1. This site uses cookies. By continuing to use this site, you are agreeing to our use of cookies. Learn More.

Problem with monoprice 5-rca component cables

Discussion in 'DIRECTV HD DVR/Receiver Discussion' started by 420benz, Feb 2, 2011.

  1. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    OK Now that i found the problem, i need to run a 50ft extension cord. What gauge should i use?
     
  2. JonW

    JonW Icon

    526
    0
    Dec 21, 2006
    You shouldn't use an AC extension cord (or a cheater plug) as a permanent solution.
     
  3. denvertrakker

    denvertrakker Legend

    136
    0
    Feb 5, 2009
    Colorado
    Yes, but sometimes ya gotta do what ya gotta do. I'd use no less than a 14 gauge heavy duty cord (preferably one of those big ugly orange things).
     
  4. MountainMan10

    MountainMan10 Icon

    607
    0
    Jan 30, 2008
    Maybe one of the power outlets is wired backwards. You can test that with a surge protector that has "protected" and "grounded" lights.
     
  5. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    That's what i was thinking. Run the wire to the outlet in the other room, cut off the plug and hard wire it to the outlet.
    .
     
  6. hasan

    hasan Well-Known Member

    5,957
    54
    Sep 22, 2006
    Ogden, IA
    There appear to be some serious errors in some of your statements:

    "The law of energy conservation means there is no "loss of power", instead that RF current just gets reflected, for the most part. a typical RG-6 connector loses about 0.1 to 0.4 dB, and that is dependent upon how well you cut the fitting more than anything else."

    That one is flat out wrong. Yes the current is reflected, but the reflected power is lost due to the attenuation of the cable. As the reflections go back and forth, the cable loss reduces the current accordingly, so power is lost every time a reflection takes place, because the reflected wave must travel down the lossy cable length all over again. In coax cables, nearly all of the losses are resistive (and cannot be avoided). Dielectric losses in coax are not worth worrying about in well designed cables.

    "Its the difference in impedance between the source (cable) and load (connector) that causes the voltage drop and reflections to happen."

    Voltage drop is generally considered a resistive issue, not an impedance issue. The resistive part of the impedance (Z is composed of R, Xc and Xl), produces attenuation (and voltage drop). The impedance bump of a connector is almost never resistive dominated (that's a really rotten connector), but rather, as you noted, caused by not properly controlling the diameter and spacings of the cable cylinder, as well as irregularities in the dielectric constant of the insulating material.

    The difference in impedance does cause the reflections. Any voltage drop is resistive in nature.

    " If the impedance is exactly the same, there is no loss at all, but that only happens theoretically."

    I can construct a simple impedance network that is perfectly matched, that has terrible losses. All you have to do is insert enough R and adjust the inductive and capacitive reactance (Xl and Xc) for a perfect match. The Z will be perfect and the losses will be atrocious. Simply matching network impedances is no surety of low losses. The reflections will be near zero, to be sure, but the resistive component of the impedance can attenuate the signal dramatically. Good matching networks don't do this. Poor ones do.

    In a well designed impedance matching network, losses are minimized (R), but there are always losses and they can be quite significant.


    Now, not much of this has anything to do with the connector issues being discussed. Most connectors are pretty well designed and don't suffer from resistive losses until corroded. Similarly, anything but the cheapest connectors are manufactured to be "constant impedance"....but there are some pretty rotten barrel audio connectors out there.

    Everything else you said was very well done. I don't mean to nit pick, but
    the conservation of energy argument was so misapplied that I couldn't let it go. The other two are still quite real, but of less importance in practical terms, with properly designed networks and connectors.
     
  7. denvertrakker

    denvertrakker Legend

    136
    0
    Feb 5, 2009
    Colorado
    Noooooooo!!!!:eek2: Danger, Will Robinson!

    You do NOT want to do that.

    Besides being a violation of almost any building code in the world, that's a very foolhardy thing to do. Just run the extension cord to the outlet and plug it in.
     
  8. hasan

    hasan Well-Known Member

    5,957
    54
    Sep 22, 2006
    Ogden, IA
    That is absolutely true. Do not cut off the grounding plug, or otherwise bypass that ground. It is pretty apparent that there is a ground loop of some sort, or the hum wouldn't be there (assuming the ground connection on the adapter is firm).

    Find the source of the difference in ground potential and fix it, otherwise, run the extension cord to the same outlet. If the hum doesn't go away then, you have a bad ground connection elsewhere.
     
  9. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    I do know it is a NONO so i will snake it down the wall,cut a hole in the Sheetrock and plug it in.
     
  10. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    Plugging the extension cord into this outlet eliminates all of the problems including the hum.
     
  11. Davenlr

    Davenlr Geek til I die

    9,136
    27
    Sep 16, 2006
    First, you seriously need to consider having an electrician check your breaker box. Since you have differential between the two circuits, either something is wired wrong, or your ground is ineffective. While he is there, he can run some ROMEX from the power box to where you need the new outlet and put in a real plug.

    In the case of an extension cord for permanent use, get a heavy duty plug (15Amp), a roll of Romex (3 conductor electrical wire designed for indoor or outdoor use), and a receptacle and box (15 Amp). Run the romex through the sheetrock where you want it. Cut to the length you need, and attach the receptacle to one end. Then attach the plug to the other end (White to silver, Black to Gold, Green to Green), and plug it in. Romex is designed for permanant use, and is much safer than an extension cord, which often uses molded on plugs which overheat and fail.
     
  12. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    Let me see if i understand.You want me to add a male plug on one end of the romex.Then run the romex through the wall as close as i can to the tV.Install a jem box and an outlet in the wall and plug the TV into that outlet. The male plug gets pluged into the live outlet. Is that correct?
     
  13. Davenlr

    Davenlr Geek til I die

    9,136
    27
    Sep 16, 2006
    Yes, that would be safer, and easier than cutting a hole in the wall big enough for an extension cord plug to go through. Of all your electronics added together are less than 15A (1800W) you could just get a 1500VA UPS unit, and plug everything into that behind the TV, and plug that into your existing outlet. It would provide about 5-10 minutes of reserve power in case of an outage, and keep everything on one circuit. FWIW, my 46" tv, 3 sat receivers, power inserter, PS3, DVD recorder, AV amp, and an emergency fluorescent light, and all my networking gear only use 550W when its all on at once. That would be the BEST solution, if you have the money for one.
     
  14. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    Sounds like a good plan. THANKS.
     
  15. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    Just a thought. What if I hard wired the Romex directly into the outlet in the other room and installed another outlet in my office.
     
  16. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    From what i have read this is not against code?????????????
     
  17. TomCat

    TomCat Broadcast Engineer

    4,153
    100
    Aug 31, 2002
    Sorry, but I stand by all of these statements 100%.

    Power is never "lost". Energy is always converted to some other form. In the case of resistive issues, which should have been able to go without saying (which is why it was not addressed in my post) it is converted to heat. The phrase "for the most part" was my disclaimer exactly referring to exactly that just in case some smart-ass decided to bring that up. Good thing that never happened, huh?

    This phenomenon is the very reason we can use connectors that are not perfectly the same impedance as the cable. Otherwise, the signal would bounce back and forth between the connectors with no attenuation, which would result in (for digital signals) a level of intersymbol interference so severe that nothing could be decoded. And that is a "loss" that is complete, if you want to parse how "significant" it is.

    The losses I stated (0.1 to 0.4 dB) are not due primarily to resistive issues, they are 99.999% due to reflection (meaning if there were no reflection there would be no loss through the connector), and the amount of power reflected is due to how imperfectly the connector matches the characteristic impedance. And naturally the signal is attenuated during the reflection and partially converted to heat energy, but never actually "lost". So no, I am not the one who is "flat out wrong".

    That would be a great answer were we talking about AC circuitry. Unfortunately for you, what we are talking about is waveguide technology, which is approached slightly differently.

    And, BTW, impedance is also resistive in nature, except that it is also subject to frequency. If you isolate any impedance-based circuit to a single frequency (thereby removing frequency as a component completely), it behaves only resistively, or as if it were only resistive in nature, and follows ohm's law perfectly. But that is not what we are discussing, we are discussing loss in cable and connectors, which is waveguide technology, and somewhat different.

    A coaxial cable is designed similar to a waveguide, and that means that characteristic impedance is everything. A perfect termination through a perfect connector will appear to the signal as if it were an infinite piece of cable--all signal will pass through unimpeded, save for the inevitable attenuation of the medium itself.

    A bad connector or one cut imperfectly will react that same way for the most part, but the change in impedance due to the bad connector or the bad connectorization will cause a portion of the RF power to be reflected. Likewise, a termination that is not perfectly the same impedance will do the same thing. All of the laws of physics and electronics still apply, and all of your formulas will still bear this out for perfect or non-perfect connections.

    And of course there are resistive losses, or attenuation over distance. I still think that is pretty obvious and can still go without saying.

    Of course you can. Anyone can. But once again, we are not talking simple AC circuits, we are talking waveguide technology, so none of that really applies, and is not a valid argument disproving anything I said. That said, any circuit you can construct probably could have a counterpart in a waveguide situation, and again, all the rules of physics would apply equally.

    I don't get what your point is. Your examples seem to support the opposite argument. Losses are only significant when the connector is damaged or corroded, or when connectorization is poor. In a system that is working properly, losses will never be significant.

    I disagree that it was misapplied by me. I do not disagree that it was misapplied by you. That I did not state the obvious regarding resistive attenuation should not be something to invalidate my statement, in my opinion.

    That said, I always value your input, and I am grateful for this chance to respond. Actually, I think there is more that we agree with than we disagree with, so I am still a little puzzled why what I said was characterized as being so far off base. But I'll survive. :)
     
  18. hasan

    hasan Well-Known Member

    5,957
    54
    Sep 22, 2006
    Ogden, IA
    I think it was the conservation of energy argument, which is a red herring. In an AC circuit no one cares about conservation of energy, because the focus is on optimum power transfer, eliminating reflections (part of same), and perhaps phase. The power is "lost" in terms of being available to maintain a signal level. Saying that energy is not lost (merely converted to heat), doesn't help in a real AC circuit. That part that is converted to heat results in a power loss at the far end, and that's all that matters. Conservation of energy is an interesting, but irrelevant side bar.

    Secondly, unless I'm completely lost in what was being discussed, waveguides and coax are two entirely different transmission lines. No one I know is using a waveguide as a transmission line in our setups. The mechanism for both propagation and loss in a wave guide are completely different than that of coax. I thought we were discussing a simple video signal over a coaxial cable. If I missed something where microwaves suddenly got involved, you have my apologies.

    Reflections do result in less power being available at the load, every time the reflection occurs. It can't be avoided. And, of course we're talking AC, or we wouldn't be even using the term impedance. There is no impedance at DC.

    Nevertheless, we can agree to disagree. I don't think you are using the appropriate engineering protocols (waveguide), and voltage drop only happens across a resistance and is not changed by reflection.

    We agree more than we disagree, and in the end two of the three variables are likely to be of little concern. Nevertheless, I wasn't trying to tweak you. I think those are errors and shouldn't be ignored. I just don't like explanations that appear to have errors in them. I pointed out what I thought they were.

    So, even though I largely agree with several of your conclusions, I wasn't comfortable with how you got there. No biggie.:)
     
  19. 420benz

    420benz Member

    635
    1
    Dec 19, 2006
    Georgia
    All i would like to know is if post #35 is a feasible solution.
     
  20. TomCat

    TomCat Broadcast Engineer

    4,153
    100
    Aug 31, 2002
    Maybe so. But this is not the Science BB at MIT where certain precepts are expected, its the internet, where freewheeling ideas are discussed regularly. If no one went off topic on a forum, forums would be pretty uninteresting. And I don't really think you are the heir apparent to any right to throw the first stone in that area, either. Remember, I am the one who DID NOT post about resistive loss until pressed, BY YOU, because I considered it beside the point. Go figure. And I would never excoriate you for going ever so slightly OT to make that analogy.

    Two hunks of cable from the same reel are "two entirely different transmission lines" as well, so your hyperbole is wasted.

    The mechanism may be different, but the physics and principles are exactly the same. Coax is certainly closer to waveguide theory than it is to the simple AC circuitry we studied in Electronics 101. And that is how it is approached and understood, or at least that is how it was approached when I studied to be a cable plant system designer, something I did for a number of years back in the day for TCI and Group W.

    Shocker. What would be do without you? In case you missed it, there is no DC in RF signal transmission, so I don't understand why that is even on the table.

    I agree with me. The only one I disagree with is you. And that is all I agree to. Voltage drop IMPLIES resistance, and there is resistance in transmission lines of all kinds. If there is a reflection, that means less of the power is presented to the load, and if power drops due to that, voltage drops, and the amount of voltage drop across the load changes. Unless we have repealed Ohm's law, how can it not? So of course it's changed by a reflection.

    I am the last guy to disagree with that, pretty obviously. I just still don't accept that there were errors there. If I thought there were, I would certainly say "Hmm, you know, you're right. I never thought about it that way." I have a long history of doing exactly that right here on this forum. You may not realize that because it really doesn't play out that way very often. But I always admit when I am wrong (I was wrong about the HR2x, thinking it would never amount to anything and would never surpass Tivo, and I came clean on that one and many more).

    So you probably won't be hearing anything like that from me any time soon. I consider your arguments to be completely without merit.
     

Share This Page