I am an electronics newbie. I am about 5 hours into the videos on this site which are very helpful.

I am building a golf ball display case that includes 27 LEDs. My initial design includes 27 white LEDs, 27 330 Ω resistors, and a 9v battery.

After watching one of the videos I learned about how to calculate the resistance properly for this type of circuit. Doing the math myself (which I hope is correct) it appears the total resistance for this circuit is 12.22 Ω. Does that mean I can put a 12.22 Ω resistor (or whatever standard resistor is closest) just off the power source and not include the 27 resistors I expected to use?

Thanks for your help.

 

Howdy, welcome to AAC, which is my home site. You probably recognize my avitar.

It can be done, but it is a terrible idea. Resistors are cheap, LEDs are expensive, and if an LED burns out the rest have to pick up the slack. The next goes because it is being pushed hard, resulting in the rest taking on more load. You can see where this is going, eventually you have a cascade reaction.

Generally one LED (or a chain of LEDs in series) and one resistor is the way to go.

This explains it in more detail, especially chapters 1 and 2...

LEDs, 555s, Flashers, and Light Chasers

 

27 white LEDs in parallel = about 1/2 A; not good for 9V alk. Can use 2 LEDs in series for 13 strings @ about 1/4 A or about 1 hour, also providing all are same Vf.; still not recommended. A higher operating V would be better, say 12V [8-AAs] giving 9 strings of 3 in series, @ 180 mA; assuming 20mA, 3.2V white LEDs, series resistor of 120 ohms, 1/4W.

 

If you are building a display case, lite by LEDs, you might want to consider something like a buck puck. It would give you a constant current source and you LEDs will last longer.

You could power everything with a wall wart instead of a battery.


A more experienced forum member might let us know if you would still want current limit resistors on top of that.

 

You won't be happy with the expense involved in replacing batteries on a daily basis.

You really need something like a wall-wart supply. They're cheap and widely available.

 

I missed the 9V battery part. Last I heard on the other forum it was a 9V wall wart.

 

Fair enough.

But, what our OP needs to decide is that they want to run this display case from a wall-wart, or from batteries.

If from batteries, it will require constant replacement of the batteries.

If from a wall-wart (plug supply) they can get somewhat reasonable efficiency and economy if they choose wisely.

That's what we're here for, right?

 

I doubt he want to go to the effort, but he could add a timer, something to push and it lights up the display for a minute or so.

 

We don't really know what our OP wants; that for them to decide.

Having a display lit up all day could cost them big money if not done efficiently.

Replacing batteries is not my idea of being efficient.

 

Wow. Thank you all for the quick responses! I really appreciate all of the help.

Happy New Year!

I expect this to be turned on for about 5 hours at a time, which will probably happen 3 or 4 times per month. You're right that I don't want to replace the batteries all of the time, but if it's once every 2 or 3 months or so I won't really mind. In fact, after 2 or 3 months I expect to get tired of it and forget to turn it on most of the time.

The buck puck looks like a good idea and is something I have never heard of (that will probably happen a lot in the near future). My questions about the buck puck are:

1) Which one would best fit my design? I saw quite a few options, but I didn't notice one that mentioned anything as high as 27 LEDs.

2) (Dumb question warning) Since the power supplied would be different than my original 9 volt battery wouldn't I have to recalculate which resistors to use?

By the way, I kind of thought through the power consumption during my initial design. In the structure of the golf ball holder (which is already done, waiting for the the LEDs to be delivered) I have the main power channels (just strips of 14-2 stapled down and ready to distribute power to all of the LEDs) connected to a coaxial power jack and 2 matching power jacks to make changes if needed. I also built a cabinet with a door to hide any power source I need to use (2.5" x 5"). My original design was to hide a power outlet behind this and use a 9v power supply I found in my electronics junk drawer.

While I would love the experience of adding a timer, I don't want to add a timer, because I would like to run the lights for an extended period of time when I'm in the room. As I mentioned above, that wouldn't be very often or for very long.

Sorry for the long response. To summarize:

- Thank you for the advice. I will use a single resistor for each LED.
- I will read chapters 1 and 2 of the LED blog linked above.
- Thank you in advance for any help with my questions about the buck puck.

 

After doing a little more research I may have made some rookie mistakes.

1) While searching ebay it looks like a wall wart is the power supply I found in my junk drawer. Is that right? Is it the same as I would have used for an old cordless phone or something similar?

2) Based on Bernard's response, it doesn't sound like this will run for 5 hours straight with a 9v battery. Maybe I'll just display it for a few minutes at a time instead of keeping it on when I'm in the room. That should resolve quite a few problems.

Thank you for any help on my questions on the previous post.

 

After doing a little more research I may have made some rookie mistakes.

1) While searching ebay it looks like a wall wart is the power supply I found in my junk drawer. Is that right? Is it the same as I would have used for an old cordless phone or something similar?


Thank you for any help on my questions on the previous post.

Yes that is correct. You might even have tos of those little black boxes, if yo are like me. But be careful, they come in all flavors. Some DC, some AC and all kind of voltages.

 

You might get 5 hours per session, but only about 2 sessions worth of power. A wall wart is much better, or even D cells. Of course, batteries tend to have a habit of leaking.

Another advantage of a wall wart, they don't discharge and change their voltage over time.

You can increase the efficiency by putting 2 LEDs in series. Problem with this is white LEDs can drop 3.5 to 3.8 volts, and 9V batteries tend to drop to 7V very quickly, so that doesn't work well.

Did you read chapters 1 and 2 of the article I pointed you to?

 

Spinnaker - Yes. I checked through a large number of leftover wall warts and found a 9v DC one. My understanding is it would be interexchangable with a 9v battery.

Bill - I don't currently have a power outlet available where I'm planning on placing this, so batteries are much more convenient. I'm not against going to a D-cell batteries if they will work better. What about placing multiple 9v batteries on this circuit? Would that double the left-span? I assume it would require different resistors.

Single 9v battery
R = (9 - 3.5) / .02
R = 275

2 9v batteries
R = (18 - 3.5) / .02
R = 725

4 D batteries
R = (6 - 3.5) / .02
R = 125

Are my calculations correct?

Thanks again for all of your help.

 

...and yes. I did read the first 2 chapters of the blog you referenced. Would you recommend that I include a LM317 with this project?

 

Maybe, but I wouldn't.

How complex are you willing to go? The LM317 has a really big drawback, it requires 3V drop to power the circuitry. It is possible to build a current regulator that drops a lot less voltage.

Overall I would stick with a resistor only. It isn't always the most efficient, but it is simple and works well.

D cells will still run down, but a lot more slowly. You could use 4 of them for a 6V power supply and do much the same thing. The resistor would be smaller, of course, but that isn't a big deal (at least from my POV).

You can get a 4 D cell battery pack from Radio Shack and many other sources.

Are you planning on using a simple switch, or a timer of some kind? My idea of the timer is something you push the button on and it stays on for however long you built it for. This would eliminate the possibility of accidentally discharging the batteries by leaving it on for an extended period of time.

The reason I mentioned the article again is it explains why you don't want to use one resistor with a bunch of parallel LEDs, and covers the basic theory of what you want to do.

 

As you may have noticed from most of my quesitons, I am not entirely sure what I'm doing. The extent of my experience is putting LEDs and resistors together to make a LED light up (and an occasional button to turn it on or off).

I have a 555 timer I bought a long time ago, but I have no idea how to use it.

I wasn't planning on having a switch on it at all. When the power is plugged in it will be on. When I'm done looking at it I would unplug it. I know switches aren't difficult to add, but I don't know of a really good place to put it. The power adapter is hidden in one of the side cabinets, which is probably where I would put the on/off switch anyway. Below is an example of the layout.

|__p|_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_x_|___|

Each x is a golf ball. The p is the hidden power supply and adapter.

Would you recommend 2 9v batteries or 4 D batteries?

 

4 D cells. Compare sizes, it is a clue as to power density.

A switch is easy, put it in line with either power wire from the battery pack. You can even get push on / push off switches, they come in huge numbers of configurations, but all are easy to use.

If you change your mind about the wall wart let us know what is written on it.

 

Maybe, but I wouldn't.

How complex are you willing to go? The LM317 has a really big drawback, it requires 3V drop to power the circuitry. It is possible to build a current regulator that drops a lot less voltage.

Overall I would stick with a resistor only. It isn't always the most efficient, but it is simple and works well.

Bill,

Could he use a Buck Puck as I mentioned? Though from what I understand from a post from Sarge, in an answer to a similar question of mine, these things also have an overhead.

But you are right a resistor would be cheap and easy.

 

Two 9V batteries in series to make 18V will last exactly as long as one 9V battery. The extra voltage will heat the resistors instead of making light in the LEDs.