Hi guys can someone tell me if i need 33 ohms or isit 3.3k ohms

Peter

 

Thanks Ron

So im looking for a 1/2 watt 3.3k or 4.7k ohms resistor

Peter

No you are looking for a 3.3 ohm or 4.7 ohm resistor. Not a 3.3 K or 4.7 K. Ohms not K Ohms.

Ron

 

This is what i found 0.6 watt 3.9 ohm metal film resistors, is this any good?

 

That should work. This is not a very exacting science. Too much added resistance and the LEDs will just be a little dim.

Ron

 

You posted an ad for an ebay three AAA cells battery holder, not for AA cells. AAA cells will last 1/3rd as long as AA cells.

Instead of guessing about the resistor value, measure the voltage across the 15 ohms resistor in one letter when it is turned on with three new button batteries and use Ohm's Law to calculate one resistor for 5 letters, powered by three AA cells in series making 4.5V.

 

I calculated wrong!

I used 20 Ohms for the resistance of the button cells, but that is for 1 cell. With three cells in series, it would be 60 Ohms. Which means the calculated series resistor should be 12 Ohms.

Bob

 

Some letters have 3 LEDs (the letter I), some have 4, some have 5 and most have 6 LEDs. Then the resistor value must be calculated for the number of paralleled LEDs in each letter.
Some letters are missing.

 

Sorry for joining this post so late. Wouldn't it just be as easy (if not easier) to wire all the negative battery terminals together, wire all the positive battery terminals together, and then connect those wires to a 3 cell AA battery holder? Forget about calculating the resistance of each individual letter, just assume each has been engineered for the proper current and only add one primary resistor between the new battery holder.

Pick any single letter and calculate the current being used for that individual letter. Assume (I know, bad practice) assume you have 10 milli amps running any given letter. Removing the button cells, your current will go up. So you'll need to replace that internal resistance. IF they are 20 Ω internal resistance each then 3 of them would be 60 Ω. I would assume that adding a 60 ohm resistor from the new battery pack probably would be safe. Given the internal resistance of a single AA battery is probably low (I'm not the expert on this) having slightly more resistance is just going to mean the letters will glow a little less brightly. Probably not even noticeable.

Turn on all the switches for each letter and use a master switch at the battery pack to turn them all on and off at once.

Am I wrong? Someone correct me if so.

 

The datasheet for an AG13 button cell says it is used for a 0.2mA tiny current, not for lighting LEDs.
If its internal resistance is 20 ohms when new then 60 ohms replaces the resistance of 3 cells in series as in one letter. But 12 ohms is needed to be in series with 5 letters.

 

I didn't look up the data sheet. I know. Don't say it. I'll shut up now.

 

Ideally each LED should have its own resistor based on the battery voltage, LED forward voltage and current. Then all the LEDs would be equally bright.

Bob

 

Lately I have been gathering up discarded LED Christmas tree strings and measuring their LEDs. I am amazed that their forward voltages in a string are nearly the same so I have paralleled many LEDs in circuits and they all look the same and work well. Even different colors. All the reds have the same voltage, all the greens have the same voltage but different from the reds of course, yellow, orange and even white LEDs.

 

Careful, you might end up dispelling a very popular myth.

 

Paralleling LEDs and blowing up the brightest ones with the lowest forward voltage is not a myth. The disposed LED Christmas tree light strings I gathered did not work. Why? Because the brightest LEDs with the lowest forward voltage have already been blown up leaving the ones with the same forward voltage for me to use.

I have a cheap Chinese flashlight with 24 LEDs directly in parallel and it works fine. Probably all the LEDs the factory gets are tested and grouped into piles with the same forward voltage. The factory making the strings of Christmas tree lights might not do that.

 

But the myth says that all of the rest will fail as well in a chain reaction, but you seem to have many surviving LEDs.

Did you do an empirical test that proves the ones that failed were because of being in parallel?

Perhaps they just failed because of other reasons.

 

Perhaps they just failed because of other reasons.

I admit that a few LEDs failed because their leads rusted away.

 

Hi guys project completed i used the 3.3 bob suggested.

Thanks guys

 

Thanks, always nice when someone shows their finished work.

 

Gotta check to see why that "u" is a different colour, i had to buy a replacement "u" as it was broken.

 

Glad it all came together.

Ron