LED Cue-light

There may well be 3V between tappings but it will take doctoring something to get it pre rectification... Besides, I'd have said that a higher voltage would be more room for voltage drop over long cable...
 
voltage drop is a function of current and is extremely low; but it works, and has for decades and is easy and cheap, why re-invent the wheel?
 
I think in this instance I'm going to have to use a Scott Adams quote: "Engineers like to solve problems. If there are no problems readliy available, they will create their own problems"
 
Any cheap multitap transformer will have 3 volts between some windings,I didn't bother with series resistors but it still works, it will probably corrode to death before it fails, but if you need to run off a higher voltage then a resistor in each line would be needed.I actually cannibalized an old intercom unit for the master station and used all the old switches, I think the total cost for a 6 way system was $15.


Yes I agree it can work but whether it is a good design is a different question.

Also in this case I would feel that 3v is to low to drive two leds in series down 20+ metres of cable. Considering Vf is typically 1.4 V rougly. So you get nearly 2.8 V drop just in the leds before considering voltage drop in the wire.

So Charcoal I would suggest for peace of mind you stick with the current limiting resistors and a 6V supply for simplicity.

Sorry Allthings I missed the bit where you said you would use resistors for higher voltages.
I follow your transformer theory but in this case it would be cheaper and easier to use a plugpack he has at hand then having to wire a transformer.
 
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100M of fig 8 cable has resistance of 6 ohms, with 20ma the voltage drop is .12v, but the main concept I was trying to get across was using ac and using the positive side for one light and the negative side for the other.
 
So I've been re-spec'ing this project from Jameco, to try and get the cost down to a more reasonable figure.
Here are the part's I've spec'ed so far:
-$4.35 x 4 281658 Male panel mount xlr
-$5.69 x 4 281640 Female panel mount xlr (These two are not the most aesthetically pleasing, but hey, whatcha gonna do?)
Jameco is taking you for a ride Charlie.
Genuine Neutrik-
D Series Male XLR Panel Jack
The females are backordered until March, but they have a generic-
3 Pin Female XLR Panel Jack
The total with shipping is under $25, a savings of $10.
Don't worry, they are the best!
 
(These are run-of-the-mill LEDs, right? So why are the pack sizes and price so dramatically different?)

-$.287 x20 1766614 Mounting hardware, 5mm led (To save on shipping I wanted to go all Jameco, but I have no clue what these look like, and how they operate. Anyone have a clue?)

More thoughts: I'd like to get some sort of strain relief on the DC in. The ME used some normal electrical strain relief, that one would see on a conduit box deal. That is an intriguing option. I'm not sure what those are called though. I also thought perhaps I should bump this up to a 9V power supply and apply a status LED in parallel with all of the cueing LEDs. So I'd need a resistor to drop it from 6V to whatever is recommended for the specific LED, probably 3V or so. Also, I have a question, can LEDs pass current if they die? Lastly, enclosure suggestions. I'm still a little shakey on what exactly I should go for, plastic/metal, what type, etc, but I just realized that Jameco sells enclosures, so I may poke around on there a little.

Different materials produce different coloured light but also cost different amounts hence the difference in price I suspect...

Mounting can also be done by drilling a 5mm hole, sticking the LED in and then applying hot glue from the rear. Can't get a pic of what you have referenced so can't comment...

You could go all out and use a *real* connector for the DC... Industry "standard" is a 4 pin XLR w/ power between pins 1 & 4.

Dead LED won't be passing current...
 
Charcoal the four units look all right in your diagram. I take it just below the switches in the control box the Red and Green refer to Leds?

In this diagram you have failed to show any resistors. I would place one current limiting resistor in the +6v leg to the battery for each of the four sections.

What I don't understand is the two leds connected across the power supply. If these are indicators that the power is on then you really only need one. Also you need a resistor there as well.

For a strain relief here's an idea. You will have to cut holes in the case so you will have a little bit of material let over. If it's big enough make a rectangular slice out of it. Then place over the DC adaptor cable and screw it to the case. I am not sure if the cutoff material will be big enough but this may work.

I have just had a look at the specs for the Leds you are buying. I would use 25 milliamps for working out the current limiting resistor. If you use this value both Led's should be bright enough but you will only need one resistor. Although the data sheet doesn't show Vf for the red one use 2v for both leds. The red led will handle the slighlty more current it will get by using this value.

I would aim for 25 milliamps current because this should be in the range for your DC adaptor. The most current you should draw is roughly 70 milliamps per each unit + 20 millamps for the on indicator = 300 milliamps. Actually each section should draw less then 50 milliamps but when you design you allow for variance.

I know I have been pushing the resistors but if you don't use them then your DC supply will probably not cope with the current drawn as well as the Leds may fail.

Using the figures I gave you if you do the resistor calculations I will check them for you.
 
cutlunch, How do you figure 50mA / section? I can only get 25mA (plus fudge factor)...
 
How about we wait for cutlunch to get back to me as to where his number comes from then we might be able to agree on things and THEN you can build it once rather than finding we were thinking different things and you've taken half of mine and half of his and ended up with something what don't work too good...
 
I am back. Charc I got your pm. I am not on any of these message systems you mentioned but I'll pm you later if this post doesn't help.

Good point Chris you are dead right. Sorry Charcoal I made a mistake.
When two leds are in series the current flows through both of them so in this case each of the four segments will only draw 25ma. I forgot and worked it out for parallel where both Leds would draw 25 millamps each.

So Charcoal apart from my slight error it doesn't effect your design.
These resistors we'll go over again. Each of the four section needs one resistor that is connected from the + voltage to the switch just like in your drawing excpet now the resistor replaces the wire you have drawn from + to the switch.

If you have that other LED you have drawn across the + and - V as a power on indicator you will have to connect one leg of a resistor to the + v supply rail. Then connect the other leg of the resistor to the anode of the led then connect the cathode of the Led to the - V rail. (just like in the circuit of the led calculator)

That Led calculator you found Charcoal you can use. In fact I am going to book mark it because it will give you values of resistors that you can buy not just a theorectical value you have to make up from different resistors.

Ok so here is your homework.
Using the calculator with these figures work the resistor for your single power on Led.
Source voltage = 6
diode forward voltage = 2
diode forward current (milliamps) = 25

Enter those figures into the calculator then write down the answer.

Now the two leds in series might seem harder but it's not.
Use the same value for source voltage = 6
Use the same value diode forward current (milliamps) = 25

Ok the difference is the diode forward voltage. Because we have two leds in series both are going to drop 2 volts each so add that together 2 + 2 =4.

So use 4 for the diode forward voltage in the calculator.

Once you have these worked out post the resistor values here and we'll check them.

PS for ease just use 1/2 watt resistors for all resistors it'll save any hassles.
 
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In reality, one can use any wattage resistor. By a quick sum, you're looking at about 100mW through the resistors so 1/4 watt will also work quite fine. Just use what you can easily get and is cheap.:mrgreen:
 
Charcoal these values will work but they are not the same as I got using the values I gave you.
But they are definitely not wrong and will work well in this circuit.

I am not sure what figure you changed - Supply voltage, V forward or the current If. If I stay with my values of Vsupply = 6V and Vforward = 2 but change Iforward to 20milliamps then I get the same values as you.
But if I use the 25 milliamps for Iforward then I get single led resistor value 180 ohms vs your 200 ohms.
For the two leds in series I get 82 ohms versus your 100 ohms.


But don't worry about it Charcoal your values will work and they are close enough that you pass the course !!!!!.


And a very good observation about the 25ma vs 30ma. I am just a cautious designer I doubt you would have any trouble using 30 ma for the current in your calculations. I tend to take what they say is the max value and do what they call derating ( take a lesser value). This just allows for a margin of error. In this case it's not a problem.

You can use 1/4 watt resistors for everything you don't need to buy 1/8 watt for the stages and 1/4 watt for the single led.
I only suggested 1/2 watt resistors because here our 1/2 watt resistors are the same size and cost as the 1/4 watt. 1/4 watts are fine in your design.

Don't forget I would like to see pictures if the finshed project.
 
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I also want to see photos of the finished product. Can one even get 125mW resistors? Even the 1/4 W are getting hard to find these days... Who wants a 5% 1/4W when you can get a 1% 1/2W for the same cost?
 
Wow, can't get 180R resistors... that's sad. To get 200R, you have to be looking at the E24 series and E24 includes 180R as well, 180 is also in E12 for those who care:mrgreen:

If you are paying more than about 4 or 5 cents for a resistor in the <=1/2W then me thinks you is getting ripped off. And even those numbers are a little pricey in my book...
 
Wiring in parallel:

I've been thinking of what the "best" way of wiring these parallel loops is. I've come up with a "twofer" idea. Holding two wires right next to eachother, and then holding the third (source) up against the other two, like this: "====------" if that makes sense? That way I could put heatshrink tubing around the connection. Perhaps I should pick up a dremel for some of the work.

It all depends on where this join is... You could put a loop through XLR on your indicator boxes, wired in parallel with the input. But your idea is essentially sound. While you didn't mention it, I'm presuming you would apply solder before heatshrinking it. You'll also want some form of mechanical protection for the join, depending on where it is. If it's inside a box, it'd be quite fine. But outside, you would probably want a layer or two of glue impregnated heatshrink around the whole thing... Not sure what you want a dremel for...
 
Is this "twofer" for the Stage Manager's Console or the Cue Stations? If for the Master Panel, just use multiple D3Fs. For each Q-Station, use one D3M and two D3Fs for the most flexibility.



D3M = panel mount, 3pin, male
D3F = panel mount, 3pin, female
A3M = inline, 3pin, male
A3F = inline, 3pin, female
A5M = inline, 5pin, male (DMX-512A approved/required)
A5F = inline, 5pin, female (DMX-512A approved/required)

Above are Switchcraft numbers, but any mfg. can substitute.
 

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