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'pic max voltage'
1998\03\30@125724 by peter

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Does anyone know the max voltage you can run a 16c84 at ?

I tried the other day, at about 8 volts it became unstable
could this have been because it was misreading the eeprom ?
I should try programming it at a higher voltage.
But before I do, has anyone else done this ?

It would be great if its possible to run direct from a 9 volt
battery

I am not bothered if the odd one gets fried in this application
also as current drain will be very low I can use a large current
limiting resistor of 330 Ohms or so in the supply line

As a last resort suppose I could put a led in the supply line
--
Peter Cousens
email: spam_OUTpeterTakeThisOuTspamcousens.her.forthnet.gr  phone: + 3081 324450, 380534
snailmail:  Folia, Agia Fotini, Karteros, Heraklion  Crete, Greece.

1998\03\30@132340 by Michael Hagberg

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just put in a 78L05 or do you want jelly on your toast?

michael

-----Original Message-----
From: Peter Cousens <.....peterKILLspamspam@spam@COUSENS.HER.FORTHNET.GR>
To: PICLISTspamKILLspamMITVMA.MIT.EDU <.....PICLISTKILLspamspam.....MITVMA.MIT.EDU>
Date: Monday, March 30, 1998 12:00 PM
Subject: pic max voltage


{Quote hidden}

1998\03\30@141101 by Mike Keitz

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On Mon, 30 Mar 1998 08:54:27 +0300 Peter Cousens
<peterspamspam_OUTcousens.her.forthnet.gr> writes:
>Does anyone know the max voltage you can run a 16c84 at ?
>
>I tried the other day, at about 8 volts it became unstable
>could this have been because it was misreading the eeprom ?
>I should try programming it at a higher voltage.
>But before I do, has anyone else done this ?

The Absolute Maximum Rating is 7V.  I suspect that the EEPROMS, both data
and program, will be mis-biased at voltages higher than that and the chip
will not function.  Also, even if the PIC doesn't burn out right away,
the high stress from excess voltage could cause it to fail very quickly.

>It would be great if its possible to run direct from a 9 volt
>battery

It is better to drop the voltage down to about 3V, the PIC will use less
current and battery life will be extended.  A battery consisting of 3 or
4 AA or AAA cells in series could be connected to the PIC directly and
run it for much longer than a 9V would.

>>I am not bothered if the odd one gets fried in this application
>also as current drain will be very low I can use a large current
>limiting resistor of 330 Ohms or so in the supply line

A simple regulator of a voltage divider with fairly large resistors
setting the voltage to the base of an NPN transistor voltage follower
could work.  Of course there are reasonably-priced micropower 3-terminal
linear regulators which would be ideal for longest battery life if that
is important.

>As a last resort suppose I could put a led in the supply line

Yes, this is practical.  Depending on the color of the LED about 1.5 to
3V will be dropped.  And the power lost by the LED isn't completely
wasted (as it would be with a linear regulator) since it could act as a
pilot light.


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1998\03\30@150839 by Hardy e/ou Rafael Pinto

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Hi everyone!!

>It is better to drop the voltage down to about 3V, the PIC will use less
>current and battery life will be extended.  A battery consisting of 3 or
>4 AA or AAA cells in series could be connected to the PIC directly and
>run it for much longer than a 9V would.
>
>>>I am not bothered if the odd one gets fried in this application
>>also as current drain will be very low I can use a large current
>>limiting resistor of 330 Ohms or so in the supply line
>
>A simple regulator of a voltage divider with fairly large resistors
>setting the voltage to the base of an NPN transistor voltage follower
>could work.  Of course there are reasonably-priced micropower 3-terminal
>linear regulators which would be ideal for longest battery life if that
>is important.

   Anyway a 9V battery has a charge of about 100mAh. Any regulator you
intend to use must have a very tiny bias current (78L05/LM317) so it would
not "suck" the battery... I'd rather better use an IC or two NPN
(2N3904/BC549C) with high hFE in Darlington configuration with resistors
around 100k. That should do it.

>
>>As a last resort suppose I could put a led in the supply line
>
>Yes, this is practical.  Depending on the color of the LED about 1.5 to
>3V will be dropped.  And the power lost by the LED isn't completely
>wasted (as it would be with a linear regulator) since it could act as a
>pilot light.
>

   Other good idea is to use two CR2032 (3V-coin type) batteries! They have
about 180mAh, are much smaller and much lighter then 9V batts and it's not
too dificult to find holders for them! Check RS Components!

http://rs-components.com



   Rafael Pinto

1998\03\30@154537 by Bob Blick

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Hi Peter,

I have noticed that the PIC starts drawing current above 7 volts, even
with the clock stopped. At 8 volts output pins start moving away from
supply rails and the chip doesn't work so well, like the complementary
output transistor is starting to come on.

Have lost pins at 9 volts, and the chip stops, but works again at normal
voltages(without the pins that burned out).

I won't trouble you with the typical off-topic replies, you already got
many of those :-)

Cheers,
Bob

1998\03\30@164017 by Alex Torres

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> From: Hardy e/ou Rafael Pinto <@spam@cppintoKILLspamspamNETRIO.COM.BR>

> >It is better to drop the voltage down to about 3V, the PIC will use less
> >current and battery life will be extended.  A battery consisting of 3 or
> >4 AA or AAA cells in series could be connected to the PIC directly and
> >run it for much longer than a 9V would.

>     Anyway a 9V battery has a charge of about 100mAh.

Yea. 9v * 0.1Ah = 0.9 Wh, but AA cells usually have  500-600mAh, 2AA :  3v
* 0.5Ah=1.5Wh - more power!

But if the device's price is not a problem - the best solution is to use
DC/DC convertor, not linear regulator,
and step-down convertor may be more effectivity then step-up.

Alex Torres, Kharkov, Ukraine (exUSSR)
KILLspamaltorKILLspamspamchat.ru
2:461/28@FidoNet
www.geocities.com/SiliconValley/Lab/6311

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