LED Array

Yes, with a separate series resistor for each array.

Yes it will, but if you want 40 LEDs you will need to supply a lot of current.

As you have already calculated, you cannot put 3 LEDs in series since that will require more than 9V.

The most power efficient solution would be to use a switched mode power supply, this will allow you to regulate the current without wasting power in the series resistor. However, it is more complicated that using a resistor and the power savings wont be that great (~25%). Probably not a good idea for a novice, but if you are interested this is the sort of thing I mean:

Another solution would be to use higher value series resistors. That will reduce the current but also brightness of the LEDs. However the LEDs will probably be visible at currents well below the maximum. It depends what you want them for - if you are going to be looking straight at then from short range indoors they will probably be too bright with

20mA. Try some higher value resistors and see what it looks like.

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gareth.harris

Actually the current will be a little less:

I = (Vsupply - Vled)/R; therefore I = (9-6.6)/150 so I = .016

20 strings would use 320 mA.

But it doesn't make sense in any event, because you won't get much life out of a wimpy 9V battery. I'd recommend using a "wall wart" supply instead. There are many alternatives - here's three:

+---+-----+----}}---+-----+-----+ | | | | | | | [LED] [LED] [LED] [LED] [LED] ------ | | | | | | | 12V +|---+ [LED] [LED] [LED] [LED] [LED] | Wall | | | | | | | Wart | [LED] [LED] [LED] | | | -|---+ | | | [270R][270R] ------ | [110R][110R] [110R] | | DCTX-1216 | | | | | | +---+-----+----}}---+-----+-----+

The part number is from Allelectronics for $6.75

That's 12 strings of 3 LEDs with a 110 ohm current limiting resistor for ~ 19 mA and 2 strigs of 2 LEDs with 270 ohm for ~ 20 mA. Total current is ~ 270 ma. You could replace the wall wart with a 12 volt gel cell for portable operation.

Here's another, with 10 strings of 4 LEDS, each with a 150 ohm resistor:

+---+-----+----}}---+ | | | | | [LED] [LED] [LED] ------ | | | | | 16V +|---+ [LED] [LED] [LED] | Wall | | | | | Wart | [LED] [LED] [LED] | -|---+ | | | ------ | [LED] [LED] [LED] DCTX-161 | | | | | [150R][150R] [150R] | | | | +---+-----+----}}---+

If you *must* use 9V batteries, how about three of them in series? Then you would have 5 strings of

8 leds, each string in series with a 33 ohm resistor, with a total current draw of ~ 91 mA. You would get more life this way than you would if you put three batteries in parallel (which is not a great idea for other reasons) owing to the lower discharge rate.

Ed

NiMh AA's would be a *FAR* better choice than a 9V if you have the space. They'd last at least 3 times longer. An Energizer 9V is rated at 625 mAh, while their AA NiMh's offer a choice of 1700, 2000, 2200 and 2500 mAh. 6 cells would give you 7.2 volts, and you could go with the 20x2 you mentioned using 33 ohm resistors to limit the current to ~18 mA. Or you could go with a 10x4 array with 56 ohm resistors at ~ 21 mA if you use 12 cells. You can get closer to 20 mA by using 1% resistors vs 10%, but it's not necessary.

I don't know which brand of NiMh is best.

Ed

If you pulse the supply to leds it is possible to exceed the normal running voltage. I am currently operating a lamp with 3 leds in parallel from a 6V supply with no series resistor. I use a 1/3 duty cycle at a frequency well above the persistence of vision (around 200Hz) No power loss in series resistor (cos there isn't one) and 2/3 power saving due to 1/3 duty cycle However some power needed to drive the 555 timer pulse circuit. overall I get close to 3 times battery life. Lamp has been in use 2 years so seems reliable. I plan to launch a kit of bits, circuit diagram and theory notes on my web site in a couple of months. Watch

Also cost wise AA batteries are much less expensive than PP3 batteries and can be used in ready made battery holders to give a wide range of voltages.

Hope this is useful Ed

Another thought or two: A PP3 alkaline battery costing about =A32-00 has about 550mAh of energy=20 and would run your proposed idea for about 1hour 15 mins.

4 AA alkaline batteries giving 6V and 2500mAh and costing about =A31-20,=20 with the battery saving circuit I propose, would run for about 30=20 hours.

Ed