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'[EE]: Voltage Compare'
2002\05\22@174154 by Tal Bejerano - AMC

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Hi All

I want to measure and get a warning if car battery is drop under 11.5v.
look at some circuits that use op-amps/comparators and read the explanations
and I wonder, how can it work? if the circuit use the battery voltage to
operate and the voltage drops, then all the adjustments are not correct. as
I understand it I need 2 voltage sources: one is stable and the other is the
battery I monitor. ???????

Regards

Tal Bejerano
AMC - ISRAEL

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2002\05\22@181719 by Jinx

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> I understand it I need 2 voltage sources: one is stable and the
> other is the battery I monitor. ??????

I was looking at one yesterday I might make up to protect my
sealed leads from being under-voltage. It uses an LM336-2.5
as the comparator reference

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2002\05\22@184100 by Bob Ammerman

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Use a voltage regulator from the car battery down to 5V. Now you have a
stable reference to which you can compare the battery voltage.

Bob Ammerman
RAm Systems

{Original Message removed}

2002\05\22@190426 by Pic Dude

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I did this recently with a PIC 16F872.  A 5V regulator drops the ~12V
down to the operating voltage of the PIC, as well as providing the reference
voltage.  The input is simply a voltage divider on the ~12V supply to limit
the voltage going into the PIC.  Actually I set it up so 25.5V at the input
will divide down to 5V.  This way, 25.5V reads 255 on the top 8 bits of
the A/D and I saved myself some math that way.  But I do lose a bit of
resolution (not a big deal).  Also have upper and lower alarms settable
with a single-button interface.

Cheers,
-Neil.



{Original Message removed}

2002\05\22@193845 by M. Adam Davis

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Radio Shack has some red LEDs which have a 2.25 typical voltage drop.
You can use five of them in series with a resister.  When they go out
the voltage of your battery will be under 11.25 volts - though the
voltage drop varies from LED to LED so you may want to buy ten and
measure the drop, selecting those that will give you the total drop you
need.

They will dim as the voltage goes down, going out completely when it's
lower than their combined voltage drop.

You would want to use a suitably large resistor and perhaps a zener,
since voltage spikes while the engine is running can destroy the LEDs.

Alternately you can reverse the whole setup.  Use two nine-volt
batteries in series to generate 18 volts.  Ground the negative to the
car ground.  Hook three LEDs and a resistor between the car battery + to
the 9 volt +.  When the car drops past 11.25, the voltage difference
between the two terminals will be 6.75 and the LEDs will light.  The
problem with this is that 9volt batteries start off with more than 9
volts, and they, of course, die over time as well.  You can use a
voltage regulator to overcome this, but at that point you may as well
throw in a few more components and make a real solution.

But to answer your question - some circuits either use limits built into
semiconductors (such as the forward voltage drop of a junction as
described above), while others use two voltages, with one as the reference.

To generate a reference voltage from your source you could simply use a
low current 5 volt regulator.  As long as the battery is above 6 volts
the regulator will always provide 5 volts out regardless of the battery
voltage.  There are special voltage regulators that are meant just for
reference, but in your case a zener diode would work just fine as a
cheap regulator.

-Adam

Tal Bejerano - AMC wrote:

{Quote hidden}

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2002\05\22@200358 by Tal (Zapta)

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Typically you need 3 voltages for the thing to work.

The first is the power to operate the comparator . You want this to be above
some minimum (e.g. 8V) such that it can power
the rest of the circuit.

The second is the reference, this is a fixed voltage (e.g. 5V) that does not
change when the power changes. You can get the
reference voltage by using a Zener diode and a resister or by a more fancy
(and accurate) voltage reference chip.

The third is the compared voltage which is relative to the power voltage and
is compared to the reference. For example, you can
have a voltage divider (two resistors) that will map your threshold voltage
11.5V to your reference voltage (e.g. 5V).

Of course, if the battery is completely dead, you will not get any
indication.

Tal



> {Original Message removed}

2002\05\22@205106 by James

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You use a zener diode as a voltage reference in whatever circuit you choose
so no matter what the supply voltage is your reference will be stable
(within reason).

----- Original Message -----
From: "Pic Dude" <spam_OUTpicdudeTakeThisOuTspamAVN-TECH.COM>
To: <.....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU>
Sent: Wednesday, May 22, 2002 4:03 PM
Subject: Re: [EE]: Voltage Compare


> I did this recently with a PIC 16F872.  A 5V regulator drops the ~12V
> down to the operating voltage of the PIC, as well as providing the
reference
> voltage.  The input is simply a voltage divider on the ~12V supply to
limit
> the voltage going into the PIC.  Actually I set it up so 25.5V at the
input
> will divide down to 5V.  This way, 25.5V reads 255 on the top 8 bits of
> the A/D and I saved myself some math that way.  But I do lose a bit of
> resolution (not a big deal).  Also have upper and lower alarms settable
> with a single-button interface.
>
> Cheers,
> -Neil.
>
>
>
> {Original Message removed}

2002\05\22@223506 by Rick C.

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What is the significance of 11.5 volts?
An automobile battery should never be allowed to discharge that low. It wouldn't
start the engine.
Here are the voltages and percents, SOC (state of charge) for normal lead acid
batteries:
12.7 - 100%
12.6 - 90%
12.5 - 80%
12.3 - 70%
12.2 - 60%
12.1 - 50%
12.0 - 40%
11.8 - 30%
11.7 - 20%
11.6 - 10%
<11.6 is considered fully discharged
These measurements of capacity should be done after the battery is idle (under
no load) for six hours.

Running a car battery to less than 10% SOC about three times will permenantly
damage the battery and it will never fully recover. Deep cycle batteries are a
little more forgiving but should not be used for automobile starting.
Rick

Tal Bejerano - AMC wrote:

{Quote hidden}

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2002\05\22@231821 by Rick C.

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Useful link.
http://www.uuhome.de/william.darden/carfaq.htm
Rick

Tal Bejerano - AMC wrote:

{Quote hidden}

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2002\05\23@005635 by Rick C.

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Let me try to give you a straight answer. All that is needed is an op amp,
voltage reference, and a voltage divider.
A piezo buzzer can be used as the alarm.

Use an LM336-2.5 as a reference to the + input to the op amp. It is powered from
the +12 volt battery source.
Use any op amp. LM741, LM1458, etc. Plus to +12 volt battery source, - to
ground.
Piezo buzzer from op amp output to +12 volt battery source.
A voltage divider from the +12 volt battery source to ground. The divider made
up of two resistors and a 1K0 trimpot.
+12 volt battery source through 8K2 resistor to end of pot. Other end of pot
through 2K2 resistor to ground. Wiper of pot to negative input of op amp.
Set trimpot to threshold where you want the alarm to sound.
If there is too much hysteresis, place a feedback resistor from the op amp
output to the - input if the op amp. Value should be around M10.
Total current draw will be about 5 mils when off. A CMOS op amp will reduce this
significantly.

I'm just doing this from memory so you might have to fudge the values slightly
but in theory, it will work.
Good luck
Rick



Tal Bejerano - AMC wrote:

{Quote hidden}

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2002\05\23@010350 by Jinx

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> http://www.uuhome.de/william.darden/carfaq.htm

What you'd call "comprehensive"

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2002\05\23@073019 by Byron A Jeff

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On Thu, May 23, 2002 at 12:07:02AM -0400, Rick C. wrote:
> Let me try to give you a straight answer. All that is needed is an op amp,

Yes.

> voltage reference,

Yes.

> and a voltage divider.

And yes.

> A piezo buzzer can be used as the alarm.
>
> Use an LM336-2.5 as a reference to the + input to the op amp.
> It is powered from
> the +12 volt battery source.

The only discussion here is the choice of part. Common parts are almost
always better. A 7805 (or even better a 78L05 if handy) could do the
job as well. No need for a specialized precision part.

> Use any op amp. LM741, LM1458, etc. Plus to +12 volt battery source, - to
> ground.

> Piezo buzzer from op amp output to +12 volt battery source.

Yup.

> A voltage divider from the +12 volt battery source to ground. The divider made
> up of two resistors and a 1K0 trimpot.

Exactly.

{Quote hidden}

Excellent work. I finally all made sense to me when Roman explained that the
control voltage (2.5V in your case, 5V in mine) doesn't have to be in the
range of the voltage to be measured.

BAJ

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2002\05\23@084034 by Rick C.

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Correction. The piezo should be across the output and ground, not the output and +
supply. I had it inverted because I originally had a transistor driver on the output
when I did it from memory. You're right, any stable reference can be used but since
Jinx brought it up (LM335-2.5), I decided to reference it.
Rick

Yup.

Exactly.

Byron A Jeff wrote:

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2002\05\23@101525 by Jinx

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> when I did it from memory. You're right, any stable reference
> can be used but since Jinx brought it up (LM335-2.5), I decided
> to reference it

The circuit I looked at was in this month's Silicon Chip. It doesn't
use a PIC so there's no need for a 5V supply. I agree that even
so it would still have been cheaper to use a 5V regulator than a
more expensive Vref IC. I've found the 78L05 is always between
4.98 and 5.02, which is good enough. It just happened that I was
thinking about battery protection soon before I picked up this
issue

Re: the Silicon Chip circuit

http://home.clear.net.nz/pages/joecolquitt/bat_guard.html

From time to time, it's asked how to implement a high-side
FET switch. SC detail the problems of driving high-side FETs
and have chosen a pulse transformer, using a toroid and a
few windings, to do it. Has this been offered as a solution
by the list before ? If not, any comments ?

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