Driving RGB LEDs with an Arduino

Wed Sep 28, 2011 6:26 pm   

On Sep 28, 5:43 pm, John Fields <jfie...@austininstruments.com> wrote:

You ran the numbers? That's useful.

The bad news is it takes 500mA @ 5V to run it
if I go with 150mA.

The booster is claimed to be 90% efficient "on
average" but we'll see...

If we assume the booster is 80% efficient that's
846mA @ 3.6V - not going to happen according
to the Duracell page.

OTOH it's software driven and I don't think all
LEDs would ever be on at full power. I could
even put a limiter in the code to keep it below
a certain output power.

If I do that then rhe only way it would ever go
to full power would be a program crash or a
faulty Arduino. Maybe I could add a thermistor
to the battery pack and cut the power if they
overheat.

Wed Sep 28, 2011 7:01 pm   

On Wed, 28 Sep 2011 09:26:25 -0700 (PDT), fungus
<openglMYSOCKS_at_artlum.com> wrote:


---
500mA * 5V = 2.5 watts
---


---
If the booster's 80% efficient, then to supply 2.5 watts it'll need
3.125 watts in.

Using three Duracell MN1500 AA cells in series will give 4.5V, and to
get 3.125 watts out of that battery, the booster will suck 625mA out
of it.

Looking at the constant current discharge curve on page 5 of:

http://www1.duracell.com/oem/Pdf/others/ATB-5.pdf

Shows that the MN1500 cells can supply 700mA for just under 2 hours
before they go "flat" at 0.8V leaving the battery with a 2.4V output.

However, your booster will keep working until the battery voltage
drops to 1.2V.

That's even more time, so it seems like you ought to get substantially
more than the hour you said you wanted out of it.

--
JF

Wed Sep 28, 2011 9:21 pm   

On Sep 28, 9:01 pm, John Fields <jfie...@austininstruments.com> wrote:

Only when they're fresh...

The 1.5V from batteries is a bit of a myth.
Batteries drop down to about 1.25V very
quickly then go down slowly from there.

See graphs on this page:

http://www.powerstream.com/AA-tests.htm

I figure they'll be down to about 3.6V after a few
minutes and that puts the current draw on the
borderline of what Alkalines can manage. They
could overheat and die very quickly...

Wed Sep 28, 2011 10:49 pm   

On Wed, 28 Sep 2011 12:21:58 -0700 (PDT), fungus
<openglMYSOCKS_at_artlum.com> wrote:


---
So you're planning on running this thing with old batteries???
---


---
Indeed.

Fresh alkaline cells start out at about 1.6V and, in order to keep
from blowing stuff up, that needs to be taken into account.

Measure one.
---


---
None of that matters if, at the end of its life, the battery has
supplied the energy needed for the time required, e.g., it's just
another dead soldier.

In your case, since the boost converter you've chosen needs 3.125
watts in to run your circuit for one hour, that energy is 3.125
watt-hours, and the question is: "Can that much energy be extracted
from a three-cell AA alkaline battery before its terminal voltage
decays to 1.2V.

If not, your project, in its current state, is doomed.
---


---
"I figure"...
"on the borderline"...
"They could"...

Got some numbers instead of just conjecture?

--
JF

Thu Sep 29, 2011 1:41 am   

On Sep 29, 12:49 am, John Fields <jfie...@austininstruments.com>
wrote:

Yep.

....except that:
a) It should never be on full power

b) The current being drawn is really under software
control. With an Arduino I could actually measure
the battery voltage and dim the LEDs as it drops.

If it turns out that it won't run at full power I can
either add a failsafe mechanism (to prevent bad
things happening in case of Arduino crashes)
or drop the current to 100mA.

With a good failsafe mechanism I could even
increase the maximum current per color and
get extra brightness when only one LED is lit.

(This is the great thing about software...)


I need to get a boost board and see how efficient
it really is under my conditions. Maybe they really
are 90% efficient as claimed.

There's a couple in the post as I write this...

Thu Sep 29, 2011 2:52 am   

fungus wrote:

Correct. The power rating is what can be dissipated over the
full length of the wire in the pot. If you set the pot such
that current flows through less than the full length of wire,
then less power can be dissipated. For example, say you
set the wiper at 1/2 the total resistance:
P=I^2*(R/2) instead of P=I^2*R

Ed