I was looking foward to powering a small Integrated Circuit using only the power provided by the ethernet outlet of my room. Is that even possible?

I've googled and I've found that the voltage that it provides is something in between 2v and 3v. Since it is not a DC voltage, but a random AC one, I believe that it would be not possible to power an IC without having to use some sort of AD conversor or a simple peak detector circuit in order to maintain the voltage.

Am I wrong? Would you have any advises for that case?

  • 6
    \$\begingroup\$ Well, there is the PoE (Power over Ethernet) standard. Maybe take a look at that? \$\endgroup\$
    – Rev
    Jan 21, 2013 at 7:29
  • 1
    \$\begingroup\$ "What IC" or more specifically "what power consumption" would be a rather critical question. \$\endgroup\$ Jan 21, 2013 at 14:40
  • 1
    \$\begingroup\$ Reverted to the original title without the limitation to scavenging imposed by the 3rd-party edit. It's up to Luis to clarify if PoE would or would not be an option now that the possibility has been raised. \$\endgroup\$ Jan 21, 2013 at 14:46
  • \$\begingroup\$ @Chris -- That isn't correct. The OP references: (1) AC, (2) data voltages not PoE voltages. PoE is an alternative to the premise of the question and is suggested by all in this role (e.g "Have you also considered...") \$\endgroup\$ Jan 21, 2013 at 21:35
  • \$\begingroup\$ @Rev1.0 PoE is not a better option in my case as I'm working with the standard networks that have in ordinary buildings. Thanks, though! \$\endgroup\$
    – Luis
    Jan 21, 2013 at 23:30

1 Answer 1


You are describing using a concept called "Energy Harvesting," but you are trying to use the data pairs of the ethernet port as your energy source.

Update: Well, let's qualify this a bit...

While extremely interesting (I did my masters work in this area), what you are describing simply will not work well in practice for a number of reasons:

  1. All versions of Ethernet over twisted pair cable specify differential data transmission over each pair with transformer coupling. That means that there is no DC power path. You have current moving in both directions through an isolating transformer. You will need circuitry to convert and condition it. A lot of the power you will acquire will be more than consumed in the quiescent power of your conversion and conditioning circuits. There will be very little, if anything, left for the load.

  2. The line is only active when data is being sent to you (or broadcast). Unless you are creating a structured environment where you control the network, data (power in your scheme) will be unreliable.

  3. If you can control the network, just install a Power-Over-Ethernet power-supply in between the network switch and your device. A PoE power source adds DC power (-48V) to the otherwise unused pairs of copper in the category 5 cable (10bT, 100bTx). It can even work with Gigabit Ethernet now by riding data on-top of the power pair (so it serves dual purposes). It's that simple. Why bother with harvesting?

Design Experiment

Here's a common Ethernet interface chip (CP2200) from Silicon Labs. enter image description here

Here is an abstraction:

enter image description here

  • The characteristic impedance of the cable system is about 100 Ohms (which is why you see the 100 Ohm termination resistor in the Silicon Labs figure).

  • The nominal peak transmission output current of the CP2200 is 15mA (page 9). It should be noted that there are high current chips available, even ones with programmable current output (such as DP83223).

  • At peak efficiency (matched impedance) the load must present the equivalent of 100 Ohms at the transmission frequency.

  • The trasmission system uses a 1:2.5 step-up transformer

Maximizing Power Transfer:

At the other end (the output of the network jack), the maximum peak current is 6mA (from 15mA / 2.5). It's delivered into an ideal load of 100 Ohms to reach an maximum instantanious power of P = I^2 R = 3.6mW or about 2.5mW,rms (not bad! and 10x higher than my original estimate).

For 15mA maximum output, the output stage of the transmitter adds about 120 Ohms in source resistance.

  1. Working backwards you have 200 Ohms on the remote side of the transformer
  2. The 2.5 turns ratio results in an impedance transformation to an apparent 32 Ohms on the primary (transmitter's) side of the transformer.
  3. That's 480mV on the primary winding.
  4. The transformer steps it up by 2.5X to 1.2V on the secondary.
  5. Half of the voltage is lost to the cable impedance resulting in 0.6V peak to the ideal load.

That's P=V^2/R = 3.6mW. It matches the ideal expectation so we're good.

Here's the Problem in Practice:

Unfortunately, power delivery isn't the total story. Now you have to be able to use it.

It's bipolar so you need to rectify, de-ripple, and (possibly) step-up (or otherwise convert/regulate). There just isn't a lot of voltage overhead for this.

You are working with 0.6V and you need to transit two diodes in the full-bridge rectifier. Even using low forward drop diode types, you are still looking at around 0.3V (per diode). That means the voltage available (and therefore power) for you to use in your load is basically nothing.

Alternate Rectifier Architectures

There are other approaches to harvesting besides the diode bridge, so it isn't impossible, but it's highly impractical to do this.

For example, you could use a half-wave rectifier (most RFID tags I looked at do this) to eliminate one of the diodes (but you lose half the waveform).

In this case, you get

  • 0.3V,peak * 6mA (ideal) = 1.8mW (peak) = 1.27mW (rms)
  • With only half the cycle generating you're down to about 640uW (micro-Watts)
  • Then you have to derate by your transmission duty cycle (the percentage of time you are keeping the transmitter active)

...and that's maximum. If you change your load away from exactly 6mA, you will get decreased efficiency and therefore much less power output that you would otherwise expect due to the impedance mismatch this introduces.

Harvesting rectifier design is an area of active research and there are more efficient ways to use a single diode. If you really are committed to pursuing this, reply and I'll go find some cites/ideas for you.

  • 3
    \$\begingroup\$ RFID works for the same basic reason as PoE: by design. The radio field strength in passive RFID applications in general is not chosen by the requirements for noise immunity but to power the RFID tag. \$\endgroup\$
    – MSalters
    Jan 21, 2013 at 15:59
  • 1
    \$\begingroup\$ @Luis -- It isn't really 100 Ohms in the resistor sense (Google "Transmission Line Theory"). It is the effect of the cable on a signal at 10MHz (or 100MHz). RF signals are a bit different in behavior (there are a few more caveats that we gloss over when teaching DC at the introductory level) than your question implies. If you don't use an ethernet cable, the power available to your device will actually be even less than estimated above... unless you keep the entire cable system to less than about 15cm. \$\endgroup\$ Jan 22, 2013 at 1:32
  • 1
    \$\begingroup\$ @Luis -- I would also add that the voltage you need at the load (within reason) isn't important. What you care about is power (voltage times current) because (with some loss) you can convert current into voltage with various circuits. Assuming a perfect converter, 1W could be expressed as 1A @ 1V, or 0.5A @ 2V... see? \$\endgroup\$ Jan 22, 2013 at 1:35
  • 1
    \$\begingroup\$ @Luis -- Cool! Glad to be helpful. 15cm is total distance between transmitter and receiver. You must include all of the relevant distances (so the wire in the wall between the jack and the Ethernet switch counts too). \$\endgroup\$ Jan 22, 2013 at 1:55
  • 2
    \$\begingroup\$ @Luis -- I'd start with another transformer: in front of the rectifier, after the jack. Try 1:10 step-up turns ratio. It isn't likely to deliver a lot of power, but you should see what you can get out of it. Perhaps you could harvest from multiple ports in parallel. \$\endgroup\$ Jan 22, 2013 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.