Rainbow Floods...ASSEMBLED!!

[ceniam]

Is this the same connector some people are using for their RJ-45s? Im trying to figure out why this one is 1 dollar vs the 5$+ connectors.

https://www.roc-noc.com/product.php?productid=78

[Steven]

ceniam wrote:

Is this the same connector some people are using for their RJ-45s? Im trying to figure out why this one is 1 dollar vs the 5$+ connectors.

https://www.roc-noc.com/product.php?productid=78

That is just the strain-relief/waterproof part. With that part you still have to open the case and plug the cable into the socket on the inside.

The $5 part is a short STP cable with a jack on one end that allows you to connect without opening the case.

[Robsaus]

I ordered 30 of the IP67 RJ45 waterproof connectors for this guy on Ebay for $4.00 each... Heres the address

http://cgi.ebay.com/IP67-Rated-Outdoor-RJ45-Connector-/370391045170?pt=LH_DefaultDomain_0



If you send him a email asking for a larger qty then he might ship them for free like he did for me.



Rob

[ceniam]

Thanks for the ebay link. Was able to get 40 of them and wrangle the free shipping

[grandad]

grandad wrote:

OK,
For all us UK enthusiasts, here is an alternative to the Rainbow Flood.
It utilises almost exactly the same components, although the construction differs.

Hi again,

I have now made 4 of these floods with components from http://www.phenoptix.com.
They are astounding in their output and at just 20GBP total per flood, including housing I think they are good value.
The housing is from Halfords, here in the UK. Stock number 301986 @ 3.99GBP
I am currently making up a custom PSU for a 30V DC supply to run these, attached to a CMB-16D (when it arrives)

Martin




Attached files

[Paul Roberson]

Bill Hoffman wrote:

Here's the way I hooked mine up, but there are different ways to do it, as previously noted. I decided not to use dmx, simply because I didn't have it at the time I put together my first Rainbow Flood.

I didn't use the RF connectors that came with the Rainbow, I took a 25' cat-5 cable and cut one end off it. I wired the bare wires into the Rainbow Flood where the orange wire controlled red, the green wire controlled green, and the blue wire controlled blue, I didn't use the brown wire (but you could to control something else if you wanted). Then I took a short piece of cat-5 and wired one bare end to my LORDC controller using 3 channels, and the other end I went to Lowes and bought a 5 pack of RJ45 females and wired one of them on this end. I put the LORDC controller and the 12v power source in a plastic enclosure, and now all I have to do is connect the 2 and I can control the Rainbow with LOR. Basically that leaves me with 13 empty channels in my LORDC, so I could technically hook up 4 more Rainbows, or 8 more Rainbows if I piggy back them. You would just have to watch the total amps of you 12v power source. Here's a pic.



Bill

I know I have seen it somewhere, but can't find it right now. What pins are what on the flood?

[Denny]

Not sure if this is what you are looking for, but in this thread, go back to around July 10-11 and see if that is the pin diagram you want.

[Paul Roberson]

Denny wrote:

Not sure if this is what you are looking for, but in this thread, go back to around July 10-11 and see if that is the pin diagram you want.

Thanks!! That is what I was looking for.

[Dan C]

I finally got around to building one of my floods. I mounted it in an weatherproof halogen light similar to the housings others have used. After getting it all wired up. I tested each light color with a bench PSU and found the red to be not quite as bright as Blue and Green. Just like some others have mentioned.

I was curious if anyone has tried different resistor values for the Red LED's? I remember someone I used to work with that was an electronics engineer and he built a RBG light controller for my computer a few years ago. He said that you cannot use the same resistor for the red leds as they require a different amperage or something along those lines.

By the way, of the 6 larger 1/2 watt resistors and the smaller 1/4 watt resistors, which ones are for the RED led's only?

[terrypowerz]

The reds are fed by the 6 half watt resistors.

You don't want to drive them any harder than they already are being driven.


It is inherent that red LEDS have a lower lux than green and blue...

The human eye is also less sensitive to the red spectrum I believe.
(I am sure someone will correct me if I am wrong on that!)

[Steven]

Dan C wrote:

I remember someone I used to work with that was an electronics engineer and he built a RBG light controller for my computer a few years ago. He said that you cannot use the same resistor for the red leds as they require a different amperage or something along those lines.

Close. Here is some basic electronics theory: Ohm's law says that the current through a device of constant resistance is proportional to the voltage across it. However, LED's (and diodes in general) are not constant resistance devices, so you can't use Ohm's law on them.

The current through a LED (or any diode) is an exponential function of the voltage across it. This means that the current will be essentially zero until the voltage reaches a value known as the forward voltage. As the voltage rises beyond this, the current rises sharply, until it gets high enough to destroy the device.

For green and blue (and white) LED's, this forward voltage is about 3.5V. For red LED's, it is about 2.0 volts. This voltage is determined by the physics of the semiconductors used in the LED, which also determines the color.

The LED's in the Rainbow Floods are designed to work at 30mA. This current is determined by how the devices are manufactured, including the size of the junction and the mass of metal used to dissipate the heat. The usual failure mode of LED's is by excess heat caused by excess current. Therefore, it is not recommended to exceed the rated current.

The Rainbow Floods are designed to work at 12 volts. If we look at green and blue, we will see that if we put 4 LED's in series, then it will need 14 volts, which we don't have. Therefore, there are 3 LED's in series, which gives a forward voltage of 10.5 volts. Since voltage adds and subtracts in series, we can see that at a supply of 12 volts, we have an extra 1.5 volts for the green and blue sets. Using Ohm's law (which works on resistors), we calculate the resistance needed as R = V / I or 1.5 volts / 30 mA = 50Ω. Thus the 12 resistors supplied for the blue and green circuits are 47Ω (the nearest standard value).

We can use the power formula to calculate how much power is being used by these resistors: Power = Voltage * Current ~= 1.5V * 30 mA = .045W. Thus 1/8 W resistors are supplied.

For the red LED's, the board could have been designed with sets of 5 red LED's in series, which gives about 10 volts forward voltage, but to keep it simple, the red is the same circuit, with 3 in series. This gives about 6 volts forward voltage, leaving 6 volts for the resistor. Using Ohm's law again, R = V / I = 6 volts = 30 mA = 200Ω. The value supplied is 180Ω. Applying the power formula gives Power = Voltage * Current ~= 6 V * 30 mA = .180W. Thus 1/8 W resistors would be too small, and 1/4 W resistors are used for these.

I tested each light color with a bench PSU and found the red to be not quite as bright as Blue and Green.

There are two reasons for this. Since the forward voltage of the red LED's is less than the green and blue, there is less power being supplied to them (and more being dissipated by the 180Ω resistors). Brightness is roughly proportional to power. Also, as Terry said, the human eye is less sensitive to light in the wavelength produced by red LED's.

[Dan C]

That is a great explanation of Resistor calculations! Thank you.

I'm still having a hard time finding the connection between volts and amperage requirements. Basically here is what I'm thinking, if the Red led's are designed for 2volts and there are 6 in series, then that would require 12volts total without any resistors. So you should be able to remove the resistors and have them at full brightness.
I know this is not the case as that would over-drive them and burn them out.

I can't seem to understand why though.
Sorry for being difficult.

[Max-Paul]

First Steven, excellent post.

Dan C.,

Ok, remember for what ever reason. Who ever made the board, only put 3 of the Red LEDs in series. If they had put 6 in series, then yes you are right there would have been no need for the resistor.

On top of Ohms law and Watts law, there is a 3rd law called Kirchoff's law.
http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

Simply it says that the voltage drop across each item in series equals the voltage applied to the circuit. In Stevens example of the red led. Each LED drops 2 volts. So 3 * 2 = 6. And the resistor drops 6 volts. So 3 LED = 6 + resistor 6 = 12 total volts. If you where to remove the resistor then each LED would have 12 / 3 = 4 volts applied to each LED. As Steven said, once the forward voltage is achieved, the current spikes up rapidly. This 4 volts is twice the voltage rating and I cant even begin to tell you how many amps that would be. But I can tell you that your LED would have a very, very short life.

Once worked at a place that got some security booths. The booths could be locked out and there was a green and red LED to indicate status of the booth. Well the yahoos who designed and built the controller and booths had a 30VDC Power supply. And the resistor was to small of a value. The LEDS lasted about 4 days and started to pop off the front half of the LED like pop corn. Thats what can happen if not like a flash bulb.

[Steven]

Dan C wrote:

if the Red led's are designed for 2volts and there are 6 in series, then that would require 12volts total without any resistors. So you should be able to remove the resistors and have them at full brightness.

The reason this won't work is based on practice, not theory.

In practice, the forward voltage of a red LED is not exactly 2 volts. It can vary depending on manufacturing variances, and also on temperature.

In practice, the resistance of the Cat5 cable we use to drive the rainbow floods is about .17 Ω/m. If you eliminate the dropping resistor in the fixture, the resistance of the power cable is now significant. This means the flood near your house would be much brighter than the flood that is far away from the power.

In practice, the voltage of the power supply can vary with load. Therefore, the brightness of these LED's without the resistor would change when other LED's on the same power supply were turned off and on.

In practice, a 12V power supply is not exactly 12V. It can range from 12 to 14 volts, typically.

Even in theory, the current through a (light emitting) diode changes wildly with small changes in voltage. In practice, we can't regulate the voltage enough to keep the current controlled. Therefore, the only practical answer is resistors.

[Ponddude]

Steven, were you in our heads when we were designing these?! haha. Excellent explanation though...absolutely perfect. Not many people completely understand Ohm's law.

Earlier somewhere in this ridiculously long thread I explained how color is really created within an LED. It has to do with the elements and chemicals inside the LED. That, along with the distance between the diode create a different arc in the current creating the color. Due to the limitations within the elements inside any red LED the output rating (MCD) is inherently lower. This goes for every red LED out there.

For instance, the MCD rating for the reds used on all the Rainbow lights is 2500. For the greens it is 4000 and the blues is 3700. You can see, it is much lower than the greens and blues which is the real reason for the reds being less bright as the blues.

Now some higher end LEDs will have 2 chips (or 2 diodes) within one LED to create a red that looks brighter. A lot of higher end RGB LEDs will have 4 chips in them, one for blue, one for green and 2 for red. This will increase the red color MCD rating.

Also, do not think that increasing the voltage to the LED will increase the MCD rating. In fact, the higher the voltage the less bright the LED will be and the shorter its life will be. Like Steven said, it is important to stay within the specs of the LED or else the LED will have a very short life span. Now, when the LEDs are being manufactured not every single one is exactly the same. A good example is the hole size for the LEDs on the Rainbow lights is to the specs on data sheet, but you will be pretty hard pressed to find one of the LEDs fit snuggly in the Rainbow lights...even the Chinese can't keep it straight.

[JonB256]

edit - appears I've gotten this in a little late, but:

The LEDs used are normally considered maxed out at "20milliamps." The amount of current is controlled by the resistor choice. You could use a lower value resistor to increase the current and light output but it will be at the expense of lifespan of the LED.

The spec sheets for Piranha Superflux LEDs show that Green gives the highest light output, then Blue. The only one giving less light per LED than Red is the Ultra-Violet (black light).

If you truly want matched brightness from Red, Green, Blue, then you need more Reds on the board. Instead of 18, there probably should be 24 Reds, 18 Greens, 18 Blues.

[Dan C]

Wow, Thanks for everyone's input on this. I'm glad I brought this up.

[terrypowerz]

Steven wrote:

Dan C wrote:

if the Red led's are designed for 2volts and there are 6 in series, then that would require 12volts total without any resistors. So you should be able to remove the resistors and have them at full brightness.

The reason this won't work is based on practice, not theory.

In practice, the forward voltage of a red LED is not exactly 2 volts. It can vary depending on manufacturing variances, and also on temperature.

In practice, the resistance of the Cat5 cable we use to drive the rainbow floods is about .17 Ω/m. If you eliminate the dropping resistor in the fixture, the resistance of the power cable is now significant. This means the flood near your house would be much brighter than the flood that is far away from the power.

In practice, the voltage of the power supply can vary with load. Therefore, the brightness of these LED's without the resistor would change when other LED's on the same power supply were turned off and on.

In practice, a 12V power supply is not exactly 12V. It can range from 12 to 14 volts, typically.

Even in theory, the current through a (light emitting) diode changes wildly with small changes in voltage. In practice, we can't regulate the voltage enough to keep the current controlled. Therefore, the only practical answer is resistors.

Or utilize a constant current driver

[Dan C]

So if they supplied led's are only 2500 MCD and 2volts. Then could we replace the Red led's with these? They are 2.1 volts and 4500 MCD. That should more closely match the output of the green/blue leds. -you would need to change the size of the 1/2 watt resistors to 64 ohm though.

http://www.mouser.com/ProductDetail/Optek/OVFSRAC8/?qs=UY4NetwcvwlN6dUy2/5NtA%3d%3d

[cenote]

terrypowerz wrote

Or utilize a constant current driver

Exactly....