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PICList Thread
'Battery consumption & LEDs'
1998\08\06@093246 by freeman.kilpatrick

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Hello All,

I am relatively new to PICs (migrating from STAMPS because I wanted
interrupts), and I did a little experiment to see how much power would be
required for a very small circuit.  I set up the standard "blinking LED"
program, and planned to let it run for a week to see what would happen.  I
had a standard energizer 9V battery driving a PIC 16F84 (through a voltage
regulator, of course), 2 LEDs, and an RC clock circuit.  I was surprised to
see that the battery was dead after only a few days.  I wasn't using
low-power LEDs, just standard Radio Shack stuff, but I still thought the
circuit would last longer.  How can I even make a battery-based application
with a "power on" LED if they drain batteries this fast?

Any advice/education welcome.

Alex

1998\08\06@100246 by Stuart Allen

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> Hello All,
>
> I am relatively new to PICs (migrating from STAMPS because I wanted
> interrupts), and I did a little experiment to see how much power would be
> required for a very small circuit.  I set up the standard "blinking LED"
> program, and planned to let it run for a week to see what would happen.  I
> had a standard energizer 9V battery driving a PIC 16F84 (through a voltage
> regulator, of course), 2 LEDs, and an RC clock circuit.  I was
> surprised to
> see that the battery was dead after only a few days.  I wasn't using
> low-power LEDs, just standard Radio Shack stuff, but I still thought the
> circuit would last longer.  How can I even make a battery-based
> application
> with a "power on" LED if they drain batteries this fast?
>
> Any advice/education welcome.
>
> Alex


Hello Alex,

I would imagine that it was the regulator that caused the battery to run
down so quickly, especially if it was standard 7905 type. The regulator was
probably drawing a large proportion of the total power from the battery;
check its spec and look at quiescent current, thats how much current it
effectively 'wastes'.

Try a special low IQ, drop regulator. MAXIM and Linear Technology have
examples, amongst others.

Also, 9v batteries don't have much capacity at all. Around 550mAH. So you
need to save power in as many places as possible. Does your PIC program go
to sleep at all? Using the watchdog timer to wake itself up? Putting the PIC
to sleep is the key to low power design. It can be asleep whether the LED is
on or off, and will use a fraction of the power it usually does, dropping
from 10s of mA to a few 1uA if you are careful.

What frequency is the clock? The slower the less power it uses. This is
documented in the data sheet.

And last of all, use a low current LED, and use the biggest series resistor
that still keeps it visible.

What else... change the flash duty cycle, ie. on for 2/10s off for 2
seconds.

Hope this helps,

Stuart.

1998\08\06@115840 by Peter L. Peres

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On Thu, 6 Aug 1998, Freeman Kilpatrick wrote:

> required for a very small circuit.  I set up the standard "blinking LED"
> program, and planned to let it run for a week to see what would happen.  I

A standard *low power* blinking LED has Ton = 0.2sec Toff ~= 2 sec. This
alone reduces power consumption from 50% (Ton = Toff) to 10% (5 times).

A LED rigged like this for low power operation, should be driven with a
L/C switching arrangement if one can afford it. A PIC output can do the
chopping (RA4). You will probably use an 'inverted' chopper and connect
the LED A to +Vbatt directly, such that this will be the common rail. Its
current limiting resistor will be calculated for a V = Vleds + 0.3 Volts
(only !) as this is what the chopper supply will output (the duty cycle
will be around 1:2.5 for 9 V input, 3.3 V out = 2 red LEDs and a R for
0.3V@15mA = 20 ohms).

This again reduces the power for the LEDs from the original 9V * 15 mA ~=
0.125W to about 3.3V * 15 mA * 1.2 = 0.06 W, almost twice (assuming
continuous use - but the same factor applies for pulsed operation).

Last, use as low a clock frequency as possible (32 kHz or 500 kHz ceramic
or RC if you can live with the drift), and put the PIC to sleep while the
LED is off by using the watchdog timer. This again induces some time
incertainty.

When you will have done all this, you will have reduced the power
requirements from:

BEFORE:

9V battery
PIC runs all the time @ 4MHz = 1.5 mA (approximate v.)
LEDs run 50% of the time = 7.5 mA
Total: 9 mA
6F22 standard 9V battery will last just 220mAh/9mA ~= 22 hours. An
alkaline will last twice as long, probably.

AFTER:

9V battery
PIC runs 10% of the time @500 kHz, sleeps in between = (a few micro-amperes)
LEDs run 10% of the time at ~ 7mA while on = 0.7 mA over time (d.c. = 10%)
Total: less than 0.8 mA
6F22 standard 9V battery will last about 220mAh/0.8mA = 275 hours, and an
alkaline may last 2.5 times as long under these conditions. The actual
time obtained may be slightly shorter.

So, you can 'lean' your application as much as you please. You can
probably squeeze it down to the power required to run one LED with a 10%
duty cycle at 80% system efficiency, which will lead to about 25
milli-watts for the led, divided by 10, or, at said 6F22
220mAh*9V/2.25milli-watts ~= 880 hours. And this is about as far as you
can go with a 6F22, unless you have patient users who can wait 4 seconds
to see the LED flash, such that you can double the number.

Incidentally, 270 hours is well in excess of 1 week, and 880 hours is just
over a month by almost a week.

hope this helps ;)

       Peter

1998\08\06@120048 by Calvin

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part 0 2457 bytes
<META content=text/html;charset=iso-8859-1 http-equiv=Content-Type>
<META content='"MSHTML 4.72.2106.6"' name=GENERATOR>
</HEAD>
<BODY bgColor=#ffffff>
<DIV>There are several thingsd you can do to save power:<BR>a)Use a low dropout
regulator (NOT! a 7805).<BR>b)Use a large resistor to limit the LED current (not
too large, obviously).<BR>c)Keep the duty-cycle of the LED as low as
possible.<BR>d)Put the PIC to sleep between cycles.<BR>e)Use the lowest clock
speed possible for the PIC.<BR>f)Use a high efficiency LED.<BR>g)Any other
suggestions someone else may have...<BR><BR><BR>Calvin<BR><BR><BR>-----Original
Message-----<BR>From: Freeman Kilpatrick &lt;<A
href="spam_OUTfreeman.kilpatrickTakeThisOuTspamafosr.af.mil">.....freeman.kilpatrickKILLspamspam@spam@afosr.af.mil</A>&gt;<BR>To:
<A href="PICLISTspamKILLspamMITVMA.MIT.EDU">.....PICLISTKILLspamspam.....MITVMA.MIT.EDU</A> &lt;<A
href="EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU">PICLISTspamspam_OUTMITVMA.MIT.EDU</A>&gt;<BR>Date:
Jueves 6 de Agosto de 1998 8:22 AM<BR>Subject: Battery consumption &amp;
LEDs<BR><BR><BR>&gt;Hello All,<BR>&gt;<BR>&gt;I am relatively new to PICs
(migrating from STAMPS because I wanted<BR>&gt;interrupts), and I did a little
experiment to see how much power would be<BR>&gt;required for a very small
circuit.&nbsp; I set up the standard &quot;blinking LED&quot;<BR>&gt;program,
and planned to let it run for a week to see what would happen.&nbsp;
I<BR>&gt;had a standard energizer 9V battery driving a PIC 16F84 (through a
voltage<BR>&gt;regulator, of course), 2 LEDs, and an RC clock circuit.&nbsp; I
was surprised to<BR>&gt;see that the battery was dead after only a few
days.&nbsp; I wasn't using<BR>&gt;low-power LEDs, just standard Radio Shack
stuff, but I still thought the<BR>&gt;circuit would last longer.&nbsp; How can I
even make a battery-based application<BR>&gt;with a &quot;power on&quot; LED if
they drain batteries this fast?<BR>&gt;<BR>&gt;Any advice/education
welcome.<BR>&gt;<BR>&gt;Alex<BR>&gt;<BR></DIV></BODY></HTML>

</x-html>

1998\08\06@121909 by org Hager

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On Thu, 6 Aug 1998, Stuart Allen wrote:

> > circuit would last longer.  How can I even make a battery-based
> > application
> > with a "power on" LED if they drain batteries this fast?
> >
> > Any advice/education welcome.
> >
> > Alex
>
>
> Hello Alex,
>
> I would imagine that it was the regulator that caused the battery to run
> down so quickly, especially if it was standard 7905 type. The regulator was
> probably drawing a large proportion of the total power from the battery;
> check its spec and look at quiescent current, thats how much current it
> effectively 'wastes'.

I've recently measured the quiescent current of an 78L05, and it was
2.5mA! The LM2936Z-5.0 from NS is much better, but needs 10uF of blocking
capacitance at the output to behave properly.

>
> And last of all, use a low current LED, and use the biggest series resistor
> that still keeps it visible.

You can also drive the LED in its `on' state in a PWM-like fashion, e.g.
at 1kHz with a 1:5 duty cycle at 10mA. You will get 2mA of mean current
flowing through the LED, but it will appear brighter to the human eye than
a LED with 2mA steady current. I don't know what the optimal duty cycle
is, but you can save a lot of power that way.

Georg.

1998\08\06@122936 by Peter L. Peres

picon face
On Thu, 6 Aug 1998, Georg Hager wrote:

> I've recently measured the quiescent current of an 78L05, and it was
> 2.5mA! The LM2936Z-5.0 from NS is much better, but needs 10uF of blocking
> capacitance at the output to behave properly.

Huh ?! There was something wrong, for sure. I see max. 0.8 mA for a 'bad'
L05. A good one is more like 0.1 mA. It's not the quiescent current that
is going to get you, it's the lousy efficiency. Drive current for the o/p
transistor comes from the power supply and is not passed to the load...
the 317L is more predictable with 120 uA quiescent and passes the series
transistor drive c. through the load...

Peter

1998\08\06@131950 by Jan Derogee

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Hello Alex,
If you are using normal led's that draw a 20mA current, then your batt. will
last for 2000mAh (batt capacity) / 20mA = 100 hours (aprox 4 days)
If you blink your led (on period = one tenth of the cycle time) then you
can decrease your power drain. But your led won't burn as bright.
If you want a better efficiency for your total project (this also can reduce
power drain) use a better voltage regulator (a switching regulator (step-down)).
If you connect LED's or other power consuming components then try avoid
connecting them to your voltage regulator. Tie your led directly to the positive
9V leads (don't forget to use a higher resistor-value) and control your led
by pulling your PIC-output-ports down (LED lights-up if output=0).

A lower operating frequency for your PIC may also reduce power drain (but
look this up in your datascheets (it may not be worth the effort)).
Try to put your PIC to sleep when it is not needed (i'm not sure if this
can be done in 100% software, perhaps some hardware is needed (check datasheets)

Good luck with this project. I hope this is of any use.
Let me know what you achief.

Jan Derogee

1998\08\06@140113 by Harold Hallikainen

picon face
       It seems that the problem is that we need to drive the LED with a
current source and all we have is a voltage source.  We convert the
constant voltage to a constant current using a series resistor, which is
not very efficient!  So, how about this?

       To avoid ascii art, here's a description of the circuit...

       From +5V connect an inductor in series with an LED in series with
a diode to +5V.  Connect the junction of the LED and the diode (cathode
of LED, anode of diode) to a PIC output pin.  Pulse the pin low.  The LED
current will ramp up as the inductor current increases.  Set the pin
high.  The LED will continue to glow as the the inductor current ramps
down.  The time the pin is high should be long enough to make absolutely
sure the inductor current has ramped down to zero so we don't start the
next ramp up from some value other than zero, eventually resulting in
infinite current (not good).  Once you've done one ramp up and down, do
it again!  The diode is a catch diode and could possibly be replaced by
the clamp diode in the PIC.  With ideal components, this SHOULD be an
efficient way of getting current thru the LED without drops across
resistors.


Thoughts?


Harold




Harold Hallikainen
@spam@haroldKILLspamspamhallikainen.com
Hallikainen & Friends, Inc.
See the FCC Rules at http://hallikainen.com/FccRules and comments filed
in LPFM proceeding at http://hallikainen.com/lpfm


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1998\08\06@144439 by Harold Hallikainen

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       On further thought, to eliminate regulator dissipation due to LED
current, we should connect the "left end" of the inductor to unregulated
+ supply, connect the cathode of the catch diode to unregulated + supply,
and use an open drain output on the PIC to pull the junction of the LED
and the catch diode low.  Also, if the LED can tolerate the unregulated
voltage as a reverse voltage, we can move it over to the catch diode
position, eliminating one part (remember, "the ideal design has zero
parts).
       In this situation, we have +supply going thru an inductor, thru
the LED back to + supply.  No current flows.  The "right side" of the
inductor is pulled low by an open drain PIC output.  Inductor current
ramps up.  The open drain output is released and the inductor current
ramps down through the LED (lighting it).  The process is repeated.


Harold


---------------------------------[previous message
below]---------------------------------

       It seems that the problem is that we need to drive the LED with a
current source and all we have is a voltage source.  We convert the
constant voltage to a constant current using a series resistor, which is
not very efficient!  So, how about this?

       To avoid ascii art, here's a description of the circuit...

       From +5V connect an inductor in series with an LED in series with
a diode to +5V.  Connect the junction of the LED and the diode (cathode
of LED, anode of diode) to a PIC output pin.  Pulse the pin low.  The LED
current will ramp up as the inductor current increases.  Set the pin
high.  The LED will continue to glow as the the inductor current ramps
down.  The time the pin is high should be long enough to make absolutely
sure the inductor current has ramped down to zero so we don't start the
next ramp up from some value other than zero, eventually resulting in
infinite current (not good).  Once you've done one ramp up and down, do
it again!  The diode is a catch diode and could possibly be replaced by
the clamp diode in the PIC.  With ideal components, this SHOULD be an
efficient way of getting current thru the LED without drops across
resistors.


Thoughts?


Harold




Harold Hallikainen
KILLspamharoldKILLspamspamhallikainen.com
Hallikainen & Friends, Inc.
See the FCC Rules at http://hallikainen.com/FccRules and comments filed
in LPFM proceeding at http://hallikainen.com/lpfm

--------- End forwarded message ----------

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1998\08\06@145504 by David VanHorn

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>        It seems that the problem is that we need to drive the LED with a
>current source and all we have is a voltage source.  We convert the
>constant voltage to a constant current using a series resistor, which is
>not very efficient!  So, how about this?


Any linear regulation is just as inefficient.
For double bonus points, use the PIC as it's own switching regulator,
controlling a pass transistor in a buck configuration.

(Also a great excersize as to why emulators often suck!)

1998\08\06@170114 by Mike Keitz

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A lot of people have posted a lot about this subject today.  I'll try to
avoid writing a lot.

First, use a high-efficiency LED.  This is a no-brainer, a tremendous
amount of bang/buck.

Second, everyone so far has taken the 9V battery/linear regulator as a
given.  For most PIC work, it isn't best.  Since a PIC can operate from
3V to 5.5V, a regulator isn't necessary unless something about the rest
of the circuit requires a stable voltage.  A set of three "penlight"
batteries such as AA or AAA size in series will put out 4.5V when new,
and be about completely run down when the voltage reaches 3V.  If a
stable voltage is required, consider 4 batteries and a micropower linear
regulator to 3.5 or 4V.  Especially the AA size have much more ma-H
capacity than a 9V, and when it is time to replace them, new ones will
cost less than the 9V.

Of course, also think about the user interface so the LEDs don't have to
be on any more than necessary.  Even if high efficiency, the LEDs can
easily use more power than the rest of the circuit.  Quick bright flashes
are much more noticeable thatn steady dim light.  Driving an LED that's
supposed to be on "constantly" with pulses too fast to see is probably
worthwhile if the processor needs to be running during that time anyway.

Many have mentioned switching techniques, but this definitely isn't a
good idea for a beginner like the original poster.  Moving the battery
voltage closer to the LED voltage will greatly narrow the advantage that
can be obtained (by increasing the efficiency of the non-switching
circuit).

If there is a lot of extra voltage, I think there are dual LEDs of the
same color that can be operated in series to get twice the brightness
from the same current.  But a battery of only 3V won't be enough to light
them.

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1998\08\07@084031 by Peter L. Peres

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On Thu, 6 Aug 1998, David VanHorn wrote:

> >        It seems that the problem is that we need to drive the LED with a
> >current source and all we have is a voltage source.  We convert the
> >constant voltage to a constant current using a series resistor, which is
> >not very efficient!  So, how about this?
>
>
> Any linear regulation is just as inefficient.
> For double bonus points, use the PIC as it's own switching regulator,
> controlling a pass transistor in a buck configuration.
>
> (Also a great excersize as to why emulators often suck!)

Good point ! But, from the previous postings, I would way that my posting
did not make it ! I had suggested a switching regulator to drive the 2
LEDs with RA4, using the + unreg rail as common rail for the switching
regulator ! Almost all follow-ups suggest the same ideas, but no-one
refers to my posting ?! Maybe it did not make it ?

Peter

1998\08\07@095743 by Claudio Rachiele IW0DZG

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1998\08\07@095956 by David VanHorn

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>> Any linear regulation is just as inefficient.
>> For double bonus points, use the PIC as it's own switching regulator,
>> controlling a pass transistor in a buck configuration.
>>
>> (Also a great excersize as to why emulators often suck!)
>
>Good point ! But, from the previous postings, I would way that my posting
>did not make it ! I had suggested a switching regulator to drive the 2
>LEDs with RA4, using the + unreg rail as common rail for the switching
>regulator ! Almost all follow-ups suggest the same ideas, but no-one
>refers to my posting ?! Maybe it did not make it ?
>
>Peter

It made it here, and yes, that woks.  The thing to note is that any time you
are limiting current with a resistor, you are wasting power.  In low power
apps, you learn to think in terms of Joules, rather than amps or volts.

1998\08\07@101413 by Claudio Rachiele IW0DZG

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1998\08\07@131406 by Claudio Rachiele IW0DZG

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1998\08\07@142415 by Claudio Rachiele IW0DZG

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         Foreign Native Name CN=Claudio Rachiele IW0DZG/OU=Italy/O=IBM\nIBMIT\n\n
         Organization        IBM
         Org Unit 1          Italy
         Last Name           IW0DZG
         First Name          Claudio



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1998\08\07@155425 by Don

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Claudio Rachiele IW0DZG wrote:
>
>                    Status Distribution August 06, 1998 18:40:28
>
> The message regarding "Re: Battery consumption & LEDs" sent on August 06, 1998
18:40:28 was sent by
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>           Organization        IBM
>           Org Unit 1          Italy
>           Last Name           IW0DZG
>           First Name          Claudio
>
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>
> X.400 Status    769
> Explanation     Router: Unable to open mailbox file D14HUBM1/14/H/IBM mail.box
: Server not responding

1998\08\09@104613 by Peter L. Peres
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On Fri, 7 Aug 1998, David VanHorn wrote:

> It made it here, and yes, that woks.  The thing to note is that any time you
> are limiting current with a resistor, you are wasting power.  In low power
> apps, you learn to think in terms of Joules, rather than amps or volts.

The point is, that I fried a 2 mA LED while testing, and I did not manage
to fry it with the R ;) So, one can remove it later...

Peter

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