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PICList Thread
'Sensing Low Battery Condition'
1997\10\13@140426 by Ricardo Seixas

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Hi PICers

       Does anyone know a quick and dirty way (no Op Amp) to measure a low
battery condition ?
       Let's me explain:
       I've built a portable camber meter using a 16F84 that's powered by
a 9V battery and regulated down to 5V with a 7805, now I'm thinking to
implement a simple low voltage indicator that drives a pin (High or Low)
when battery reachs 5.5V.
       This pin must be shared, since all pins are already used :(
       It's no funny when you're in the middle of a camber adjustment and
the battery dies ...
       Any sugestions ?

Thanks


Ricardo Seixas

-------------------------------------------------------------------------
Ricardo Seixas
Mechanical Engineer
spam_OUTrseixasTakeThisOuTspamciclone.com.br

... Smile, tomorrow will be worst ...

-------------------------------------------------------------------------

1997\10\13@142914 by Jean Mercier

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Your 7805 regulator needs at least 7 volts to operate properly !
Use a low drop out regulator with an on board low input indicator. See
Maxim !
Jean

Ricardo Seixas wrote:
{Quote hidden}

1997\10\13@162205 by Roger L Stevens

picon face
Try the MAXIM MAX 666. It is an adjustable low drop out
regulator with a low battery indicator output  (LBO).
The LBO is open drain, low for low battery, thus may
be shared with another function.

Thanks to those who posted the HOLTEK info for me.

Roger
-----------------------------------------
What is a JOWETT JUPITER ?
Are there really Jowett Nuts ?
rogerlstevensspamKILLspamjuno.com
-----------------------------------------

1997\10\14@013025 by Mike Keitz

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On Mon, 13 Oct 1997 16:02:14 -0300 Ricardo Seixas
<.....rseixasKILLspamspam.....CICLONE.COM.BR> writes:
>Hi PICers
>
>        Does anyone know a quick and dirty way (no Op Amp) to measure
>a low
>battery condition ?

Use one of the Panasonic voltage detectors in the Digi-Key catalog with a
voltage divider on the supply pin.  The current drawn from this pin is
only 5uA or so so the voltage divider resistors can be large.


>        This pin must be shared, since all pins are already used :(

It depends on what sheme you're using for pin sharing.  They have 3 kinds
of outputs, I think CMOS, inverted CMOS, and open-drain.  You want the
open-drain one so the output current doesn't flow through the voltage
divider and upset the comparison.  If you cut the power to the open-drain
one (pull Vdd pin low through a diode), the output should stay at
high-impedance.

1997\10\14@015048 by tjaart

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Ricardo Seixas wrote:
>
> Hi PICers
>
>         Does anyone know a quick and dirty way (no Op Amp) to measure a low
> battery condition ?
>         Let's me explain:
>         I've built a portable camber meter using a 16F84 that's powered by
> a 9V battery and regulated down to 5V with a 7805, now I'm thinking to
> implement a simple low voltage indicator that drives a pin (High or Low)
> when battery reachs 5.5V.
>         This pin must be shared, since all pins are already used :(
>         It's no funny when you're in the middle of a camber adjustment and
> the battery dies ...
>         Any sugestions ?
>
> Thanks
>
> Ricardo Seixas
>

Two resistors as a voltage divider on an input. Choose the values so the
input will 'trip' from high to low when the battery goes low. It won't
be
very accurate, but it will be cheap & simple.

--
Friendly Regards

Tjaart van der Walt
EraseMEtjaartspam_OUTspamTakeThisOuTwasp.co.za
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1997\10\14@054507 by Mike Smith

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-----Original Message-----
From: Ricardo Seixas <rseixasspamspam_OUTCICLONE.COM.BR>
To: @spam@PICLISTKILLspamspamMITVMA.MIT.EDU <KILLspamPICLISTKILLspamspamMITVMA.MIT.EDU>
Date: Tuesday, 14 October 1997 3:35
Subject: Sensing Low Battery Condition


{Quote hidden}

Measuring terminal volts may not be a good way of indicating low battery,
becuase you will get so little warning of impending dead battery that it is
meaningless (nicads are a good example).
How about storing 'battery used' time?  This works well with rechargables.
Microchip make a dedicated chip for battery management that includes this -
you interrogate it for battery life.

MikeS
<RemoveMEmikesmith_ozTakeThisOuTspamrelaymail.net>

1997\10\14@084214 by Larry G. Nelson Sr.

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You can use a comparitor. Set the input voltages using a divider network on
the raw DC and  the other side off the regulated DC. When the raw DC
divided value goes below the regulated value the comparitor will change
state and you will know the battery is below the threshold. I use a similar
technique with a 16C73 using an A/D channel to monitor battery voltage. Set
the divider networks so the output voltage from each is equal when the
battery voltage is at the minimum you find acceptable.


At 04:02 PM 10/13/97 -0300, you wrote:
{Quote hidden}

Larry G. Nelson Sr.
TakeThisOuTL.NelsonEraseMEspamspam_OUTieee.org
http://www.ultranet.com/~nr

1997\10\14@140004 by Steve Smith

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Use a TL431CLP programable zener needs 2.5v ref therefore can set to turn on
at 4.8V with just 2 resistors and a cap output is low when voltage is ok any
port b line with pull ups turned on works fine. Using a 4k7 it can be twined
with something like an LCD enable line (active high) set to output read pin
as input the TL431 takes over cost about #0.30 total.

ascii circuit (poor)

5v-----2k7-----I      TL431,o----4k7-------pic port
0v-----3k0-----I-----TL431,c
0v--------------------TL431,g

Pic port pin may need a pull up r depending on use (use on an output active
high pin)

Cheers Steve.....

1997\10\16@022625 by Dwayne Reid

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>        Does anyone know a quick and dirty way (no Op Amp) to measure a low
>battery condition ?
>        Let's me explain:
>        I've built a portable camber meter using a 16F84 that's powered by
>a 9V battery and regulated down to 5V with a 7805, now I'm thinking to
>implement a simple low voltage indicator that drives a pin (High or Low)
>when battery reachs 5.5V.
>        This pin must be shared, since all pins are already used :(

I have a simple circuit using a few resistors and almost any PNP transistor.
It works well only when you are using a low dropout regulator however.

I generally use a National LP2950 regulator (0.2 V dropout, loads less than
50 mA) or a LM2931 (similar dropout, good for several hundred mA, but a real
pig for current when the battery gets low).  My best recomendation is the
LP2950 if your current requirements will allow it.  NOTE: any of these low
dropout regulators MUST have a large output filter cap (10 uF) or they will
oscillate.

The actual low V dropout detect is based upon the 0.6 V emitter-base turn on
voltage of the transistor.  When the battery voltage drops to less than
about 5.7V, the transistor turns off.  The dropout voltage of the regulators
mentioned earlier is about 0.2V, well below the turn on voltage of the
transistor.  It gives solid, reliable low voltage warning from while the
battery is still able to keep the regulator in regulation down to the point
where the micro won't function any more.



           -----------    ------/\/----+----  to PIC pin
          |         E \  / C    47K    |      HI=batt OK
          |          ------            \      LO=batt low
          |             |              /
          |             /         100K \
          |             \ 47K          |
          |             /             gnd
          |   -------   |
  from ---+---| reg |---+-- +5v to circuit
  batt    |   ---+---   |
     i/p ---     |     --- o/p
     cap ---     |     --- cap
          |      |      |
         gnd    gnd    gnd


notes:
1) the voltage at the collector of the transistor nearly the same as the
battery voltage while the battery is above the dropout voltage.  The 47K
series resistor at collector limits the current into the port pin protection
diodes.  The 100K pulldown resistor pull the port pin to gnd when the
transistor turns off.  Note that the port pin must be configured for TTL
input levels (don't use pin RA4 unless you change the 47K resistor).

2) The circuit that you are powering with this must draw a certain minimum
amount of current or the regulated voltage will rise above the desired
value.  This is because both 47K resistors are supplying current from the
unregulated input to the regulated output.  For example, if using a 9V
battery, the minimum current the circuit must draw is (9-5) / 47K + (9-5) /
47K = 0.17 mA.  Note that almost none of the current is wasted - most of it
goes to the 5 V rail.

3) I have used this when I am really short of IO pins by using the output
from the transitor as a pullup for a data line or pushbutton input and
adding a 1 uF cap from the collector to gnd.  Software then checks to see if
the line is low for a longer period than normal; if it is, then the battery
is too low and I can warn the user of the fact.  The added cap at the
collector ensures that short "blips" of current that might pull a weak
battery below the dropout threshold do not corrupt the normal function of
the IO line.

If you don't need to warn the user of a low battery and just want the PIC to
play dead, then use the output to drive MCLR.  You'll need to change the
output series resistor to about 10K, the pulldown resistor to about 39K, add
a signal diode from MCLR to VDD (anode to MCLR), and another 1 uF cap from
MCLR to either VDD or gnd.  The diode ensures that MCLR stays within
allowable limits (it doesn't have a protect diode to VDD), the cap keeps
noise from resetting the PIC inadvertantly.  I personally don't use this
method for user operated devices, however, a variation of this is what I use
for all unattended (embedded) type projects.

I hope this helps.

Dwayne Reid   <RemoveMEdwaynerspamTakeThisOuTplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, Alberta, CANADA
(403) 489-3199 voice     (403) 487-6397 fax

1997\10\16@130423 by Mike Keitz

picon face
On Thu, 16 Oct 1997 00:30:11 -0600 Dwayne Reid <dwaynerEraseMEspam.....PLANET.EON.NET>
writes:

>I have a simple circuit using a few resistors and almost any PNP
>transistor.

 When the battery voltage drops to less
{Quote hidden}

Rearrange the circuit so the resistor in series with the emitter instead,
and connect the base directly to the 5V line.  The collector voltage will
then rise only to about 5.6V regardless of the battery voltage.   A 100K
resistor from the collector to the PIC pin will limit flow into the
protection diode to a very small amount.

The circuit becomes a common-base voltage amplifier then.  When the
transistor comes out of saturation (because the input voltage is
falling), the emitter and collector currents are essentially the same.
If you use a 1M load resistor, 2.5 uA of collector current are needed to
reach 2.5V on the PIC pin.  For a 6V minimum input (1V minimum difference
across regulator), the emitter resistor would need to be (1 - 0.7) / 2.5u
= 120K.  With this setup the PIC pin will reach 5V when the input voltage
rises to 6.3V.  When the input voltage is well above the threshold, 5.6
uA of the emitter current goes to ground through the load resitor, and
the rest into the 5V line through the base.  Also, it is safe to short
the transistor's collector to ground.  This may be useful driving MCLR
for example and providing a manual reset button.

The sensing threshold could be raised more predictably by putting diodes,
Zeners, etc. in the emitter circuit.  This could be useful if the
regulator is not the low-dropout type, or advance warning of power-off is
required to backup into EEPROM, etc.

 Note that the port pin must be configured for
>TTL
>input levels (don't use pin RA4 unless you change the 47K resistor).

Don't use RA4 with the original circuit (unless adding a diode), since
the open-circuit voltage can rise above 5V (changing the 47K resistor to
prevent this would make the PIC pin voltage go below 2.5V even when the
battery was well above 6V).  But, using the emitter-resistor circuit, RA4
may be the best choice, since it can withstand 5.6V without flowing any
current.  Applying significantly more than 5V to RA4 could put the PIC in
a test mode.  The emitter-resistor circuit should also be suitable to
drive MCLR directly.

>
>2) The circuit that you are powering with this must draw a certain
>minimum
>amount of current or the regulated voltage will rise above the desired
>value.  This is because both 47K resistors are supplying current from
>the
>unregulated input to the regulated output.

True.  If the circuit is to go to sleep, the 5V line will rise above 5V.
There doesn't seem to be a simple way to disable the voltage-detector
circuit to prevent it from leaking current into the 5V line.

1997\10\16@160744 by Dwayne Reid

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 <snip>
>>
>>            -----------    ------/\/----+----  to PIC pin
>>           |         E \  / C    47K    |      HI=batt OK
>>           |          ------            \      LO=batt low
>>           |             |              /
>>           |             /         100K \
>>           |             \ 47K          |
>>           |             /             gnd
>>           |   -------   |
>>   from ---+---| reg |---+-- +5v to circuit
>>   batt    |   ---+---   |
>>      i/p ---     |     --- o/p
>>      cap ---     |     --- cap
>>           |      |      |
>>          gnd    gnd    gnd
>>
>>

Mile Keitz suggested:

>Rearrange the circuit so the resistor in series with the emitter instead,
>and connect the base directly to the 5V line.  The collector voltage will
>then rise only to about 5.6V regardless of the battery voltage.   A 100K
>resistor from the collector to the PIC pin will limit flow into the
>protection diode to a very small amount.

good point, good call.  There was a good reason that I didn't do that on my
original version (but I can't think of it right now, dammit), and inertia
just kept me using the circuit.  Another circuit that I posted here a couple
of years back used a 4.3V zener and the Vbe drop to make a simple 5V zener
regulator.  The output of that one directly drives MCLR, and the transistor
for that one IS used in common base configuration.  Many thousands of that
one out there so far (part of a furnace controller).

<more snip>

>The sensing threshold could be raised more predictably by putting diodes,
>Zeners, etc. in the emitter circuit.  This could be useful if the
>regulator is not the low-dropout type, or advance warning of power-off is
>required to backup into EEPROM, etc.

I suppose . . . but would you really want to?

Personally, I want to get as much life out of the battery as possible.  That
means using the low dropout reg.  If I have to worry about updating EEPROM,
I use a port pin into the PIC (not MCLR, which would shut the PIC down
immediately) and power the product with alkaline batteries.  So far, the
0.4V difference between signalling a low battery and the regulator reaching
dropout voltage has allowed enough time to complete any pending writes
(usually just a single byte).

<more snip>

But, using the emitter-resistor circuit, RA4
>may be the best choice, since it can withstand 5.6V without flowing any
>current.  Applying significantly more than 5V to RA4 could put the PIC in
>a test mode.  The emitter-resistor circuit should also be suitable to
>drive MCLR directly.

Your suggested change means that RA4 is indeed the optimal input pin.  Only
3 components implements the dropout detector: the transistor, with B tied
directly to VDD, the emitter resistor to Vin, and the pulldown resistor from
C to gnd.

<more snip>

>True.  If the circuit is to go to sleep, the 5V line will rise above 5V.
>There doesn't seem to be a simple way to disable the voltage-detector
>circuit to prevent it from leaking current into the 5V line.

I haven't done any projects where I have had to put the PIC to sleep yet.
The most that I have done is power down things like displays, sensors, etc
while still leaving the PIC operating.

Dwayne Reid   <EraseMEdwaynerspamplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, Alberta, CANADA
(403) 489-3199 voice     (403) 487-6397 fax

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