I kinda thought it would change because doesn't current change I a parallel circuit whereas in series it doesn't?! Then kinda guessed that if you calculate a curent for one circuit then, use that value to calculate the overall current as in the case of parallel current??

 

Okay, the drawing helps immensely. A few questions:

1) What is the value of the resistors used for each LED color and quantity?

Five Red LEDs in series:

Three Red LEDs in series:

Three Blue LEDs in series:

Three White LEDs in series:

2) Can't read the other portion of the drawing that mentions circuit quantities. You said you had 543 LEDs total, is this correct? Your drawing shows 33 red LEDs correctly, but 18 x 3 = 54 blue LEDs (qty of 124 stated) and 15 x 3 = 45 white LEDs (qty 88 stated). I can't find a combination of the stated values that gives 543. Also, if 88 and 124 are used in series of 3, you'll have one LED in blue and white that needs its own resistor which is not shown. If this is the case, we'll need the value of those resistors as well.

3) You don't necessarily have to draw all 500+ LEDs, so let's try this:

a) What is the total number of LEDs are you using and how many of each color are you using?

b) How many sets of five red LEDs in series are there?

c) How many sets of three red LEDs in series are there?

d) Are there any red LEDs in other sets (one, two, etc.)? If yes, how many?

e) How many sets of three blue LEDs in series are there?

f) How many sets of one blue LED are there (single blue LED with its own resistor? What is the resistor value?

g) Are there any blue LEDs in other sets (two, etc.)? If yes, how many and hat is the resistor value?

h) How many sets of three white LEDs in series are there?

i) How many sets of one white LED are there (single white LED with its own resistor)? What is the resistor value?

j) Are there any white LEDs in other sets (two, etc.)? If yes, how many and what is the resistor value?

Perhaps drawing everything out would be a better option after all . . .

So literally if you have 3 LED's in series drawing 60mA then

Wait, wait, wait - if you're putting 60mA continuously across three LEDs in series that are rated for 20mA, you've just cooked them. If the LEDs each require 20mA and they are in series, then 20mA is passing through the string, not 60mA. You would, however, need triple the forward voltage required. If the LEDs have a forward voltage of say 3.2V, three in series would require 9.6V but only 20mA. Because LEDs require current limiting, you want to supply at least 1-2V extra and drop it across a current limiting resistor. This is what you're doing now.

If you put 100 LEDs in series, you'd need over 320V(!) but you'd still only pull 20mA from your power supply.

When you put LEDs in parallel, the current adds up but the voltage does not. So if you put 3 LEDs in parallel, each would require at least 3.2V (again more so you could use a resistor for each one) and 20mA. You could use a 5V power supply, but now the current draw is 60mA. Again, if you used 100 in parallel, you'd still only need 3.2V+, but now 2A!

Because it is often more cost effective to get a power supply with a larger voltage than a huge current, you generally want to put as many LEDs in series as you can. Additionally, when LEDs are in series, they are each pulling the same amount of current, so they will all be at the same brightness. If you put all LEDs in parallel, the value of each resistor could be slightly different from one to another resulting in some LEDs appearing brighter than others.

Whew. Sorry, it's late, so if this diatribe doesn't make sense, let me know and I'll try to explain more clearly later.

 

Since the LEDs are all rated for 20mA, I'm assuming you have size the resistor for each string to nominally result in the same current (hopefully less than 20mA) to flow in each string. Or perhaps you have decided on a different current for each color in order to get a desired relative brightness. So let's call the current Ired, Iwhite, Iblue.

Your total current is

Itotal = 7*Ired + 15*Iwhite + 18*Iblue.

 

Alas!! We are singing from the same hymn sheet. Ill give you all the data for each circuit and hopefully it will make sense.

Red:
6 x
100Ω for 5 LED's in series

1x
330Ω For 3 LED's in series

Drawing 140mA in total

Blue:
13 x
150Ω For 3 LED's

Drawing: 260 mA totalling 39 LED's




41 x
150Ω for 3 LED's

1 x
470Ω for 1 LED

Drawing: 840mA totalling 124LED's



39 x
150Ω for 3 LED's

1 x
470Ω for 1 LED

Drawing: 800mA totalling 118LED's






1 x
470Ω for 1 LED

Drawing: 20mA totalling 1LED




White:
16 x
120Ω For 3LED's

1 x
270Ω For 2LED's

Drawing: 340mA totalling 50LED's






29x
120Ω For 3LED's

1x
470Ω For 1LED

Drawing: 600mA totalling 88LED's






29x
120Ω For 3LED's

1x
470Ω For 1LED

Drawing: 600mA totalling 88LED's

 

My drawing albeit incredibly crude was not - by a long shot precise it was just to clarify the ambiguity surrounding how its wired up. There are a total of 8 separate circuits each consisting of a standard series/parallel array and each one is connected in parallel to the main circuit. That's all it was meant to show really.

The spacing in the previous post is to add emphasis as to what I am classing as 1 circuit. In the instance for example where there are 124 blue, the reason why I've included the single one within the circuit is purely to do with tidiness and spacing on the board so yes - that LED is included in the circuit along with 41 groups of 3LED's

Just a few other points to clarify:
I must re-emphasise I've never done anything like this before so if anything looks a little obscure its because I have tried to be as cautious as possible. For example why 4 blue circuits instead of just 1?? I figured that it is the current which will wreak havoc in terms of overheating and potential fire risks so I have limited circuits to no more than 150LED's to try and eliminate this.

Same for the voltage choice - instead of messing with high voltages when I don't essentially know what I'm doing could be dangerous.

All series LED's should be resisted appropriately as I used an online calculator to suggest appropriate resistors for the following parameters:
Number of LED's in circuit
LED forward voltage
LED forward current
supply voltage.

The LED's actually have an operational range I.e off the top of my head the whites illuminate between 3.1 - 3.5V but I chose the median values as this would allow for unexpected voltage generation in the circuit or as the is more likely the case, because of the size of the circuit it may not generate enough power to illuminate the LED's

The circuit is totally finished and working perfectly - all balanced etc so current overload I'm guessing isn't really an issue right??

Would you still prefer me to draw a full schematic or have you got enough information to advise me on the correct formula to calculate the current??

As previously mentioned the current is my main area of concern because that is what will be responsible for safety issues right??

 

My drawing albeit incredibly crude was not - by a long shot precise it was just to clarify the ambiguity surrounding how its wired up. There are a total of 8 separate circuits each consisting of a standard series/parallel array and each one is connected in parallel to the main circuit. That's all it was meant to show really.

The spacing in the previous post is to add emphasis as to what I am classing as 1 circuit. In the instance for example where there are 124 blue, the reason why I've included the single one within the circuit is purely to do with tidiness and spacing on the board so yes - that LED is included in the circuit along with 41 groups of 3LED's

Just a few other points to clarify:
I must re-emphasise I've never done anything like this before so if anything looks a little obscure its because I have tried to be as cautious as possible. For example why 4 blue circuits instead of just 1?? I figured that it is the current which will wreak havoc in terms of overheating and potential fire risks so I have limited circuits to no more than 150LED's to try and eliminate this.

Same for the voltage choice - instead of messing with high voltages when I don't essentially know what I'm doing could be dangerous.

All series LED's should be resisted appropriately as I used an online calculator to suggest appropriate resistors for the following parameters:
Number of LED's in circuit
LED forward voltage
LED forward current
supply voltage.

The LED's actually have an operational range I.e off the top of my head the whites illuminate between 3.1 - 3.5V but I chose the median values as this would allow for unexpected voltage generation in the circuit or as the is more likely the case, because of the size of the circuit it may not generate enough power to illuminate the LED's

The circuit is totally finished and working perfectly - all balanced etc so current overload I'm guessing isn't really an issue right??

Would you still prefer me to draw a full schematic or have you got enough information to advise me on the correct formula to calculate the current??

As previously mentioned the current is my main area of concern because that is what will be responsible for safety issues right??

 

Alas!! We are singing from the same hymn sheet. Ill give you all the data for each circuit and hopefully it will make sense.

Reorganizing your data a bit:

Group|Color|Strings|I(mA)||LEDs|Vled(V)|Vr(V)|R(Ω)|I(mA)
1|Red|6|120||5|10|2|100|20.0
2|Red|1|18||3|6|6|330|18.2
TOT | Red | 7 | 138 || 33
3|Blue|13|234||3|9.3|2.7|150|18.0
TOT | Blue | 13 | 234 || 39
4|Blue|41|738||3|9.3|2.7|150|18.0
5|Blue|1|19||1|3.1|8.9|470|18.9
TOT | Blue | 42 | 757 || 124
6|Blue|39|702||3|9.3|2.7|150|18.0
7|Blue|1|19||1|3.1|8.9|470|18.9
TOT | Blue | 40 | 721 || 118
8|Blue|1|19||1|3.1|8.9|470|18.9
TOT | Blue | 1 | 19 || 1
9|White|16|280||3|9.9|2.1|120|17.5
10|White|1|20||2|6.6|5.4|270|20.0
TOT | White | 17 | 300 || 50
11|White|29|508||3|9.9|2.1|120|17.5
12|White|1|29||1|3.3|8.7|470|18.5
TOT | White | 30 | 537 || 88
13|White|29|508||3|9.9|2.1|120|17.5
14|White|1|19||1|3.3|8.7|470|18.5
TOT | White | 30 | 537 || 88 TOTAL | | 180 | 3243 || 540

You were shooting for 20mA per string. You have 180 strings so that would nominally be 3.6A. Most of your strings should be pulling a little bit under 20mA, so your estimated total current draw is about 3.25A. If you supply is rated for 5A, you should be just fine, even allowing for variations in LED forward voltage.

Another thing to check is the power dissipation in the resistors. Since all have about the same current of 20mA, the worst case will the largest resistor, which is 470Ω. Calling this 500Ω for simplicity, your I^2*R power will be 200mW. So be sure that you are using at least 0.25W resistors. I'd be happier with 0.5W since the rule of thumb is to only put half the power into a resistor as what it is rated for. One thing you could do is replace your 470Ω resistors with two 1kΩ resistors in parallel. Your next highest power consumption are your 330Ω resistors and they should be a bit under 125mW so 0.25W resistors are fine.

 

That's fantastic!! Thanks ever so much for taking the time to calculate this. Just out of interest did you calculate it using the formula you gave previously?? I.e

16*Ired + 17*Iwhite ... Etc

??

Also now we've calculated the current draw how do you add the values of resistance for the wiring? I decided last night I'd LIKE to add an extra 42' of wiring between the driver and the display. So the total wiring will be 144' which according to the sums of another will generate 1.725Ω Of resistance? Ill get those resistors changed all others are 0.25W

 

That's fantastic!! Thanks ever so much for taking the time to calculate this. Just out of interest did you calculate it using the formula you gave previously?? I.e

16*Ired + 17*Iwhite ... Etc

No. You gave enough information to calculate the nominal current for each string independently.

Let's pick the first row. You have 5 red LEDs, each of which drops 2V. So the LEDs will drop a total of

Vled = (5redLED)*(2V/redLED) = 10V

You have 12V total which must be dropped across the series connected resistor and LEDs. So the voltage across the resistor is

Vr = 12V - Vled = 12V - 10V = 2V

The current in the resistor, which is the current in the LEDs, is

Ir = Vr/R = 2V/100Ω = 20mA

Now, this doesn't allow for the range of variation of the forward LED drop. For doing things properly, you should allow for the minimum voltage drop across the LED since that will result in the maximum voltage across the resistor and, hence the maximum current. This will manifest itself most significantly when you have the highest number of LEDs in series because you will have the greatest change in voltage and the smalled voltage across the resistor, so the net effect can be pretty high.

For instance, you said that the white LEDs could be as low as 3.1V, which would make the three in series drop 9.3V instead of 9.9V. That would add an additional 0.8V across the resistor, which only has 2.1V across it to begin with. This would result in an increase in current of more than a third. In addition, we should allow for the fact that the resistor has a tolerance on it (probably 5%), so it might only be 95% of it's nominal value. The current in that string could then be as high as

I = 2.7V/(0.95*120Ω) = 23.7mA instead of the nominal 17.5mA. That a factor of 1.35. The variation in the other strings should be less than that. But if they were all that, then the total current draw would rise to 4.4A.

Another thing that should be allowed for is the power supply voltage variation. If it is 13V instead of 12V, then that directly places an additional 1V across all the current limiting resistors. If the supply is only regulated to ±10%, it could conceivably by as high as 13.2V and still be in spec. So the current in the string of three white LEDs could be as high as 34mA, or right at twice what you expect. That would run you over your 5V limit, but it relies on everything being at the edge of its spec envelope in just the wrong way, which is very unlikely.

On the other hand, the output voltage could be as low as 10.8V and be in spec. If your LEDs also happened to be at the high end of their range, 3.5V, and your resistors were high by 5%, then you would only end up with

I = (0.9*12V - 3*3.5V)/(1.05*120Ω) = 2.4mA

The problem is that the uncertainty in the voltage drop across the current limiting resistor is a large fraction of the nominal voltage drop across it. To reduce this range of possible operation, you either have to reduce the uncertainty or you have to increase the nominal voltage drop. Let's consider the string that only has a single while LED in it.

Imin = (0.9*12V - 1*3.5V)/(1.05*470Ω) = 14.8mA
Inom = (1.0*12V - 1*3.3V)/(1.00*470Ω) = 18.5mA
Imax = (1.1*12V - 1*3.1V)/(0.95*470Ω) = 22.6mA

So the same variations that can produce nearly a 100% variation in the string of three white LEDs can't even produce a variation of 25% in the strings with just one white LED.

But everything is a compromise. If you use just single-LED strings, you will have nice, stable light levels but you will need to supply a total of around

Itot = (20mA/LED)*(540LED) = 10.8A

Also now we've calculated the current draw how do you add the values of resistance for the wiring? I decided last night I'd LIKE to add an extra 42' of wiring between the driver and the display. So the total wiring will be 144' which according to the sums of another will generate 1.725Ω Of resistance? Ill get those resistors changed all others are 0.25W

This is something you need to be careful about. We just saw above how a drop of 1V in the power supply voltage could be devastating to your three-LED strings. With 3.25A of total current, it only takes 0.31Ω of total leadwire resistance (so twice the distance from the supply to the LEDs) to cause a 1V drop.

Is the 144' the total distance from the driver to the display, or is that 72'?

Assuming it is the later and that you have a total lead wire length out and back of 144', then you need the wire resistance to be 0.31Ω/144' or 2.1Ω/kft for a 1V drop at the display. If you want only 0.5V drop, then you need the wire to be under 1Ω/kft.

http://en.wikipedia.org/wiki/American_wire_gauge

So you would need #10 AWG wire to stay below 0.5V drop while #14 AWG would give you a about a 1.2V drop.

It sounds like you are planning to use #20 or #22 AWG guage, which seems too small to carry several amps of current any distance.

 

Basically what's happened is that I have recently got a model of the truck I drive for work and would like to make a small display case with 6LED's in it.
I figured being as we are within the operational parameters (because nothing has blown up yet and it has been switched on for like half an hour and not showing signs of heating up yet) - that it would be a good idea to power these LED's off the driver that's powering the current display instead of buying a separate one.
The display that I've made sits at the foot of my bed and the display case will be on my bedside cabinet (a distance of 7 meters from the LED display)
So what I intend to do is this:
Over here in Britain we use 3 core wire with one lead going to earth (from my American travels I seem to remember you use just 2 cores) so because the output isn't dangerously high from the LED driver, I would have the negative lead going into a junction, one end going straight into the 543LED display and the other travelling to the display cabinet. The positive lead would go straight down the 7m lead up to the display cabinet and then I would tap it into the earth wire with a switch so it would become the positive wire of the 543 LED display and I can control it independent from my bedside. So what you would have in a 3 core wire I 2positives and 1 negative lead. See diagram below.

The display is currently using 100' of 22 gauge wire to link everything and the 3 core wire i intend to use I believe is 12 gauge wire or standard electrical appliance 13A wire if that makes sense?

 

Here is another very crude diagram to show what my intention is.

 

Anybody please cast some light :-(

 

Anybody please cast some light :-(

Being just a little impatient, aren't we? It's been less than four hours since you posted your diagram and already you appear miffed because your request for free assistance hasn't been satisfied? We have lives, too, you know.

 

Nooooo no impatience here man just didn't want the thread to get swallowed up into obscurity never to be seen again so I thought I'd pull it back to the front

 

Hell, my first post on the forum went about three weeks before getting it's first response. Threads often sit for a day or so before getting any notice, if for no other reason than that people have to work and sleep.

The general rule is to not ping a thread to bounce it up to the top until it has dropped off the first page of the thread list (or at least is about to).

Also, when you do, be upfront about it. Don't ask whether anybody can shed some light with a frowny face -- that just screams impatience and more than a little bit of a sense of entitlement (which I know you didn't mean). Instead, just say something like, "Pinging the thread to keep it on the first page. I'm still here and very interesting in hearing what anyone has to say."

 

Ok that's fair enough I never really use forums so forgive me for my inadequate social protocol via cyberspace.

Please can you tell me whether this wiring plan will work or not then my project is done??

 

We can't tell because you haven't given enough detail about how the thing is actually wired. You are using #22 wire for most of the hookup, but you aren't asking much, if any, of it to carry ALL the current.

So you need to put together a schematic or a list that breaks out the major interconnects and indicates how long the connecting wires are and what guage.

 

Would you be willing to use a relay at the junction? With just a couple volts to play with you need to be careful about running that much current through any long lengths of wire.

So you use a relay to send power to the big display and use the power coming back from your bedside switch to energize the relay coil.

 

Ok well the driver has two wires coming out of it which will logically carry all the current it needs otherwise its a design flaw. It then goes into 12 guage wire (if that's standard wire for carrying 13A for mains appliances) the

Negative splits into two seperate wires of the same guage after 1'. One wire is 22' and the other is 6' before leading onto the 22 guage wire where it powers all the big display.

The positive again - attaches to 12 guage. It then travels 50ft with a switch at 25' before being attached to 22 guage wire where it powers the big display.

I have literally no idea what the various lengths of wire are In the display because due to a couple of errors in bulb placement, some individual bulbs ended up on a length of wire 10" long - also a lot is longer to get over wooden spars etc all I know is I used near enough exactly 4 rolls of 22 guage wire including wastage etc.

Hmmmm relay - well I don't actually know what one is so if you still think it would require one - then let me know and ill research it and acquire one

 

What you will need to do is draw a schematic that is broken down into enough detail to let you calculate the voltage drops across the sections of wire, or at least put upper bounds on it. You need to treat each section of wire as a resistor and calculate the voltage drop based on that resistance and the current that flows through it.

It might seem odd that we need to be concerned about running a current of less than 4A through a cord that is normally expected to handle 13A. The thing that we have to keep in mind is that if the circuit that the cord is designed for is normally a 240V circuit and that a drop of 2.4V represents a reduction of only 1% of the circuit's nominal voltage, while in your circuit that is more than enough of a drop to extinguish all but one of your red LEDs, all but four of your white LEDs, and very possibly all but three of your blue LEDs as well.

But, if I understand things correctly, you presently have a display that works and that you are satisfied with and now you just want to add a branch that goes over to the head of your bed to (1) power a much smaller display, and (2) allow you to install a switch to controll the main display.

One approach you can take is just to try it. Add the proposed length of wire (no need for a switch or anything else) and see if it still works.

If it doesn't, then if you wanted to be a bit more systematic you could get some resistors to emulate the wire drop and insert them until you find out how much resistance you can tolerate. If you get a few 5W 0.5Ω resistors, then you can put one of them in series with the supply output and see what happens. Then put a second in parallel with it and see what happens. Keep doing that until you get a display that is acceptable. The resistance you have at that point is 0.5Ω/N where N is the number of resistors you have in parallel. Multiply that by about 75% to give yourself a bit of margin and the result is your allowable lead wire resistance for the additional wire.

If you are talking about #12AWG wire, then the ampacity is typically 25A to 30A and the resistance is 5.2mΩ/m. If you are talking about something rated for 13A, then #18AWG wire -- often used in light duty extension cords -- has a typical ampacity of 14A and the resistance is 21mΩ/m, so a factor of four more.

You are wanting to add 14m of wire carrying about 3.5A of current. If you want to hold the additional voltage drop to under 0.5V, then you need a wire resistance of less than 143mΩ which means a resistivity of under 10mΩ/m. So you can see that #12AWG is a factor of two better than you need which #18AWG is a factor of two worse than what you need. With #18AWG you would pick up another 1V of voltage drop, which may well cause you problems.

But try it and see. You aren't going to cause a dangerous situation -- at worst some of your LEDs may get dim or go out completely.

Think of a relay as a remote switch. Basically, it is an electromagnet that you control with a small current (with the switch at your bedside) and the magnet either closes or opens a set of contacts (a switch) that controls a potentially much larger current. When you start your car, you are using a key switch to energize a solenoid (which is nothing more than a special-purpose relay) to control the hundreds of amps that are needed by your starter.

A suitable relay can be had for just a few dollars.