Video Camera - POE - splitter/injector problem.. need some assistance

Discussion in 'The Projects Forum' started by vbtalent, Mar 10, 2014.

  1. vbtalent

    Thread Starter Active Member

    Feb 20, 2010
    34
    0
    Its been a while since I posted on the forum.. "Hello" again.

    I started a relatively easy project to convert a video camera to use POE using an injector and splitter but for some reason I don't understand why I can't get enough voltage to the camera and could use some suggestions/help.

    The camera uses a 5v 2000ma walwart and has been disassembled.

    I bought a splitter and injector from EBAY, the injector seems fairly straight forward and I'm using a 12v 1000ma walwart to power it.

    The injector seems like its doing its job correctly and I am able to get POE of 12+ on the spare pair of the ethernet cable.

    Supplying the POE voltage to the splitter, I am able to adjust down the output voltage to 5.28v the exact voltage of the camera walwart.

    All good so far..

    So here is the strange part, when I hook up the output of the splitter to the power input of the camera the voltage is 2.45v considerably less than what is required for operation and much different than the measurement when not connected.

    Is this a problem with the splitter or injector or both, any ideas?

    Thanks.

    VB,
     
  2. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,135
    200
    I have to ask what the lenght of the wires are. Unfortunately POE is a real mess and your not using a .af compatable POE system.

    The injector your using generally just uses the spare pairs and injects a DC voltage. SO the line lenght and all the resistances like contact resistances all cost you, It would be better putting the regulator at the camera side and inject the 12 V @ 1A. So, you need all 8 conductors and must use 100bT or 10bt. They may double the pairs to get a lower voltage drop.

    Taking 12 *1 and 5*2 that means 12V in to power a 10 W out system. That assumes NO losses.
    Also some systems may take more current to start,

    You are walking on the edge and you would be better off injecting 12 V.

    Real POE systems also get wierd, but for the most part use approximate 54 V on one of the 4 pairs.

    I am running one real POE system at home for my DSL modem. The modem uses 5V and the POE system I selected operates on two pairs or 4 wires total.

    There does look like there is significant resistance. You could put a say 1/2 load at 5V or R=5/0.5 or 10 ohm resistor. P=V*V/R or power = 25/10 or 2.5 or greater watts.

    Measuring the drops across the actual RJ45 contacts might help.

    Using this http://www.powerstream.com/Wire_Size.htm calculator 24 AWG wire, 12 V, 1A, 100 feet and you get 6.7 V calulated at the end. Then the regulator input probably has to be 3V or so (not checked) bigger than the input, you get 6.7-3 or 3.7 Volts.

    For long lengths, the final regulation needs to be at the point of use end.
     
  3. vbtalent

    Thread Starter Active Member

    Feb 20, 2010
    34
    0
    Thanks for the reply KISS.

    So, voltage at the input side of the splitter does have some loss, actually a lot 5+, but at the output side of the splitter it appears to be regulated at 5+, the needed voltage for the camera to operate. When I actually attach the wires to the input connector then the voltage drops to around 2.4v, definitely my problem.

    If I measure the same voltage across the input connector on the camera using the supplied wallwart I get the 5+, as anticipated.

    I'm really confused why the splitter that appears to be supplying 5+ regulated and the wallwart supplying 5+ have different impacts on the input connector of the camera, 5+ volts is 5+volts... the only differences that I see is one wallwart supply's 1000ma and the other 2000ma, the 2000ma works..

    VB,
     
  4. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,135
    200
    Everything is so marginal, that it likely won;t work. You house may be rated for 200 A at 120 VAC, but if you only have a 10 W light bulb on, that;s what your house draws.

    12 W available and 10 W out is too close for comfort with a switching power supply in there too.

    "Measuring across terminals": I didn't explain that because it's probably counterintuitive and difficult to measure. The test might be take a voltmeter and measure across say pin 8 at the PCB injector to pin 8 on the camera. The would include two connector drops and the wire between them. Use a short cable (6") and it;s mostly the connector.

    Your would try to measure th voltage across what you might think is a wire, but it;s modeled as a number of resistors in series.
    PCB/solder
    The contact resistance of connector
    The contact resistance of the cable to wire
    The wire resistance
    The contact resistance of the cable to connector
    The contact resistance of the connector
    The PCB/solder contact resistance

    AT every place there can be a loss. The higher the current through these items, the higher the voltage loss.

    If you try to draw too much current from the 12 V supply @ 1 Amp, may fold back it's output or reduce it.

    A 12 V 2 A adapter MIGHT work.

    Putting the regulator on the camera side is a better option as well.

    It could even be a bad cable. The power source (12 V @ 1A) needs upgrading, 24 at 1A is preferable to 12 V at 2A.

    Also ne careful that the Ethernet cable is not a crossover cable. The camera may mot mind, but the crossover cable reverses the polarity, Won't happen with real POE.
     
  5. vbtalent

    Thread Starter Active Member

    Feb 20, 2010
    34
    0
    All good information and points to check.. thanks.

    I'll do a little more sleuthing this evening and post back.

    VB,
     
  6. vbtalent

    Thread Starter Active Member

    Feb 20, 2010
    34
    0
    So, I did some more poking around yesterday and it turns out to be user error, as usual.. DOH!!

    I had wired the splitter incorrectly into the camera, flipped thew wires around and it was good to go.

    Really appreciate the suggestions and ideas it was extremely helpful in diagnosing the problem.

    VB,
     
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