LED Array

[[email protected]]

I am interested in building an LED array with about 40 LEDs. They take
3.3 volts at 20mA.

Does it make sense to make 20 arrays of 2 in series with a 150 ohm
resistor, and connect all of those in parallel to a 9 volt battery?

In this way they will pull 400 mA from the battery. I don't think this
is the best solution as it will deplete my 9v battery too quickly.

I am a novice so any help is greatly appreciated!

Thanks!

 

[Gareth]

I am interested in building an LED array with about 40 LEDs. They take
3.3 volts at 20mA.

Does it make sense to make 20 arrays of 2 in series with a 150 ohm
resistor, and connect all of those in parallel to a 9 volt battery?

Yes, with a separate series resistor for each array.

In this way they will pull 400 mA from the battery. I don't think this
is the best solution as it will deplete my 9v battery too quickly.

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:

http://www.linear.com/pc/productDetail.do?navId=H0,C1,C1003,C1042,C1031,C1061,P2114

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.


--

 

[ehsjr]

I am interested in building an LED array with about 40 LEDs. They take
3.3 volts at 20mA.

Does it make sense to make 20 arrays of 2 in series with a 150 ohm
resistor, and connect all of those in parallel to a 9 volt battery?

In this way they will pull 400 mA from the battery. I don't think this
is the best solution as it will deplete my 9v battery too quickly.

I am a novice so any help is greatly appreciated!

Thanks!

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
http://www.allelectronics.com/
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

 

[[email protected]]

Thanks so much! I *must* use batteries but the system only needs to
run for about 30 minutes or so... what about NIMH AAs? Any
recommendations on which would be best? I'd like to be able to
recharge them rather than constantly have to get new batteries

I am interested in building an LED array with about 40 LEDs. They take
3.3 volts at 20mA.

Does it make sense to make 20 arrays of 2 in series with a 150 ohm
resistor, and connect all of those in parallel to a 9 volt battery?

In this way they will pull 400 mA from the battery. I don't think this
is the best solution as it will deplete my 9v battery too quickly.

I am a novice so any help is greatly appreciated!

Thanks! 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 $6.75http://www.allelectronics.com/
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

 

[ehsjr]

Thanks so much! I *must* use batteries but the system only needs to
run for about 30 minutes or so... what about NIMH AAs? Any
recommendations on which would be best? I'd like to be able to
recharge them rather than constantly have to get new batteries

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

I am interested in building an LED array with about 40 LEDs. They take
3.3 volts at 20mA.

Does it make sense to make 20 arrays of 2 in series with a 150 ohm
resistor, and connect all of those in parallel to a 9 volt battery?

In this way they will pull 400 mA from the battery. I don't think this
is the best solution as it will deplete my 9v battery too quickly.

I am a novice so any help is greatly appreciated!

Thanks! 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 $6.75http://www.allelectronics.com/
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

 

[[email protected]]

I am interested in building an LED array with about 40 LEDs. They take
3.3 volts at 20mA.

Does it make sense to make 20 arrays of 2 in series with a 150 ohm
resistor, and connect all of those in parallel to a 9 volt battery?

In this way they will pull 400 mA from the battery. I don't think this
is the best solution as it will deplete my 9v battery too quickly.

I am a novice so any help is greatly appreciated!

Thanks!

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 http://www.geocities.com/sirkituk/index.htm
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

 

[[email protected]]

I am interested in building an LED array with about 40 LEDs. They take
3.3 volts at 20mA.

Does it make sense to make 20 arrays of 2 in series with a 150 ohm
resistor, and connect all of those in parallel to a 9 volt battery?

In this way they will pull 400 mA from the battery. I don't think this
is the best solution as it will deplete my 9v battery too quickly.

I am a novice so any help is greatly appreciated!

Thanks!

Another thought or two:
A PP3 alkaline battery costing about £2-00 has about 550mAh of energy
and would run your proposed idea for about 1hour 15 mins.
4 AA alkaline batteries giving 6V and 2500mAh and costing about £1-20,
with the battery saving circuit I propose, would run for about 30
hours.

Ed