Coax Wire Gauge vs Watt Capacity (

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  • Опубликовано: 14 янв 2025

Комментарии • 18

  • @carolmartin7042
    @carolmartin7042 18 дней назад

    Thank you.
    A simple uncomplicated way of viewing the situation.
    Have a Happy Hanukkah and Happy New Year too. N0QFT Glen

  • @SteadArcFab_Ministries
    @SteadArcFab_Ministries 18 дней назад

    Great explanation! 👌👊

  • @timburdick6207
    @timburdick6207 18 дней назад

    Hey Dave, great video. I found that Belden 9913F7 cable works great for an antenna system. The LMR series is also a very good coax cable to use for antennas. Both of the 9913F7 and the LMR-400UF cables are the flexible type cables and work very well for an antenna that rotates. Again, thanks for a very informative video. 73’s….....

  • @marcelleuenberger4648
    @marcelleuenberger4648 18 дней назад

    The counter-intuitive fact that aircraft wiring prefers higher current per gauge to reduce weight is really interesting, it makes sense but I would not have expected it either.
    Thanks for throwing that in! :)

    • @petergrhill
      @petergrhill 18 дней назад

      Don't aircraft use 400 HZ instead of 50/60 HZ. I do know that you still get a shock, that will wake you up, from the voltage that the aircraft use.

    • @user-fr3hy9uh6y
      @user-fr3hy9uh6y 18 дней назад

      The losses in the cable determine the heat. The size of the cable bundle determines the ability of a cable in the center of the bundle to get rid of the heat. If you are designing an airplane, you also have the time to consider this along with signal integrity at the other end of the wire. For house wiring, it is just a value that always works. No calculations are required.

  • @user-fr3hy9uh6y
    @user-fr3hy9uh6y 18 дней назад +2

    I have never considered using DC theory on an RF calculation. I would have gone with cable losses, heat, compaired with the ability of the cable to dissipate the heat.

  • @garryhammond3117
    @garryhammond3117 18 дней назад

    Thanks Dave - Happy New Year! - Cheers - VE3GHP

  • @brianm1696
    @brianm1696 18 дней назад

    Does the length of the run factor into the calculation when selecting the wire gauge?

    • @user-fr3hy9uh6y
      @user-fr3hy9uh6y 18 дней назад

      Cable losses are greater for longer cables. For DC and low frequencies, like house wireing, yes. If your 12v power is only 5v when it gets to the other end of the wire, it may not work. For RF, look at the cable loss of your cable at the frequency that you are using. It will be given per a particular length. With that, you can determine the max length you can use for your application.

  • @MarcusPocus
    @MarcusPocus 18 дней назад +1

    what about coax cables in RF situation where the skin effect rise with the frequency.. most of us start with the cheap RG-58 but nobody know the Frequency vs Power limits (we only see the power loss tables vs frequencies).. also i never saw the power limit of a BNC connector often saw in the field.. anyway, thanks for this subject!!

    • @user-fr3hy9uh6y
      @user-fr3hy9uh6y 18 дней назад

      Good question, I had to look it up. For BNC, I found 3.5 amp typical, but there are some that are rated at 5 amps. The losses are dependent on frequency, and that will affect how hot they get. You don't want to melt the dielectric.

  • @NehemiahShelley-w7e
    @NehemiahShelley-w7e 18 дней назад

    Great content, as always! I have a quick question: My OKX wallet holds some USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). What's the best way to send them to Binance?

  • @ke4tmt91
    @ke4tmt91 18 дней назад +2

    I expectecd some mention of skin rffect.

  • @geirha75
    @geirha75 15 дней назад

    1500 W into Z=50ohm is about 5,5A

  • @richardl6751
    @richardl6751 17 дней назад

    1,500 watts into 50 ohms its a little less then 5.5amps.

  • @johnbarber7157
    @johnbarber7157 16 дней назад

    WHAT? find a way to dissipate the heat?
    LOL... also note rf voltages are peaks. Insulation has a lot to do with the power level. DC current is the same as rf and ac. Current is current.

  • @robertbenedict1225
    @robertbenedict1225 18 дней назад

    Hmmmm. If P/I=Z=50 then multiply both sides by 50/I to get I = P/Z = 100/50 =2. I think P = V*I or P/I = V.