1

For RF applications, we are usually considering of impedance matching to choose load impedance as the conjugate of the source input impedance for the maximum transferred power. But how about design the source having zero input impedance? Is it even possible or good? A naive thought is that all the power will be on the load if so. I might think it is true for DC case and for amplifier design to have a small output impedance and big input impedance. I lack the knowledge to find the missed chain to put all the knowledge together. How do you think? Thanks!

george andrew
  • 129
  • if the source has zero impedance then any reflection from the load will be completely reflected back again creating nasty standing waves on feed line – hyportnex Nov 27 '17 at 19:12
  • "But how about design the source having zero input impedance?" - How? How do you propose to eliminate all parasitic inductance and/or capacitance? – Alfred Centauri Nov 27 '17 at 19:23
  • There is an amateur radio stack exchange too, where it might be good to post this. From a practical standpoint: the centerpoint impedance of a half-wave dipole (~50 ohms) is easily matched to either a tube or transistor-powered RF amplifier circuit; designing a very low impedance RF amp would require designing a very low impedance antenna. Physics conspires against you in this case (see the amateur radio SE for why). – niels nielsen Nov 27 '17 at 19:26
  • Thank you guys, but anyone give me some equations? – george andrew Nov 27 '17 at 19:29
  • @Alfred_Centauri Since high power RF amplifiers are always made by parallel connection of many (very many) transistors (except the ones using GaN ) the natural output impedance is indeed very low and it takes a lot of effort to match it to $50\Omega$. One of the main source of inefficiency is the matching circuit itself, still in practice it is never a good idea to drive the antenna from a low impedance directly because the uncertainties of the actual load. The reactive mismatch of the source can be resonated out (narrow band) but the resistive one still needs a transformer. – hyportnex Nov 27 '17 at 19:42
  • Yes it can be done (approximately). For example, in ECL logic systems, it's very common to use matched termination on the load and low-impedance (~5 ohm) termination at the source. – The Photon Nov 27 '17 at 19:42
  • @hyportnex Can I consider case1: AC voltage source 1v, input impedance 50ohms, load 50ohms, so the circuit is matched. case2: AC voltage source 1v, input impedance zero ohm, load 50ohms and 50ohms serially connected. Does the voltage source hurt in this case? – george andrew Nov 27 '17 at 19:44
  • no it does not, but the input impedance of a transmission line off which there is an antenna hanging is uncontrolled in practice, and the higher the frequency is the less you know about that actual input impedance. Not only the length is often variable but the neighborhood of the antenna will also influence its radiation impedance. – hyportnex Nov 27 '17 at 19:52
  • 1
    @hyportnex, very low is not zero. – Alfred Centauri Nov 27 '17 at 20:03
  • @hyportnex, thank you for the answers. I understand now that to select 50ohm mainly is for practical easiness. But 1. why not matching to a low impedance value? using transmission line or other lumped element, basically we can get any wanted value for the load side, right? 2. I really don't understand the case1 and case2 I mentioned, so for case 2, do we still have the stand wave problem or we don't? – george andrew Nov 27 '17 at 20:05
  • as I tried to say above practical power amplifiers do employ impedance matching to low value because their output impedance is indeed very small, about $1$to$5\Omega$ or even less. That matching circuit is there to raise the output impedance of the amplifier to that of the transmission line that it is to drive. Again, if the amplifier is matched then any reflection from the line/load it drives will be absorbed by the source impedance. Note too that an RF transistor is unlike a low frequency one e.g., it likes to drive a very specific load impedance to be able to deliver maximum power! – hyportnex Nov 27 '17 at 20:27
  • Thank you very much@hyportnex. I like your comments. I think I have an answer to my confusion now. I hope to vote for your comments, but I didn't see where I can click the up-triangle. Also thank @ niels nielsen and @ The Photon. All of your helps make the question more clear, from practical point of view and from theory. – george andrew Nov 28 '17 at 15:02
  • Possible duplicate of Condition for Maximum Power in the circuit – sammy gerbil Dec 01 '17 at 14:45

1 Answers1

1

The aim of impedance matching is (usually) to achieve the maximum transfer of power to the load. This is often found to be not an intuitive solution.

A simplistic view is that in a closed system with a fixed voltage and a fixed total impedance, the power is dissipated/radiated/applied in both the supply and the load.

There is a limiting factor in how much current the supply can deliver, based on its internal impedance. If the load impedance is zero then all of the power is dissipated in the supply because P=I2R and the only R is in the supply.

As the load impedance is increased, the total current will drop but the power dissipated at the load increases, because V stays constant and so I decreases (inversely) proportionally as R increases. If you were to graph this it would show the increasing power at the load, and what happens as the impedance of the load becomes higher than the impedance of the supply, and that the "sweet spot" is when the load impedance matches the supply impedance.

Source and load circuit (By Cjp24 - Own work based on: Source and load circuit.png  by Omegatron, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=57112455) Maximum power transfer (By Inductiveload - Self Made in Inkscape and Mathematica, Public Domain, https://commons.wikimedia.org/w/index.php?curid=2081927)

In this, it looks like having zero impedance in the source would mean all the power is delivered to the load. In a DC circuit nobody cares much (joke) :-) but otherwise what happens is that only the load dissipates power, but not all the power will be transferred to the load.

The reflection coefficient also needs to be considered, which is calculated by Z2 - Z1 / (Z2 + Z1). When Z2 = Z1 then the reflection coefficient is zero and all the power is transferred. You could also take the transmission coefficient as 1 - reflection coefficient. But when either the source or the load is zero then all the power is reflected and doesn't get dissipated in the load.

And not to forget, but only the resistive component of the impedance dissipates power (not the reactive component) but the reactive component contributes to the impedance and is frequency-dependent.

** edited to expand answer

Mick
  • 955
  • 8
  • 15
  • I understand this very well. Instead, if you look into the discussion about what I was confused, you will see I am interested in something else.But thank you very much! – george andrew Nov 28 '17 at 14:56
  • My bad, I was distracted halfway through – Mick Nov 29 '17 at 01:02