- Dec 28, 2019
Hi Terry and do not apologize, I appreciate you taking the time to elaborate.It would depend upon the hardware you choose to use in your display. Power injection cable length will depend upon the number of pixels in a prop (or the number of pixels that are being driven by a particular cable if multiple injection cables are implemented), type of pixels (regulated or resistor), how hard the pixels are driven, and cable type (ohms per unit length, foot, meter, or whatever). I may have forgotten one of two but that's most of them I think.
Number of pixels and their drive level will determine the prop's amperage requirements. The type of pixel will determine allowable voltage drop although drive level plays into this as well. From voltage drop and amperage requirements the maximum length of a power injection cable can be determined based on a cable's resistance per foot rating.
An example may help. I'll fudge a few numbers as I do not have everything at hand presently. I have a Coro GE icicle, 3 actually but I'll use only 1 here. 150 pixels and I plan on using 5 volt WS2811s. Most individuals use .06A per pixel at 100% white for those but from testing I find them closer to .05A (I'll use .05A). From testing I also know that I cannot see anything above 90% which is still exceedingly bright so I'll limit drive for the prop at 80%. Amp requirement is then 150 * .05 * .8 = 6A.
Voltage drop is a bit more difficult to know but again from testing 5 volt WS2811s seem to start dropping in the 3.8 volt range (I'll use 4 volt). At this point the power injection cable needs to supply 6 amp with a voltage drop of no more than 1 (5 minus 4) volt. Ohms law can be used to determine the cable's maximum resistance (R = E / I) so 1 / 6 = 0.16667 ohm.
There is a 'got-cha' here that I feel a number of individuals overlook. Wire charts are not giving the ohms per foot of a cable but rather a single conductor within the cable. Voltage drop within the cable for the supply's negative run to the prop needs to be considered with the supply's positive run to the prop. Thus the 0.16667 above is negative and positive combined and needs to be halved. 0.08333 ohm is the value that needs used in the following.
I plan on using 14/2 which according to standardized charts has a resistance of 0.002525 ohm per foot. Maximum power injection cable distance is then the 0.08333 / .002525 = 32.6 feet. This would be for a single power injection cable. If I would run a 2nd cable and inject at the end then each cable essentially powers 75 pixels rather than 150 so the distance would double.
Apologies for the length; got carried away. I think there are no mistakes in all this. Cheers.
I am new to this and I love detailed information.
Thank you, thank you!