LF: Re: VLF - vertical probe measurements

["James Moritz" <james.moritz@btopenworld.com>]

Dear Chris, LF Group,

Assuming the antenna impedance seen by the TX stays the same (i.e. you are 
not changing output power by altering the antenna matching), the DC current 
drawn by the amplifier will be, to a first approximation, proportional to 
the output current and the output voltage. The E and H fields produced by 
the antenna are also proportional to the currents and voltages at the TX 
output. So overall, direct proportionality between Idc and E is what you 
would expect.
The apparent disparity between voltage/current/power relationships does not 
really exist - the change in DC input power relative to RF output power 
depends on the type of PA. If a class AB type fed with a constant supply 
voltage, as the output is reduced, efficiency becomes lower, i.e. a greater 
proportion of the DC input is dissipated by the amplifier. RF output power 
is approximately proportional to the square of the DC input power and the 
square of the DC current in this case. If a class D or E type, in order to 
reduce the supply current, you have to reduce supply voltage in the same 
proportion also. In this case, RF output power is nearly directly 
proportional to DC input power, the square of  DC supply current, and the 
square of DC supply voltage.
Cheers, Jim Moritz
73 de M0BMU

----- Original Message ----- 
From: "Chris Osborn" <g3xiz@yahoo.co.uk>
To: "LF Group" <rsgb_lf_group@blacksheep.org>
Sent: Friday, April 15, 2011 1:40 PM
Subject: LF: VLF - vertical probe measurements


I noticed that the probe's pick-up voltage is directly proportional to my 
PA's supply current.
Assuming that the PA efficiency doesn't change drastically with current then 
would I be correct in assuming that the probe's voltage is directly 
proportional to the radiated power?