Newbie Q. re: solar light/battery

[CDESC]

"JeffM" <jeffm_@email.com> wrote in message
news:01f59c1b-937c-4934-b7d0-08721fa9f9c5@d19g2000prm.googlegroups.com...

Kris Krieger wrote:
I'm trying to figure out how to [build] a solar light
(I have a market for stained-glass solar lights and,[...]
(note that I'm a crafts person, not an electronics person

Like most people who find themselves
attacked from behind by a wild hare,
it appears you haven't actually thought this through.
The first step in a project is writing a specification.

Don't take it like if I would think this place is meant for peoples to laugh
their ass off on peoples who literally started their post with "I'm a newbie
and I need help" but I tend to take for this poor guy who, luckilly wasn't
stuck his head down in a wrecked car on the roadside providing the kind of
introduction you gave him.


-- Now to that guy. The best way to learn not to put salt in your eyes is to
know how bad it hurts -- Google your way to some electronic plans on solar
NiCD chargers, light sensors and how to do that switching. You'll get parts
for about $5 to $10 off eBay. Get a few different solar panels, a few
different kinds of leds and do the tests. Doing the maths is cool, but some
peoples did them and they're selling their product to walmart for about 3$
each because it literally sucks. Some think that they can get it right on
paper, never test anything, start the production based on theories and get
poor results.

Do your tests, find the right leds type and quantity for the illumination
you want and start from there.

;-)

Will the lights come on automatically?
Are you going to use *multiple* LEDs in each one?
Have you selected a LED to use?
Do you understand the arithmetic involved with LEDs?

Once you understand how *those* part work
and can put some actual NUMBERS to your requirements,
then you can move on the *battery* part and the arithmetic there.

On that front, it seems foolish for a complete rookie
to try to *build* something that he can buy as a complete unit.
http://www.google.com/products?q=intitle:Solar+intitle:Battery-Charger&scoring=p&price=between&price1=1&num=100
You certainly won't be able to do it cheaper
than the pros who build them in large quantities.

 

[JeffM]

Kris Krieger wrote:

a solar light
(I have a market for stained-glass solar lights and,[...]
(note that I'm a crafts person, not an electronics person

Like most people who find themselves
attacked from behind by a wild hare,
it appears you haven't actually thought this through.
The first step in a project is writing a specification.

Will the lights come on automatically?
Are you going to use *multiple* LEDs in each one?
Have you selected a LED to use?
Do you understand the arithmetic involved with LEDs?

Once you understand how *those* part work
and can put some actual NUMBERS to your requirements,
then you can move on the *battery* part and the arithmetic there.

On that front, it seems foolish for a complete rookie
to try to *build* something that he can buy as a complete unit.
http://www.google.com/products?q=intitle:Solar+intitle:Battery-Charger&scoring=p&price=between&price1=1&num=100
You certainly won't be able to do it cheaper
than the pros who build them in large quantities.

 

[Michael Black]

On Sun, 1 Jun 2008, Kris Krieger wrote:

On that front, it seems foolish for a complete rookie
to try to *build* something that he can buy as a complete unit.

I stated the reason. It's not relevant, tho', why I want to do it - IMO,
it's sufficient that I *do* want to do it.

But. the relevance of the comment is because of a "cold" post.

People post here all the time, and they leave out details. They ask
for something very specific, but then it turns out that's because they've
based that on a false assumption. Without detail, we have to interpret,
and many times it's not obvious if the poster asking the question knows
what he's doing, or doesn't.

There's nothing wrong with wanting to build rather than buy.

But, lots of times people ask about how to build something because they
don't realize it's available off the shelf. They can still proceed to
build it, but it should be tempered by an awareness that such things
are available.

So someone asks how to build an alarm clock. They want to put it
in a box and use it for some specific purpose. Do they really need
to build one, or can they get by with buying a used clock radio and
stripping out the clock? It really depends on whether their first
goal is getting results or building something. It usually isn't clear
if they've even considered an existing clock radio (and even though
salvaging parts is a time honored thing, the beginner often seems
to overlook scrap electronics as a source of parts, rushing to the
parts store with the list of parts out of the article and paying
premium prices as a result).

And forget about the building versus buying, buying some existing
piece of equipment can often be faster and maybe cheaper means of
getting some parts. If you have to track down a multi-pin IC to
build that clock, the closest and cheapest source will be the clock
radio, and once you've got that, one has to rethink whether it's
really valid to extract the IC and build around it, or just use
the clock radio as a source of a the "clock module".

Sometimes you can't get parts via consumer electronics, because
the parts are too specific and don't show up in such equipment. But
consumer electronics can often be a great source of components
that are hard to track down even if they are commonly used in consumer
electronics equipment.

This is especially so when talking about buying things at rummage or
garage sales. Then the price often drops to almost nothing, which can
really beat buying the parts new.

We had some of those garden lights, they all broke during this winter's
heavey snowfall. So I got a set of solar cells, and some reaonably
bright white LEDs, the nicads and even the circuitry to make them
all work. Maybe it wouldn't be the best choice to buy them to get
the solar cells, but I suddenly have some without having to fuss at
all. If I was building something in single quantities, I would have
to evaluate the ease of getting them that way compared to finding
a place that sells them and will sell to me and the cost of
the full order to get them.

Michael

 

[Peter Bennett]

On Sun, 01 Jun 2008 16:41:10 -0500, Kris Krieger <me@dowmuff.in>
wrote:

<much snippage>

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.

The voltage across a lit LED depends primarily on the LED's colour -
red LEDs are about 1.7 volts, yellow 1.9 volts and green 2.1 volts.
Blue and white LEDs are 3.3 - 3.6 volts, I believe. (a white LED is
really a blue led with white phosphor, since a single LED chip can
only produce a single colour.) If you see an LED rated at 1.2 volts,
it will be infra-red - probably not much use to you (and it won't
work reliably off a 1.2 volt battery anyway, as the battery voltage
will drop as it discharges.

You will have to connect two or more batteries in series to operate
the LEDs, or have some electronics to step the 1.2 volts up to
something that will run the LEDs.

Since the voltage across an LED is determined by its chemistry/colour,
LEDs are normally run from a higher voltage than their rating, with a
series resistor to limit the current.

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.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[John Fields]

On Sun, 01 Jun 2008 16:41:23 -0500, Kris Krieger <me@dowmuff.in>
wrote:

"Tom Biasi" <tombiasi@optonline.net> wrote in
news:Yd-dnQBiscg2YN_VnZ2dnUVZ_qvinZ2d@giganews.com:


"Kris Krieger" <me@dowmuff.in> wrote in message
news:jtadnZ4sbrOzZN_VnZ2dnUVZ_t_inZ2d@earthlink.com...
Michael Black <et472@ncf.ca> wrote in
newsine.LNX.4.64.0806011152470.27894@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...

JF

 

[John Fields]

On Mon, 02 Jun 2008 08:36:57 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 01 Jun 2008 16:41:23 -0500, Kris Krieger <me@dowmuff.in
wrote:

"Tom Biasi" <tombiasi@optonline.net> wrote in
news:Yd-dnQBiscg2YN_VnZ2dnUVZ_qvinZ2d@giganews.com:


"Kris Krieger" <me@dowmuff.in> wrote in message
news:jtadnZ4sbrOzZN_VnZ2dnUVZ_t_inZ2d@earthlink.com...
Michael Black <et472@ncf.ca> wrote in
newsine.LNX.4.64.0806011152470.27894@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...

---
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,P37356,D24889

Either will allow you to drive the number of LEDs you need, and
driving 4 LEDs at about 80% efficiency means that the LEDs will be
dissipating:

P = IE = If * Vf = 0.02A * (4 * 3.5V) = 0.280 watts.

If you chose to use three 1500mAh NiMH cells that'll give you:

W = 3.6V * 1.5AH = 5.4 watt-hours

available to drive the LEDs, but you're going to waste 20% of that in
the driver, so you have left, to drive the LEDs:

W = 0.8 * 5.4WH = 4.32 watt hours,

which means that the LEDs will illuminate for:

4.32WH
T = -------- ~ 15.43 hours
0.28W

until the battery voltage falls to 3V.

I think most NiMh batteries are rated at 0.2C in order to get their
full capacity during discharge, so a 1500mAh battery should be
discharged at 300mA or less in order to get the full 1.5Ah out of the
battery.

Since the 4 LED array is dissipating 0.28 watt and the LED driver is
80% efficient, the battery must supply:

0.28W
P = ------- = 0.35 watts
0.8

into the driver.

Then, since the battery is providing 3.6 volts to the driver, and the
driver is dissipating 0.35 watts, The current into the driver must be:

P 0.35W
I = --- = ------- = 0.0972 amperes = 97.2mA,
E 3.6V

which is well below the battery's 0.2C limit of 300mA.

That means that with 4 LEDs in series and 20mA going through them they
should stay lit for at least 15-3/4 hours until the battery voltage
falls to 3V.

In order to keep from damaging the battery, the battery should be
disconnected from the driver when its (the battery's) voltage falls to
3V.

More on that tomorrow, and also on how to charge the battery properly
from a PV array.

JF

 

[John Fields]

On Tue, 03 Jun 2008 00:16:28 -0500, Kris Krieger <me@dowmuff.in>
wrote:

John Fields <jfields@austininstruments.com> wrote in
news:2nq844llenim5vb0vmmoc5c43ehnap0fmr@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.

JF