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The Main Problem
The problem with POE is voltage loss in thecable. CAT5 conductors were never designed to carry significantcurrent, and are very thin. Consequently, the resistance of (say) a 20mPOE cable is significant; 1.8 ohms per conductor to be precise. As wepass the current necessary to power the AP though such a cable, thereis an associated voltage loss. And if the voltage at the AP end of thecable falls too low, the AP does not work at all, or performs at lowefficency.One common symptom of a poorly engineered POE cable is:
The reason: As the AP starts transmitting, itdraws more current over the POE cable. That means a bigger voltagedrop, and the voltage at the AP is no longer sufficient. So, itcrashes, reboots, and starts over.An Example
- AP at the end of a POE cable appears to work fine when idle (i.e. little or no wireless traffic)
- AP continually reboots as soon as the AP starts a high-speedfile transfer, where the file transfer data is being sent from the APto a 'client'.
It all comes back to that basic equation known as Ohm's Law, which states:
(Voltage Drop in cable) = (Cable resistance) x (Current)
V = I x RLets say we have the following:
So, we can calculate the voltage loss in the cable as follows:
- 20m POE cable. The round-trip resistance totals 1.8 ohms.
- An AP that:
- draws 200 mA (idle)
- draws 500 mA (transmitting full speed)
- requires 5.0V (+/- 0.5V) for proper operation
- comes with a 5.0V regulated power supply
V = I x R
Idle: V = 0.200 x 1.8 = 0.36
Transmitting: V = 0.500 x 1.8 = 0.90
That means the voltage at the AP is:
Idle: 5.0 - 0.36 = 4.63 V (OK - as it above 4.5V)
Transmitting: 5.0 - 0.90 = 4.10 V (Not OK - as it is below 4.5V)
Engineering Your Own POE Cable
You need to do the following:
- Operate your AP as normal (i.e. no POE cable)in such a way that it is transmitting full speed. Typically you willsetup a simple wireless network (AP and Client), connecting two PCswith high speed ethernet interfaces (NICs), and start a long filetransfer using File Transfer Protocol (ftp) or similar.
- Measure these values:
- DC Voltage at the AP (Va)
- DC Current into the AP (Ia)
- Note the required AP voltage (Vr). This is often written on the AP, or in the manual.
- Calculate the maximum length of CAT5 POE cable using this equation:
Max Length = (Va - Vr) / (Ia x 0.094). [Note: 0.094 is the resistance per meter of CAT5e cable]
Vr = 12 Volts (written on AP)
Va = 13 Volts (measured with AP transmitting)
Ia = 400 mA (measured with AP transmitting)
Max Cable Length = (13 - 12)/(0.400 x 0.094) = 26 meters.Where does the 0.094 ohms/m come from?
CAT5e has a maximum DC resistance specification of 0.0938/m. (See this link.) We wire the 4 unused conductors in a POE cable like this:
where each pair of wires is shorted at each end (ie each pair is wired in parallel). So considering a 1m length of cable:
Thats All folks.
- Resistance of one conductor = 0.094 ohms
- Resistance of two conductors wired in parallel = (0.094/2) = 0.047 ohms
- Round Trip Resistance for both conductor pairs = (2 x 0.047) = 0.094 ohms