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LF: Re: VLF - vertical probe measurements

To: <[email protected]>
Subject: LF: Re: VLF - vertical probe measurements
From: "James Moritz" <[email protected]>
Date: Fri, 15 Apr 2011 14:18:50 +0100
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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" <[email protected]>
To: "LF Group" <[email protected]>
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?



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