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I want to match RF power amp to aerial. There is no 50 ohm or 75 ohm transmission line inbetween.

So, I consider my aim to be matching certain load-reflection coefficient (Γload) to a short circuit (no transmission line).

Can I do it using a Smith chart? I'd appreciate an advice.

UPDATE

I guess I do not need to do any matching here whatsoever. What I need is to calculate load-reflection coefficient, mount proper components and then connect an antenna. Am I right?

RF circuit

ivan
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  • Use a "Dummy Load" on the RF Amp output, that matches the antennas designed impedance. Once matched to the "Dummy Load", connect to the antenna for the final fine trimming. – Optionparty Jan 21 '14 at 19:58
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    What is up with the close votes? This is a perfectly valid question if you took a basic RF course. – Matt Young Jan 21 '14 at 20:17
  • What is between the matching network and antenna? – Matt Young Jan 21 '14 at 20:59
  • A short wire, l< – ivan Jan 21 '14 at 21:02
  • There is no such thing as a 'direct connection' when talking about an RF current path. Any signal path carrying a current has a characteristic impedance, it is only the length of this compared to the frequency that matters. If the impedance does not match both the source and load you get mismatch and power loss. It certainly is not Zo=0 ohms which would be a short to GND producing a 100% anti-phase reflected current – Jay M Jan 21 '14 at 22:53
  • Sorry, must be R=0 ohm, not Z=0 ohm. What I meant was mounting proper components to create desired load-reflection coefficient and then connecting the antenna. – ivan Jan 21 '14 at 23:03
  • @ivan what do you think the desired load reflection coefficient might be? Not 0? – Phil Frost Jan 21 '14 at 23:36
  • According to RF amp calculations Γload = 0.248e^(14.5 deg) – ivan Jan 21 '14 at 23:42
  • @ivan by what calculation did you determine that is the *desired* value? – Phil Frost Jan 21 '14 at 23:47
  • @PhilFrost I use 'RF Circuit Design' by C. Bowick(1982). There is an algorithm for RF amp design. I got S-parameters from transistor datasheet and used equations from 6-14 to 6-21. – ivan Jan 21 '14 at 23:59

1 Answers1

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If your antenna and device would work in the ordinary case with a transmission line, it will work without a transmission line, with the antenna connected directly to the device also.

The reason is simple, and could be explained a number of ways. Here's one: does it work if the transmission line is 100 feet long? 50 feet? 5 feet? 5 inches? 0.5 inches? Does it just suddenly stop working at some length? I think not.

Here's another: the whole reason we design our output stages to drive a 50 ohm load, and use 50 ohm transmission line, and make the antenna feedpoint impedance 50 ohms, is so that the transmission line is irrelevant to the output stage. Energy is emitted from the transmitter, travels down the feedline, and is delivered to the load, all without any reflections, regardless of the transmission line length. If nothing ever reflects back, how can the output stage ever know anything about the transmission line? As far as the output stage can tell, it's driving a 50Ω resistor, or a 50Ω transmission line on infinite length. It can't tell the difference. So if the transmission line is gone, but you never knew it was there, does it matter?

You also seem to have some misunderstanding about what a reflection coefficient is, and why we care. The reflection coefficient describes what fraction of the voltage is reflected back at the source when a wave encounters an impedance discontinuity. We engineer everything to have the same impedance so that there are no such reflections, and the reflection coefficient is 0. We do not match devices to a reflection coefficient: we match device impedances so that we can make the reflection coefficient zero.

Phil Frost
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