In some notes on transmitters I have found the statement that "inductive coupling between transmitter amplifier stages reduces the generation of harmonics", with no further explanation. Can anyone please explain why this is so, or convincingly refute it if it isn't?

  • $\begingroup$ Can you cite such a source please? $\endgroup$ – Phil Frost - W8II Oct 4 '16 at 16:56
  • $\begingroup$ I would rather not reveal the source for reasons of privacy of the author. I don't think it is relevant, the assertion, if correct, should be part of transmitter theory whoever said it. $\endgroup$ – Harry Weston Oct 4 '16 at 17:56
  • 1
    $\begingroup$ "Inductive coupling" could mean a lot of things. I can think of ways to reduce harmonics, as well as ways to create them with inductive coupling. So without any context it's a bit of a guess what the author was getting at. $\endgroup$ – Phil Frost - W8II Oct 4 '16 at 20:56
  • $\begingroup$ Thanks for your comments Phil Frost, but there is no other context, it is in a list of methods of reducing harmonics. It is this lack of any explanation that prompted this question. $\endgroup$ – Harry Weston Oct 4 '16 at 21:52

Absent any other context, I have a couple guesses.

A ferrite core in a transformer has losses that increase with frequency, so consequently the higher frequency energy in the harmonics is dissipated as heat in the transformer. Provided of course that the core is not saturated to the point where its non-linear properties introduce more harmonics than are removed.

Additionally, any real inductor, of any core material, or a transformer winding, has a parasitic capacitance due to the proximity of the windings. The inductance and this capacitance define a self-resonant frequency for the inductor, and above this frequency the inductor presents a capacitive reactance instead of an inductive one. If the transformer is designed such that the harmonics are above the self-resonant frequency, those harmonics will then see a shunt capacitance to ground, effectively low-pass filtering them.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.