Hello, First time poster, excuse my lack of detail here.

I'm working on creating a tower mounted digitizer/tuner enclosed in a NEMA box. The box contains the radio, preamps, and a fiber media converter(so i can just run one fiber down the tower), fans, and a custom made power distribution unit that has a 6 pin canon connector input on the box. It functions very much like a PCI-E connector on a graphics card. 3 power cables, 3 ground cables, all 20AWG in the box.

I have a bench DC power supply that maxes out at 15VDC and 48A. I can use this power supply to power my box with no problems with a custom made 3ft long 6 conductor cable (18AWG). I draw approximately 6amps when the unit is functioning fully at 15VDC. (Power distribution can handle 20volts). Box needs at least 13 Volts to function.

My issue is that I'll need a cable run of at most 300ft, avg 150ft, to reach the box when it's on the tower. I've made a 300ft cable with 6 conductors (18AWG) but I cannot get the box to power on. I've verified the voltage at the power distribution unit pins so I know its reaching all the way there. I've tried to take into account the cable loss of around 4 ohms round trip but it still doesnt work. I have higher voltage power supplies that I tested with the box and 300ft cable and they still dont work. One is a 24Volt, 6Amp psu. The other is a 48v 15amp psu.

We are trying to keep the cable small, hence the parallel conductors.

I feel like this is a small issue that I'm overlooking. Any advice or help would be greatly appreciated.

 

You are looking at a pretty long run. This cable chart should help.

Your current draw is about 6 amps you mention. What you need to do is start calculating the I*R loss. Take your cable and measure the resistance of your power lead going and coming (two leads). With just the cable laying flat short the power leads on one end. Now measure the resistance as accurately as possible on the other end. Multiply the resistance you read in Ohms times 6 Amps and that should tell you what your voltage drop is round trip. Subtract that voltage from your supply voltage and that is about what your load is seeing. Be it parallel conductors or a single large conductor the total area of the conductor is what it's all about. Actually using a single conductor is going to result in a smaller cable anyway as multiple conductors will have multiple insulation's making for a larger bundle to get the same wire area. I think if you do the math you will be surprised.

Your 4 ohms is pretty close. 300 feet of AWG 18 is about 2 Ohms so round trip is about 4 Ohms. With a single run of 6 Amps I get 24 volts of loss or I*R drop. Awg 18 has a diameter of about 0.0403.

The idea is not to apply a higher voltage at the start of the run, the idea is to use a large enough wire.

Ron

 

Have you tried installing about a 10000μF/63VDC capacitor at the load?

 

Have you tried installing about a 10000μF/63VDC capacitor at the load?

That's a good suggestion! Yopu may want to try that.

Ron

 

why not supply 120 vac to the top of the tower and put the supply in the box at the top of the tower? the capacitor will only help a little during surges, not steady supply. also, does your supply have sense lines? if so, run the sense lines to the top of the tower and connect them to the dc lines, they are there to compensate for voltae drop .

 

why not supply 120 vac to the top of the tower and put the supply in the box at the top of the tower? the capacitor will only help a little during surges, not steady supply. also, does your supply have sense lines? if so, run the sense lines to the top of the tower and connect them to the dc lines, they are there to compensate for voltae drop .

That's how it's normally done. Running 120 VAC for example up the tower would get the current to less than an amp. So if is practical or possible to move the DC supply to the top of the tower things may go better.

Ron

 

If you could also run an extra small gauge pair, you could use one of the automatic regulating sensing power supplies made by Hammond and others for this very application, the small gauge pair sense the voltage at the load end and continuously adjust the source so that the volt drop is automatically counteracted.
Max.

 

If you could also run an extra small gauge pair, you could use one of the automatic regulating sensing power supplies made by Hammond and others for this very application, the small gauge pair sense the voltage at the load end and continuously adjust the source so that the volt drop is automatically counteracted.
Max.

That's actually a good idea, using a power supply with external sense lines. You may want to consider that.

Ron

 

I'm taking in all the great suggestions all.

The 2 standalone power supplies I'm using do have sense ports. I'll see about running an extra pair of wire to see what it does (if it works, we'd need to order some 8 conductor cable and change the cannon connector). I personally think having an AC/DC converter in the box is the way to go but it might pose a heat issue (we are still going to pursue it.) I also think the 18AWG wire might be too small for the current at the distance we want. I'm going to try using some 10AWG to see what happens, just for the hell of it.

 

Let us know how this progresses and works out.

Ron