0
$\begingroup$

I am IT person not Ham operator and I guess it is difficult for you to understand me and my different thinking so this is a very simple question that will probably answer the dilemma at Coaxial cable loss, dbm, db correlations as nobody is able to understand the puzzles that I'm having.

So Imagine that you put 40 dBm (10 Watts output) amplifier to boost 2Ghz signal and hope to travel good distance.

This cable is having 30dB attenuation at 100 meters at that frequency. As I understood 30dB attenuation is reducing the original power 1000 times. Ok, ok.

But boy I am feeding the line with 10Watts right ? That could be possibly translated as 10Volts X 1 Amp !

Instead of antenna at the distant end (at 100 meters) if I put a light bulb by soldering the copper core for + and shielding for - will the 8 volts rated bulb (drawing current of 100mA) will it light or not ? By the way the electrical properties for the cable are 3 ohm resistance per 100 meters.

What's your answer ?

For

Thanks

$\endgroup$
1
  • $\begingroup$ Are you asking about feeding one end of a 100m transmission line with 10V DC? or with 10W 2GHz?. It matters: DC will give some light, RF very little if any. $\endgroup$
    – glen_geek
    Jun 18 at 14:29

2 Answers 2

2
$\begingroup$

If we assume from "8 volts rated bulb (drawing 100mA)" that your bulb is meant to be an 80-ohm equivalent load, and it's at the end of 100 meters of coax with a loss of 30dB/100m, then:

  • The RMS voltage across the light bulb will be 0.896 V.
  • The RMS current through the light bulb will be 11.2mA.
  • The power delivered to the light bulb will be 10mW (10dBm).
  • The power dissipated as heat in the coax will be 9.99W.

You tell me whether 10mW means the light bulb "lights" or not.

Side note, not really relevant to your question: usually we would want to specify the characteristic impedance of the coax (50 ohms, 75 ohms, or rarely something else) and consider the impedance mismatch between that and the bulb, but in this case it doesn't really matter as long as the coax is matched to the amplifier. Because 99.9% of the power is dissipated in the coax, the "light bulb" makes almost no difference to the SWR seen by the amplifier — we can make it a dead short, a 1000-ohm resistance, capacitive, inductive, whatever we want, and the SWR at the amplifier would still be no worse than 1.003:1.

$\endgroup$
3
  • $\begingroup$ In a previous comment of yours, I concluded the following: A DC at 0mhz will reach the bulb without a problem and charge it, it will light at 100m. A DC at 2Ghz will reach only 10mW and it won't light. This is in contrary with this thread where they say ohms law don't care much about the DC frequency in almost all cases (if I read it right). physics.stackexchange.com/questions/320159/… By the way thank you for your help so far! $\endgroup$
    – Svetoslav
    Jun 18 at 10:06
  • $\begingroup$ Please look at the attached image. So DC @ 0mhz says only 6% loss (even for 10V bulb at 100mA) while at DC @ 2Ghz 99% loss? $\endgroup$
    – Svetoslav
    Jun 18 at 10:27
  • 1
    $\begingroup$ @Svetoslav there is no such thing as "DC @ 2GHz". 0MHz is the only kind of DC. But yes, the loss will be much much lower at DC. At DC the total resistance will be 80+3 = 83 ohms. If you put in 10 watts (28.8V @ 347mA), 9.64W would go to the bulb and 0.36W would be lost in the cable. $\endgroup$ Jun 18 at 19:45
2
$\begingroup$

Short answer is no, the bulb will not glow. The maximum available 10mW at the end of the wire is not enough to make a 0.8W bulb glow.

The impedance of the lightbulb will matter. The maximum available power at the end of the wire is 10mW, but that is only available for load with an impedance that matches the wire. If the impedance of the load is not ideal then most of the available power will be reflected back into the wire and the load will receive even less than the 10mW that is available.

The stated lightbulb has 80 ohms of resistance at DC. Between skin effect and inductance, it is likely to have much higher impedance at 2GHz. This means it will have a much higher impedance than most coax, and so will receive only a small amount of available power.

You might even find that because of the impedance of the lightbulb, it doesn’t light up even when connected directly to the output of your amplifier! (Depending on the exact geometry of the bulb, and how you make the connection)

$\endgroup$
1
  • $\begingroup$ Thank you for your response on this. I appreciate it ! $\endgroup$
    – Svetoslav
    Jun 18 at 10:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .