Moxa showed a VDSL link over barbed wire...I thought that was interesting. On Mar 9, 2017 4:46 PM, "Chuck McCown" <[email protected]> wrote:
> This all started with the VDSL2 ethernet line extenders. They are a > constant load of 6 watts. I want to power them over the pair of wires that > the VDSL signal is going over. So the question is, if I put 48 volts on > the pair at the house through a VDSL filter, can I get enough power out at > the far end to power that VDSL unit. 100 ohm loop. About 2000 feet of 24 > gauge. The answer is yes. > > Current is 0.12953 A > Voltage at the load is 46.322 V > > You can brute force it with trial and error, that is how I got my first > estimate, but I wanted to know exactly. There is a formula. > > Trying to visualize how I would use Thevenin. The battery is a short, > right? So from the load’s perspective you are seeing 100 ohms. Not sure > what the Thevenin equivalent of a constant power load is. > > *From:* Chuck McCown > *Sent:* Thursday, March 09, 2017 4:28 PM > *To:* [email protected] > *Subject:* Re: [AFMUG] Ohms law > > But what is the formula? > > *From:* Dave > *Sent:* Thursday, March 09, 2017 4:20 PM > *To:* [email protected] > *Subject:* Re: [AFMUG] Ohms law > > Current =.125A at load > Voltage=35.5 at load > > If my current is correct then I should be on point. > Otherwise I would use Thevenins Therom to get closer. > > > On 03/09/2017 05:08 PM, Chuck McCown wrote: > > The questions are: > What is the current and voltage on the load.� > � > *From:* Chuck McCown > *Sent:* Thursday, March 09, 2017 4:05 PM > *To:* [email protected] > *Subject:* [AFMUG] Ohms law > � > Had a fun afternoon.� > � > Solve this.... and give the general formula... > � > 48 volt power supply > 100 ohm wire resistance to the load. > 6 watt load. > � > Took me some time.� Not trivial.� > > > -- >
