Missed the orginal post - but a PIC should onil draw nano amps in sleep mode
and ma as required by the circuit otherwise.

The following is a reply to the same problem discussed in another newsgroup.
pick a resistor that you believe will drop a tolerable amount of voltage at
the average current. For example, if you believe the average current is
150uA, you might pick 1K 1% which will drop 150mV at that current. Then put
a capacitor across it to smooth the reading at the repetition rate of
whatever pulses etc. are being drawn from the supply. Maybe you pick a time
constant of 1 second, so 1000uF electrolytic. This is a single pole LPF. You
use the meter in voltage mode, of course. In stubborn cases, you may wish to
add a second pole, say a 100K resistor and 10uF tantalum. That won't affect
the reading with a >10M input Z, the average current is still 1mV/uA within
a couple of percent.

If the voltage turns out to be too large or small - so it drops too much
voltage to the circuit, or the voltage is too small to measure easily, then
just change it (and the capacitors)

Note that if current is drawn in short pulses you have to make sure that the
instantaneous decrease in the voltage is not too high or the reading may be
inaccurate or the circuit may malfunction. A 'scope is useful in this
regard. You can also calculate it to bracket the problem.

Eg. Supply the 150uA is essentially all composed of pulses: 150mA for
200usec with a 200msec period. A 100uF capacitor would see a voltage change
at the terminals of delta-V = I * Ton/C = 300mV, which is a bit high. The
1000uF would yield a more reasonable 30mV, so the troughs would be 180mV
down from nominal, not 450mV down. The 30mV/300mV p-p of ripple will yield
jumpy readings as the 5Hz frequency "beats" with the measurement cycle of
the meter. This is where the second pole filter comes in handy.

Roy

CA Zuke wrote:
> 
> Hi,
> 
> Can someone PLEASE tell me what the current (Amperes) should be for
> powering a pic?
> 
> I use a L7805 voltage regulator but it still seems like I can get a
> whole range of different levels of current depending on the batteries
> use. (also you get different Amps for the L7805. Some are 0.1A and
> others are 1A or even 2A)
> 
> I have 4 1.2V 1800mA batteries but I guess that current is far too
> high. The 9V rechargable I have is 150mAh. How is it possible that the
> 9V has so little curent compared to he 1.2V??? I aways thought 9V is
> stronger?
> 
> Anyway.. I really want to power my pic off a battery. So please let me
> know what the current and voltage should be.
> 
> Thanks in advance
> CE AUKE


I run a PIC16F628, four EEPROMs, and one LED (LED is ON for only 100ms
ever sec.) from a 9v (actually 7+ volt) NiCd battery.  LED anode is
connected to the 78L05 output through several hundred ohms resistor, LED
cathode is connected to a PIC pin which pulls it low to turn the LED
on.  I have used this circuit for about a year and it works fine.

As long as the total current of your circuit never exceeds the
current-supplying capability of the battery or the 78L05, your circuit
should work fine too.

"Just do it."

Michael

Spehro Pefhany wrote:

> On Mon, 18 Oct 2004 10:48:01 +0100, the renowned Ian Bell
> <ruffrecords@yahoo.com> wrote:
> 
>>CA Zuke wrote:
>>
>>>> So, you need to pick a power source, either battery or regulator that:
>>>> 
>>>> a) provides a voltage suitable for the pic
>>>> 
>>>> b) is capable of providing at least the amount of current the pic
>>>> requires.
>>>> 
>>>> AFAIK pics require only a few tens of milliamps current so a 78L05 or
>>>> your batteries should be fine.
>>>> 
>>>> HTH
>>>> 
>>>> Ian
>>> 
>>> So does that mean that the current of the power source can never break
>>> a pic? Because the pic only takes the amount of power it needs? That
>>> sounds strange... Does that mean if I have a pic connected to one led,
>>> and I give it 10A that everything will work properly if the Voltage is
>>> within the specifications of the pic?
>>
>>Yes.  You cannot 'give it 10A'.
> 
> 
> Well, you *could*, with a CC supply...
> 
> A little inaccuracy sometimes saves tons of explanation.
>     --Saki (1870 - 1916)

I was quite conscious of the several inaccuracies in my reply but I felt
that, given the OPs obvious lack of basic electronics knowledge, further
detail would only confuse him.  I'm with you and Saki on this one.

Ian
> 
> 
> 
> Best regards,
> Spehro Pefhany

-- 
Ian Bell

Meindert Sprang wrote:
> "Hans-Bernhard Broeker" <broeker@physik.rwth-aachen.de> wrote in message
>
>> <Nitpick> Or you apply a current, and the device determines how much
>> voltage you have to supply to get that current through it.  Devices
>> known as "constant current sources" aren't exactly the typical case,
>> sure, but they do exist.  Many households have one these days, known
>> to the laypersons as a (simplistic version of a) battery
>> charger. </Nitpick>
> 
> I know they exist, but I didn't want to throw that on him too, because
> it would have contributed heavily to the confusion, I think :-)

Don't forget Van de Graffe machines, which have a much higher
voltage compliance range than the common tufted battery charger :-)

-- 
 "I support the Red Sox and any team that beats the Yankees"
 
 "Any baby snookums can be a Yankee fan, it takes real moral
  fiber to be a Red Sox fan"

On Mon, 18 Oct 2004 13:09:39 +0200, the renowned "Frank Bemelman"
<f.bemelmanx@xs4all.invalid.nl> wrote:

>"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> schreef in bericht
>news:fc57n0lnlab11ddtkocvdvm4cbglm79072@4ax.com...
>> On Mon, 18 Oct 2004 10:48:01 +0100, the renowned Ian Bell
>> <ruffrecords@yahoo.com> wrote:
>>
>> >CA Zuke wrote:
>> >
>> >>> So, you need to pick a power source, either battery or regulator that:
>> >>>
>> >>> a) provides a voltage suitable for the pic
>> >>>
>> >>> b) is capable of providing at least the amount of current the pic
>> >>> requires.
>> >>>
>> >>> AFAIK pics require only a few tens of milliamps current so a 78L05 or
>> >>> your batteries should be fine.
>> >>>
>> >>> HTH
>> >>>
>> >>> Ian
>> >>
>> >> So does that mean that the current of the power source can never break
>> >> a pic? Because the pic only takes the amount of power it needs? That
>> >> sounds strange... Does that mean if I have a pic connected to one led,
>> >> and I give it 10A that everything will work properly if the Voltage is
>> >> within the specifications of the pic?
>> >
>> >Yes.  You cannot 'give it 10A'.
>>
>>
>> Well, you *could*, with a CC supply...
>
>Sort of last hack, if it won't run?
>
>I am ashamed to admit that I actually did that once ;)

Or accidentally turn the voltage up on a bench supply until it hits
the CC limit. Stuff happens. 


Best regards, 
Spehro Pefhany
-- 
"it's the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> schreef in bericht
news:fc57n0lnlab11ddtkocvdvm4cbglm79072@4ax.com...
> On Mon, 18 Oct 2004 10:48:01 +0100, the renowned Ian Bell
> <ruffrecords@yahoo.com> wrote:
>
> >CA Zuke wrote:
> >
> >>> So, you need to pick a power source, either battery or regulator that:
> >>>
> >>> a) provides a voltage suitable for the pic
> >>>
> >>> b) is capable of providing at least the amount of current the pic
> >>> requires.
> >>>
> >>> AFAIK pics require only a few tens of milliamps current so a 78L05 or
> >>> your batteries should be fine.
> >>>
> >>> HTH
> >>>
> >>> Ian
> >>
> >> So does that mean that the current of the power source can never break
> >> a pic? Because the pic only takes the amount of power it needs? That
> >> sounds strange... Does that mean if I have a pic connected to one led,
> >> and I give it 10A that everything will work properly if the Voltage is
> >> within the specifications of the pic?
> >
> >Yes.  You cannot 'give it 10A'.
>
>
> Well, you *could*, with a CC supply...

Sort of last hack, if it won't run?

I am ashamed to admit that I actually did that once ;)

-- 
Thanks, Frank.
(remove 'x' and 'invalid' when replying by email)

CA Zuke wrote:
>>So, you need to pick a power source, either battery or regulator that:
>>
>>a) provides a voltage suitable for the pic
>>
>>b) is capable of providing at least the amount of current the pic requires.
>>
>>AFAIK pics require only a few tens of milliamps current so a 78L05 or your
>>batteries should be fine.
>>
>>HTH
>>
>>Ian
> 
> 
> So does that mean that the current of the power source can never break
> a pic? Because the pic only takes the amount of power it needs? That
> sounds strange... Does that mean if I have a pic connected to one led,
> and I give it 10A that everything will work properly if the Voltage is
> within the specifications of the pic?

Think of electricity as water.  If you raise a bucket of water 2 meters, 
then that's voltage.  If you attach a hose to the bottom of the bucket, 
that the hose is your PIC and the water running through it is the 
current.  As you can probably picture in this (flawed) analogy, if you 
have a much wider pipe, more "current" can flow.  As with the bucket, 
you "apply" a voltage (the height to which you raise it) and the current 
(the amount of water running through it) is a function of the size of 
the pipe (impedence).

You might find this http://www.ibiblio.org/obp/electricCircuits/ 
interesting and useful.  It's a free online textbook about electric 
circuits and electricity.  You'll find that your PIC experiments will be 
much more sucessful and rewarding if you pause for a bit to get some of 
the basics.  Even if you only read through chapters 1-3, you'll get a lot.

Ed

"Hans-Bernhard Broeker" <broeker@physik.rwth-aachen.de> wrote in message
news:2thlbdF1urnn1U5@uni-berlin.de...
> <Nitpick> Or you apply a current, and the device determines how much
> voltage you have to supply to get that current through it.  Devices
> known as "constant current sources" aren't exactly the typical case,
> sure, but they do exist.  Many households have one these days, known
> to the laypersons as a (simplistic version of a) battery
> charger. </Nitpick>

I know they exist, but I didn't want to throw that on him too, because it
would have contributed heavily to the confusion, I think :-)

Meindert

On Mon, 18 Oct 2004 10:48:01 +0100, the renowned Ian Bell
<ruffrecords@yahoo.com> wrote:

>CA Zuke wrote:
>
>>> So, you need to pick a power source, either battery or regulator that:
>>> 
>>> a) provides a voltage suitable for the pic
>>> 
>>> b) is capable of providing at least the amount of current the pic
>>> requires.
>>> 
>>> AFAIK pics require only a few tens of milliamps current so a 78L05 or
>>> your batteries should be fine.
>>> 
>>> HTH
>>> 
>>> Ian
>> 
>> So does that mean that the current of the power source can never break
>> a pic? Because the pic only takes the amount of power it needs? That
>> sounds strange... Does that mean if I have a pic connected to one led,
>> and I give it 10A that everything will work properly if the Voltage is
>> within the specifications of the pic?
>
>Yes.  You cannot 'give it 10A'. 


Well, you *could*, with a CC supply... 

A little inaccuracy sometimes saves tons of explanation.
    --Saki (1870 - 1916)



Best regards, 
Spehro Pefhany
-- 
"it's the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

Meindert Sprang <mhsprang@nocustomspamware.nl> wrote:
> "CA Zuke" <ceauke@yahoo.com> wrote in message
> news:ff8b1683.0410180038.1e8c54b3@posting.google.com...
> > So does that mean that the current of the power source can never break
> > a pic? Because the pic only takes the amount of power it needs? That
> > sounds strange... Does that mean if I have a pic connected to one led,
> > and I give it 10A that everything will work properly if the Voltage is
> > within the specifications of the pic?

> I think you don't quite understand how this works. You don't *give*
> something a current, you apply a voltage. And the devices that gets the
> voltage determines how much current it draws from the voltage source.

<Nitpick> Or you apply a current, and the device determines how much
voltage you have to supply to get that current through it.  Devices
known as "constant current sources" aren't exactly the typical case,
sure, but they do exist.  Many households have one these days, known
to the laypersons as a (simplistic version of a) battery
charger. </Nitpick>


-- 
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.