Feeling sorry for those poor souls who moved out of the wrong side of the track neighborhoods and left their towers and antenna farms to go live in gated communities where they are trying to figure o
how much loss are you willing to tolerate. A kilowatt into 50 ohms is 1000=I^2*50 -> I = sqrt(1000/50)=sqrt(20), about 4.5 Amps (RMS). 30 ga wire is 0.1 ohm/ft, a 40m dipole is 66 ft of wire (althoug
Chas, A little math shows that 5.5 Amps rms into 50 ohms is 1.5 kW. My handbook says the DC "fusing current" of #26 copper is 20 Amps. RF skin effects increase the resistance (perhaps by 5X at 10 MHz
Good point (although I don't think the dissipation goes up as the square of resistance) Handy skin depth calculator: http://www.microwaves101.com/encyclopedia/calsdepth.cfm AWG 30 is 10 mils in diame
P=I^2 R 73 Roger (K8RI) _______________________________________________ _______________________________________________ TowerTalk mailing list TowerTalk@contesting.com http://lists.contesting.com/mai
You're right, of course. But I was only off by one in the exponent... :-( -mse -- Martin Ewing, AA6E Branford, CT _______________________________________________ _____________________________________
Nope - the current will be much higher in an electrically short antenna. What matters is not its high reactance - that will get tuned out by the matching circuitry - but the low radiation resistance
I'm not buying that! Current flow depends on impedance, R+jX, not radiation resistance. Your short whip runs at high voltage and low current. There is lots of current in the matching network, but lit
Can any of you guys say the conclusion in plain English. I am new at this and still learning electrical theory. So... good idea or not? feasible or not? Call the fire dept or not? I am studying for t
..."I'm not buying that! Current flow depends on impedance, R+jX, not radiation resistance."..... Ah, but it's confusing. For a short vertical that is not tuned to resonance, if the source is a volta
When I said a short vertical, I meant electrically short, i.e. << one wavelength (and also << 1/4 wavelength). Such a monopole above a ground plane always presents a high impedance, and it's mainly c
Martin wrote: ..."When I said a short vertical, I meant electrically short, i.e. << one wavelength (and also << 1/4 wavelength). Such a monopole above a ground plane always presents a high impedance,
Martin, with respect this is completely wrong. Take a simple example of a shortened dipole that's only a quarter-wavelength overall length. It has a feedpoint impedance of about 13-j750. If this ante
To further illustrate what is happening here, I took my "quarter-wave" dipole example, matched it with a simple L network at 30 MHz (series 4.1uH inductor and shunt 180pF capacitor) and looked at the
Dave, I don't know of any programme that will calculate the voltages and currents - I just did it the "hard way" with Smith Chart and a pocket calculator :) If you simply want to see what sort of com
Apologies - that was the local link on my PC! The Tuner simulator is at: http://fermi.la.asu.edu/w9cf/tuner/tuner.html 73, Steve G3TXQ _______________________________________________ ________________
G3TXQ wrote: I don't know of any programme that will calculate the voltages and currents - I just did it the "hard way" with Smith Chart and a pocket calculator EZNEC will give it to you. To get the
Taking your 13-j750 as the feedpoint impedance (including loss resistance within the antenna), if you're actually dissipating 1kW (either by radiation or IR loss), the inphase component of the curren
Jim, Nope! [13-j750] is the equivalent SERIES representation of the feedpoint impedance, so the same 8.7 Amps flows through the resistive part AND the reactive part; there is no 500 Amps flowing anyw
Bud, There's no mystery - the other 860 Watts gets dissipated in the 2.4 Ohms loading coil loss resistance and the 4 Ohms Ground loss assumed by HI-Q. The maths all hangs together: Efficiency = Rrad/