# Pic Currrent

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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.

CE AUKE

Re: Pic Currrent

The regulators you mention output a given voltage.  The amount of current
they give depends on the load attached to them e.g. a pic plus peripheral
circuits.  They regulate the voltage so it stays the same no matter what
the load current (within certain limits).  The various current you mention
for 7805 regulators are the maximum current load you can draw from each one
whilst it still regulates the voltage.

Similarly with batteries, the currents given are the maximum load the
battery can sustain.  (Actually this is not true but it will do for the
purposes of this explanantion).

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

--
Ian Bell

Re: Pic Currrent

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?

Re: Pic Currrent

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.

So if you have a power supply of say 5V/1A it means that the output voltage
is 5V and it *can* supply a maximum current of 1A. The attached device is
the only thing that determines how much current is really needed.

Observe Ohm's Law: U = I * R. You set U by the voltage of the power source
and the PIC has a certain resistance (over simplified) that determines how
much current actually flows.

I case of the batteries you mentioned, the "current" of 1800mAh is actually
the capacity of the battery and it is the product of current and time. To
put it simple: a 1800mAh cell can deliver 1800mA during 1 hour, or 180mA
over 10 hours and so on.

a bit about the basics of voltage, current and resistance etc. It will clear
up things a bit.

Meindert

Re: Pic Currrent

<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 ( snipped-for-privacy@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.

Re: Pic Currrent

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

Re: Pic Currrent

Correct. If you have, say, a huge 4V "car" battery (don't know if you
can get them any more), you can happily run a PIC (assuming it'll work
at 4V) of it, and it'll run for about a year(?); it will take whatever
current it needs. This is true of all electronic circuits.

Yes. You do not "give" it 10A; you "offer" it 10A: it will take whatever it
needs.

However, an "electronic circuit" can include a short circuit. That will take
10A (and more). It will probably melt and, in extreme cases, cause a fire.
It may or may not damage the PIC, depending on where the short is.

Richard [in PE12]

Re: Pic Currrent

Yes.  You cannot 'give it 10A'.  If you connect the correct voltage to it
then it will take whatever current it needs.  The current rating of a
regulated voltage power supply for example only tells you how much current
it *can* supply if the load demands it.  This is not strange but a simple
consequence of ohms law which can be expressed as:

I = V/R

Where I is the load current, V is the applied voltage and R the load
resistance.  The PIC effectively has a certain value of R.  When you apply
a voltage V it is the above formula gives the current.  The PIC will only
take more current if you increase the voltage.

Ian
--
Ian Bell

Re: Pic Currrent
On Mon, 18 Oct 2004 10:48:01 +0100, the renowned Ian Bell

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"
snipped-for-privacy@interlog.com             Info for manufacturers: http://www.trexon.com
We've slightly trimmed the long signature. Click to see the full one.
Re: Pic Currrent

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)

Re: Pic Currrent
On Mon, 18 Oct 2004 13:09:39 +0200, the renowned "Frank Bemelman"

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"
snipped-for-privacy@interlog.com             Info for manufacturers: http://www.trexon.com
We've slightly trimmed the long signature. Click to see the full one.
Re: Pic Currrent

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

--
Ian Bell

Re: Pic Currrent
[...]

Whoa, Silver!  Let's stop right here, for a little consideration.

Now, how do I put this politely...: you're in *desparate* need of some
you stop thinking about microprocessors for a while and go back to
lamps, batteries, switches, and maybe some resistors.  Learn how they
actually behave, learn Ohm's law, Kirchhoff's node and loop rules, and
what the symbols in an electronics schematic actually mean.

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

Re: Pic Currrent

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

Re: Pic Currrent

As a hardware dunce, I thank you for the link!

Re: Pic Currrent

The current consumption depends on the clock frequency. Look it
up in the datasheet (or measure it). It (may) also depend on what you
hang on the outputs of the PIC. If the pic sources current to, for
instance, leds, the current consumption goes up.

Yes, there are several choices. First find out how much current you need.

Think in terms of power and physical battery size. 1.2 x 1800 -> 2160.
9 x 150 -> 1350. 9V batteries have less energy density, because a lot
of space is wasted by the insulation of the individual stacked cells
inside the package.

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

Re: Pic Currrent

Your problem is not current.  It is drop-out voltage.  Voltage
regulators like the 7805 require that the input voltage be a certain
amount higher than the output voltage (at least several volts in the
case of the 7805).  There are special Low Dropout (LDO) Voltage
Regulators that can work with difference of only 200 mV or so.  For
battery powered applications, look into LDO regulators.

-Robert Scott
Ypsilanti, Michigan
(Reply through this forum, not by direct e-mail to me, as automatic reply

Re: Pic Currrent
On Fri, 15 Oct 2004 13:28:45 GMT, the renowned no-one@dont-mail-me.com

Exactly. And you probably want the circuit to work down to 4.0V
battery voltage or so if there are 4 of them, so a Vdd of 3.3V and an
LDO regulator would be about right.

Don't forget to study the data sheet on the LDO regulator. They are

Best regards,
Spehro Pefhany
--
"it's the network..."                          "The Journey is the reward"
snipped-for-privacy@interlog.com             Info for manufacturers: http://www.trexon.com
We've slightly trimmed the long signature. Click to see the full one.
Re: Pic Currrent

Aren't there any PICS that run on 1.8 to 5.5V? Then, with 4 1.2V cells you
don't need a regulator at all....
You could probably get away with 3 or even just 2 cells.

Meindert

Re: Pic Currrent

The batteries are rated in milli-ampere-hours, which is the capacity, not the
current.  The actual current is determined
by the circuit that the battery is connected to.  If your circuit draws 100 mA
and you use 4 cells in series, the life
is approximately 1800 mAh/100 mA = 18 hours.

The 9 volt battery has a higher voltage.  The energy capacity is 9 x 150 mAh, or
1350 milli-Watt-hours, which is less
than the energy density of the AA cell (1.2 x 1800).

A battery approximates a constant voltage source, providing the nominal voltage
to the load, as long as the current
isn't too high.  The resistance of the load determines how much current is drawn

I = V / R

where I is the usual symbol for current.  Knowing the current and the battery
capacity in ampere hours or
milli-ampere-hours, you can determine the approximate life of the battery.  The
formula is not exact because most
batteries decrease the output voltage as they are discharged.  At some point the
voltage becomes too low to power the
circuit properly.  Also, the capacity of a battery in mAh depends on current
being drawn and temperature.   Most
batteries lose capacity when they are discharged at low temperatures.

It would help to find a tutorial on electrical units of volts, amperes, watts,
ohms.

The voltage requirements of the processor varies with the model, but is in the
general range of 3 to 6 volts.  Some
models can run on slightly less voltage.  When run on the lowest voltages, they
cannot run as fast.  You also need to
consider the voltage and current requirements for the peripherals that are
connected to the processor.