby state. This should leave enough reserve for even an older battery
> to start the car after sitting for potentially weeks. Therefore, if a car
> battery is, say 48 A-H, I could consume 4.8 A-H over 24 hours, or 200 mA
> (2.4 W @ 12 V).

Was the time hard-coded, or were you measuring the voltage to determine the
state of charge?

So what would happen in the standby state? And how did you exit from it?

> I also toyed around with ideas like having the system "learn" the usage
> patterns of the car over the course of multiple weeks, and have it power
> itself up in time for the next anticipated drive.

Hmm, I like the idea, but I think the implementation can get messy.

Vitaliy

>> A scheme I've employed is to monitor OCV when the car is
>> not running - State-of-charge in the most common lead-acid
>> chemistry is approximately

Hereunder, 1. 2. & 3. are formula references.

1.    >> ( OpenCircuitVoltage - 11 ) / 2.8

> This is clearly wrong.

Only for a limited range of values of 'clearly' and 'wrong'.

ie
(i) The formula is understandable and useful to most people as is, and

(ii) While it will be obvious to any person liable to be interested in this
thread that the formula 'does not quite scan' as written, it is very
unlikely to tax anyone's thinking power in any significant way to understand
what is being said. While it is technically correct that every formula or
equation should be dimensionally correct and have all units shown, doing
this can on occasion detract from the understandability of what is being
said. And, that is arguably probably the case here.

"Properly written" the equation should read something like,

2.       State of charge %  = (OpenCircuitVoltage Volt - 11 Volt) / (0.028
%_full_Charge/Volt) %_of_full_charge %

I use % rather than a fraction here as that is (obviously) clearer and more
correct than using a dimensionless number in the range 0 - 1. to indicate a
state of charge.

"Obviously" any formula which uses mixed units is dimensionally incorrect.
ie in this case substracting a dimensionless number from a voltage is
"incorrect".
Any practitioner skilled in thre art, and about 99.5 % of people liable to
be at all interested in this threrad, is almost certainly going to relaise
that the '11' indicates a voltage. And also that there is an implied
conversion for delta-voltage to state of charge when the factor of 2.8 is
invoked. And that the state of charge range is from 0 (or 0.0 or 0.000 or
...) where 0 indicates 0 and 1 indicates 100%.

My point in writing this is to note that none of us always write things 100%
correctly 100% of the time (including both myself and the questioner), that
often it is more understandable to allow of a degree of 'short hand' when
discussing such things amongst consenting adults in PICList, and that
striving endlessly to correctly express things  can lead to less rather than
more clarity.

What was originally being said here was both 'obvious, and useful, and maybe
also modertaely correct :-).

Viz.

3.       A useful rule of thumb for determining the state of charge of a
lead acid car battery is to subtract 11 Volt from the battery voltage and
todivide the result by 0.028. The resultant figure is the approximate
battery charge in percent.

I think the original formula (1.) did a much better job of explaining that
to interested parties than did the "more correct" 2. or the correct but
somewhat obfuscated 3. .
No?

Russell

David Bearrow wrote:
> Car batteries are rated by "Reserve Capacity". To convert from
> reserve capacity to amp hours one must first understand the
> definition of "Reserve Capacity" which is the time in minutes that a
> new fully charged battery will deliver 25A at 80 degrees Fahrenheit
> and until it discharges to 1.75V per cell.

So for a car battery, that means draining it to a voltage of 1.75V * 6 =
10.5V?

This corresponds to "voltage loaded at full discharge":

I'm assuming this means that after the battery discharges down to 10.5V, it
won't be able to start the engine.

{Quote hidden}

It looks like the "reserve capacity" value is not available for all car
batteries (I just checked a few, on sears.com). Is it used interchangeably
with ampere-hours, using the formula above?

Vitaliy

Russell McMahon
> 3.       A useful rule of thumb for determining the state of charge of a
> lead acid car battery is to subtract 11 Volt from the battery voltage and
> todivide the result by 0.028. The resultant figure is the approximate
> battery charge in percent.

Open circuit voltage = 12.8V

12.8V - 11V = 1.8V
1.8V / 0.028 = 64.29%

This doesn't sound right. It assumes that at 100% charge, the battery
voltage should be 13.8V.

What gives?

Vitaliy

Vitaliy wrote:

>
> What about the impact of this small drain on the battery life? Say, if you
> drain the battery to 50% capacity on a regular basis?

It will have a noticeable effect. Offhand, I'd say lifetime will be 75%
of normal if operated under mild temperatures. But in a rough
environment it will be worse.

Cheerful regards,

Bob
Russell McMahon wrote:
> Only for a limited range of values of 'clearly' and 'wrong'.

The units were missing to the point of some confusion, particularly what
exactly the result of the equation was supposed to be.

> (ii) While it will be obvious to any person liable to be interested in
> this thread that the formula 'does not quite scan' as written, it is
> very unlikely to tax anyone's thinking power in any significant way to
> understand what is being said.

So how far into sloppiness do you want to decend?  Sometimes it's OK to let
a few minor details be understood from context when you're really sure
everyone is in the same context.  However, basic sloppiness like leaving off
units is just plain wrong, and frankly inexcusable in technical circles.
You'd never get away with it in freshman physics class, but it's somehow OK
in the real world when it might actually matter but there's no grade
attached?

> "Properly written" the equation should read something like,
>
> 2.       State of charge %  = (OpenCircuitVoltage Volt - 11 Volt) /
> (0.028 %_full_Charge/Volt) %_of_full_charge %

You are making it deliberately too messy.  All that would have been required
would have been simply:

FractionFull = (OpenVolts - 11V) / 2.8V

********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.
> This doesn't sound right. It assumes that at 100% charge, the battery
> voltage should be 13.8V. What gives?

That's exactly what it assumes, corresponding to a full-chg terminal
voltage of 2.3V per cell. Some charging regimes use higher values
for EOC, in order to agitate the electrolyte, but such behavior in autos
usually means the regulator is pau...

IME the other common value, usually encountered in general aviation
applications, is 13.4V.

Jack
Vitaliy wrote:
> Dave Tweed wrote:
> > I once was doing some system-level design of a car entertainment system,
> > part of which needed to be running even when the engine was off. On the
> > assumption that most cars are started at least once a day, I decided that
> > I could consume 10% of the nominal battery capacity over the course of 24
> > hours, after which, the equipment would put itself into a very low power
> > standby state. This should leave enough reserve for even an older battery
> > to start the car after sitting for potentially weeks. Therefore, if a car
> > battery is, say 48 A-H, I could consume 4.8 A-H over 24 hours, or 200 mA
> > (2.4 W @ 12 V).
>
> Was the time hard-coded, or were you measuring the voltage to determine the
> state of charge?

We never actually built the system, but the idea was that it would simply
be hard-coded.

The actual requirement was that under most circumstances, the equipment
wanted to be in its "active" state for at least one hour *before* the next
drive. This requirement was to be met by simply leaving it active for up to
24 hours after the previous drive.

> So what would happen in the standby state? And how did you exit from it?

Pretty much nothing. The system would become active again the next time
the ignition was turned on, as indicated by the "switched" power lead from
switched and unswitched power.

> > I also toyed around with ideas like having the system "learn" the usage
> > patterns of the car over the course of multiple weeks, and have it power
> > itself up in time for the next anticipated drive.
>
> Hmm, I like the idea, but I think the implementation can get messy.

Yeah, we never really pursued this.

-- Dave Tweed
John Gardner wrote:
>> This doesn't sound right. It assumes that at 100% charge, the battery
>> voltage should be 13.8V. What gives?
>
> That's exactly what it assumes, corresponding to a full-chg terminal
> voltage of 2.3V per cell. Some charging regimes use higher values
> for EOC, in order to agitate the electrolyte, but such behavior in autos
> usually means the regulator is pau...
>
> IME the other common value, usually encountered in general aviation
> applications, is 13.4V.

So you're saying the Wikipedia figures are wrong?

Vitaliy
>> 3.       A useful rule of thumb for determining the state of charge of a
>> lead acid car battery is to subtract 11 Volt from the battery voltage and
>> todivide the result by 0.028. The resultant figure is the approximate
>> battery charge in percent.

> So if we were to trust the figures from this page:

> Open circuit voltage = 12.8V
>
> 12.8V - 11V = 1.8V
> 1.8V / 0.028 = 64.29%
>
> This doesn't sound right. It assumes that at 100% charge, the battery
> voltage should be 13.8V.
>
> What gives?

As I said about John's formula: " ... and possibly also moderately correct."
:-)

ie John's formula is an attempt to take a parameter (O/C Voltage) that is
known to NOT accurately reflect state of charge, and to use it to
APPROXIMATELY determine state of charge. The result is necessarily
approximate and will vary, possibly widely, with circumstance.

At the end of this email I've added information, from the above Wikipedia
page, that relates to the voltage ranges for 6 cerll 12V LA batteries. As
will be seen, the voltages in a given situation vary by battery sub-type,
temperature and recent past conditions. A standard "float" voltage for
automotive lead acid batteries is 13.8 V. But, none of the non-charging
voltages reflect the full charge voltage that John's formula implies
(13.8V), the closest situation being the line "After full charge the
terminal voltage will drop quickly to 13.2 V and then slowly to 12.6 V."
This doesn't make his formula wrong per se - just reflects the inexact
nature of the assumptions that it is based on.

ie using the Wikipedia data the formula would perhaps be

SOC% = (Vbattery V - 11 V ) / 0.022 %/V   % in Olin nomenclature or

SOC% = (Vbattery - 11) / 0.022  % in John nomenclature

John's formula imples a 11V potential for full discharge whereas the
"Reserve Capacity" one that you were discussing implies a 10.5V end of
discharge potential.
Adding 13.8 and 13.2V full charge points  leass to formulae of (using John's
Russell modified nomenclature)(for clarity :-) ):
E&OE.

13.8, 11              SOC% = (Vbattery - 11) / 0.028 %
13.8, 10.5           SOC% = (Vbattery - 10.5) / 0.033 %
13.2, 11              SOC% = (Vbattery - 11) / 0.022 %
13.2, 10.5           SOC% = (Vbattery - 10.5) / 0.027 %

To these you can add the Wikipedia observations that O/C voltage at full
charge is 13.2V initially, slowly reducing to 12.6V and thatv there is a
temperature effect, as ?John? noted and Wikipedia elucidated -0.022%
V/degree_C wrt 20 degrees C. So time since charge termination may be
usefully factored in if the situation allows or demands.

*** ie *** the original formula was an approximation, YMWV, and you need to
apply whatever corrections are required to make the result as accurate as
possible and/or required for your given application.

In situations requiring extreme precision you may require some or all of
"gas gauging", historical records and trends, full cell & environmental
data - including acid specific gravity* and temperature. Whereas, for
approximate "am I running the battery dangerously low" consumer electronics
applications, some version of the simple formula above may suffice.

* Real women (and men) even now use battery acid specific gravity as part of
the 'gold standard' method of cell state of charge, cell balance and overall
battery condition measurements - especially so for large and expensive
batteries such as are used for telecommunication sites or larger battery
backup UPS systems. In the good old days it was common practice to use a
hydrometer to measure per cell  battery acid specific gravity in automotive
batteries. I suspect that in this age of sealed-for-death batteries, many
modern auto-shops would not even own a hydrometer, let alone know how to use
one.

Russell
_________________________

Wikipedia usefully says:

These are general voltage ranges for six-cell lead-acid batteries:

a.. Open-circuit (quiescent) at full charge: 12.6 V to 12.8 V (2.10-2.13V
per cell)
b.. Open-circuit at full discharge: 11.8 V to 12.0 V
c.. Loaded at full discharge: 10.5 V.
d.. Continuous-preservation (float) charging: 13.4 V for gelled
electrolyte; 13.5 V for AGM (absorbed glass mat) and 13.8 V for flooded
cells
1.. All voltages are at 20 °C, and must be adjusted -0.022V/°C for
temperature changes.
2.. Float voltage recommendations vary, according to the manufacturer's
recommendation.
3.. Precise (±0.05 V) float voltage is critical to longevity; too low
(sulfation) is almost as bad as too high (corrosion and electrolyte loss)
a.. Typical (daily) charging: 14.2 V to 14.5 V (depending on
manufacturer's recommendation)
b.. Equalization charging (for flooded lead acids): 15 V for no more than
2 hours. Battery temperature must be monitored.
c.. Gassing threshold: 14.4 V
d.. After full charge the terminal voltage will drop quickly to 13.2 V and
then slowly to 12.6 V.
Russell McMahon wrote:
>Adding 13.8 and 13.2V full charge points  leass to formulae of (using
>John's
Russell modified nomenclature)(for clarity :-) ):
E&OE.<

What's "leass"?

Vitaliy

> Russell McMahon wrote:
>>Adding 13.8 and 13.2V full charge points  leass to formulae of (using
>>John's
> Russell modified nomenclature)(for clarity :-) ):
> E&OE.<
>
> What's "leass"?

So basically, the formula is enough to get a very rough estimate, and for
anything more precise you have to factor in many more variables, which make
the calculation impractical.

I do have a useful answer to my original question, though. Thanks, everyone.

Vitaliy

> So you're saying the Wikipedia figures are wrong?

"All models are wrong. Some models are useful."
All absolute data is always wrong :-).
How wrong?" is the question.
By using an analog of some other condition (in this case battery voltage for
charge state) we trade or known inexactitude in exchange for convenience and
ease of use. Too much convenience can lead to unnaceptable results :-).

Re my " ...leass ..."
Dunno what I meant to put there - my spilling chucker is dead and my read
through missed it.
I think it may have been several part words run together during editing and
not deleted.

Maybe if I install IE8 my spilling chucker will leap back into life.
Thenagain ... .

Russell

> These are general voltage ranges for six-cell lead-acid batteries...

Good info, IME. My choice of 11V rather than 10.5V reflects the
fact that cell longevity is (roughly) inversely proportional to depth
of discharge. A cell "never" discharged below 80%, and otherwise
cosseted,  may well give acceptable service for 10 years.

The same cell regularly discharged to 50% will likely fail in less
than 4 years, and 20% less than 18 months.

Capacity & longevity degrade rapidly with falling temperature.

I own & use a hydrometer, but it's a messy interface to a micro...

Jack

p.s. -  For the bottom line on PbSO4 battery tech, check with a
submarine engineer ( Not me - I evaded military service by joining
the USAF, but I'm related to one... )
> John's formula is an attempt to take a parameter (O/C Voltage) that is
known to NOT accurately reflect state of charge, and to use it to
APPROXIMATELY determine state of charge. The result is necessarily
approximate and will vary, possibly widely, with circumstance.

Yep, especially temperature. This approach has the virtue of tending to
reflect reality more accurately the longer the battery sits, corresponding
well with many vehicle applications. But YMMV.

Jack
Actually there are lots of fact to affect the durability of a battery. Some of the well designed car has a good trade-offs/balances between loads and alternators. The environment also plays an important role.

Funny N.
Au Group Electronics, http://www.AuElectronics.com

________________________________
From: Gerhard Fiedler <listsconnectionbrazil.com>
To: Microcontroller discussion list - Public. <piclistmit.edu>
Sent: Sunday, June 14, 2009 5:15:33 PM
Subject: Re: [EE] How long does it take to drain a car battery?

Bob Blick wrote:

{Quote hidden}

I think that's a realistic best-case scenario. As a battery runs down,
it seems to have more and more trouble providing those 800A. You can
still listen to the radio, but the starter motor won't crank :)

There's also the state of the engine (how often and how long do you have
to crank for it to start?), the temperature (battery capacity goes down
quickly as it gets cold -- and I don't mean Brazil-cold, I mean
Canada-cold :), economy-sized batteries and other factors.

Gerhard
submarine uses expensive battery normal person won't be able to afford. :)

Funny N.
Au Group Electronics, http://www.AuElectronics.com

________________________________
From: John Gardner <goflo3gmail.com>
To: Microcontroller discussion list - Public. <piclistmit.edu>
Sent: Sunday, June 14, 2009 10:21:47 PM
Subject: Re: [EE] How long does it take to drain a car battery?

> These are general voltage ranges for six-cell lead-acid batteries...

Good info, IME. My choice of 11V rather than 10.5V reflects the
fact that cell longevity is (roughly) inversely proportional to depth
of discharge. A cell "never" discharged below 80%, and otherwise
cosseted,  may well give acceptable service for 10 years.

The same cell regularly discharged to 50% will likely fail in less
than 4 years, and 20% less than 18 months.

Capacity & longevity degrade rapidly with falling temperature.

I own & use a hydrometer, but it's a messy interface to a micro...

Jack

p.s. -  For the bottom line on PbSO4 battery tech, check with a
submarine engineer ( Not me - I evaded military service by joining
the USAF, but I'm related to one... )
I'm sure it's expensive :)

Auto electric systems are slowly, grudgingly, getting better. Hear Hear.

Jack
On Sun, 2009-06-14 at 15:09 -0300, Isaac Marino Bavaresco wrote:
> It takes much less than half the charge of the battery to start the engine.
> Usually you can start the engine lots of times in a short period without
> time for recharging.

Perhaps, in the best of conditions.

That said, cars are in pretty horrible environments. I guarantee that a
car battery in -30C weather will not do a very good job of starting the
engine with a similar charge level.

Unfortunately there are simply too many variables at play here. A big
one in todays cars is what the cars drain on the battery is just sitting
there, it can be surprisingly high with some cars.

TTYL

Plus, they are made of expensive metals, silver, gold.
So, if you got one from the sub, you got a lot of \$.

The auto industry doesn' change that much due to lower cost requirement and cycles of cost reduction  process.

Thanks to the material industry, the cars are still drivable. :)

Funny N.
Au Group Electronics, http://www.AuElectronics.com

________________________________
From: John Gardner <goflo3gmail.com>
To: Microcontroller discussion list - Public. <piclistmit.edu>
Sent: Sunday, June 14, 2009 11:16:40 PM
Subject: Re: [EE] How long does it take to drain a car battery?

I'm sure it's expensive :)

Auto electric systems are slowly, grudgingly, getting better. Hear Hear.

Jack
On Sun, 2009-06-14 at 14:16 -0700, Bob Blick wrote:
> A long long way back(1999) I built a computer for my car to play mp3's
> and I made it shut down after the car had been off for a half hour, so
> it wouldn't need to reboot if I was just running a short errand. I

Amazing Bob, I did the exact same thing, although a couple years after
that.

It was a Pentium 233 running 98 IIRC sitting in the trunk. I used a
serial IR receiver and IR remote to control winamp and a character LCD
display connected to the parallel port for feedback (along with a volume
control and the power button).

It worked quite well, but the inverter always rebooted when I started
the car, never got quite as far with it as you did, still, you brought
back alot of memories.

My friends still mention my car PC from time to time... :)

TTYL

Herbert Graf wrote:
>> A long long way back(1999) I built a computer for my car to play mp3's
>> and I made it shut down after the car had been off for a half hour, so
>> it wouldn't need to reboot if I was just running a short errand. I
>
> Amazing Bob, I did the exact same thing, although a couple years after
> that.
>
> It was a Pentium 233 running 98 IIRC sitting in the trunk. I used a
> serial IR receiver and IR remote to control winamp and a character LCD
> display connected to the parallel port for feedback (along with a volume
> control and the power button).

Sounds cool. Where was the IR sensor physically located?

Did you have to write a custom app to output song info to the LCD?

> It worked quite well, but the inverter always rebooted when I started
> the car, never got quite as far with it as you did, still, you brought
> back alot of memories.
>
> My friends still mention my car PC from time to time... :)

I've just read a post on MP3Car.com where someone said they're using a
Sheeva plug as their CarPC:

http://www.mp3car.com/vbulletin/1321159-post8.html

Vitaliy

Herbert Graf wrote:
> On Sun, 2009-06-14 at 14:16 -0700, Bob Blick wrote:
>> A long long way back(1999) I built a computer for my car to play mp3's
>> and I made it shut down after the car had been off for a half hour, so
>> it wouldn't need to reboot if I was just running a short errand. I
>
> Amazing Bob, I did the exact same thing, although a couple years after
> that.

When it's time to build something, it's build time everywhere :)

http://bobblick.com/techref/projects/yammp3/yammp3.html

Cheers,

Bob
>> Amazing Bob, I did the exact same thing, although a couple years after
>> that.

> When it's time to build something, it's build time everywhere :)

Approximately citing the (non verbatim) quote:

"It steam engines, come steam engine time".

Russell
>> This doesn't sound right. It assumes that at 100% charge, the battery
>> voltage should be 13.8V. What gives?
...
> IME the other common value, usually encountered in general aviation
> applications, is 13.4V.

When I was involved with developing radio telephone units, we worked on
13.6V for a fully charged battery on float charge (i.e. the vehicle had the
motor running).

On Sun, 2009-06-14 at 21:56 -0700, Vitaliy wrote:
> > It was a Pentium 233 running 98 IIRC sitting in the trunk. I used a
> > serial IR receiver and IR remote to control winamp and a character LCD
> > display connected to the parallel port for feedback (along with a volume
> > control and the power button).
>
> Sounds cool. Where was the IR sensor physically located?

I believe I sat it on the dash so anybody could control it.

> Did you have to write a custom app to output song info to the LCD?

Actually it was a plugin for winamp, very easy to get going.

TTYL

In the automotive industry, car companies won't add a supplier module
to the car if it draws more than 2mA when the car isn't running
(except for a short period before going into sleep), except for some
modules that must be always on (such as RF receivers) - but even those
have a very low power budget - certainly not as high as 100mA
constantly.

100mA will drain the battery far enough to be noticable by a user
after several days, especially if the battery is not new.

However, after the engine and body controllers go to sleep, the CAN
(or vpw, or other vehicle bus) bus should go into a bus off state, at
which point you should be able to put the OBD-II device into sleep
mode.  Why must you draw 100mA when there's no bus traffic to monitor?
The only thing I can think of is perhaps for running a radio
receiver, but in that case you can turn it on once a second or so for
a few mS and see if anyone is sending, rather than running the
receiver all the time.  Otherwise I can't imagine a reason to draw
100mA constantly...

If you frequently drain the battery below 50% you will lower the
battery's ability to keep a charge.  If you have to draw 100mA, put a
timer on there and shut it off after 48 hours, or at minimum sense the
battery voltage and shut down if it drops below 11.5 to 12V and the
bus isn't active.

On Sun, Jun 14, 2009 at 12:10 PM, Vitaliy<spammaksimov.org> wrote:
{Quote hidden}

> -
{Quote hidden}

Thank you very much for the info. I did not know about the 2 mA limit. Is
there a document that spells this out?

We are considering different power saving options for our future devices,
but my question pertained to our existing products. When we designed them,
we did not consider that someone would want to leave the device inside their
vehicle.

Thought it sounds like it should be easy, in fact it isn't always easy to
know when the OBD bus is off. It is a common misconception that the data is
always available on an active bus (this is not true in a lot of cases).
Moreover, some ECUs respond to requests for data even when ignition is in
the OFF position (this is the worst possible situation, the ECU actuators
can completely drain the battery in a matter of hours).

Vitaliy

On Fri, Jun 19, 2009 at 5:41 PM, Vitaliy<spammaksimov.org> wrote:
>> In the automotive industry, car companies won't add a supplier module
>> to the car if it draws more than 2mA
>
>
> Thank you very much for the info. I did not know about the 2 mA limit. Is
> there a document that spells this out?

I'm not aware of a 'generic' document that covers this - I should have
worded that differently to indicate that all of the modules I've
worked on have upper limits in sleep that are lower than 2mA - the
modules I've worked on have been the larger (ie more functionality)
modules, so my assumption is that the smaller modules have equal or
more strict requirements.  2mA is still on the high side for such a
module.  A high-end vehicle has 25+ electronic modules on the various
CAN busses, and many on the LIN busses, etc.  Even 50mA during sleep
is too much for the whole vehicle.  A large vehicle gets a large
battery not just because it takes more energy to start the larger
engine, but to keep all the creature feature electronic devices happy
without draining the battery.

Keeping the customer happy is important, but a big reason why the
current draw must be so low is so the manufacturer doesn't have to
jumpstart hundreds of cars in inventory every day they get shipped out
simply because they've been sitting around for more than a few months,
or perform battery connection/disconnection.

So you don't need to confine yourself to 2mA, unless your customers
demand months of operation while off, in which case you might suggest
a battery maintainer...

> We are considering different power saving options for our future devices,
> but my question pertained to our existing products. When we designed them,
> we did not consider that someone would want to leave the device inside their
> vehicle.

The ongoing problem of what to do when your customers use your product
in an unintended way...

If you have the ability to do so (hardware supports it, can upgrade
the software) the simplest solution would likely be to monitor the
voltage on the connector and shut the device down if it stays below
11.5V for more than 30 seconds (to avoid shutting down during engine
starts, etc).  If not... "This product was not designed to remain
attached to the vehicle for long periods of use in the engine off
state.  We recommend customers install a battery maintainer for these
situations."

Alternately, make a cable that plugs inbetween the vehicle and your
device which provides power to your device via rechargeable battery,
and perhaps solar panel.

Or the cable can merely check the battery voltage and under low
voltage conditions cuts power to the device.

> Thought it sounds like it should be easy, in fact it isn't always easy to
> know when the OBD bus is off. It is a common misconception that the data is
> always available on an active bus (this is not true in a lot of cases).
> Moreover, some ECUs respond to requests for data even when ignition is in
> the OFF position (this is the worst possible situation, the ECU actuators
> can completely drain the battery in a matter of hours).

That's true.  Life is full of wonderful challenges...  ;-D