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PICList Thread
'12c508 fed by 9V battery'
1998\07\16@035217 by Leo van Loon

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Dear friends,

I am producing an electronic dice with a PIC12C508 as a kit for children.
For simplicity I must use a 9V block battery, but the PIC does not like 9V.
How can I reduce 9V to 5V with a minimum of components?
Does only a serial 4,7 V zener work?
I feed the seven LED's of the display directly from the battery, that can be
no problem.

Leo van Loon
SBB simpeltronics
Netherlands
tel +31 (0481) 450034
fax+31 (0481) 450051
mail spam_OUTsbb.simpeltronTakeThisOuTspamtip.nl
url http://www.sbb-simpeltronics.nl
SBB simpeltronics ontwikkelt technische projecten voor basisschool en
basisvorming.
SBB simpeltronics develops technical projects for children in primary and
secondary education.

1998\07\16@035637 by David BALDWIN

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Leo van Loon wrote:
{Quote hidden}

--
       Use 78L05 instead of zener

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1998\07\16@090344 by paulb

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Leo van Loon wrote:

> For simplicity I must use a 9V block battery, but the PIC does not
> like 9V.

 Dead right there, recently mentioned on this list.

> How can I reduce 9V to 5V with a minimum of components?
> Does only a serial 4,7 V zener work?

 As I see it, you go either of two ways.  For robustness, you use a
78L05 (as previously advised), because this allows you to maintain
optimum performance over a wide battery voltage.  Preferably use a low-
dropout regulator version.  And these are almost as cheap as
transistors.

 Alternately, consider the range of workable voltage of the PIC (about
3 to 6V is it not?) as a resource.  Connect the two LEDs which must
operate together (the diagonal ones) in series and use them to feed the
PIC and simultaneously drop 3V from the battery.  To drop another 0.6V
just to get it within ratings, use a series diode.

 These LEDs may barely light (but not in sleep mode) especially if you
use high- efficiency ones (but you wouldn«t, would you?).  To truly
light them, switch a resistor across the PIC using one of its outputs.

 Similarly the other two pairs are in series plus a silicon diode, from
the +9V supply to a PIC output each but *via* a resistor of the same
size.  The single LED must needs be driven separately, either by a
transistor or using a series "ballast" LED and diode as the others.

 The big concern with electronic dice (you meant you were producing an
electronic "die") is that switching the display LED current can warp the
randomisation algorithm.  With a PIC you have the opportunity to
actually run the randomisation very quickly (timing the button press in
µS, mod 6) then knowing the desired result, implement a "run-down"
display which eventually stops on the already-determined value!

 Actually, on reflection, a single silicon power diode in series with
all three LED pairs will both reduce the voltage to bring it into the 6V
rating and provide reverse-battery protection.

 Cheers,
       Paul B.

1998\07\16@113132 by Peter L. Peres

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On Thu, 16 Jul 1998, Leo van Loon wrote:

> Dear friends,
>
> I am producing an electronic dice with a PIC12C508 as a kit for children.
> For simplicity I must use a 9V block battery, but the PIC does not like 9V.
> How can I reduce 9V to 5V with a minimum of components?
> Does only a serial 4,7 V zener work?
> I feed the seven LED's of the display directly from the battery, that can be
> no problem.

If the PIC does not do anything peculiarly complex, a simple series
resistor should be enough, once you know how much current it needs when
working, and making sure that this is fairly constant. Of course you get
to decouple the PIC. Also, make sure that the PIC does not sleep. A power
switch is mandatory.

hope this helps,

       Peter

1998\07\16@162218 by Luis Gonzalez

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Leo van Loon wrote:
>
> Dear friends,
>
> I am producing an electronic dice with a PIC12C508 as a kit for
children.
> For simplicity I must use a 9V block battery, but the PIC does not
like 9V.
> How can I reduce 9V to 5V with a minimum of components?
> Does only a serial 4,7 V zener work?
> I feed the seven LED's of the display directly from the battery, that
can be
> no problem.
>
> Leo

Hello Leo:
There are many aspects that affects the choice for DC-regulation needs
such as:
- the voltage range within your circuit must operate (I4m used 9v
batteries as low as 7 volts for some applications and here all linear
regulators as the 78L05 fails because the dropout can be 2volts or more)
- the quisquent current that you can (loss( in the regulator (I4m
actually working in telephone digital circuits and here the zener
regulator is the worst solution for regulate 5v to one PIC16C65 and some
peripherals because they need some apreciable amount of current trough
it to give the specified voltage drop)
- the operation current of your circuit because many parameters depend
on it (dropout of linear regulators, time response of Low-Dropout
Regulators LDOs, load and line regulation of zener regulators and many
others.

My actual application includes PIC16C65A, LCD display 2*16, I2C memory,
tone encoder and decoder and RS-485 interface taking all the supply
voltage from the telephone line that can vary from 5 to 15 volts
depending on the tipe of line, the RENs connected, the weather, etc. and
finally I decide design a transistor-zener based regulator with 2 BJTs,
1 zener and 3 resistors, and this solution was better than the 78L05 and
simple zener option.

If you have more details about your needs please don4t doubt in say it
to me.


Luis Gonzalez
Bogota, COLOMBIA
.....lucho554KILLspamspam.....hotmail.com

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1998\07\16@190423 by Russell McMahon

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Note that when you turn the displays off they will rise to the
voltage that they are supplied from - this will be 9 volts if
you feed them directly and the PIC may not like it. If you use
drive resistors (as you should) then the PIC protection diodes
will conduct and MAY save you. Better design is to run the
displays from the 5 volt supply that the PIC is using.

A 7805 regulator in a TO220 package will supply the PIC and a
display happily and costs very little more than a 78L05 - the
78L05 is capable of lower power and will probably not have
enough capacity for a bright LED display.

LM7805s (same as LM340T5) or similar can be purchased from most
electronic suppliers who will also be able to supply a simple
circuit to use them.

----------
> From: Leo van Loon <EraseMEsbb.simpeltronspam_OUTspamTakeThisOuTTIP.NL>
> To: PICLISTspamspam_OUTMITVMA.MIT.EDU
> Subject: 12c508 fed by 9V battery
> Date: Thursday, 16 July 1998 19:47
>
> Dear friends,
>
> I am producing an electronic dice with a PIC12C508 as a kit
for children.
> For simplicity I must use a 9V block battery, but the PIC does
not like 9V.
> How can I reduce 9V to 5V with a minimum of components?
> Does only a serial 4,7 V zener work?
> I feed the seven LED's of the display directly from the
battery, that can be
> no problem.
>
> Leo van Loon
> SBB simpeltronics
> Netherlands
> tel +31 (0481) 450034
> fax+31 (0481) 450051
> mail @spam@sbb.simpeltronKILLspamspamtip.nl
> url http://www.sbb-simpeltronics.nl
> SBB simpeltronics ontwikkelt technische projecten voor
basisschool en
> basisvorming.
> SBB simpeltronics develops technical projects for children in
primary and
> secondary education.

1998\07\17@005417 by tjaart

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Luis Gonzalez wrote:

> Leo van Loon wrote:
> >
> > Dear friends,
> >
> > I am producing an electronic dice with a PIC12C508 as a kit for
> children.
> > For simplicity I must use a 9V block battery, but the PIC does not
> like 9V.
> > How can I reduce 9V to 5V with a minimum of components?
> > Does only a serial 4,7 V zener work?
> > I feed the seven LED's of the display directly from the battery, that
> can be
> > no problem.
> >
> > Leo

Your enemies are drop-out and quiescent current.

Use a low-drop regulator for starters. JRC makes a cheap range.

To solve the second problem, use some mechanical contact that
will give some sort of a pulse when the die is activated (I suppose
it could be a pushbutton), and a 4011 flip-flip to switch the
regulator on. The PIC can switch the regulator off again.

--
Friendly Regards

Tjaart van der Walt
KILLspamtjaartKILLspamspamwasp.co.za

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1998\07\17@034528 by William Chops Westfield

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Doesn't the 12C50x run down to 3.something volts?  Increase your
"battery life" by using a 3.something + epsilon regulator instead
of aiming at 5V.  Of course, that'll be more expensive than the
ubiquitous 5V regulators (which tend to be cheaper in the 1A TO220
cases than in the Lower power to-92 cases, anyway, at least from
many of the "hobbyist" suppliers.  A similar argument applies to
zener regulators (of the parallel sort, anyway.)

BillW

1998\07\17@231608 by J Nagy

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>Dear friends,
>
>I am producing an electronic dice with a PIC12C508 as a kit for children.
>For simplicity I must use a 9V block battery, but the PIC does not like 9V.
>How can I reduce 9V to 5V with a minimum of components?
>Does only a serial 4,7 V zener work?
>I feed the seven LED's of the display directly from the battery, that can be
>no problem.
>

I've been quite successful at driving LEDs directly from a '508 with a 3V
supply, and no series resistor. The LED current is typically about 10mA.
Food for thought...

       Jim

1998\07\19@045043 by paulb

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Russell McMahon wrote:

> Note that when you turn the displays off they will rise to the
> voltage that they are supplied from - this will be 9 volts if
> you feed them directly and the PIC may not like it.

 Hmmm.  Why would they do that?  Are they not diodes with a certain
threshold voltage, about 1.3 to 1.4 Volts?  It is true that Zeners have
a soft "knee", but that is not to say they have outright leakage, else
all our rectification circuits would have bad problems!  A LED is a
different device, but AFAIK it does not leak appreciably.

> A 7805 regulator in a TO220 package will supply the PIC and a
> display happily and costs very little more than a 78L05 - the
> 78L05 is capable of lower power and will probably not have
> enough capacity for a bright LED display.

 The TO-220 package 7805 has a typical quiescent current of 3mA.
Probably more than a running PIC and enough to light a LED on its own
(very nicely - standard method of "adjusting" regulators in 1.4V steps
is to put a LED in the "ground" line and you get a free pilot that also
indicates shutdown!

--
 Cheers,
       Paul B.

1998\07\20@091233 by RHS Linux User

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On Thu, 16 Jul 1998, Leo van Loon wrote:

> I am producing an electronic dice with a PIC12C508 as a kit for children.
> For simplicity I must use a 9V block battery, but the PIC does not like 9V.
> How can I reduce 9V to 5V with a minimum of components?

 LM2931Z-5.0  -- a low-dropout voltage regulator specifically made for
battery-powered applications.  (National Semiconductor, if the 'LM' wasn't
a dead giveaway)

 --Crow

/**/

1998\07\21@003031 by Mike Hamilton

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You could use a low drop out regulator from Maxim part number Max667.  This
has a nice feature that can tell the pic when the battery is too low.  Two
of these chips can be obtained for free as samples if you want. Just connect
to their web site. I think it is http://www.maxim.com.
{Original Message removed}

1998\07\21@004135 by Eric Fixler

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actually, http://www.maxim-ic.com

I use them fairly frequently; they have a wide input range and source 250 mA.

e






>You could use a low drop out regulator from Maxim part number Max667.  This
>has a nice feature that can tell the pic when the battery is too low.  Two
>of these chips can be obtained for free as samples if you want. Just connect
>to their web site. I think it is http://www.maxim.com.
>{Original Message removed}

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