LEDs on Batteries: How to Measure Voltage?

[Meseria]

Hello everybody,

I tried to make a special LED I <3 you gift for my girlfriend and it seems that I failed to take into account the battery discharge curve. I was using 50 red LEDs that I originally measured with my voltmeters LED function at 1.7v.

With measuring LED voltage, is it better to create a small circuit with say, 9v battery, 1 LED and a 1k ohm resistor. Then measure the voltage on the anode and cathode of the LED? I did it this way and it seems like they were running between 2.0 and 2.2 volts at a pretty good brightness. The voltage should be constant and it is the mA value that determines how bright or dim it is, right? So, is this the best way to figure out typ forward voltage for an LED?

Anyways, i setup my light array with 10 parallel series of 5. I had a 22 ohm resistor. The whole ohms = 9v - 1.7(5) over .020 shot back 25 ohms. So I wire it all together and it isn't bright at all, its very dim! Now, this was using my voltmeters LED return of 1.7v, which I think is incorrect.

So did I mess up in that I need to measure the typical voltage of the LED at a load? That gave me a bit higher in the volts dept.

My Energizer 9v's discharge curve seems to go from 9v to about 8.5 quickly, and then lowers to about 7.5 for quite a while before it is very close to dying. This is an alkaline battery. I have some NiMH rechargeable AA's and AAA's and they seem to start about 1.3, drop to 1.2 after a little bit and then stay at 1.2 until they are almost out. LEDs seem to have a very strict need for a constant source of voltage, so do I need to use batteries that have a discharge curve that is quite horizontal, like the NiMH?

 

[vk6kro]

You measured good brightness at 2.2 volts, so use this figure for your calculations.

Allow 25% of the total voltage for the resistor. So this leaves about 6.75 V for the LEDs, so you would have 3 of them using 6.6 volts.

In your test, the voltage across the 1 K resistor was (9 -2.2) or 6.8 volts, so the current was 6.8 mA.

Assume 7 mA for the LEDs so the resistor would be (9 - 6.6 ) volts / 0.007 amps or 342 ohms.
You can get 330 ohm resistors.

If you want 50 LEDs, you will need 16 sets of (one 330 ohm resistor and 3 LEDs) and 1 set of (one 680 ohm resistor and 2 LEDs) and this will draw about 119 mA from the battery. This is OK for a short time, but expensive if you want it to keep doing it.

 

[Fish4Fun]

Meseria,If we examine your circuit we initially find:

5 * (1.7V) = 8.5V ==>
9V - 8.5V = .5V ==>
.5V = I * 22ohms = 22.7mA

In general 22.7mA should make most standard LEDs light fairly brightly. However, you are attempting to share the 22.7mA with 5 parallel strings of LEDs, so each string is only seeing 22.7mA/5 = 4.54mA per string. To ensure each string is getting 22.7mA through a single resistor, you would calculate R as follows:

.5V = 5 * 22.7mA R = 4.4ohms

and a total current through the resistor of 113.5mA; however, this is generally bad practice as it can lead to current runaway. If one of the LEDs fails then the current through the remaining strings increases. Instead you should use 5 * 22ohm resistors, one in series with each string.

At a drain of 113mA your strings will drain an Alkaline battery in ~ 5 hours. (Rated capacity for an Alkaline PPP3 => 565mAh).

Fish

 

[Meseria]

Allow 25% of the total voltage for the resistor. So this leaves about 6.75 V for the LEDs, so you would have 3 of them using 6.6 volts.

Why would I want to allow 25% to go to the resistor? Is it just good practice to not be extremely close in regards to voltage requirement and voltage available? Seems this may also allow the LED to remain brighter for longer as the battery begins to die (lowers in voltage)?

In general 22.7mA should make most standard LEDs light fairly brightly. However, you are attempting to share the 22.7mA with 5 parallel strings of LEDs, so each string is only seeing 22.7mA/5 = 4.54mA per string.
Fish

So, for example, if I have 4 LEDs in a series, running 9v source, 2.2v per LED and 20mA, Ill be at ohm = (9-8.8)/.02 = 10 ohm resistor. So if I doubled it and put 8 LEDs in two parallel strings with 4 LEDs in a series per string, each LED now would only get 10 mA rather than 20, and thus be quite a bit dimmer?

 

[Wetmelon]

So if I doubled it and put 8 LEDs in two parallel strings with 4 LEDs in a series per string, each LED now would only get 10 mA rather than 20, and thus be quite a bit dimmer?

Yes. Assuming you're not doubling the resistance as well (which would half again the current to the strings).

 

[vk6kro]

Why would I want to allow 25% to go to the resistor? Is it just good practice to not be extremely close in regards to voltage requirement and voltage available? Seems this may also allow the LED to remain brighter for longer as the battery begins to die (lowers in voltage)?
So, for example, if I have 4 LEDs in a series, running 9v source, 2.2v per LED and 20mA, Ill be at ohm = (9-8.8)/.02 = 10 ohm resistor. So if I doubled it and put 8 LEDs in two parallel strings with 4 LEDs in a series per string, each LED now would only get 10 mA rather than 20, and thus be quite a bit dimmer?

The 25% figure is a good guide. If you have the sum of the LED voltages too close to the supply voltage, then the current will be very dependent on the supply voltage. If it were to drop below 8.5 volts as in your case, then the LEDs would go dim or dark.

If you have parallel strings of LEDs, each string with its own resistor, and the battery does not change in voltage, then each string will draw whatever it drew alone.