On November 22, 2011 at 12:36, charris said...
Back to the basics...
Definitely. This project might be the first one where you have to challenge your assumptions and understanding of Xantech products. I'm sticking with Xantech for this because they have had less squirrely products over the years than other manufacturers.
For various reasons we have decided that we would like to have an IR distribution "back-up" or "redundancy" system on all our projects where we have HD video distriibution. This mean an IR receiver on every TV.
I'm assuming that you are not going to control ANY TVs with this IR system, but you're only controlling sources. Also that all sources are in the same zone, as you don't define any zones at all. Moving on --
Well, okay. Let's say you have sixteen IR sensors, and somehow a tiny bit of IR interference -- don't ask me how -- pops up in two places in the building. These will add together. You'll have a much greater possibility of system-crippling interference with sixteen sensors than with two or four. Think this through strategically, because often Xantech sensors will pick up remote signals from forty feet away! (Garbage, too!)
My main question is a very basic one: How do I calculate how many IR recievers an IR distribution block can take? e.g a basic one like the Xantech 78944?
[Link: xantech.com]Let's start by talking about the basic IR system. An IR sensor is powered by 12 volts and ground. Its output can be as much as maybe 50 mA... whatever it its, it's too much to feed directly to an LED. Therefore, a resistor is placed in series between the sensor output (78944 input), which we'll call the bus, and each LED. Now the surprise -- the 78944
does not require any current at all! It's just jacks and screw terminals interconnected to route signals according to the labels, and four 470 ohm resistors inside the 78944 that feed the proper amount of current to the LEDs.
Similar solutions are sold by almost everybody e.g Snapav, Niles, e.t.c, the problem is that they only take 1 IR receiver and I need several. They might be able to accept more in parallel but where is the limit?
You're making an assumption here. The diagrams
only show one sensor connected to each input. The very fact that the CB18 exists, and its description, "Easily connects home run wires from multiple IR Receivers," tells us that Xantech is quite comfortable with eight sensors being wired in parallel. The ninth set of connectors on that piece is used to jumper to the other components in the system. I say if you can connect 8, you can connect 16 just by working out the screw terminals by getting another CB18.
Most of my projects will require 3-16 IR Receivers. A better solution than going parallel on a basic IR block and hoping it works
It will, but if you have a problem with one receiver, you'll wish you had separate Phoenix connectors. Buy a CB18 and cut it in half!
is the combination of Xantech 79520 and 79620 which can do a total of 9 zones.
[Link: xantech.com][Link: xantech.com]Why is this better? To discuss what you're trying to do, please suggest your reasons for your conclusions.
9 zones means, for instance, 9 satellite receivers that you want to control independently. This might be what you want to do, but you haven't told us a thing about that. Describe the sources you want to control, if that's the case. Maybe for 16 TVs you want one zone. Maybe you want eight. Tell us. It will completely change what you do with IR, and it's all completely manageable.
When I look at the 795 and 796, I do not see the focus you seem to be looking for, which is input to ONE ZONE from multiple sensors. Instead these have multiple IR output zones, each of which
shows one sensor. Read the instructions there; they focus on getting LED signals to lots of places, not taking them in as one zone from many places. Again, the CB18 (kudos to Fred for this) suggests says that you can treat many sensors as through they are just one.
Has anybody seen something similar from other manufacturers? Or something similar that can do more IR receivers e.g 16?
I'd bet you could get two of Fred's terminal strip and just parallel the whole shootin' match. In fact, you'd
want to do that, because that way if there's interference you can just pull one Phoenix connector at a time to locate the interfering sensor. Or one terminal strip to divide the system in half.
On November 22, 2011 at 13:14, Fred Harding said...
One, you want to make sure you have adequate current to drive your targets. Since a typical dinky link will use between 10 and 20 ma, it should be simple to do the math.
Absolutely. One reason to stick with Xantech is that they publish the required current for active devices. Two years ago I tried to get the not-published required current for a Niles sensor and the Niles techs said a) they could ask engineering but they didn't have any idea how long it would take to get an answer and b) they would not guarantee that the answer would be correct. Nobody from Niles has ever contradicted this. I did the measurements myself and moved on without them. Don't build complicated systems with Niles unless you're ready to do your own characterization of required current.
Do be aware that wire gauge and length will impact your current draw.
When we think of impacting the current draw, what usually comes to mind is drawing more current, not less, but the resistance of longer wire runs make for less current draw. Smaller wire gauge, in long enough runs, will starve the sensors for current, perhaps so much that they don't work, so, well, that could be a bad idea. However, I've run 160 feet with doubled-up conductors of a CAT5 with perfect results. Se calculations below.
As a rule, I specify the Xantech 782 power supply (1 amp rating) when I have more than three targets, because I like the comfort factor that having adequate current gives.
Well, yeah, but if one dinky link requires 20 mA, you can do 50 of them with only a half amp! On the other hand, you have to also add up the amounts of current drawn by all the LEDs. My rule is a bit more rigidly stated than Fred's, but will get you more bang per power supply: add up the current specs for all devices including LEDs, and use one 782 per 500 mA of required current.
Be aware that you certainly can deploy multiple power supplies through crafty wiring if the number of targets gets too big.
The "crafty wiring" part means that if you need more than one power supply, you wire things up like this: you designate sections; the grounds of all sections and the sensor outputs of all sections are tied together, but the +12 volts of each section only feeds to the devices in their own section.
And, for instance, you only need to feed 12 volts to a 78944 if that 78944 is going to supply current to some sensor. If it's only being used to feed LEDs, then IR IN (our bus) and ground is all you need.
Second, you want to organize your wiring. A terminal strip like the Xantech CB18 will allow you to put 8 targets into it, with one set of outputs. You certainly can deploy several of these as job size dictates.
That's what I'm talking about! This is the device that has Xantech saying that 8 sensors, keyboards, whatever, in parallel is just peachy.
Third, regarding the 795 series; those give you the ability to have dedicated and common sources controlled by physical connection on the blocks themselves. If you wish to have some sources controlled by all targets, but other sources by one target only, these work great.
I don't see their use in your application, though, because you have not stated that you want the different sources to be in actually different zones.
Fourth, remember that ir systems are cumulative in terms of how they deal with interference. If every target has a small amount of noise, the main emitter will have a lot of noise.
Yup.
On November 22, 2011 at 15:39, charris said...
The problem is, even if you know that a dinky link receiver requires 20ma, how do you calculate how much it needs overs 150 feet of cat5/6?
When a device requires 12 volts at 20 mA, that means it needs 12 volts at 20 mA
at the device. You must create a wiring system that delivers 12 volts at 20 mA
to the device. If the wire is so long that its resistance lowers the voltage substantially, then it won't work.
Theoretically -- don't even think about doing this -- you could supply a higher voltage, then drop it to twelve volts at each sensor, and thereby work out much longer wire runs. Don't do it because it's not worth the time.
Since we're using regulated supplies, we've got the right voltage to start with. Therefore we want to use wire large enough that its resistance doesn't drop the voltage over the length of the run.
Let's take a look at CAT5 wire. It's 24 gauge. Now's the time for you to look up a chart of wire gauges and resistances on the internet and store that on your computer somewhere near your list of TV and CATV channel assignment frequencies.... but I digress.
24 gauge wire has a resistance of 25.67 ohms per thousand feet. Let's look at a run of 100 feet. Let's double up the CAT5 wires.
Doubling up the wire halves the resistance, and the run is 1/10 of a thousand feet, so the hot lead has a resistance of 1.2835 ohms -- call it 1.5 ohms for no good reason -- and the ground wire has the same resistance.
The dinky link requires 20 mA, and it's sending a bus signal out to, say, four LEDs, each of which getd 5 mA. (The 470 ohm resistor calculates out to 2.5 mA, but let's say 5 mA for safety.) That's 40 mA out on the power and back on the ground, and 20 mA used by the dinky link and 20 mA coming back on the IR bus.
Since the LEDs have resistors in series with them, we can ignore the calculations here. The 78944 has 470 ohm resistors inside, so wire resistance in the one ohm range is negligible.
Now, about the voltage drop on the power and ground:
The hot lead has a resistance of 1.5 ohms and carries 40 mA out to the sensor. The voltage drop across that wire is
E=IR, or 1.5 ohms x 0.040 Amps = 0.060 volts.
The ground has a resistance of 1.5 ohms and carries 20 mA back from the sensor. Its voltage drop is then half that of the power lead, or 0.030 volts.
Thus if we start with 12 volts at the power supply, a doubled-up run of 24 gauge wire results in a voltage drop of 0.090 volts, delivering 11.91 volts, a reduction of 3/4%. There's no reason at all that this should be a problem. Remember that I exaggerated the resistances, too.
This is a lot of thinking and calculating to do, but learn it! Once you get what this is about, it will make lots of other things easier to sense, which you should always do, and calculate if you must.
And if if you calculate this can you use any power supply with the 78944 without frying it?
The 78944 is a bunch of connectors and four resistors. Don't worry about frying it. But don't EVER use anything but a regulated 12 VDC supply with Xantech devices!
How? By directly powering the IR reciever through the CB10?
The CB10
DOES NOT do anything at all to power an IR receiver (sensor, target). Look at the illustration -- there's the usual power, IR, ground, crap at the bottom, and ten LED outputs at the top. The amplification part allows one sensor to feed ten LEDs. It does not affect the sensor AT ALL. The power leads are just terminal strip connections.
Maybe a key point to state here, after all this, is that the Xantech IR sensors are made so that when you connect two of them together, the output of one does not feed into the output of another. When one outputs a voltage (and current), that output is not affected at all by any other sensors in parallel with it. That's why the CB18 works.
Thanks for asking. This was fun. If any of this is wrong in calculations or other details, it's still been helpful to me to know it, so it's probably worth just coping with any mistakes I might have made. But do ask questions!
