Ok i thought i was keeping it simple but now it is reall simple i was going to use the two 1/8W resistors like you showed : )

here is a picture:

I don't know why I missed this, but I did - and I can't see your image of your circuit board.

It looks like it's posted via IP address. Could you try again please?

 

OK, I see the image now - the connection from me to that hosting site must've been down earlier. Yes, that will work just fine When you install the LEDs, make sure that the shorter lead is towards the '-' power connection.

 

A 2.2V LED in series with a 68 ohm resistor and powered from 5.0V has a current of 41ma which is much too high for your old LEDs.
Two 2.2V LEDs in series with a 68 ohn resitor and powered from a 5.0V supply will have a current of 8.8mA which is low and they will be dim.

Don't connect LEDs in parallel. They are all different and will not share the current well.

 

Audioguru,
The LEDs go in the two upper locations, cathode up, so they will be in series.
This will leave 0.6v to drop from the 5v USB supply.
The two lower locations will be occupied by one 100 Ohm and one 150 Ohm resistor, resulting in 60 Ohms, and 10mA current flow.

 

How would you make it have more than one led in a series wouln't you need to have more volts than just 5. i did one of those equations you gave me and just three leds made it go into the negitives even the led calculator said the same thing. what would you use to make it work.

 

How would you make it have more than one led in a series wouln't you need to have more volts than just 5. i did one of those equations you gave me and just three leds made it go into the negitives even the led calculator said the same thing. what would you use to make it work.

That's correct. Since you stated that your LED's Vf is 2.2v, and your supply is 5V, then the maximum number of LEDs that you can connect in series is 2.
MaximumLEDsInSeries = Integer((VoltageSupply - 0.5)/VfLED)
MaximumLEDsInSeries = Integer((5 - 0.5)/2.2)
MaximumLEDsInSeries = Integer(4.5/2.2)
MaximumLEDsInSeries = Integer(2.04545...)
MaximumLEDsInSeries = 2

The reason for the 0.5v is to ensure that the current limiting resistor is dropping at least 0.5v so that regulation is pretty stable for LEDs that you have measured for Vf. If you have NOT measured them for Vf, you should increase that number to 1.5 or 2. Otherwise, the brightness of your LEDs will vary considerably, and you risk burning them out due to providing them with too much current.

If you wish to connect more LEDs in series, then you would need to increase the voltage being supplied to the circuit.
For example, if instead of a 5V supply, you had a 12V supply:
MaximumLEDsInSeries = Integer((VoltageSupply - 0.5)/VfLED)
MaximumLEDsInSeries = Integer((12 - 0.5)/2.2)
MaximumLEDsInSeries = Integer(11.5/2.2)
MaximumLEDsInSeries = Integer(5.227272...)
MaximumLEDsInSeries = 5

Then you would calculate the current limiting resistance as before:
Rlimit = (VoltageSupply - VfLED(total)) / ILED
Rlimit = (12 - (5*2.2)) / 10mA
Rlimit = (12 - 11) / 0.01 A
Rlimit = 1 / 0.01
Rlimit = 100

Make sense?

 

so would a 270 ohm work for three leds if i have a power supply of 9 volts

 

Yes, but you'll only get 8.9mA (approx) through them.

Remember how to calculate the current?
9-(3*2.2)=2.4V remaining to drop across the resistor.
I = E/R
I = 2.4 / 270
I = 8.888....mA
You could also use two 120 Ohm resistors in series, which would give you 240 Ohms and 10mA current. Of course, that's up to you.

 

i made a quick and small circuit it board with only one led to see how it works.I put a 220 ohm resitor with the 5v and the led and it works good. I think it is geting 12ma so it shouldn't be that bad for it.

 

i made a quick and small circuit it board with only one led to see how it works.I put a 220 ohm resitor with the 5v and the led and it works good. I think it is geting 12ma so it shouldn't be that bad for it.

Roughly 12.7mA.

If your LEDs are truly rated for 10mA max, you're overstressing them.

 

wow if that is to mutch then they are so dim you can barley see them no more of those for me lol

 

Are you SURE that their maximum current rating is 10mA?

One good thing about them - you won't go blind looking at them like you can the new super-brights.

 

lol lol Thats what the thing on radio shack said that they where only 10 ma which i did not know what that ment at the time

 

Ahh, ok.

Do you still have the radio shack stock number that's on the package?
Or did you get one of those LED assortment packs?

 

For years, Radioshack didn't know that LEDs produce light so they never spec'd the brightness. The weight and the size were spec'd but not the brightness.
I can't remember if they even said the color of the LEDs.
Their catalog had a page full of LEDs for you to select one by its weight.
DUMB!

 

Audioguru,
Radio Shack has made so many horrific blunders in the last 30 years, we could discuss it forever, and agree upon a great many points.

My Radio Shack Model I computer is now over 30 years old. My Radio Shack Model 4P is now about 20 years old. One day I'll dig them out and fire them up for posterity.

I don't have a clue when R.S. pulled out of Canada. Perhaps they should've stopped trying to sell cell phones in places where there were no towers, and get back to their roots of selling radios and radio equipment.

However, I have sitting next to me the cardboard backing for R.S. Cat No 276-041a that I bought about a decade ago, which I'll now copy verbatim:
-------------------------------------------
RED
Light-Emitting Diodes (2)
T-1/34 size
-------------------------(flipping it over)--------------------------
ABSOLUTE MAXIMUM RATINGS (25°C)

Power Dissipation: 60mW
Forward Current: 20mA

Optoelectrical Characteristics
(at 10mA)

Forward Voltage: 2.0V
Luminous Intensity: 6.3mcd
Peak Wavelength: 650nm

Short lead is Cathode (-) minus.

(barcode - no, I'm not going to try to replicate it. Ok, skinny bar, large space, skinnybar, fat bar....)
276-041
Made in Japan or China
Custom Packaged in USA for Radio Shack
-------------------------------------------
There you have it. Yes, 6.3mcd is pitifully dim by today's standards.

B3ans,
I think you got confused by the 10ma mentioned on the "Optoelectrical Characteristics" specification.

I'll suggest that you can safely run these LEDs at up to around 17mA or so, but don't go over that - unless you measure the Vf of each one.

These LEDs are perfectly suited for experimentation purposes, power-on indicators, and for such purposes as you're experimenting with now. I'll bet you know more about LEDs than when you started on this project, right?

You can buy super-bright LED's online for cheap ... but beware; they can actually cause severe damage to your vision if you look at them for more than a few moments. The light they put out is very concentrated; your eyes' iris only adjusts to the average of ambient light. Those bright little LEDs will burn permanent "dark spots" in your retina.

 

lol thank you for your help. I learned there is alot more to leds then i thought. i need to continue to read the tutorial that this site has. here is a link to the leds i got that i probaly will not get agaqin lol.

http://www.radioshack.com/product/i...igkw=green+led&kw=green+led&parentPage=search


down in the subscription its says 10ma max but i do not know verry mutch about circuits trying to learn more though.

 

$0.75 each for 20mcd 10ma LEDs - wow!

One way to find an LED's current limit is to start at a minimal level (say, 5mA) and slowly increase it until the LED starts to change color, then reduce the current until the color is back to normal.

I just tried the above experiment with some green LED's I obtained from RS and other sources a decade or so ago. Even when I exceeded 40mA, the color didn't change.

It's likely that Radio Shack received so many complaints about the LEDs burning out, that they lowered the max current ratings for their LEDs. Usually what happens is a new experimenter will try to directly connect an LED directly across a couple of batteries without a current limiting resistor, resulting in the LED trying to conduct all of the current the batteries can generate. It works, for perhaps a few milliseconds (1 mS = 1/1000 of a second.) Then the frustrated experimenter tends to blame the LEDs as being junk, rather than understanding what they did was not correct.

When batteries are fresh, their internal resistance is quite low, and increases as the battery is discharged. You can find out what the internal resistance of a battery is by using a digital multimeter and an external resistor of between 1k and 3k Ohms.

Measure the voltage of the battery without the external resistor connected.
Then connect the resistor across the battery terminals, and measure the voltage again.
Let's say your first reading was exactly 1.5v, and the 2nd reading was 1.49v.
InternalResistance = (1-(LowestVReading/HighestVReading))x ExternalResistorValue
InternalResistance = (1-(1.49v/1.50v)) x 1000 0.99333....
InternalResistance = (1-0.99333....) x 1000
InternalResistance = 0.00666... x 1000
InternalResistance = 6.66... Ohms (Resistor of the Beast)