Sandeep wrote:
> Guys, any way can we know voltages approx across 20M traps ?
> excited with 25 watt...
I'll assume you have a 40m dipole, with 20m traps halfway down each leg?
to prevent 20m energy from passing, so it is effectively a shorter dipole on 20m.
When operating on 20m, those traps are at the ends of the 20m dipole,
those ends have the same 3000 ohm impedance that we would have at?
the feedpoint of an end fed halfwave.? Actual impedance will vary depending on
trap construction, how close the wires are cut for resonance, nearby objects,?
height from ground, ...? ? But I'd expect this to be correct within a factor of two.
Power is Volts*Amps, or Volts*(Volts/Ohms).
25watts = Volts * Volts / 3000ohms,? ?so Volts = square_root(25 * 3000) = 274 Volts rms of 20m RF
Jerry, KE7ER
On Sat, May 18, 2019 at 03:13 PM, Sandeep Lohia wrote:
On Mon 15 Apr, 2019, 8:42 PM Jerry Gaffke via Groups.Io, <jgaffke=
[email protected]> wrote:
The transformer of an EFHW? matching network matches the 50 ohm transmitter
to an expected impedance into the wire of between 2500 and 3000 ohms.
Power is Volts*Amps, or Volts*(Volts/Ohms).
At 5 Watts of RF power,? ? 5watts = Volts*Volts/3000ohms,? ?therefore? Volts = sqrt(3000*5) = 122 Volts RMS.
At 100 Watts, it's still around 548 Volts RMS, not yet kilovolts
?
Thanks Jerry, was helpful...
& here's online calculator too :
?
?
Guys, any way can we know voltages approx across 20M traps ?
excited with 25 watt...
?
