I require instructions to build an electrical circuit.

CIRCUITS PURPOSE 
The purpose of the circuit is to string as many LED's off a standard 9
Volt battery as possible.

ANSWER REQUIREMENTS
1) I cannot read circuit diagrams, but I have a basic understanding of
electronics, answer will need to be descriptive for beginners use.

2) Each components for the circuit must be individually identified
with a 'Cat. No.' from the Dick Smith electronics catalogue:

http://www.dse.com.au/cgi-bin/dse.filereader?4153930c103b5338273fc0a87f9c06c6+EN/catalogs/CTG0000606

3) The LED's must be non-coloured (ie white light) and bright is preferred.

Not having done electronics in a while, wouldnt just stringing them up
in paralel be the easiest way to get the most LED with least energy
waste?



|-----|----|---|---|- .etc ---|---|
V     *    *   *   *          *   |
|-----|----|---|---|----------|---|

* lights, V voltage source

Erm, paralleling LEDs straight off 9 volts will be ?interesting? for a
few milliseconds. Great way to fry LEDs. (sorry nanoalchemist).

I?ll try to find time to get back to you in a day or two for ?guidance?.

Best

The first problem I have is with the Dick Smith electronics website. I
can see a Super Bright White LED part no Z3800 but the website will
not show me the details for that item; it just goes back to the
homepage.

The circuit is quite simple. You need a resistor in series with the
LED to limit the current flowing through the LED. The value of the
resistor is found by the formula

   R = (Vs - Vf) / If

where Vs is the supply voltage (9V in this case), Vf is the forward
foltage drop across the LED, and If is the forward current through the
LED. Vf and If must be obtained from the seller's technical data,
which I cannot get from their website.

Using another supplier (maplin.co.uk) for a 3mm white LED, which
probably has similar characteristics, Vf is 3.6V and If is 30mA. The
formula thus becomes

  R = (9-3.6) / 0.030 = 180 ohms.

Unless you are using lots of LEDs, the cheapest (quarter-watt)
resistors should be fine.

The circuit for one LED is thus: positive of battery to resistor,
resistor to anode of LED, cathode of LED to negative of battery. The
cathode is normally indicated by the shorter of the two leads and/or a
flat on the body, or the letter k.

  V(+)----------Res-----aLEDk-----V(-)

For multiple LEDs, you can either use a separate resistor for each
LED, or one resistor common to all LEDs. Separate resistor puts your
component cost up, but means your LEDs are more likely to be the same
brightness. Also, if some of your LEDs got disconnected (intentionally
or accidentally), the current might be too high for the remaining LEDs
which would burn out. Resistors are cheaper than LEDs.

For separate resistors, use the same value you have calculated by the
formula above. Wire each LED-and-resistor in parallel across the
battery terminals.

V(+)-------|---Res-----aLEDk---|---V(-)
           |                   |
           |                   |
           |---Res-----aLEDk---|
           |                   |
           |                   |
           |---Res-----aLEDk---|
           |                   |
           |                   |
           |---Res-----aLEDk---|


For a common resistor, divide the resistor value from the formula by
the number of LEDs, so eg 180/10 = 18 ohms for 10 LEDs. Then wire from
the battery to the resistor, from the resistor to the anodes of all
the LEDs, and from the cathode of all the LEDs back to the battery.

V(+)---Res---|----------aLEDk---|---V(-)
             |                  |
             |                  |
             |----------aLEDk---|
             |                  |
             |                  |
             |----------aLEDk---|
             |                  |
             |                  |
             |----------aLEDk---|

Hope that helps. 

By the way, an alkaline 9V PP3 battery has about 500mA/h, so will run
10 LEDs for about one and a half hours, roughly.

Owain

There are many issues with your apparently simple requirement --
second nature to a design engineer but everyone?s got to start
somewhere :-)

You say a standard 9 volt battery. Do you mean the little PP3 type?
This is an unfortunate choice for cost effectiveness -- but if you
have to. Problem is that though alkaline types are specified as being
500mA hours and 9 volts, reality is substantially different. A new
battery without load will be 9.5 volts or above, initially dropping
rapidly with discharge, then levelling off down to a bit over 6 volts
at which point very little is left in it and the voltage plummets. So
to get the most out of your battery it has to be able to supply the
required load over quite a range of voltage. This largely makes a
nonsense of resistor calculations. Furthermore, the internal impedance
of a PP3 is relatively high and they are not designed for heavy load
-- which as far as a PP3 is concerned is anything over 50mA.
Essentially though, the heavier the demand which is placed upon the
battery, the less will be the capacity realised. The upshot of all
this is that ten LEDs in parallel running at 30mA will give nothing
like an hour and a half of decent brightness (sorry Owain). They?ll
start of bright for a minute or two then rapidly fall to glowing level
for an hour or two. I doubt this is what is required.

Next aspect is do you just want lots of lights giving out as much
light as possible, or do you want intense beams? All things being
equal (which they are not) the narrower the beam angle, the brighter
will be the beam. Think of a 60 watt car headlamp. If you merely want
lots of lights go for wide angle, otherwise go for narrow. You?ll find
though that 3mm devices do not focus as hard as 5mm ones so for high
intensity, go for 5mm. There are other devices such as ?Luxeon
Lumileds? (not available from your chosen supplier) which are
*mega-powerful*. Getting cheaper but a PP3 would balk at the power
requirements. Manufacturers (lamentably) tend not to spec both the
luminance (total output as visibly perceived) and candela rating
(brightness) so direct comparison is difficult. It?s easy to translate
from one to the other but I doubt you want to get involved in
arctangents calculations so I?ll leave it out.

There would appear to be three LED contenders on the supplier?s site.
Links didn?t work right but if you search by part number they do come
up.

http://www.dse.com.au/cgi-bin/dse.storefront/4156d0b1006571a0273fc0a87f9c0746/Product/View/Z3800

http://www.dse.com.au/cgi-bin/dse.storefront/4156d0b1006571a0273fc0a87f9c0746/Product/View/Z3980

http://www.dse.com.au/cgi-bin/dse.storefront/4156d0b1006571a0273fc0a87f9c0746/Product/View/Z3982

The last one is the best, but ?orribly expensive. None of them are
wide angle though. Personally I get mine shipped from the states, paid
with Paypal, and even including shipping ($2 USA) they come out much
cheaper :-

http://www.lsdiodes.com/

Or alternatively, very good, though shipping is a bit more :-

http://www.ultraleds.co.uk/

The issue of how to drive the beasts. Bear in mind that LEDs are
diodes and the current through them is non-linear. At, say, 2.5 volts,
very little current will flow. Say you had a guaranteed 9 volts, the
voltage drop of a white LED at best working current (generally about
25mA) is roughly 3.6 volts. So if you run two in series with a
resistor you?ll be using the same current for both and halving your
power requirements. So that?s 1.8 volts at 25mA through 72 ohms,
dissipating 45mW. That?s all in an ideal world though. Dropping
battery voltage would soon dim the LED very severely. Better
brightness retention will be obtained by having a single LED and
resistor in a chain because the voltage across the resistor will
change proportionally less. But to get the same brightness as two in
series you?d have to use twice the current so you?re back where you
started.
Another mooted arrangement is parallel LEDs with one resistor. This is
to be avoided because it doesn?t work reliably. What happens is that
as a diode heats up its voltage drop falls, and although there is a
dependence of voltage drop rising with current, you still end up with
one LED hogging the current and blowing up. One proviso is that if you
incorporate a smallish value resistor (say 10 or twenty ohms) in
series with each LED, all fed from a higher value resistor, this small
resistors largely swamp out the voltage change or variations in the
individual LEDs.

This brings us back to how to drive lots of LEDs from a PP3. Basically
a bad idea unless you just want lots of little ones running at low
current. Rechargeable cells have far lower source impedance and
flatter voltage curves so can be used more effectively. But they are
initially expensive and possibly unsuitable for your requirements.
However, if you can be flexible with your power supply, four AA cells
in a holder driving a switching regulator is the obvious recourse.
Gonna have to learn to read circuit diagrams though. Don?t worry, if
you keep at it you?ll wake up one day and suddenly realise that these
squiggly lines make sense.

The subject of switching regulators is undeniably complex but you
don?t have to get too involved, merely follow the application notes
provided by manufacturers. Not expensive either, and would repay
itself after one set of batteries! There are many varieties on the
theme, but ideally you would want one which delivers a specified
current through a bunch of LEDs in series. The switcher would take the
initial voltage and raise it with up to 90% efficiency, but more
usually ~ 75%. Amongst others, check out :-

http://www.zetex.com/ 
http://www.maxim-ic.com/ 
http://www.linear-tech.com/ 

Your main problem might be getting hold of devices -- none available
from ?your? web site, but it?s the way to go if you want to do it
cheaply and efficiently.

There is another alternative. Just depends on what your prime
requirements are. Do you want it bright, cheap, efficient, one unit or
many etc? If you want a stable light output level which doesn?t dim
from the moment you switch on, you could use an adjustable current
regulator for each LED. These are not altogether readily available but
a standard regulator can be adapted by connecting a resistor (whose
value you would calculate based upon the regulator?s regulating
voltage and the current you require) between the output and common
terminals. You would then connect the LED between the common terminal
and zero volts. If you want precise details of how to do this and what
regulators would be suitable, then post a note.

Sorry for such a long reply -- I could have made it tem times as long
without even trying :-) But as I said, anything more you want, just
ask.

Best

try this website for help 

http://wolfstone.halloweenhost.com/TechBase/litlpo_PoweringLEDs.html

moffo

I understand you are a beginner to electonic circuits. But you seem
to very inquisitive. So let me try to solve your puzzle. Your question
reads:

"The purpose of the circuit is to string as many LED's OFF a standard 9
Volt battery as possible."

I am taking OFF as OF.

Method 1: (if you want o glow all LEDs simultaniously)
In order to connect maximum number or LEDs to a 9Volts supply you have
to choose the minimum resistance (= R) LEDs. Also check for the
current n(= I)

Now you know how much is the Voltage (V= 9) and how much is the
current (I of the LED). So now you can find the total resistance you
require for the circuit.
as

    V = IR
(http://www.grc.nasa.gov/WWW/K-12/Sample_Projects/Ohms_Law/ohmslaw.html)

    so R = V/I

 From the data specifications of the LED you can find R and I. Since
I and R are fixed for LED. Once R is found divide
          R/(R of LED) = total number of LEDs you can connect to the
circuit in SERIES. Let us say the value as n. then your circuit will
look like this.

    (+)  --^^^^-----^^^^----
...................................----^^^^^^^^^------ (-)
          LED1 resistance                                LEDN resistance.
   The (+) sign shows the positive end of the battery and (-) shows
the negetive end of batter. This connection will work with proper
calculation.

Method 2: Using a multiplexer.
   This is more complicated. If you understand the multiplexer logic
let me know.

Your best bet would probably be to run the LEDs in parallel using
resistors to lower the power going to each LED. Depending on the
resistor you can use two or three LEDs for each resistor. I have tried
it out on a breadboard and that would probably work best to get the
most out of the 9 volt, although you may need a lot of resistors
because you can power many LEDs with a 9 volt battery. any more
questions or comments about this and you can e-mail me at:


                                                              ajrojee@gmail.com

Bye the way a very good site for purchasing electronics is 

              http://jameco.com/webapp/wcs/stores/servlet/StoreCatalogDisplay?langId=-1&storeId=10001&catalogId=10001