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Default OT: Car battery volt drop

All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?
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Default OT: Car battery volt drop

Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


At that sort of current it depends more on the battery's self discharge
characteristics than anything else.

In other words there's no simple answer, it depends on so many things:-

Ambient temperature
Battery condition (part or fully charged)
Battery capacity
Type of battery (wet cell, calcium, etc.)

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Default OT: Car battery volt drop

On 20/04/2021 20:02, Cliff Topp wrote:

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


12.5V

If the temperature drops by 10C overnight you may see 12.45V

After another 200 hours (at a constant temperature) you may see 12.45V



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Default OT: Car battery volt drop



"Cliff Topp" wrote in message ...
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


You can't calculate that.
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Default OT: Car battery volt drop

On 20/04/2021 20:17, Chris Green wrote:
Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


At that sort of current it depends more on the battery's self discharge
characteristics than anything else.

In other words there's no simple answer, it depends on so many things:-

Ambient temperature
Battery condition (part or fully charged)
Battery capacity
Type of battery (wet cell, calcium, etc.)


And quite commonly (from experience of my own and neighbours), it'll
still be high enough to start after a fortnight, but likely not after 3
weeks.




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Default OT: Car battery volt drop

Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


Part of the problem with batteries, is determining what is going on,
and exactly how healthy the car battery is.

I have a lot of trouble with that.

For example, I've had many batteries, where I get out in the
drive, hook up the dumb charger, and after two minutes the stupid thing
measures 18V. And you just know that's wrong, the battery is not
"stiff enough", it's only got a fraction of the normal ampere-hour (Ah)
capacity. Yet, for maybe several weeks, it continues to turn over the
car. But you know its days are numbered. And how do you tell
one 18V charging session from another 18V charging session
(where it's failed and won't start the car) ?

There are a ton of articles to entertain you.

https://batteryuniversity.com/index....attery_runtime

https://batteryuniversity.com/learn/..._is_the_c_rate

OK, so the second article says the end-of-discharge voltage
is 1.75V. Six times that is 10.5V .

https://batteryuniversity.com/learn/...d_acid_battery

A cell voltage of 2.10V at room temperature
reveals a charge of about 90 percent.

Six times that is 12.6V and that's after you've disconnected the charger
and let it stabilize for 12+ hours or so. Right after it is charged, it
reads higher than that, and taking an open circuit voltage reading
is then "not representative" of battery state. Your smart charger
does its best to terminate charge at the right condition ("topping"),
but the actual working voltage is not visible until the electrolyte has
settled and equilibrated (sulphuric acid the same composition, right at the plate
surface, as 5mm away from it). Using Specific Gravity (SG) with a
liquid sampling device, is about as useful as a voltage reading,
and even SG can be deceiving if only half of the plate (down deep)
is working.

For practical discharge ("running a golf cart"), you limit the
depth of discharge to about 25% on lead acid. The 10.5V value,
would be if you were designing a "cutoff circuit", to disconnect
the battery from the load and stop all additional discharge.
25% depth of discharge is preferred, to maximize the number of charge
cycles you can get from it.

Lead Acid batteries are inefficient at high load. This
is captured in some of the discharge curves shown for
UPS (uninterruptable power supplies, sealed lead acid SLA).
If you run a UPS at the rated load, maybe the battery lasts
for 2 minutes. But the relationship is non-linear, and
if drawing 50mA, then you get the ampere-hour rating of
the battery back.

50Ah is 50000 mAh, divided by 50mA is 1000 hours.
Whereas if you drew 50000mA, you might be expecting
1 hour (60 minutes), but the stupid lead acid battery only lasts
for 20 minutes.

Summary: At the low load you describe, if the battery
is brand new, mint condition, fully charged, it might last
1000 hours before it hits 10.5V (and can't crank the car).

*******

To crank a car, some percentage of charge is needed, because
the battery impedance matters and it gets "weak in the knees"
at low charge. For example, I did a test using two metering
devices here. A clamp-on DC ammeter with peak hold. A voltmeter
with peak hold. And when the car starts, the values I got
were 9V @ 150A . The battery voltage then, was 12.6 internally,
but Ohms law saw to it that 3.6V dropped across the "resistor"
inside the battery. 3.6V/150A is 24 milliohms. When drawing
50mA, there's practically no voltage drop at all, from the
ideal internal value.

I would shoot for 25% discharge, so if the battery is
60Ah, that gives you 15Ah to work with, 50mA load is
300 hours or 12.5 days. And then, perhaps the car will
still start.

There are some examples here of "pulse-challenging" a battery.
And starting a car is a rather large pulse. Car batteries
are designed to provide CCA, they're not a stationary
storage system, like keeping your house warm for 10 days.
They're actually intended for short term use, cranking the
starter. Whereas a golf cart battery is "motive storage"
and could consist of many batteries and many amp-hours.
And it might take all day, to make a dent in the bank of
batteries.

https://batteryuniversity.com/learn/...ct_performance

*******

Getting back to the voltage, here are some values.

https://batteryuniversity.com/learn/...tate_of_charge

SG
100% 1.265 12.65V === 26C (78F) after a 24h rest
75% 1.225 12.45V Voltage is a strong function of temperature,
50% 1.190 12.24V and temperature compensation is required
25% 1.155 12.06V in the real world, like when it is -20C in the drive.
0% 1.120 11.89V

10.5V is cutoff (hard on the battery, could reverse bias something).
10.5V is not a useful value for our home maintenance plan.

We're more interested in voltages that will start a car.

11.9 or 12.0V is about as low as you should go, before
turning the key and starting. The battery needs some juice
in it, to be useful. You can run a 20mA LED all the way down
to 10.5V because the battery impedance doesn't matter for
ridiculously low loads. We need that "stiff 24 milliohms"
value, when pumping 2 horsepower into the starter motor.

When I had a car battery fully charged, and a suspected starter
problem (on a four cylinder car), the measurements during starting
were 9V @ 150A. A four cylinder starter should probably have been
drawing 100A, in which case the battery voltage would not have
dropped so low. That's at the point where the starter is heating
significantly. And you'd give it a lot of time between "trials" :-)
Cars which start easily on the fourth crank, might not be so bad.

For max life then, I could run the battery, day after day,
from 100% full to 75% full. This would (perhaps) maximize the
number of charge cycles. Like, if I was running a golf cart,
that would be a useful operating range.

If I run the battery from 100% ro 25%, that's 75% charge difference,
if the battery is 50Ah times 0.75, I have 37.5Ah or 37500mAh,
divided by 50mA, 750 hours. Round and call it a month. That means,
if I do that for a month, the car will barely start, the starter
will be getting a bit warmer than normal.

But then, real batteries (and you probably have some idea how
poorly the current one is doing), If I was connecting my charger
and seeing 18V while it charged, I would be out of my fricken
mind to be waiting a month to test. I'd be lucky to get three
days our of it at 50mA. When they're on their last legs,
the vampire load really hurts them.

Ya see, it's like predicting when a relative is going to die :-)
You're hoping hoping... and before you know it, they're 92 :-)
You can sorta tell from some of the symptoms, that the battery
"is not well", but it's damn hard to tell "only three days left"
or "won't be able to crank the car a second time". That's why
we don't try to run them right down to the wire.

You can see in my worked results, I'm mostly interested in
the ampere-hours and what percentage of the battery
I've kicked the snot out of. While the voltage gives
some indication of the remaining charge, you must allow
the battery to stabilize before reading it. Whether
charging or discharging, that disturbs the acid and plate
surface conditions.

The 11.9V or 12.0V minimum starting voltages, that is at 25C,
and temperature corrections must be applied at temperatures
other than the "room temperature". If you don't apply
temperature correction, you can be off by 50% of battery
capacity! (Mistake a reading of 100% full, for actual
50% full.) Good smart chargers have a thermistor, and
measure temperature and attempt to do at least a little
bit of correction when charging. Equipment at 5X to 10X the
price, comes with a thermistor on a cable, to be clamped
to one of the electrodes in an attempt to measure temperature.

Even hydrometers, for measuring SG, you need to run your
finger across the chart, for the right temperature.

Paul
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Default OT: Car battery volt drop

Cliff Topp expressed precisely :
All modern cars will have an amount of quiescent current draw to power things
like the alarm, the clock, the radio presets and so on when the car is parked
up and switched off. I've seen it written somewhere that around 50mA can be
considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing 12.5V,
with a 50mA draw overnight what will the voltage be at, say, 9am?


OK, thanks everybody.
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Default OT: Car battery volt drop

Paul was thinking very hard :
Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the car
is parked up and switched off. I've seen it written somewhere that around
50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


Part of the problem with batteries, is determining what is going on,
and exactly how healthy the car battery is.

I have a lot of trouble with that.

For example, I've had many batteries, where I get out in the
drive, hook up the dumb charger, and after two minutes the stupid thing
measures 18V. And you just know that's wrong, the battery is not
"stiff enough", it's only got a fraction of the normal ampere-hour (Ah)
capacity. Yet, for maybe several weeks, it continues to turn over the
car. But you know its days are numbered. And how do you tell
one 18V charging session from another 18V charging session
(where it's failed and won't start the car) ?

There are a ton of articles to entertain you.

https://batteryuniversity.com/index....attery_runtime

https://batteryuniversity.com/learn/..._is_the_c_rate

OK, so the second article says the end-of-discharge voltage
is 1.75V. Six times that is 10.5V .

https://batteryuniversity.com/learn/...d_acid_battery

A cell voltage of 2.10V at room temperature
reveals a charge of about 90 percent.

Six times that is 12.6V and that's after you've disconnected the charger
and let it stabilize for 12+ hours or so. Right after it is charged, it
reads higher than that, and taking an open circuit voltage reading
is then "not representative" of battery state. Your smart charger
does its best to terminate charge at the right condition ("topping"),
but the actual working voltage is not visible until the electrolyte has
settled and equilibrated (sulphuric acid the same composition, right at the
plate
surface, as 5mm away from it). Using Specific Gravity (SG) with a
liquid sampling device, is about as useful as a voltage reading,
and even SG can be deceiving if only half of the plate (down deep)
is working.

For practical discharge ("running a golf cart"), you limit the
depth of discharge to about 25% on lead acid. The 10.5V value,
would be if you were designing a "cutoff circuit", to disconnect
the battery from the load and stop all additional discharge.
25% depth of discharge is preferred, to maximize the number of charge
cycles you can get from it.

Lead Acid batteries are inefficient at high load. This
is captured in some of the discharge curves shown for
UPS (uninterruptable power supplies, sealed lead acid SLA).
If you run a UPS at the rated load, maybe the battery lasts
for 2 minutes. But the relationship is non-linear, and
if drawing 50mA, then you get the ampere-hour rating of
the battery back.

50Ah is 50000 mAh, divided by 50mA is 1000 hours.
Whereas if you drew 50000mA, you might be expecting
1 hour (60 minutes), but the stupid lead acid battery only lasts
for 20 minutes.

Summary: At the low load you describe, if the battery
is brand new, mint condition, fully charged, it might last
1000 hours before it hits 10.5V (and can't crank the car).

*******

To crank a car, some percentage of charge is needed, because
the battery impedance matters and it gets "weak in the knees"
at low charge. For example, I did a test using two metering
devices here. A clamp-on DC ammeter with peak hold. A voltmeter
with peak hold. And when the car starts, the values I got
were 9V @ 150A . The battery voltage then, was 12.6 internally,
but Ohms law saw to it that 3.6V dropped across the "resistor"
inside the battery. 3.6V/150A is 24 milliohms. When drawing
50mA, there's practically no voltage drop at all, from the
ideal internal value.

I would shoot for 25% discharge, so if the battery is
60Ah, that gives you 15Ah to work with, 50mA load is
300 hours or 12.5 days. And then, perhaps the car will
still start.

There are some examples here of "pulse-challenging" a battery.
And starting a car is a rather large pulse. Car batteries
are designed to provide CCA, they're not a stationary
storage system, like keeping your house warm for 10 days.
They're actually intended for short term use, cranking the
starter. Whereas a golf cart battery is "motive storage"
and could consist of many batteries and many amp-hours.
And it might take all day, to make a dent in the bank of
batteries.

https://batteryuniversity.com/learn/...ct_performance

*******

Getting back to the voltage, here are some values.

https://batteryuniversity.com/learn/...tate_of_charge

SG
100% 1.265 12.65V === 26C (78F) after a 24h rest
75% 1.225 12.45V Voltage is a strong function of temperature,
50% 1.190 12.24V and temperature compensation is required
25% 1.155 12.06V in the real world, like when it is -20C in the
drive.
0% 1.120 11.89V

10.5V is cutoff (hard on the battery, could reverse bias something).
10.5V is not a useful value for our home maintenance plan.

We're more interested in voltages that will start a car.

11.9 or 12.0V is about as low as you should go, before
turning the key and starting. The battery needs some juice
in it, to be useful. You can run a 20mA LED all the way down
to 10.5V because the battery impedance doesn't matter for
ridiculously low loads. We need that "stiff 24 milliohms"
value, when pumping 2 horsepower into the starter motor.

When I had a car battery fully charged, and a suspected starter
problem (on a four cylinder car), the measurements during starting
were 9V @ 150A. A four cylinder starter should probably have been
drawing 100A, in which case the battery voltage would not have
dropped so low. That's at the point where the starter is heating
significantly. And you'd give it a lot of time between "trials" :-)
Cars which start easily on the fourth crank, might not be so bad.

For max life then, I could run the battery, day after day,
from 100% full to 75% full. This would (perhaps) maximize the
number of charge cycles. Like, if I was running a golf cart,
that would be a useful operating range.

If I run the battery from 100% ro 25%, that's 75% charge difference,
if the battery is 50Ah times 0.75, I have 37.5Ah or 37500mAh,
divided by 50mA, 750 hours. Round and call it a month. That means,
if I do that for a month, the car will barely start, the starter
will be getting a bit warmer than normal.

But then, real batteries (and you probably have some idea how
poorly the current one is doing), If I was connecting my charger
and seeing 18V while it charged, I would be out of my fricken
mind to be waiting a month to test. I'd be lucky to get three
days our of it at 50mA. When they're on their last legs,
the vampire load really hurts them.

Ya see, it's like predicting when a relative is going to die :-)
You're hoping hoping... and before you know it, they're 92 :-)
You can sorta tell from some of the symptoms, that the battery
"is not well", but it's damn hard to tell "only three days left"
or "won't be able to crank the car a second time". That's why
we don't try to run them right down to the wire.

You can see in my worked results, I'm mostly interested in
the ampere-hours and what percentage of the battery
I've kicked the snot out of. While the voltage gives
some indication of the remaining charge, you must allow
the battery to stabilize before reading it. Whether
charging or discharging, that disturbs the acid and plate
surface conditions.

The 11.9V or 12.0V minimum starting voltages, that is at 25C,
and temperature corrections must be applied at temperatures
other than the "room temperature". If you don't apply
temperature correction, you can be off by 50% of battery
capacity! (Mistake a reading of 100% full, for actual
50% full.) Good smart chargers have a thermistor, and
measure temperature and attempt to do at least a little
bit of correction when charging. Equipment at 5X to 10X the
price, comes with a thermistor on a cable, to be clamped
to one of the electrodes in an attempt to measure temperature.

Even hydrometers, for measuring SG, you need to run your
finger across the chart, for the right temperature.

Paul


Wow! Thanks for all that Paul, that'll take me a bit to digest lol
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Default OT: Car battery volt drop

On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


at 70Ah exactly the same. It takes a week or tow to drain a starter battery

--
"First, find out who are the people you can not criticise. They are your
oppressors."
- George Orwell


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Default OT: Car battery volt drop

In article ,
Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.


My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?


For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


Just what a battery reads voltage wise (to decimal points) has a lot to do
with the individual battery. You can work out how long the battery will
last by its capacity in amp.hrs. And knowing the quiescent load. Using
that, discharging by half seems to be the maximum that will allow the car
to start OK. As a very rough guide.

But the capacity of a battery deteriorates with age.

--
*Letting a cat out of the bag is easier than putting it back in *

Dave Plowman London SW
To e-mail, change noise into sound.
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Default OT: Car battery volt drop

On 21/04/2021 11:19, Dave Plowman (News) wrote:
In article ,
Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.


My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?


For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


Just what a battery reads voltage wise (to decimal points) has a lot to do
with the individual battery. You can work out how long the battery will
last by its capacity in amp.hrs. And knowing the quiescent load. Using
that, discharging by half seems to be the maximum that will allow the car
to start OK. As a very rough guide.

But the capacity of a battery deteriorates with age.


All true, and it also depends on how the voltage is measured, and the
type of battery.

My car has automatic engine stop when the footbrake is on and the car is
stationary. My understanding is that the battery is specifically
designed to cope with repeated starts. There were a number of conditions
where that would not apply (such as A/C running), but also where the
battery capacity and/or voltage was not deemed sufficient (by software?)
to restart the engine when the brake was released.

Over the last year the "battery insufficient" symbol has been appearing
more regularly, and now the automatic engine stop will not work at all.
The car (and battery) is 5.5 years old, so I guess it's battery age
which is preventing the autostop working. The car always starts
perfectly first time, even when it hasn't been used for a week or so.
It would be interesting to see what parameters are required to be met
for the "battery insufficient" symbol to not appear.

--

Jeff
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Default OT: Car battery volt drop

On 21/04/2021 13:22, Jeff Layman wrote:


Over the last year the "battery insufficient" symbol has been appearing
more regularly, and now the automatic engine stop will not work at all.
The car (and battery) is 5.5 years old, so I guess it's battery age
which is preventing the autostop working. The car always starts
perfectly first time, even when it hasn't been used for a week or so. It
would be interesting to see what parameters are required to be met for
the "battery insufficient" symbol to not appear.


During the first lockdown my car had little use although on fortnightly
trips to the supermarket I took it for a long drive around the block to
get it up to temperature and to exercise the A/C. During this time the
stop/start function was always (automatically) disabled. It took 100+
mile non-stop journey before it started working again.
--
mailto : news {at} admac {dot} myzen {dot} co {dot} uk
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Default Car battery volt drop

You cannot do it like that. The voltage curve on a battery drops very little
for ages, then if any of the cells are lower capacity than others there is a
quite sudden drop by 2v or so, and at that time, the low cell is being
reverse charged by the current passing through the load. Not a good thing to
occur.

Brian

--

This newsgroup posting comes to you directly from...
The Sofa of Brian Gaff...

Blind user, so no pictures please
Note this Signature is meaningless.!
"Cliff Topp" wrote in message ...
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the car
is parked up and switched off. I've seen it written somewhere that around
50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?



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What somebody electronically minded did on a battery where you can actually
get at the cell interconnects, few these days, sadly, he put a monitor so it
looked at every cell on its own and then he could tell the weak cell or
cells, but as to what to do with such info, who knows? I'm surprised modern
car electronics do not already allow this like they tend to do on Lithium
cells these days.
Brian

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On 20/04/2021 20:17, Chris Green wrote:
Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


At that sort of current it depends more on the battery's self discharge
characteristics than anything else.

In other words there's no simple answer, it depends on so many things:-

Ambient temperature
Battery condition (part or fully charged)
Battery capacity
Type of battery (wet cell, calcium, etc.)


And quite commonly (from experience of my own and neighbours), it'll still
be high enough to start after a fortnight, but likely not after 3 weeks.






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Default OT: Car battery volt drop

In article ,
Brian Gaff \(Sofa\) wrote:
What somebody electronically minded did on a battery where you can
actually get at the cell interconnects, few these days, sadly, he put a
monitor so it looked at every cell on its own and then he could tell
the weak cell or cells, but as to what to do with such info, who knows?
I'm surprised modern car electronics do not already allow this like
they tend to do on Lithium cells these days.


Given there's nothing you can do about a faulty cell in a car battery, not
much point?

--
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Dave Plowman London SW
To e-mail, change noise into sound.
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Default OT: Car battery volt drop

On 22/04/2021 11:14, Dave Plowman (News) wrote:
In article ,
Brian Gaff \(Sofa\) wrote:
What somebody electronically minded did on a battery where you can
actually get at the cell interconnects, few these days, sadly, he put a
monitor so it looked at every cell on its own and then he could tell
the weak cell or cells, but as to what to do with such info, who knows?
I'm surprised modern car electronics do not already allow this like
they tend to do on Lithium cells these days.


Given there's nothing you can do about a faulty cell in a car battery, not
much point?


As we used to say in industrial automation

"Never check for an error condition you can't handle"

Good advice.
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On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


This is an article I feel might be useful for you.

https://batteryuniversity.com/learn/...tate_of_charge
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On 22 Apr 2021 at 12:18:49 BST, "Adrian Brentnall"
wrote:

On 22/04/2021 11:14, Dave Plowman (News) wrote:
In article ,
Brian Gaff \(Sofa\) wrote:
What somebody electronically minded did on a battery where you can
actually get at the cell interconnects, few these days, sadly, he put a
monitor so it looked at every cell on its own and then he could tell
the weak cell or cells, but as to what to do with such info, who knows?
I'm surprised modern car electronics do not already allow this like
they tend to do on Lithium cells these days.


Given there's nothing you can do about a faulty cell in a car battery, not
much point?


As we used to say in industrial automation

"Never check for an error condition you can't handle"

Good advice.


It is a sound reason to replace a battery if you are not sure whether to do
so.

--
Roger Hayter


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In article ,
Adrian Brentnall wrote:
On 22/04/2021 11:14, Dave Plowman (News) wrote:
In article ,
Brian Gaff \(Sofa\) wrote:
What somebody electronically minded did on a battery where you can
actually get at the cell interconnects, few these days, sadly, he put a
monitor so it looked at every cell on its own and then he could tell
the weak cell or cells, but as to what to do with such info, who knows?
I'm surprised modern car electronics do not already allow this like
they tend to do on Lithium cells these days.


Given there's nothing you can do about a faulty cell in a car battery, not
much point?


As we used to say in industrial automation


"Never check for an error condition you can't handle"


Good advice.


It's why they no longer fit oil pressure gauges, ammeters and accurate
temperature gauges to cars these days. All they did was cause worry to
most.

--
*Whatever kind of look you were going for, you missed.

Dave Plowman London SW
To e-mail, change noise into sound.


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In article ,
Fredxx wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when
the car is parked up and switched off. I've seen it written somewhere
that around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


This is an article I feel might be useful for you.

https://batteryuniversity.com/learn/...tate_of_charge

Good article that, Fred. I knew by experience taking a spot reading of
lead acid voltage of little use. But didn't know the makers reckoned it
had to be left unused for 24 hours.

--
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Dave Plowman London SW
To e-mail, change noise into sound.
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In article , Roger Hayter
wrote:
On 22 Apr 2021 at 12:18:49 BST, "Adrian Brentnall"
wrote:


On 22/04/2021 11:14, Dave Plowman (News) wrote:
In article , Brian Gaff \(Sofa\)
wrote:
What somebody electronically minded did on a battery where you can
actually get at the cell interconnects, few these days, sadly, he
put a monitor so it looked at every cell on its own and then he
could tell the weak cell or cells, but as to what to do with such
info, who knows? I'm surprised modern car electronics do not
already allow this like they tend to do on Lithium cells these days.

Given there's nothing you can do about a faulty cell in a car
battery, not much point?


As we used to say in industrial automation

"Never check for an error condition you can't handle"

Good advice.


It is a sound reason to replace a battery if you are not sure whether to
do so.


Given the cost, makes sense to have it tested first? There are
sophisticated testers that give an instant readout of the condition. A bit
too expensive for DIY, but a decent spares place should have one. ACT is
one such.

However, if you charge the battery with one of those 14 Lidl chargers,
and it struggles to start the car, there's a very good chance it is faulty.

--
*I tried to catch some fog, but I mist.*

Dave Plowman London SW
To e-mail, change noise into sound.
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On 22/04/2021 14:37, Dave Plowman (News) wrote:
In article , Roger Hayter
wrote:
On 22 Apr 2021 at 12:18:49 BST, "Adrian Brentnall"
wrote:


On 22/04/2021 11:14, Dave Plowman (News) wrote:
In article , Brian Gaff \(Sofa\)
wrote:
What somebody electronically minded did on a battery where you can
actually get at the cell interconnects, few these days, sadly, he
put a monitor so it looked at every cell on its own and then he
could tell the weak cell or cells, but as to what to do with such
info, who knows? I'm surprised modern car electronics do not
already allow this like they tend to do on Lithium cells these days.

Given there's nothing you can do about a faulty cell in a car
battery, not much point?


As we used to say in industrial automation

"Never check for an error condition you can't handle"

Good advice.


It is a sound reason to replace a battery if you are not sure whether to
do so.


Given the cost, makes sense to have it tested first? There are
sophisticated testers that give an instant readout of the condition. A bit
too expensive for DIY, but a decent spares place should have one. ACT is
one such.

However, if you charge the battery with one of those £14 Lidl chargers,
and it struggles to start the car, there's a very good chance it is faulty.


I once went through a couple of batteries until I sussed it was the
starter motor!


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On 22/04/2021 14:33, Dave Plowman (News) wrote:
In article ,
Fredxx wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when
the car is parked up and switched off. I've seen it written somewhere
that around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


This is an article I feel might be useful for you.

https://batteryuniversity.com/learn/...tate_of_charge

Good article that, Fred. I knew by experience taking a spot reading of
lead acid voltage of little use. But didn't know the makers reckoned it
had to be left unused for 24 hours.


Thanks, I knew there was a relationship but thought it was flatter than
the graph in the link.

You can either leave it for a while or apply a load for a short time.

The best way of checking is to pulse a known current and measure
internal resistance. Though that tells you the stored charge rather than
state of charge.


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Dave Plowman (News) wrote:
In article ,
Fredxx wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when
the car is parked up and switched off. I've seen it written somewhere
that around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


This is an article I feel might be useful for you.

https://batteryuniversity.com/learn/...tate_of_charge

Good article that, Fred. I knew by experience taking a spot reading of
lead acid voltage of little use. But didn't know the makers reckoned it
had to be left unused for 24 hours.


But that doesn't tell you enough about the situation,
to be celebrating a single reading of OC voltage.

It's comforting reading the voltage and seeing a
"normal" value after it's settled. But that's not
a diagnostic as such.

The Smart Charger has more means at its disposal
of determining battery health.

Your car doesn't charge it chock-full. And if you
didn't find it full, what would you conclude ?
(Battery ? Alternator ? Voltage regulator ?)

The Smart Charger has more opportunities for making
measurements, than your car does. The car doesn't
have a desulphation cycle, it doesn't use pulse
charging (where the relaxation can be measured
after each pulse).

Paul


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On 22/04/2021 17:21, Paul wrote:
Dave Plowman (News) wrote:
In article ,
** Fredxx wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when
the car is parked up and switched off. I've seen it written somewhere
that around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say,
9am?


This is an article I feel might be useful for you.

*https://batteryuniversity.com/learn/...tate_of_charge


Good article that, Fred. I knew by experience taking a spot reading of
lead acid voltage of little use. But didn't know the makers reckoned it
had to be left unused for 24 hours.


But that doesn't tell you enough about the situation,
to be celebrating a single reading of OC voltage.

It's comforting reading the voltage and seeing a
"normal" value after it's settled. But that's not
a diagnostic as such.


It does give an indication of the state of charge. That is better than
nothing.

The Smart Charger has more means at its disposal
of determining battery health.

Your car doesn't charge it chock-full. And if you
didn't find it full, what would you conclude ?
(Battery ? Alternator ? Voltage regulator ?)


Most smart alternators only charge up to 90% in normal use.

The usual criterion is the car won't start.

The Smart Charger has more opportunities for making
measurements, than your car does. The car doesn't
have a desulphation cycle,


It's normally called an equalisation charge, and is more suited to
stationary use, and used to stir up the electrolyte to prevent
stratification.

it doesn't use pulse
charging (where the relaxation can be measured
after each pulse).


If you know the capacity of battery then charge current at constant
voltage is a good way to determine a specified state of charge.

Pulse charging is generally used to increase battery life and decrease
charging time.
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On 22/04/2021 14:27, Dave Plowman (News) wrote:

It's why they no longer fit oil pressure gauges, ammeters and accurate
temperature gauges to cars these days. All they did was cause worry to
most.


But these parameters are being monitored and if you connect a OBDII
device to the diagnostic port you can see such items on a smart phone or
tablet.

--
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On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


Couldn't the manufacturers fit a separate small battery dedicated to
supplying the quiescent items? One that would last maybe ten days. With
a user option to decide whether it should steal power from the main
battery when it became depleted?

Bill
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On 22/04/2021 16:42, Fredxx wrote:


I once went through a couple of batteries until I sussed it was the
starter motor!


I had that problem too with a Mk 4 Cortina in the 1980's. I used to do
all my own maintenance at the time, eventually took it to an outfit with
one of those Fruit-Machine sized "computerised" testers, IIRC it was
taking 600 plus amps (presumably because of shorted windings).
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On 22 Apr 2021 at 20:55:11 BST, "newshound"
wrote:

On 22/04/2021 16:42, Fredxx wrote:


I once went through a couple of batteries until I sussed it was the
starter motor!


I had that problem too with a Mk 4 Cortina in the 1980's. I used to do
all my own maintenance at the time, eventually took it to an outfit with
one of those Fruit-Machine sized "computerised" testers, IIRC it was
taking 600 plus amps (presumably because of shorted windings).


Solved a similar problem recently in an old (but only 20 years) mini with a
cheap Chinese clamp meter. 400A and turning over very slowly.

--
Roger Hayter




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On 22/04/2021 19:41, williamwright wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do
I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


Couldn't the manufacturers fit a separate small battery dedicated to
supplying the quiescent items? One that would last maybe ten days. With
a user option to decide whether it should steal power from the main
battery when it became depleted?


At 50mA that would be less than the self discharge rate of a lead acid
battery.

I therefore don't see the point, it would also make a car even more
complex than it they are already. And you've have to replace two
batteries rather than the one. For some cars a battery change is already
a dealer operation.
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In message , Fredxx
writes
On 22/04/2021 19:41, williamwright wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to
power things like the alarm, the clock, the radio presets and so on
when the car is parked up and switched off. I've seen it written
somewhere that around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how
do I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is
showing 12.5V, with a 50mA draw overnight what will the voltage be
at, say, 9am?

Couldn't the manufacturers fit a separate small battery dedicated to
supplying the quiescent items? One that would last maybe ten days.
With a user option to decide whether it should steal power from the
main battery when it became depleted?


At 50mA that would be less than the self discharge rate of a lead acid
battery.

I therefore don't see the point, it would also make a car even more
complex than it they are already. And you've have to replace two
batteries rather than the one. For some cars a battery change is
already a dealer operation.


Last year, during the first lockdown, after my new
full-of-eltronic-gizzmos Fiesta had been left undriven for four weeks, I
had great trouble with its battery (which eventually found had fallen to
8V). Essentially, the car was dead. The doors would not unlock, and I
had to consult then manual to find where the hidden door-handle keyhole
was. When I got it open, the alarm went off, and wouldn't stop.

I only have 'simple' battery chargers, and even with a homemade beast
that can deliver over 20A I couldn't get the voltage to rise. I guess
that at least two - and maybe three - cells had gone reverse-charged.
After many hours of charging, and a rather too-hot charger, I did get
the voltage up about 9.5V, after which I was able to start the car - and
the on-charge voltage rapidly rose to around 14V.

In contrast, I had absolutely no trouble with my wife's Citroen C1
which, being essentially a 2CV in a modern body, has none of the
electronic fripperies of my Fiesta. The battery was around 12V, and it
started instantly.

So I guess that the Fiesta's battery was discharging at a lot more than
the C1's. So yes - it certainly would be a good design feature if a
car's electronics were powered from a separate battery, so that the main
battery could always be available to start the car.
--
Ian
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In article ,
Fredxx wrote:
The best way of checking is to pulse a known current and measure
internal resistance. Though that tells you the stored charge rather than
state of charge.


I have an ACT tester. Expensive bit of kit. But calculates the actual
battery capacity. It's never let me down. Although I can generally tell
when my own battery is past it anyway.

--
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Dave Plowman London SW
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In article ,
alan_m wrote:
On 22/04/2021 14:27, Dave Plowman (News) wrote:


It's why they no longer fit oil pressure gauges, ammeters and accurate
temperature gauges to cars these days. All they did was cause worry to
most.


But these parameters are being monitored and if you connect a OBDII
device to the diagnostic port you can see such items on a smart phone or
tablet.


One would hope that if you go to those lengths, you'd know what the
readings mean. Unlike the average motorist. But such gauges disappearing
started long before code readers.

--
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Dave Plowman London SW
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In article ,
williamwright wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


Couldn't the manufacturers fit a separate small battery dedicated to
supplying the quiescent items? One that would last maybe ten days. With
a user option to decide whether it should steal power from the main
battery when it became depleted?


Not too many would be keen on a car which still sort of starts, but has to
be taken to a garage to have all the things that rely on a memory reset?
Nor would it be a small battery. Up to 50 mA is a common quiescent drain.
Work out the size of battery needed to supply that for any length of time.

If you know the car is not going to be used for some time, disconnect the
battery. At least then you won't need to buy a new one when you eventually
want to use it.

--
*Why do we say something is out of whack? What is a whack?

Dave Plowman London SW
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In article ,
Fredxx wrote:
At 50mA that would be less than the self discharge rate of a lead acid
battery.


Not so. A lead acid in good condition has a very low self discharge rate.

--
*Proofread carefully to see if you any words out or mispeld something *

Dave Plowman London SW
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On 23/04/2021 13:44, Dave Plowman (News) wrote:
In article ,
Fredxx wrote:
At 50mA that would be less than the self discharge rate of a lead acid
battery.


Not so. A lead acid in good condition has a very low self discharge rate.


More than a 1,000 hours?


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On 23/04/2021 10:50, Ian Jackson wrote:
In message , Fredxx
writes
On 22/04/2021 19:41, williamwright wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to
power* things like the alarm, the clock, the radio presets and so on
when the* car is parked up and switched off. I've seen it written
somewhere that* around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how
do* I calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is
showing* 12.5V, with a 50mA draw overnight what will the voltage be
at, say, 9am?
*Couldn't the manufacturers fit a separate small battery dedicated to
supplying the quiescent items? One that would last maybe ten days.
With* a user option to decide whether it should steal power from the
main* battery when it became depleted?


At 50mA that would be less than the self discharge rate of a lead acid
battery.

I therefore don't see the point, it would also make a car even more
complex than it they are already. And you've have to replace two
batteries rather than the one. For some cars a battery change is
already a dealer operation.


Last year, during the first lockdown, after my new
full-of-eltronic-gizzmos Fiesta had been left undriven for four weeks, I
had great trouble with its battery (which eventually found had fallen to
8V). Essentially, the car was dead. The doors would not unlock, and I
had to consult then manual to find where the hidden door-handle keyhole
was. When I got it open, the alarm went off, and wouldn't stop.

I only have 'simple' battery chargers, and even with a homemade beast
that can deliver over 20A I couldn't get the voltage to rise. I guess
that at least two - and maybe three - cells had gone reverse-charged.
After many hours of charging, and a rather too-hot charger, I did get
the voltage up about 9.5V, after which I was able to start the car - and
the on-charge voltage rapidly rose to around 14V.

In contrast, I had absolutely no trouble with my wife's Citroen C1
which, being essentially a 2CV in a modern body, has none of the
electronic fripperies of my Fiesta. The battery was around 12V, and it
started instantly.

So I guess that the Fiesta's battery was discharging at a lot more than
the C1's. So yes - it certainly would be a good design feature if a
car's electronics were powered from a separate battery, so that the main
battery could always be available to start the car.


I empathise with your issue, but this is down to design. I am currently
using a microcontroller that takes sub microAmp when in sleep mode but
still providing a date and time function. It has to when working off a
lithium button battery.

I also know a acquaintance that had a similar issue to yours. It was a
Peugeot, and the discharge rate was solved after performing a firmware
update for his radio.

I'm more disheartened that your battery was allowed to go to 8V. Most
12V battery items I have ever come across tend to turn off at 11V to
save the battery from being destroyed. I guess there is the reduced
voltage during starting issue to contend with.

It's not a very good advert for Ford!
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Default OT: Car battery volt drop

On 23/04/2021 13:43, Dave Plowman (News) wrote:
In article ,
williamwright wrote:
On 20/04/2021 20:02, Cliff Topp wrote:
All modern cars will have an amount of quiescent current draw to power
things like the alarm, the clock, the radio presets and so on when the
car is parked up and switched off. I've seen it written somewhere that
around 50mA can be considered 'normal'.

My question is - if the quiescent current draw is 50mA (0.05A), how do I
calculate voltage drop per hour?

For instance, if I park the car up at 10pm and the battery is showing
12.5V, with a 50mA draw overnight what will the voltage be at, say, 9am?


Couldn't the manufacturers fit a separate small battery dedicated to
supplying the quiescent items? One that would last maybe ten days. With
a user option to decide whether it should steal power from the main
battery when it became depleted?


Not too many would be keen on a car which still sort of starts, but has to
be taken to a garage to have all the things that rely on a memory reset?
Nor would it be a small battery. Up to 50 mA is a common quiescent drain.
Work out the size of battery needed to supply that for any length of time.

If you know the car is not going to be used for some time, disconnect the
battery. At least then you won't need to buy a new one when you eventually
want to use it.


The last Transit I owned had two lead acid batteries of equal size.
During cranking and running they were connected together. At other times
one was isolated from any gizzmos and so zero drain.

I guess that is one solution to the problem.


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Default OT: Car battery volt drop

In article ,
Fredxx wrote:
On 23/04/2021 13:44, Dave Plowman (News) wrote:
In article ,
Fredxx wrote:
At 50mA that would be less than the self discharge rate of a lead acid
battery.


Not so. A lead acid in good condition has a very low self discharge
rate.


More than a 1,000 hours?


Yes.

A pal has a place in Spain. In better days goes there for at least 2
months at a time, leaving a car here. With the battery left connected,
totally flat on return. And toast. Disconnected, it will start the car
after re-connecting.

--
*When you get a bladder infection urine trouble.*

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