Power Loss Over Cable Length

I'm at a complete loss here, and I'm hopeful someone here with more knowledge than I can lend some assistance.

I've got a power over ethernet setup, using a passive power injector.... The device's DC adapter is rated at 5 volt DC @ 3 amps, and adequately powers the device... a multimeter reads the actual power output at about 5.5 volts.

The cable run is 50 ft.

At the other end of the cable, the multimeter is reading the voltage at 5.4 volts, so it seems that there is minimal loss... since the device only requires 5 volts, this should be adequate.... but its not... there's not sufficient power to run the device... it doesn't even try to power on with this power source....

I'm presuming I somehow need to increase the amps?

I've run into some POE calculators that tell me that for 5 volts at 3 amps over a 50 ft cable run, I would need to increase to 6.9 volts (but with a warning that the 3 amps is likely not what the device actually draws and that I could fry stuff)... these calculators also say that the loss is 1.9 volts.... but I'm not seeing much of any drop in voltage at all in real world testing.... I don't want to plug in a 7 volt power supply and fry the equipment that is expecting 5 volts....

Now, I understand that Watts = Amps * Volts, so 5V @ 3 amps = 15 Watts specified available at the business end of the power supply.... in reality it tests out 5.5V @ 3 amps = 16.5 Watts... but at the end of the cable run, I'm reading 5.4v at 1.4 amps so about 7.56 Watts.... is that correct?

So as long as I'm getting at least 15 Watts at the end of the 50 ft cable, I should be fine... Is this basically correct, my end goal should be to get the same wattage? ie is 15 volts at 1 amp the same as 5 volts at 3 amps?

If I plug a 6 volt 3 amp power supply into this system, then I'm presuming I'll see loss along the order of 0.11 Watts per foot, or about 5.5 Watts, so the end of the run should come in at something around 12.5 Watts.... and a 7 volt 3 Amp power supply should come in at something like 15.5 Watts... Hooking a "Kill-A-Watt" up to the power supply shows its actually drawing about 13.2 - 13.6 Watts.... so am I safe to pump 7 volts into a device that's spec'd for 5 volts, so long as the wattage is ~ 15 Watts?

I'm just confused.... Anyone who could help point me in the right direction?

This post has been edited by Azuretech on October 11, 2011 01:21 am

http://blog.fosketts.net/toolbox/power-ethernet-calculator/


I'm not sure how to read your post.

the dc supply is rated to 3 amps, but you only need 1.4 amps at the other end of the cable?

also, if there's 3 amps going into the cable there will be 3 amps flowing out of the cable, so when you "read 5.4 volts at 1.4 amps" how did you make the measurement?

the simple way to test this is use a 3 foot cable and see if it works.
if yes then you will have to run a separate power cord

Are you using legitimate poe equipment, our hacking something together?

If you are hacking it, there are two unused pairs in a run, use one pair for negative and one pair for positive.

Real poe equipment usually runs low current at 48v to reduce ohmic losses.

hmmm ...

did you measure the voltage at the other end under load ?
theoretically speaking , you could have a 300 foot run and shove 5 volts into it and measure right at 5 volts on the other end if you use a high impedance meter (an average dmm usually has around 1 meg impedance as a rule of thumb) .
put a 100 ohm resistor across the line and measure the voltage ... there will be a drop in the voltage as the line has a greater load on it than with the 1 meg of the meter .

also ... if the device is designed to work with a regulated 5 volt input (on the end of a short line) then the device may simply not function if the input drops below 4.5 volts .

does the device you are attempting to power up have it's own 5 volt regulator circuit built in ?
if it does you need to supply it with more than 5 volts in order for the regulator to work ... in the case of a LM7805 regulator , about 7 volts input is needed for "overhead" .

you will still have to allow for the voltage drop in the line ... so the actual inserted voltage could be around 9 volts .

also ... "increase the amps" ... a misnomer .
as long as the supply is capable of delivering the required current ...
increasing the voltage and/or lowering the resistance is the way to actually "increase the amps" .
once you visualize this , it becomes easier to understand things like the resistive losses associated with cable/wire runs , etc .

What CWB said. But rather than to put a 100ohm in series just connect your load, strip the cable a bit and measure while the load is attached.

"put a 100 ohm resistor across the line and measure the voltage ... "

Yes, it's cable resistance that is causing your problem. This will cause a voltage drop, the exact amount depending on the current being drawn.
The only way around this is to use bigger wire, or more conductors, to carry the power.
It is NOT safe to increase the supply voltage, unless the powered device always draws the same amount of current, or you add a regulator. Otherwise varying current will result in varying voltage.
One way out would be to increase the supply voltage to 10-12 volts, & use a LM7805 three terminal 5 volt voltage regulator at the far end to regulate the voltage back down where it should be. The regulator will deal with the voltage differences as long as the supply voltage is high enough that maximum current draw doesn't pull the voltage down too far for the regulator.

Even bumping the source voltage and dropping it linearly at the far end is kind of a wash... you're still burning up all of the power along the cable, plus some at the far end. You're turning the wire into a large heating element, and using the small fraction of power left over.

Cat-5 can have conductors anywhere from 18 gauge down to 24 gauge. Larger sizes may work with even one pair, two pairs in parallel should work. But at the smaller sizes, even two pairs in parallel are going to drop too much voltage over 100'. (50' down, 50' back.)

OP: Get yourself a proper PoE injector/extractor pair like this one. Sure, it's $50, but it will also fix your problem for you.

Out of curiosity, what equipment is this that's doing PoE with just 5V?

Sorry, I guess I should have given a bit more information....

-Equipment being powered: Actiontec MI424WR wireless router

-I'm using a passive POE injector, (specifically, this: Passive POE injector) which for short cable runs can just use the stock device power supply, but for longer cable runs obviously requires something with a little more juice. (understood the more expensive stuff is better, but hey, I'm on a budget.... and its a one time setup, so the extra cost is not really an option especially if I can do it cheaper with just a different power supply... heck, it wouldn't suprise me to learn that I've got one that'll work lying around the house somewhere from some old dead equipment)

-the stock wall wart power supply that comes with the router is marked 5VDC @ 3A it works if plugged into the router directly, it works for short cable runs (so I know the injectors work), but not for the longer cable run.

-The 50ft cable run is Cat 6 24 AWG ethernet cable, the injector is using 2 pairs

That being said... I'm basically trying to figure out what Volt and Amp rating power supply I should get to be able to provide equivalent power at the end of the cable run as the wall wart provides when plugged into the router directly.

This post has been edited by Azuretech on October 12, 2011 04:22 pm

Wiring resistince will be up and about 2,5 ohm. So if your devices consumes the full three amps it will drop about 7.5V, which is more than it began with

To correct this, you could start out with a 12.5V PSU. Unfortunately, that would mean that at a time your device is *not* consuming 3amps (because no devices, in sleep mode, what not) but say 1 amp that would mean your device gets 12.5V - 2,5 * 1 = 10V which will probably send your 5V router up in smoke.

You coult try with putting a second pair in parallel or try to find some middle ground between too much and too little but that is a bit shaky at best...

The wires in your cable are too small to feasibly pass that much current. Simply increasing supply voltage to compensate will fry your equipement sooner or later, and probably sooner. That injector/extractor isn't going to work.

Is your power supply regulated with feedback? sense the output voltage at the end of the cable then the power supply might be able to bump up the voltage under load to maintain voltage at the other end... not the most power efficient solution though.

I'm guessing that if he doesn't know how to calculate voltage drop over a cable we shouldn't send him searching for the feedback pin in an offline flyback..

EDIT: not to mention that wallwarts I know that are designed for 3A@5V will react poorly and explosively to 3A@12.5V

It is quite possible that it is safe to supply the router with 12 volts, which would fix the supply problem.

sounds crazy but most of the ones i have dissected drop the 5v supply down to 3.3 volts. the input power is typically buffered with a 10v cap. (replace with 16v) then fed into a single chip buck supply (You will have to look up the datasheet to determine if it can take the increased I^r losses.) and you may have to replace the inductor.

*I would run 120vac down the line and put the wall wart at the end of the cable.

You can't do it without some form of regulation. The reason being your router doesn't present a constant load...the amount of power it actually uses varies with it's activity level.
This means that the voltage drop along the cable will also vary with router activity. So, any fixed voltage supply will lead to a varying voltage at the far end of the cable.
Thus the need for a regulator at the router end. POE equipment has the regulators built in.
Don't worry about any power loss in the cable & regulator...the amount will be tiny, no more than a few watts.

"*I would run 120vac down the line and put the wall wart at the end of the cable."

120vac down the cat6 ? i hope this is not what you meant .

about that wall wart ...
that sounds like a regulated swmps type ...
so , the router is looking for a regulated 5 volt input .
as was mentioned , in this case , the internal power supply of the router might not be able to handle an increase of input voltage (marginal engineering) .

bottom line :
the only way to get around the IČR losses in the cable run is to raise the input voltage and lower the current requirements in order to ship enough power (did you ever wonder why the electric companies use voltage over current for distance transmission ?) .
going with shoving ac down the cat6 and staying within the electrical code for low voltage wiring (in the states) :
(1)grab a couple of 110vac to 24vac transformers good for 1.5-2 amperes (1 ampere will squeak by , i like to engineer a bit on the heavy side) on the secondaries . these are commonly available transformers used in heating and A/C work , etc . be selective when shopping ... some outfits think they have high dollar stuff .
(2) hook one transformer up "normally" on the "power insertion end" of the cat6 run .
(3) hook up the second transformer "bassackwards" on the other end of the run .
(4) you will now have about 110vac available to hook the original wall wart to .
(5) treat this as you would any "normal" 110 circuit ... correct insulation , an enclosure , etc .

mind you , the best way to do this is to actually run a second cable that is either four or two conductor that is easily identifiable as being different from the cat6 run (and marked clearly on both ends as to it's function) .
the reason for this is that you may know what is going on but the next poor schmuck may not be hip and possible problems of 60Hz injection on/to the signal lines .

ps ... could someone edit my post and change that "IIR" to the better expression ?
i'm too lazy to look it up .
pps ... i believe that you can run up to 48 volts on a code LV circuit ... so much the better (but those 24 volt transformers are fairly common) . as long as the transformers match you could use any voltage between 24 and 48 .

This post has been edited by tekwiz on October 14, 2011 08:48 pm

Heck, at that point, I might as well run outdoor romex and wire a GFCI outlet in a weatherproof enclosure.... kind of defeats the purpose of POE.

I did try a 7.5V 1A power supply and the router at least tries to power on, but not quite enough juice to do the job, with no ill effects.... Assuming that the 5v 3A wall wart is probably designed for way more amperage than the device would ever actually draw (I'm guessing it probably draws maybe 50-75% of that at most, which makes sense since the Kill-a-watt says its drawing between 13.2 and 13.6 watts of the 15 watts the power supply is rated for).

I'm kind of thinking a 7 Volts / 2 Amp adapter would probably be about right... From what I'm seeing at a POE calculator at http://blog.fosketts.net/toolbox/power-ethernet-calculator/ its saying that 5V 3A into 2 pair of 24 AWG at 50 ft would be a cable resistance of 0.6 ohms, and a voltage drop of 1.9 volts... so 6.9 volts at 3 amps to compensate... but with a warning that I might fry the equipment.... but....

From doing a little more research on the specs of the router, I found the following...

Input 5V/3A (Voltage regulator LTC3736 http://pdf1.alldatasheet.com/datasheet-pdf...ER/LTC3736.html accepts up to 9.8V input voltage)

It appears that it can accept up to 9.8v input voltage, so I in theory should be able to safely pump 7 Volts at 3 Amps into the line, and the LTC3736 voltage regulator on the router itself should be able to handle the excess voltage without a problem, no?

This post has been edited by Azuretech on October 13, 2011 03:07 am

I would speculate that the reason for the 9.8v limit is because to get the minimum Vin down to 2.75 volts, they had to use a rail to rail gate driver with minimum resistance and no provisions for over voltage. --Meaning if you go over 15 volts the fets get fried, and/or if you go over 10 volts the chip gets too hot because its running at ~550Khz (!) and the mosfet gate drive power requirement follows Vin squared.

You're probably going to have to find an 8 or 9 volt power supply to get it running.


@CWB Cat 5 or 6 breaks down at 6-8Kvdc when dry

"@CWB Cat 5 or 6 breaks down at 6-8Kvdc when dry"

i was thinking from the electrical code side .

horror story :
owner of a station let a wireless internet company rent space on the STL tower .
several days later a frantic call from owner "there's smoke and fire on the tower !" .
me climbs tower and finds a chunk of cat *something* doing the 60Hz buzz burn .
me feels the cat below the new box on tower --- hot --- !
me goes down tower and finds the wireless equipment .
me discovers a cat cable with a 110 plug on it stabbed into a wall outlet .
me pulls plug and tells owner .
owner tells me "get their s*** of off my tower ... now !" .
me complies with orders as owner calls lawyer .

If this chip is used, then you might get away with it. The LTC3736 has integral overvoltage protection...IF it's enabled in that particular product.
I wouldn't go much higher in the voltage, though, as it's never a good idea to run anything at it's maximum voltage. Safety factors are cheap insurance.

another solution is to jack up the voltage and reduce the current. the power loss is due to large current going through wire resistance.

if you increase the voltage, you have reduced the current going through the wire.

on the receiving end, all you need to do is to down-convert it to your desired voltage.

I would use a 48vdc for that.

obviously, you need to investigate if your router can sustain a 48vdc source on the line.

maybe these books to be useful to you -http://www.sapub.org/journal/journalsList.aspx?subjectid=107

I just said the heck with it and wired an outdoor 110v GFI outlet in a weatherproof enclosure... now I just plug the wall wart in and it all works fine... in the end, it was the most cost effective, and simplest solution.

good deal ... and you can sleep more soundly at night .

did you use that "UF" type of romex ?
that stuff has the outer insulation molded right to the wires ... not a sheath or sleeve like regular romex .
it is not my favorite stuff to work with .

I'm sure Johansen would say "You're welcome"..