As mentioned, since the batteries are 1.2V and 1500mA, I assumet aht (1)
it's simplest to stick with a 1.2V LED which, if rted at 15mA, should
stay lit for the night.

I don't know what solar arrangement ebst acheives the charging; as
mentioned, I don't knwo whetehr I need to match the voltage of the solar
cellto the battery, what maximum mA the SC should have, or if it has too
many, how to damp charging once the battery is charged.

Again, tho', I'm slowly answering my own questions adn *will* eventually
figure it out. I also decided against posting any links, it seems that
the info would be far too elementary for this NG.

As per my other post, sorry to have been a bother.

On Sun, 01 Jun 2008 16:41:23 -0500, Kris Krieger <me at (no spam) dowmuff.in
wrote:

"Tom Biasi" <tombiasi at (no spam) optonline.net> wrote in
news:Yd-dnQBiscg2YN_VnZ2dnUVZ_qvinZ2d at (no spam) giganews.com:


"Kris Krieger" <me at (no spam) dowmuff.in> wrote in message
news:jtadnZ4sbrOzZN_VnZ2dnUVZ_t_inZ2d at (no spam) earthlink.com...
Michael Black <et472 at (no spam) ncf.ca> wrote in
news:Pine.LNX.4.64.0806011152470.27894 at (no spam) darkstar.example.org:
OK, but that whole bit about "attacked by a hare" (IOW, obvious
implication being "idiot") was silly.
SNIP

I have said much in all the sub paragraphs of this thread but need to
say this:
Perhaps you are not from the USA.
The reference to the hare is not to imply that you are an idiot.
It is a cleaver play on words to imply that you had an impulsive idea.
Tom




I'm from the USA but I never heard anything about hares except for "hare-
brained", and the mockery that Pres. Carter got re: the rabbit incident.
Guess I lived in the wrong part of the US or perhaps the wrong side of the
tracks (or petroleum cracking plant as 't'were).

In any event, I seem to be slowly working out the answers to my questions
myself.

Sorry for having been a bother.

---
No bother at all. :)

The 'hare' crack was, I believe, a misspelled attempt at implying a
"wild _hair_ up your ass", and this _is_ sci.electronics.basics, where
you're right: There are no stupid questions here.


As far as the rest of it goes, first you have the LEDs, and assuming
they're common, garden-variety white, then they'll have a forward
voltage (Vf) of about 4V max at a forward current of 20mA.

Next are the cells, which will output 1.2V and, being rated at
1500mAH, will output 20mA for:

mAH 1500mAH
T = ----- = --------- = 20 hours
mA 20mA

Until the cell discharges to its terminal voltage, 1.0 to 1.1 volt.

Finally, there's the photovoltaic (PV) panel, which is rated to supply
some current into a load with some voltage across the load.

That load resistance can be determined from:

E
R = ---
I

Where R is the load resistance, in ohms,
E is the voltage across the load, in volts, and
I is the current through the load, in amperes.


Since the LEDs have a Vf(max) of 4V, then you'll need at least four
NiMh cells to build a 4.8V battery which can drive the LEDs.

For the single LED case" (View in Courier)

VBAT
/
+---------+
|+ |
[NiMH] |
| [Rs]
[NiMH] |
| |
[NiMH] |A
| [LED]
[NiMH] |
|- |
+---------+

and Rs will be equal to:

VBAT - Vf(min) 4.8V - 3.5V
Rs = ---------------- = ------------- = 65 ohms
If 0.02A

Note that Vf(min) was used for the LED's forward voltage.

The reason for that is to keep the current through the LED from
exceeding 20mA if its Vf is less than the maximum specified on the
data sheet.

Another caveat is that NiMH cells, when fully charged, will output
about 1.4V for a while, until they settle down to 1.2V, so it would be
possible for Vbat to be 5.6V initially.

Under those conditions, Rs should be equal to:

VBAT(max) - Vf(min) 5.6V - 3.5V
Rs = --------------------- = ------------- = 105 ohms
If 0.02A

100 ohms is a standard 5% value and would force the initial current,
under worst case conditions, through the LED to be:


VBAT(max) - Vf(min) 5.6V - 3.5V
If = --------------------- = ------------- = 0.021 amperes
Rs 100R


which would be fine, and the resistor would dissipate:


P = EI

=(VBAT(max) - Vf(min)) * If = 2.1V * 0.021A = 44 milliwatts


A standard 100 ohm +/- 5% 1/4 watt carbon film resistor would be fine.

More later on today...

John Fields <jfields at (no spam) austininstruments.com> wrote in
news:2nq844llenim5vb0vmmoc5c43ehnap0fmr at (no spam) 4ax.com:

[...]


---
The next thing to consider is how many LEDs you want to use and how
you want to drive them.

Instead of using discretes to build a linear constant-current
regulator (and wasting battery power (and probably money) in the
process) I suggest you go the route of an integrated LED driver such
as:

http://focus.ti.com/lit/ds/symlink/tps61061.pdf

or

http://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1003,C1094,P3
7356,D24889

I saw some of those on-line while searching LEDs, but to be honest, I don't
understand how to use them.

On Tue, 03 Jun 2008 00:16:28 -0500, Kris Krieger <me at (no spam) dowmuff.in
wrote:

John Fields <jfields at (no spam) austininstruments.com> wrote in
news:2nq844llenim5vb0vmmoc5c43ehnap0fmr at (no spam) 4ax.com:

[...]


---
The next thing to consider is how many LEDs you want to use and how
you want to drive them.

Instead of using discretes to build a linear constant-current
regulator (and wasting battery power (and probably money) in the
process) I suggest you go the route of an integrated LED driver such
as:

http://focus.ti.com/lit/ds/symlink/tps61061.pdf

or

http://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1003,C1094,
P3 7356,D24889

I saw some of those on-line while searching LEDs, but to be honest, I
don't understand how to use them.

---
They're what are called "boost converters" which convert a source
voltage into whatever's required to drive a string of LEDs at a
constant current, which is what LEDs like in an application like
yours.

In order to use them properly, just follow the instructions in the
applications shown on the data sheet.