**Now I'm stuck...**
Hi Folks,

I keep coming back to this unit as a possible solution if the Cabling is marginal. It appears to be a unit used in multi-residence applications in High Rises or Apartments (Condos)

The 300 foot solution look intriguing...

I'd love to get a simple explanation as to how to do this type of install (maybe a handbook or on-line course). Just for educational purposes though. I'm not looking for a new career :laugh.

I found the following while checking out the SD PI-6S module. It's about DIRECTV® but I'm thinking the same standards / information would pertain to Shaw Direct. My comments in Blue

DIRECTV® recommends that the distance between receiver and dish be less than 150 feet of solid copper RG-6. Here is why.

The 18 volts originating at the SWM16 must arrive at the SL5 above 16 volts for the SL5 (the DIRECTV® LNB) to operate correctly. Up to (2) volts can be lost in the coax. (Jim's edit => this means the minimum switching voltage is 16 or 11 VDC depending on what polarity is being sought on our Shaw Direct Systems.)

Model SL5 LNBs employ current management to minimize the current carried per coax. (400 mA total) Rev3 SL5 LNBs splits the 400 mA current equally at 100 mA per coax.

Current x Resistance = Voltage loss

Solid copper RG-6 has a typical loop resistance of 4 ohms per 100 feet. At 200 feet the resistance is 8 ohms. 0.1 Amp x 8 ohms = 0.8 volts loss

18 V - 0.8 V = 17.2 volts to the SL5

Model SDPI6S-T starts with 19 Volts to provided extended dish to Main Point of Entry distances. The extra voltage provides an additional 150 feet RG-6 distance. Model LA146R off sets the signal loss of 150 feet of RG-6. The LA146R located at the polarity locker does not affect the current loss in the cable to the dish. What is the voltage loss in the 19V/22k coax? Assume 100 mA per coax.

At 300 feet the resistance is 12 ohms. 0.1 Amp x 12 ohms = 1.2 volts loss 19 V - 1.2 V = 17.8 volts to the SL5

This is quite different than the Spec in my post above ~20% less resistance => ANSI/SCTE 44, "at 68°F (20°C), the maximum DC loop resistance shall be 103 ohms per 1000 ft" Or about 10 ohms per 100 ft.

Assuming the former 4 ohms per 100 ft., for the Shaw Direct system at the cabin, the lines have a resistance of 3.55 X 4 = 14.2 Ω

14.2 Ω x 0.1 Amp =1.42 volts loss 18 V - 1.42 V = 16.58 volts to the Xku LNB if solid copper core RG6 were used. This should be fine as it's above the 16 volt minimum spec for switching.

Assuming the latter 10.3 ohms per 100 ft for the Shaw Direct system at the cabin, the lines have a resistance of 3.55 X 10.3 = 36.565 Ω

36.565 Ω x 0.1 Amp =3.6565 volts loss 18 V - 3.6565 V = 14.3435 volts to the Xku LNB if solid copper core RG6 were used. This should not work (or not work well) as it's significantly below the 16 volt minimum spec for switching.

If we calculate using copper-clad steel inner conductor, according to maximum spec, the maximum DC loop resistance shall be 201 ohms per 1000 ft. So the math becomes even crazier.

So obviously some thing is wrong with my data and/or methodology. We know Dual Shield RG6 copper-clad steel inner conductor works for the 355 foot run to the dish.

I've assumed the old Quad LNB had more output than the Xku, as when the Xku was installed, the run to the neighbor's needed a boost (in line amplifiers) to overcome the 277 feet to his place. The loss for those 277 feet should be less than that the 355 to the dish, so it shouldn't need the boost - but it does. My explanation was the multiswitches pushed more voltage to overcome insertion losses to the LNB Feeds (like the Model SDPI6S-T above which starts with 19 Volts) and that benefit overcame some of the line loss as "The extra voltage provides an additional 150 feet RG-6 distance." If that's the case then:

the lines have a resistance of 2.05 X 10.3 = 21.115 Ω

21.115 Ω x 0.1 Amp =2.1115 volts loss 18 V - 2.1115 V = 15.8885 volts to the Xku LNB if solid copper core RG6 were used. Even with the multiswitch boost this would allow for marginal switching.

RG6 copper-clad steel inner conductor, if calculated in this manner, gives even worse numbers as it's more lossy than solid copper.

Now I'm stuck...