# LED Array Questions (beginner)

Discussion in 'The Projects Forum' started by Nexum, Mar 25, 2008.

1. ### Nexum Thread Starter New Member

Mar 25, 2008
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0
Hi,

I am after a little advice about building an LED array project which I would like to have somewhere on the order of 100 LEDs laid out on a grid.

First of all, a question about power consumption. From my reading a single LED consumes around 20mA of power (am I right in thinking all different colours of LED are roughly equivalent, including more exotic kinds like the blue ones?). So If I have 2 AA batteries, which gives me ~5700mAh of power, I should be able to light a single LED for 285 hours. Or 100 LEDs for a little under 3 hours.

Just want to make sure I've done my sums correctly there.

Also, if this is the case, I'm probably not going to want to run this array from a battery. Even a bunch of D batteries won't give me a whole day's worth. And so, question 2; would it be a good idea to power something such as this by the mains? If so, how is this usually done? My only experience so far is with DC and simple battery powered circuits. AC scares me somewhat, especially the idea of plugging things into a 230VAC socket. Any pointers on sorting things like that out would be a great help.

Thirdly and finally, with big arrays of LEDs like this, I have read that the battery is unable to supply a high enough current, and so transistors should be used to step that up. I barely have an idea of what a transistor is, and can't quite appreciate how a transistor has this effect, but if someone could let me know the pertinent issues around current supply to an array of LEDs, I'd be very grateful.

Thanks all for your help and time!

- N

2. ### Audioguru New Member

Dec 20, 2007
9,411
896
An AA alkaline battery cell is about 2850mAh when its current is only 25mA and the end voltage is 0.8V. Two still have a capacity of only 2850mAh at 25mA but their starting voltage is 3.0V and the ending voltage is 1.6V.

A red LED needs about 1.8V to 2.0V. A blue or white LED needs about 3.4V to 3.8V so two AA alkaline cells won't work.

If you had 100 red LEDs then the current is 2A and two AA alkaline battery cells would light them for about 15 minutes.

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3. ### SgtWookie Expert

Jul 17, 2007
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That is do-able.

LED's actually come in quite a variety of ratings and configuratons.
But if you're talking about your typical 5mm LED, they are usually rated by I (current) @ Vf (forward voltage). Maximums and typicals are usually given, but sometimes only maximums.

For example, a red LED might be rated:
20mA @ 2.0V Max, 1.8V Typ
which is saying that while the maximum Vf could go as high as 2.0V for a current of 20mA, a typical Vf at that current would be 1.8V. Even in the same lot (batch), Vf's tend to vary several tenths of a volt.

Red generally has the lowest Vf (1.7v-1.9v), then yellows (2v-2.3v), then greens (2.1v-2.5v). Blues and whites tend to have significantly higher Vf's; ranging from 2.9v-4.1v. There are lots of different colors available nowadays; each has a different chemistry to create the particular spectrum of emitted light.

You would not want to try running such a large array of LED's from a couple of penlight cells.

You should use a power supply that has an isolation transformer. Running things directly from the mains isn't safe, nor recommended.

You can convert a power supply from an old ATX-form factor computer to get a very handy "bench" power supply, for just a few dollars worth of parts.
...will lead you to quite a number of plans for converting these supplies. I converted a 250W supply from an old Compaq, which gives me +5V @25A, +3.3V@14A, +12V@8A, and -5V,-12V @0.8A max. They're quite efficient for the power they put out.

Well, there ARE DC-DC "boost" converters to step up voltage, but they are not 100% efficient (although some come close). You can boost voltage, but that reduces the available current.

LED's need to either have their current limited using a resistor, or a current limiter circuit. If you have a high enough supply voltage, you can connect several LED's together in a series "string", and use just one resistor to limit the current at some point in the circuit.

To calculate the size of the resistor you'll need to limit the current:
Rlimit = (VoltageSource - Vf(total) ) / LEDCurrentRating

Example: you have a 12V power supply, and want to light a single blue LED that is rated for 20mA @ 3.2Vf
Rlimit = (12V-3.2V) / 20mA
Rlimit = (12-3.2) / 0.02
Rlimit = 8.8/0.02
Rlimit = 440
470 Ohms is the next highest standard resistor value. Figuring what actual current will be:
I = E/R
I = 8.8/470
I ~= 18.7mA

You can use the same formula with multiple LED's - just add up all of the Vf's, and use that number for Vf(tot). The Vf(total) should be at most 0.7v less than your supply voltage, so that your Rlimit will be at least 35 Ohms. Having a resistor smaller than that may cause your LED current to fluctuate wildly with small changes in the power supply voltage. This problem gets worse for LEDs that are used in an auto or truck, as the voltage may vary from a high of around 14.5 to a low of around 11.

Another way is to use a constant current source. You can use an LM317 or LM317L to do this, by using a 60 Ohm resistor between the output and adj pins, and you get about 20mA current from the adj pin. However, your voltage supply needs to be at least 1.8v higher than Vf(total).

BTW, I use an LM317 with the 60 Ohm resistor with a 7VDC supply to get the Vf reading from the LED's I use. (Hint: you can get a 7VDC difference in potential from your modified ATX bench power supply by using +5V as the more negative source and +12V as more positive source.)

Another method of controlling current is by using a PWM source. This is getting a tad advanced though. Try some of the things I've suggested so far.

4. ### Nexum Thread Starter New Member

Mar 25, 2008
4
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Thanks, that's a great help. Obviously I hadn't considered the effects on capacity vs. voltage of using the cells in series properly.

I'm still not 100% though - if the cell has a capacity of 2.8Ah (2850mAh). And a load of 2A, I should get 1.4 hours of work out of the battery? Where am I going wrong there?

Thanks for your help, it's much appreciated,

- N

5. ### Nexum Thread Starter New Member

Mar 25, 2008
4
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SgtWookie - thanks for taking the time to produce such a comprehensive reply. It's much appreciated, and you've given me a lot of material to absorb.

Thanks!

- N

6. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
A battery is not simply a current source; it also has a certain amount of internal resistance.

At low output current levels, it isn't very significant. However, as you increase the output current, the internal resistance becomes more and more significant. The current flow through this resistance causes heat, which is effectively wasted power. In severe cases (such as the battery terminals being shorted together) the heat can build up more quickly than the battery can dissipate, and can cause the battery to explode.

7. ### Audioguru New Member

Dec 20, 2007
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896
Look at the graph I posted for an AA alkaline cell with a capacity of 2850mAh when its current is only 25ma.

At a current of 500mA, its capacity is 0.5A x 3.2h= 1600mAh.
At a current of 1A, its capacity is only 1A x 1h= 1000mAh.
So I guessed that its capacity at 2A would be about 500mAh. It might be less.

8. ### Nexum Thread Starter New Member

Mar 25, 2008
4
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Thanks for explaining - both of you - that's very interesting, I had not anticipated that batteries would behave like that when providing higher levels of current.

One thing appears to be clear - using batteries is not going to be a good way to power this thing if I need it to be on for any length of time (I do - I'm trying to make a little grow lamp).

Thanks again,

- N

9. ### DC_Kid Distinguished Member

Feb 25, 2008
640
9
if you drive them say at 50Hz @40% duty cycle then battery life should get extended.

10. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
What you are proposing will extend the battery life, but will also reduce the light output from the LED's.

For the number of LED's the OP is planning on driving, it would really be impractical to power them from batteries.