Wiring Large Amounts of LEDS

Could you précis that a bit?

My time is valuable to me - John.

I'm a little rusty on this but here goes.

1. It doesn't matter where the resistor is, either before or after the LEDs,

2. Its there to limit the current that flows through the LEDs to stop them burning out.

3. In series, all the LEDs will see the same current (which is controlled by the resistor). There will be a voltage drop of 3V across each LED, so for 9V, no more than 3 in series. You can put any amount of the 3 LED series strings in parallel with each other.

As for the resistor, you can either put one in series with each series string of LEDS, or just one in series with ALL the LED strings, in which case you need to add the current draw of all the strings in order to correctly size the resisitor.

Hope this helps!

beautifully and funamentally explained

You are going to run into problems with load sharing when you start to use large quantities of series and parallel connected LEDs. As for the resistor, it doesn't matter if you put it at the positive or negative end. I would recommend either using a constant current LED driver chip (removes the need for resistors and many will also incorporate a voltage stepup circuitry, but may be a little beyond your current level) or using a smaller number of series-connected LED circuits (with their own series resistor), and then parallel as many up as you need. All the info is on the web.

Oh, I am guessing your battery capacity required is a typo.

Slight manufacturing differences between individual LED's can cause them to have slightly different forward voltages. This will cause problems if you hook the LED's in parallel, since they will not share the current evenly. You will see this as some LEDs being dimmer than others, and the LEDs will burn out in less time than they normally would. If you don't have a problem with this, you can hook them up any way you want.

If you want the LEDs to last a long time, hook up as many in series as your battery voltage will allow, then add a resistor in series to set the current to the rated current of the LEDs. This constitutes one string of LEDs. You can then hook as many of these strings in parallel as you want.

Also, note that a 9V battery is usually only 700mAH. This means that if you have 7 parallel strings of LED's, each drawing 20 mA (7 * 20 = 140 mA), your 9V battery will last less than 5 hours ( 700 mAH / 140 mA = 5H ). In practice, you will probably only get 2 or 3 hours of use before the LEDs dim significantly.

Okay so I guess I gave a little to much information for some of you but others who responded helped me a lot to get me on the right path...

To simplify things, I want to hook up say 22 leds to a 9v battery using blue leds at 3.0 forward voltage and run at 20ma... Somebody mentioned a constant current source instead of using resistors..

So I did some searching and found this nice little pdf http://www.constant-current.com/0804-04_KONREGe.pdf that shows a simple series circuit set up with it in place of the resistor.. But it gave me some questions. If i do a combo of 3 leds set up in series paralleled to another 3 etc and then have one more added on to get to 22 leds, do I need to use more than one of these current regulators or will it regulate throughout the entire circuit? Also each series voltage will be reduced to 3volts because of the 3 leds, but how to i limit the voltage of the last led? do i need to use a resistor? I find the whole amps/volts/watts thing a bit confusing....

I also saw mention of capacitors to double a batteries voltage... So wouldn't it be best to use a capacitor to get up to 18 volts so I could series more leds and have to parallel less therefore reducing the mA/hr and get longer life (again the whole amp/volts/watts thing confuses me, i do know you don't get anything for free in life, so you doubled the voltage what am i losing??)

do I need to use more than one of these current regulators or will it regulate throughout the entire circuit?

• You would need to use more then 1 as the max output of the current source is 20mA, using your 9V battery you would need 1 per string. If you could up the voltage, the max voltage on that device is 40V so you would still need at least 2. I think you would be fine with multiple parallel LED-Resistor strings.

Also each series voltage will be reduced to 3volts because of the 3 leds, but how to i limit the voltage of the last led? do i need to use a resistor?

• You know the source voltage 9V, and you know your forward voltage drop 3V, and you know your desired current 20mA. So just start with your 9V and subtract your forward voltage drops (in this case 3) to leave you with 6V. So you will have 6V across your resistor. Next calculate resistance by R=V/I or R=6/.020 which is 300 Ohms.

i do know you don't get anything for free in life, so you doubled the voltage what am i losing??)

• There are inefficies with any circuit so there would be some losses which can not be avoided, also it is more involved then just adding a capacitor. It would require additional circuitry which takes space, money, and power to run. In certain cases it can be very useful or even required but may be a bit of overkill for your needs. If 1-9 volt battery doesn't last long enough add a second and double your life. Good luck!

Shawn

There is another method you might try. Use two 9V batteries in series for 18V.

Put a quantity of 6 each LEDs in series, no resistor(s). You may need to buy two fistfulls of them and by trial and error "grade" them visually for similar brilliance that suits you. Similarly, you can expand with additional strings of quality 6 each, but where the total would be 18 or 24, etc. This would be more energy efficient (no resistor power dissipation). I'm making the assumption that these are not blinking LED's which will not work in series configuration.

You could also use 9V, but with 3-each x 7 strings in series where you would end up with 21. Don't know if you are fixed on 22 or not.

My suggestion, after going through the good ideas already offered, is to hook up 7 strings of three LEDs in series. You want each string to draw 20mA. Subtract the series voltage drop of those 3 LEDs in series from the battery voltage and divide that answer by the 20mA (.02A) current to get the value for each of the seven resistors.

I wouldn't worry about the slight differences in the voltage drop characteristics too much. Placing the three in series increases the possibility that they will cancel out due to random chance. I don't think you will notice problems from it unless you are using a battery with less than 9V output or a low current output type battery like a carbon-zinc.

If you don't mind taking a chance about getting slight variations in brightness from LED string to LED string, you might consider using an 8.2V rechargeable battery (NiCad or NiMH) and hooking up each of the strings of 3 LEDs without current limiting resistors. You could always use trial and error to match the LEDs in each string for brightness.

You are talking about using a dropping resistor to handle LED #22. This dropping resistor will do nothing other than generate heat. Why not replace it with two more LED's for a total of 24. Then you would be getting more light (desired) rather than useless heat... or the other option would be to degrade back to 21 LEDs.

It would be nice if you could drive each string with a 20 ma current source, but this would require a higher voltage than 9 volts. I would have to look at the specs on the current driver mentioned above. In lines of KISS (keep it simple stupid) you might want to try breadboarding 7 (or 8) strings of 3 LEDs, connecting a 9 volt battery, and see what happens. It might do the job with no resistors. Also I would recommend putting the LEDs in sockets if possible. The things do burn out.

Back Later

Bill

I just want to thank everyone for taking the time out to give me some advice... With that being said I'm still stuck on some stuff.... If this looks to long just skip it ;)

I totally understand how to get the basic setup with a straight parallel circuit (highly not recomended), or a combo series/parallel circuit both of which use resistors.. I am also aware that I am limited in the amount of leds i can hook up in series at 9volts, the more I want to hook up in series the more voltage I will need.

This is a site that explains all of that in detail just like all of you have done for me

http://www.theledlight.com/ledcircuits.html

and is where I got a lot of info from and many other sites quote from here and use their pictures... Here is a quote from the site

"However, if you take that same 12V power source and put 4 LEDs in series, there would be 3V going to each LED and (assuming the LEDs are made to run off 3V) each would be powered and just dandy. Check out this illustration:"

In this quote he says that each would be powered and just dandy if you where to divide the 12 volts across the 4 leds assuming the leds are made to run off of 3v... This is what made me say hey I am using 3.0 forward voltage leds and a 9 volt battery so I should be able to hook them up in series and not require a resistor....Which also led me to beleive i could then just parallel the other series up to the desired amount of leds still requireing no resistors because each led is getting the proper voltage... But what about the mA?!?! nobody mentions the mA when hooking up the leds this way.. Is each led getting 20mA or 30mA or 50mA (which i don't want) I have no clue and can't find any answers anywhere, when I use the resistor calculator it cant be done because forward voltage and supply voltage cant be the same (and rightfully so because when you divide the two you get 0 and that makes everything 0)..

I think I will just wire it up that way and see if it melts or not ;) if not I will suck it up and use resistors...

Which brings me to my last question guys which everyone seems to have avoided.. I have a 9 volt battery, I know there are things called capacitors that I can double that to 18 volts, thereby allowing me to use a lot less strings and consume a lot less energy (unless capacitors end up using a lot of energy).. I'd rather use a capacitor than using 2 9vs in series because capacitors are small and I can hide them somewhere while 2 9volts wired together would require me to rout out another cavity which I would LOVE to avoid.... I saw somebody mention that yes I could use them and in certain situations might be necessary but i would require other circuitry and advised me to do something else.. I kinda feel like this situation requires me to use a capacitor due to my size constraints... Could anyone give me a heads up on how to incorporate a capacitor into the series parallel scheme???

Just a crazy thought.

I'm assuming that this is an electric guitar, so your output jack is mono. If you were instead to use a stereo jack and 2 core shielded cable, you can use the third conductor to supply the power to run you leds, (the shielding being the common earth for both the guitar pickup and the leds).

In this way, you don't need any batteries or capacitors inside the body of the guitar, and you can use a low (or high!) voltage dc plug in power source to run the leds. The source can be put next to the amp and hooked to the third conductor and shielding of the guitar lead (via a spliced cable). Because you are supplying DC to the leds, there should be no interference with the guitar pickup's signal to the amp, and no concerns of batteries and leds dying in the middle of the greatest solo ever played!

JUST MAKE SURE THE AMP IS PROPERLY WIRED AND EARTHED !!!!!

As i said, just a thought.

A nine volt battery is nominally 9V but actually slightly more or less. Just take your 9V battery and 7 strings of three LEDs. Use trial and error to connect a resistor in series with the 7 sets of series of LEDs.

If you had a power rheostat like one from an old automobile dash board dimmer switch, you could place it between the 7 sets of LEDs and the battery terminals. You'd start with it set to maximum resistance, then lower it until one of the stings reaches 20ma. Then measure the resistance value of the rheostat and substitute a resistor for it if its value is greater than 0. This procedure would apply in the case of an alkaline battery. See my earlier thoughts if you want to use a NiCad or NiMH battery.

The idea of the capacitor is not easy like that. It requires a minimum of several more components to boost the voltage that the capacitor would need to store.

Okay guys I found what I'm looking for I think, but I need somebody to check it out to make sure it is what I need...

"When running off a Lithium-Ion battery (typically 3.7V output voltage), each low-power LED requires up to 4V at 30 mA. To operate more than one LED for a lighting solution, an LED driver is needed to boost the voltage and regulate the current to optimize LED output."

http://www.national.com/appinfo/power/led.html#

This webpage is set up so you can input what you are looking to light up and it will recomend the led driver to use...

In the led workbench on the top I inputted the values I intend to use (would anyone be able to check if i did it properly?)

Under enter Custom Values

Vf = 3.0 V If = 0.20 Amps

Then Under Configurations and Options

Vin Min 9.0V Vin Max 9.0V (not sure if this part is right)

Number of LEDS

Series 8

Parallel 3

It then shows you how many volts out and how many amps out at 24 Volts .60 Amps

and reccommends a led driver to use, in this case it recommended this http://www.national.com/pf//LM/LM5022.html ...

So what do you guys think?

Me thinks you are on the right track. I have played with this workbench, but I have not actually built any hardware with it. You might want to run it once more with a max voltage of 9.5 volts and a min voltage of 7.0 and see if you get the same part.

I would now download the spec sheet for the LM5052, and any applications notes they have on the part. In fact I will do that too and see if there are any problems using it in your application.

Bill

I ran the problem on Nationals Web bench using the broadened battery voltage range and it still came up with the same regulator (LM 5022). The bad news is that that device requires a whole lot of passive components to work, and most of them are surface mount devices, which implies mounting on a printed circuit board. That is a whole new problem in itself.

Just for general information to all, it looks like the National design software (as written) allows a maximum of three parallel strings of LEDs.

If you breadboard up the 3 series by 8 parallel matrix of LEDs, I would recommend putting a fuse between the battery and the LEDs. Drawing 240 mA of current, I would probably use a 1/2 amp fuse. 1/4 amp might work though.

Sincerely

Bill

Yeah the cool thing is they list all the components needed and give you a schematic of the entire circuit, the bad part is needing the circuit board and the other bad part is that they can make the custom boards for you but they dont make them for the lm 5022... So now I have to learn how to make a circuit board...

Question for ya, After looking at the circuit diagram they give you, will it be as easy as ordering all the parts they list and setting it up on a circuit board the way they have it in the diagram? or will i need to add a fuse like you said? Or is there like 20 million unseen things that I am just not aware of.. I will be using the 8 series 3 parallel schematic making a total of 24 leds (i will just bury the two uneeded ones to where they are unseen or find another use for them)...

OK!!

If you are going to do your own PC board, I use EAGLE (Easily Applicable Graphics Layout Editor). It is a freebies which you can find at www.cadsoftusa.com . It has a pretty good library of parts, but I am sure that you will have to enter some of these parts (from Nationals circuit) into the library. This means going to the individual manufacturers and getting "footprints" of the individual components. Most of this can probably be found on line. It also will take a little bit of "book learning". The Eagle documentation is about 250 pages long.

For fabrication of the PC board, there is a method where you laser print the layout onto photo paper. I emphasize laser because inkjet does not work. If you don't have a laser printer, print it in ink and take it down to a copy center and have it copied with a laser copier. Once you have the layout on photo paper, you iron (use your wife's clothes iron when she is not home) the layout onto the PC board, wash the paper off, and are ready to etch the board. For reference to this method try www.fullnet.com/u/tomg/gooteepc.htm .

Other helpful information can be found at www.thinktink.com/stack/volumes/volvi/substrat.htm

The above is one sheet out of many, but you should be able to get back to their home page from there. All the above information is approximately 3 years old, and things may have changed. Worst case you can Google "Printed Circuit Board Fabrication" and get thousands of bits of information about the process.

Regarding the fuse, if you are using the National part, you can probably get away without it.

Further help?? Just ask.

Bill

Just to clarify a little, gunna be wiring 8 leds in series paralleled 3 times for total of 24 leds(no need to do 3 series by 8 parralel anymore since the lm 5022 boosts the voltage allowing to series 8 leds), so should only be 60mA of current total correct? So would the fuse be necessary?

Rule of Thumb I use.

R=(V_s-V_L)/I_L

R = Series Resistance

V_s = Supply Voltage

V_L = LED Voltage

I_L = LED Current

So, for 1x 3V(10mA) @ 12VDC, the Resistor would be (12-3)/0.01 = 900R

If 3 LED's in series, then (12-(3x3))/0.01 = 300R

If connecting multiple LED's using series parallel connection is safe, I mean, parallel connection of several series connected LED's

------------------- VCC

| | | |

R R R R

| | | |

LED LED LED LED

LED LED LED LED

LED LED LED LED

| | | |

------------------- GND

Bring back ASCII art ;o)