Power your access point without a separate power cable by using free pairs on CAT5.
A number of access point manufacturers (Proxim, Symbol, and D-Link, to name just three) are now offering Power over Ethernet (PoE) add-ons for their access points. PoE modules insert DC voltage into a standard Ethernet cable. The idea is to supply the AP's power and UTP Ethernet connectivity requirements via a single Ethernet cable.
This works great in areas where you might not have power easily accessible, such as a roof. This also allows you to more easily place the AP closer to the antenna, thus reducing signal loss over antenna cabling. Ethernet signal travels well over CAT5 cable; a 2.4 GHz signal doesn't do as well over antenna cabling. Also, Ethernet cabling is much cheaper than antenna cable such as LMR400. This hack demonstrates how to build a simple PoE module pair.
In June 2003, the IEEE ratified the 802.3af standard for Power over Ethernet, which has spurred the release of standards-compliant PoE products. 802.3af defines two types of power source equipment: end-span and mid-span devices. An end-span device is an Ethernet switch with embedded PoE technology. These switches deliver data and power over the same wiring pairs: 1/2 and 3/6.
We're going to build a pair of mid-span devices, which in the 802.3af specification can be placed between a legacy switch and the device to be powered. A mid-span device has an RJ-45 data input and a power input, and it sends the data on pairs 1/2and 3/6, while sending power on the unused 4/5 and 7/8 pairs.
4.6.1. Step by Step
If you have a device such as a VoIP phone or a Soekris PC that will accept Power over Ethernet without a secondary adapter, you will only need to build the power injector in steps 13.
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Figure 4-6. The completed power plug
Figure 4-7. The DC plug mounted in the housing
Input jack |
Output jack |
DC plug |
---|---|---|
Pin 1 |
<-> Pin 1 |
|
Pin 2 |
<-> Pin 2 |
|
Pin 3 |
<-> Pin 3 |
|
Pin 4 <-> |
DC Positive Wire 1 Center Connector |
|
Pin 5 <-> |
DC Positive Wire 2 Center Connector |
|
Pin 6 |
<-> Pin 6 |
|
Pin 7 |
<-> DC Negative Wire 1 Outer Connector |
|
Pin 8 |
<-> DC Negative Wire 2 Outer Connector |
Output plug
Input jack
DC plug
Pin 1
<-> Pin 1
Pin 2
<-> Pin 2
Pin 3
<-> Pin 3
Pin 4
<-> DC Positive Wire 1 Center Connector
Pin 5
<-> DC Positive Wire 2 Center Connector
Pin 6<-> Pin 6
Pin 7
<-> DC Negative Wire 1 Outer Connector
Pin 8
<-> DC Negative Wire 2 Outer Connector
4.6.2. Resistance Is Futile
The DC resistance of CAT5 is about 3 ohms per 100 feet per conductor, so a 250-foot cable has at least 7 ohms resistance. Most of the time, an AP draws much less than 0.8A, so you would still be above 6V at the AP. In fact, the access points typically use linear regulators to drop the voltage 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.
There is a good calculator online at http://www.gweep.net/~sfoskett/tech/poecalc.html that calculates the voltage drop for a given length of CAT5. Use it to estimate how much power you need to provide at one end of your cable run in order to power your access point.
Terry Schmidt
Bluetooth, Mobile Phones, and GPS
Network Discovery and Monitoring
Wireless Security
Hardware Hacks
Software Hacks
Do-It-Yourself Antennas
Wireless Network Design
Appendix A. Wireless Standards
Appendix B. Wireless Hardware Guide