Basically what I am trying to do is run 4 LEDs (in parallel?) off of a
battery power source. The LEDs will be in a box, one LED in each
corner, about 1 or 2 feet apart.
The parts in question are:
10mm LED (Radio Shack #276-216) - Forward Voltage 2.0V, Forward
Current 40mA, Luminous 250mcd, Wave 585nm, max values: Forward Voltage
2.8V, Rev. Voltage 5V, Forward Current Peak 100mA, continuous 60mA,
Power dissipation 168mA
there are two of these #276-216 LEDs
10mm LED (Radio Shack #276-214) - Forward Voltage 2.0V, Forward
Current 40mA, Luminous 300mcd, Wave 635nm, max values: Forward Voltage
2.8V, Rev. Voltage 5V, Forward Current Peak 100mA, continuous 60mA,
Power dissipation 168mA
there are also two of these #276-214 LEDs
so, (2) #276-216 and (2) #276-214 = 4
4 LED x 2.0V forward voltage = 8.0V requirement
I wanted to use a battery tray (radio shack #270-387B) that holds 8 AA
batteries. this tray, depending upon battery make, will "produce"
12.8V or more with new AA batteries actually putting out approx 1.6V
and not the 1.5V they are marked with.
now comes my confusion. in figuring my resistor requirements
("learned" ohms law last night) do i:
12.8 (1.6v x 8AA) - 8 (4 LED x 2.0V) = 4.8
4.8 / .04 = 120 ohm resistor on each LED?
or since there are 4 LEDs do I have a mA req. of 40mA x 4 (160mA)?
12.8 (1.6v x 8AA) - 8 (4 LED x 2.0V) = 4.8
4.8 / .160 = 30 ohm resistor on each LED? I think it's this one.
also, i would like the LEDs to run as bright as "normally" possible
without affecting their lifetime (slow burn out?) this is why i went
with the 8 AA battery tray. thought a few "extra" volts would extend
the run time of the four lights with a power req. of 8V.
further, i would LOVE to hear any suggestions or ideas that anyone may
have regarding my project in regards to extending the runtime of the
LEDs. basically want it to run as long as possible off portable
battery power. would like to stay at or under (due to size & weight
issues) 8 AA batteries.
is there any way to estimate the run time of this setup? 8v req.
pulling 40mA (or is it 160mA?) powered by controlled 12.8v source
provides x minutes of illumination?
also using a SPST switch (radio shack #275-612 3A at 125VAC / 1A at
250VAC) as on/off but this probably is irrelevant?
any thoughts, comments, suggestions are welcome! i love this stuff i
just don't know what i am doing!
THANKS AGAIN!!!
Organization: blueyonder (post doesn't reflect views of blueyonder)
Lot of questions and I`m short on time at sec,sure others will be able to
guide you, but in meantime couple links of interest:
http://www.candlepowerforums.com/
http://ledmuseum.home.att.net/
Adam
<---- 4V ---->+<-------- 8.0V --------->
+--------\/\/\/\----|>|---|>|---|>|---|>|--+
| |
--- 20 ma thru LEDs and resistor |
- (12V-8V)/20ma = 200 ohms |
--- The resistor is 500 ohms |
- Total battery current = 4*20ma = 20ma |
| Power waste = 4 * 0.02 = .08 Watts |
| |
+------------------------------------------+
If you're wiring the LED's in parallel, you can do it either of two ways.
Wire the LEDs in parallel and then wire the limiting resistor is series with
the LEDs. Like this:
<---- 10V ---->+<- 2.0V ->
| | |
+----+---\/\/\/\----+ |
| | |
| 125 ohms +---|>|---+
| | |
--- +---|>|---+ 20 ma thru each LED
- 12V | | = 80 ma total
--- battery +---|>|---+
- | | (12V-2V)/80ma - 125 ohms
| +---|>|---+ Total battery current = 80 ma
| | Power waste = 10V * 0.08 = .8 Watts
+-----------------------------+
OR, wire each LED with its own limiting resistor, then wire all in parallel,
like this:
<---- 10V ---->+<- 2.0V ->
|----+---\/\/\/\--------|>|---+
| | |
--- +---\/\/\/\--------|>|---+ 20 ma thru each LED
- | | (12V-2V)/20ma = 500 ohms
--- +---\/\/\/\--------|>|---+ Each resistor is 500 ohms
- | | Total battery current = 80ma
| +---\/\/\/\--------|>|---+ Power waste = 10 * 0.08 = 0.8 Watts
| |
+-----------------------------+
Think of your battery lifetime this way... the more current drained from it,
the shorter its life. Therefore, if you look at the circuit with the least
current drain, that would be the first example. Also less power is wasted
in that circuit as well, so more of the battery's power is going to produce
light from the LEDs rather than disappearing as heat in the resistor(s).
The current rating of your switch is irrelevant at this low current.
You can estimate the battery life by multiplying the number of amperes that
your circuit draws by the number of hours you run the circuit. Compare this
to the Ampere-Hour (AH) rating of your battery. Typical rating of an AA
alkaline battery is 2.7 AH. However, all AA batteries aren't created equal,
so how often you use the circuit, how long it is used each time,
temperature, and a few other factors will cause the actual lifetime of the
batteries to vary. But this will get you in the ballpark.
--
Dave M
Remove the xx for my correct email address
ab...@yahoo.com; ab...@lycos.com; ab...@hotmail.com
"Zap" <zapne...@yahoo.com> wrote in message
news:a4a03916.02050...@posting.google.com...
>I am just getting into electronics and have a few questions and
>concerns that I was hoping to get some help with.
>
>Basically what I am trying to do is run 4 LEDs (in parallel?) off of a
>battery power source. The LEDs will be in a box, one LED in each
>corner, about 1 or 2 feet apart.
>
>The parts in question are:
>
>10mm LED (Radio Shack #276-216) - Forward Voltage 2.0V, Forward
>Current 40mA, Luminous 250mcd, Wave 585nm, max values: Forward Voltage
>2.8V, Rev. Voltage 5V, Forward Current Peak 100mA, continuous 60mA,
>Power dissipation 168mA
>
>there are two of these #276-216 LEDs
>
>10mm LED (Radio Shack #276-214) - Forward Voltage 2.0V, Forward
Same specs, different part number - did you make a mistake copying the
specs?
>4 LED x 2.0V forward voltage = 8.0V requirement
>
>I wanted to use a battery tray (radio shack #270-387B) that holds 8 AA
>batteries. this tray, depending upon battery make, will "produce"
>12.8V or more with new AA batteries actually putting out approx 1.6V
>and not the 1.5V they are marked with.
The battery voltage will soon drop when you start using them - use 12
volts for easy calculation.
>
>now comes my confusion. in figuring my resistor requirements
>("learned" ohms law last night) do i:
>
>12.8 (1.6v x 8AA) - 8 (4 LED x 2.0V) = 4.8
>4.8 / .04 = 120 ohm resistor on each LED?
If you connect the four LEDs in series (positive of one to negative of
the next, so all the current flows through each LED in turn), you only
need 40 mA, and a single resistor. (Actually, I find most LEDs are
bright enough at 10 mA - to maximize battery life you may want to try
lower currents to see what you really need for your application.)
For 40 mA, you would need 100 - 120 ohms (but bear in mind that the
battery voltage will drop with use, reducing the LED current.)
>
>or since there are 4 LEDs do I have a mA req. of 40mA x 4 (160mA)?
No - if they are in series, all the current flows through each LED in
turn, so you only need 40 mA. (Some pedants will likely object to me
saying "flows through each in turn", but it gets the point across...)
>is there any way to estimate the run time of this setup? 8v req.
>pulling 40mA (or is it 160mA?) powered by controlled 12.8v source
>provides x minutes of illumination?
Radio Shack has (or had) a Battery Handbook that has information on
battery capacities. This book indicates that a "general Purpose" AA
cell can deliver 20 mA for four hours a day, taking a total of 40
hours to reach 0.9 volts. Since you will be drawing 40 mA, we might
expect a total of 20 hours, but the faster you discharge a battery,
the less total energy you get out of it - so I'll make a wild guess of
16 hours, or so.
Their "Heavy Duty" AA cell gives 65 hours at 20 mA/4 hours per day,
and the Long Lasting Alkaline AA gives 154 hours.
--
Peter Bennett, VE7CEI
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver-webpages.com/van-ps
That's the one, but it's one 120 ohm R in series with
the series string of all 4 LEDs.
> also, i would like the LEDs to run as bright as "normally" possible
> without affecting their lifetime (slow burn out?) this is why i went
> with the 8 AA battery tray. thought a few "extra" volts would extend
> the run time of the four lights with a power req. of 8V.
>
> further, i would LOVE to hear any suggestions or ideas that anyone may
> have regarding my project in regards to extending the runtime of the
> LEDs. basically want it to run as long as possible off portable
> battery power. would like to stay at or under (due to size & weight
> issues) 8 AA batteries.
These are conflicting requirements. More current = less battery
life. Find the amp-hr capacity of your batteries, divide that by
the current, and that will tell you the run time.
I recently found that a nominally 20 mA LED, with 10 mA through
it, was imperceptibly less bright than at 20 mA. They're very
nonlinear.
To increase battery life, use a larger value resistor to limit
the current to 20 or even 10 mA. Do some experiments to see what
kind of brightness you get.
>
> also using a SPST switch (radio shack #275-612 3A at 125VAC / 1A at
> 250VAC) as on/off but this probably is irrelevant?
The switch will be fine. :-)
>
> any thoughts, comments, suggestions are welcome! i love this stuff i
> just don't know what i am doing!
>
> THANKS AGAIN!!!
Cheers!
Rich
More than you ever wanted to know about batteries can be found at
http://data.energizer.com/
-- Foo!
"Foobar T. Clown" <fu...@gazonk.del> wrote in message
news:3CD498B7...@gazonk.del...
my question is in your second parallel example:
Total battery current = 80ma
but
(12V-2V)/20ma = 500 ohms
shouldn't it be
(12V-2V)/80ma = 125 ohms
i thought you added the values of all components? (4 LED x 2.0v = 8.0v
and 4 LED x 20mA = 80mA)?
thanks for your killer schematics! and thanks to all who replied...
<---- 10V ---->+<- 2.0V ->
|----+---\/\/\/\--------|>|---+
| | |
--- +---\/\/\/\--------|>|---+ 20 ma thru each LED
- | | (12V-2V)/20ma = 500 ohms
--- +---\/\/\/\--------|>|---+ Each resistor is 500 ohms
- | | Total battery current = 80ma
| +---\/\/\/\--------|>|---+ Power waste = 10 * 0.08 = 0.8 Watts
| |
+-----------------------------+
"Dave M" <dgma...@atxxt.net> wrote in message news:<%oWA8.5425
-SNIP-
> If you're wiring the LED's in parallel, you can do it either of two ways.
> Wire the LEDs in parallel and then wire the limiting resistor is series with
> the LEDs. Like this:
>
> <---- 10V ---->+<- 2.0V ->
> | | |
> +----+---\/\/\/\----+ |
> | | |
> | 125 ohms +---|>|---+
> | | |
> --- +---|>|---+ 20 ma thru each LED
> - 12V | | = 80 ma total
> --- battery +---|>|---+
> - | | (12V-2V)/80ma - 125 ohms
> | +---|>|---+ Total battery current = 80 ma
> | | Power waste = 10V * 0.08 = .8 Watts
> +-----------------------------+
>
> OR, wire each LED with its own limiting resistor, then wire all in parallel,
> like this:
> <---- 10V ---->+<- 2.0V ->
>
> |----+---\/\/\/\--------|>|---+
> | | |
> --- +---\/\/\/\--------|>|---+ 20 ma thru each LED
> - | | (12V-2V)/20ma = 500 ohms
> --- +---\/\/\/\--------|>|---+ Each resistor is 500 ohms
> - | | Total battery current = 80ma
> | +---\/\/\/\--------|>|---+ Power waste = 10 * 0.08 = 0.8 Watts
> | |
> +-----------------------------+
>
> --
>I think i want to go parallel with a resistor on each led
That is the WORST solution for battery life. You need the series
connection and take the advice to try 10 mA first; the LEDs are likely
to be quite bright enough **and won't get much brighter if you run at a
higher current**.
> (i was told
>with one resistor if one led goes they all go as opposed to each being
>protected individually?)
The chances of one LED failing short-circuit are approximately zilch.
But if it did, in the series circuit the others just get a bit more
current. If they were running at 10 mA, they would get about 12 mA with
one shorted. No big deal. Get a better adviser.
At 10 mA, the diode forward voltage is likely to be about 1.6 V, so you
need 560 ohms in series with the four diodes.
>
>my question is in your second parallel example:
>Total battery current = 80ma
>but
>(12V-2V)/20ma = 500 ohms
>shouldn't it be
>(12V-2V)/80ma = 125 ohms
>
>i thought you added the values of all components? (4 LED x 2.0v = 8.0v
>and 4 LED x 20mA = 80mA)?
You are totally confused. Look at the schematics. You add the VOLTAGES
for series connection but the CURRENTS for parallel connection.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!
Missed one trick so far. Chuck the LEDs and buy some ultrabright ones.
Then you can run them at a small fraction of the rated currect to do
what you need, and so increase battery life many many times. Go for
LEDS with say a 3000mcd rating.. As long as you run them below 20mA
they will live out their full life.
Regards, NT
Why is it that people have such a hard time with these elemetary basics? One
wonders if those such as the OP wouldn't be a little better if they spent
their time looking at the circuit and deducing the obvious, rather than
spending their time writing questions and replies to get others to show them
what nearly any high school kid could figure out if he spent a few minutes
in thought about the problem.
Bob.
"Bob Wilson" <rfwi...@intergate.nospam.bc.ca> wrote in message
news:udbdbab...@corp.supernews.com...
I, too believe in assisting those new to the game. But I (personally) expect
some effort at the other end as well. The OP seemed to be having a problem
understanding that in a series string of LEDs, the current remains
unchanged, but the voltage "adds". This has nothing to do with ohms law, but
I venture to say that it DOES have a lot to do with common sense, and would
have only required a moment's reflection to see the answer.
Conversely, if this wasn't obvious, then it would seem that he is not even
clear in the concept of voltage and current, in which case there isn;t much
we can do to assist.
Bob.
That SHOULD be spelled "elementary"... how "elementary" is that...?
;)
And although they may be basics for someone (you that's been into this
for x months / years / whatever) they may not necessarily be basics
for others (me that got into this just last week. never had a class,
just reading the internet and stuff. makes it hard when you don't
know a resistor is called a resistor. how would you do that search.
i thought i "learned enough" on my own to make a somewhat logical
quest for help in a place called .basics when i got stuck)
Further, wouldn't common sense (on your part) dictate that if i said i
just "learned" ohm's law last night that it would be safe to assume
there may be other things I may not be familiar with? Especially
since some say knowing ohm's law is one of the "first basics" to
learn.
To all those that gave a reply, THANKS A MILLION!
Bob, I am SORRY my post annoyed you so much that you actually took the
effort to make it publicly known instead of sharing the knowledge that
this forum is actually about. SORRY SORRY SORRY! No sarcasm
intended! Really, Sorry!
Zap, you posted to s.e.basics, where there are, by definition, no
questions too elementary to be asked. But you also posted to s.e.design,
where only sophisticated mistakes are permitted. Stick with it, and
carry on asking on s.e.basics.