Doom wrote: > > ... �n data de 15 Aug 2002 Andrei Vlad Pascal a povestit: > > >>sau daca eshti jmeker pui si tu un adaptor care sa scoata tensiune ceva >>mai mare decat iti trebe tie si asa compensezi pierderile pe cablu. Au >>facut unii o faza d-asta si au alimentat ditai modemu radio. Care a mers >>vreo 3 ani fara nici o pb.! (Modemu si antena erau la vreo 50 m de >>compu' cu pricina - da, se alimenta de la 12 V din computer, scula >>mergand la 6 V. vreo 4-6 V se pierdeau pe cablu :)) ) > > > LOL :) Asta da calcul precis!!! Made in .ro!!! Si eu recunosc ca am > incercat pe sarma aia de 120m mai multe tensiuni, dar la nici una nu a > raspuns cum trebuie , asa ca am rugat pe cineva la o casa din apropierea > hub-ului sa ma lase sa trag curent de la ei ... se rezolva lunar cu o > plasa de fructesi ciocolati pentru copilasi galbeni de grasi! :) > > Cosmin
Hai sa ma dau si eu rotund: normal sa "se piarda" ceva voltzi pe cablu, ca doar rezulta o rezistenta alimentata care alta treaba decat sa "manance" curent si sa produca caldura nu are :-)))) la o plimbare mica pe net, am gasit si teste+numere exacte, scuze de dimensiunea mailului: The DC resistance of Cat5 is about three ohms per hundred feet per conductor, so for 250 foot cable had at least seven (7) ohms resistance. Most of the time the APs draw much less than 0.8A, so you were still above 6V at the AP. In fact, the access points are probably just using linear regulators down to 5V on their insides, so as long as you're giving them something better than 6V at the terminals they're likely to work, but at 500 feet they didn't because the voltage drop was too high once they started turning on. -- Scott Carter Conclusion- The maximum segment distance for Ethernet 10BaseT is 330 feet (100 meters) (specified in IEEE 802.3), so this PoE setup should work at the maximum segment distance with the stock voltage outputs of the DC transformers (depending upon your access point and DC transformer). 500 Feet of Cable Testing Addendum (2001 May 05)- This setup was additionally tested with 500 feet of CAT 5 cabling, 170 feet more than is allowed for ethernet specifications. The ethernet signal worked over the 500 feet of cabling at both 10baseT speed (Apple Airport) and 100BaseT (Ugate 3300) with no problems. The DC power however did have problems. Both access points would power up (i.e. the lights would come on), but they would not fully initialize and function. I don't really consider this a problem with this hack since your not supposed to use 500 feet of cabling for ethernet anyway. Check the packaging, documentation, or the bottom of the device to find out how much current it draws. Usually it'll be between 50 and 90% of what the transformer can supply. (If the device ships with a 1500ma transformer, expect it to draw around 1200 at peak, and normal operating current should be around 1000.) These numbers are based on my experience alone. You can fix this by using a higher voltage transformer to overcome the drop. Be careful! This is why I said to find the actual power draw of the device. You want to do all your testing using a large power resistor which draws approximately as much power as the device. When you've found the correct transformer voltage, doublecheck your polarity and try it with the device again. The caveat here is that your new transformer rating assumes the resistance that you tested it with. If you find a transformer that works over 500 feet of cable, and you run it over 100 feet, you've changed the resistance. Current draw (the device) is the same, but resistance has gone down, so voltage drop goes down. Ergo, the actual voltage reaching the device will go UP, and you might cook something. Electrical Math (thanks to Ryan Cole for emailing this) Wire gauge / Ohms per 1000ft 20 / 10 22 / 16 24 / 25 28 / 65 The TIA-568 spec which defines Cat5 (cat 3 as well) specifies that either one shall be no more than 9.38 ohms per 100 meters. Our testing of name brand Cat 5 says more like seven ohms per 100 meters, so 3 ohms per hundred feet is safe and conservative. -- Scott Carter Ohms Law E = IR (Volts = Amps x Resistance) We'll say 22 Guage wire at 250 Ft under a 1 Amp load.... 22 Gauge @ 250 Ft = 4 Ohms E = 1 x 4 or E = 4 Volts So basically you should expect 4 Volts consumed leaving you 8 volts at the AP end. Lets do another situation... 28 Gauge over 50 ft at 2 amps.... 28 Guage @ 50 ft = 3 Ohms E = 2 x 3 or E = 6 Volts So in this case you dissipate half your volts... could be bad, or if I calculate correctly, possibly worse... 6 Volts x 2 Amps = 12 Watts dissipated over the wire 12 Watts / 50 Ft = .24 Watts per foot I would not expect that to start a fire outright, but you could use it to keep your pipes from freezing in the winter. So sunt bune iarna :-)))) C --- Pentru dezabonare, trimiteti mail la [EMAIL PROTECTED] cu subiectul 'unsubscribe rlug'. REGULI, arhive si alte informatii: http://www.lug.ro/mlist/
