50 ohm load

I'm studying op-amp circuit combined with rf mixer. While I was reading articles on the circuit,I found that the term like "50 ohm load" or "50 ohm ( )" occurs frequently. I also remember that one of the input options of Digital oscilloscope was 50 ohm load.

Because I'm a newcomer to analog electronics, I don't know significance of that term.

Why should we think about the specific value 50 ohm? How do we get such value?

Reply to
sperelat
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RF circuits are designed for 50 ohms, because the coaxial cable (used most of the time) has a 50 ohm characteristic impedance. The exception is TV antennas, cable TV and sat TV equipment.

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Reply to
Michael A. Terrell

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A 1/4 wave whip exhibits about a 50 ohm resistive impedance at
resonance so, in order to keep from having reflections in the
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Reply to
John Fields

Who are you replying to?

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Reply to
Michael A. Terrell

Actually, a coax cable can have pretty much any value for characteristic impedance, it's just that the most common values are 50 ohms and 75 ohms. (The characteristic impedance of a coax line is proportional to the natural log of the outer radius of the coax to the inner radius. See any undergraduate physics or EE text on electromagnetics. If you want to change the characteristic impedance of a coax, just change the ratio, assuming you're keeping the coax's dielectric material unchanged.)

"It can be shown" that 75 ohm coax gives minimum loss, and 50 ohm coax is a good compromise between minimum loss and maximum power handling capability. See, for example, Thomas Lee, "The Design of CMOS Radio-Frequency Integrated Circuits", 2nd edition, pp. 229 - 231.

Bob Pownall

Reply to
Bob Pownall

Sorry to follow up my own post, but I just realized I owe Michael an apology.

My first reading of his post made me think he was saying all coax had 50 ohm impedance. A second, more careful, reading caused me to realize he was just referring to the coax cable used most of the time for RF purposes.

The rest of my post still stands, of course.

Bob Pownall

Reply to
Bob Pownall

Thank you. The largest 50 ohm coax I've worked with was 6" copper and brass hardline on a 25 KW UHF TV transmitter I moved. ;-)

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Reply to
Michael A. Terrell

Because it is conveniently close to "the impedance of free space" - a universal constant.

To get as much energy into and out of an aerial, it needs to closely match "air" which is almost the same as a vacuum i.e. the impedance of free space.

So the 50 ohm value is a sort of "RF thing" (as opposed to "audio thing").

See this rather good link about it:-

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Robin

Reply to
robin.pain

On Tue, 24 Jul 2007 01:36:13 -0700, in message , snipped-for-privacy@hanmail.net scribed:

I'll add to the already excellent answers, and we'll assume 50 ohms as the Value of the Gods. Transmitters and other signal sources will have an output impedance of 50 ohms. It is expected that they will drive a 50 ohm load, as, after all, that is the Value of the Gods. As John Fields hinted, when the source impedance matches the load impedance, maximum power is transferred. There is an engineering theorem addressing that phenomenon, and is called, appropriately, the Maximum Power Transfer theorem.

Another topic that was hinted at was "reflections." A reflection is a return of the transmitted signal back into the transmitter. If a reflection happens, it indicates that the power transfer is not perfect, and hence that source and load impedances are not perfectly matched. Reflections can be measured and compensated for.

Here's an interesting phenomenon, and one that bears learning: if you disconnect the 50 ohm load and drive an RF or audio source into practical infinity, the observed voltage at the output will double. That might seem like a good thing on the surface, but it's impractical and potentially damaging to the source. One thing that happens with such a driven source is that reflections are maximized (this in fact is one part of a reference test for a transmission line and antenna - driving an open line, and driving a shorted line). When reflections are maximized, all sorts of harmonic distortion can be introduced on the waveform.

When testing RF transmitters, a "dummy load" is always attached to the output if there is no antenna path into free space, in order to prevent reflections and possible damage to the transmitter.

The concept of matching source to load applies at lower audio frequencies as well, such as when fitting loudspeakers to a stereo system. The best arrangement is to match source and load as precisely as possible. The common example is 4-ohm auto systems vs. 8-ohm home systems - ideally, one shouldn't mix components between the two.

Reply to
Charlie Siegrist

Well, a couple of additional thoughts.

50 ohms is not the impedance of free space! More like 377 ohms.

The answers given tend to focus on why 50 ohms, rather than, say, 40 or

  1. An obvious approach for experienced techs. But the OP may have been wondering why not 10,000 ohms, for example? And what does coax have to do with it? No really short answers to those questions.

Robin, what did you have in mind?

Chuck

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Reply to
Chuck

Sorry Robin, I mean to address that question to the OP.

Chuck

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Reply to
Chuck

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Not true with output transformerless amplifiers, since the amplifier
looks like a voltage source with an output impedance on the order of
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Reply to
John Fields

Right - "free space" had nothing to do with this. But the choice of standard transmission line impedances wasn't arbitrary, either.

It came, as might be expected, from what was required in RF communications work, and specifically what was needed to match standard antennas. The feedpoint impedance of a half-wave dipole (which is among the most basic antenna designs, and the basis for many more complex practical designs) in free space is around

73 ohms. Hence the common 72-75 ohm standards for many coax and twinlead feedlines, esp. those still common in video systems today (which, of course, were built around an existing standard for convenience's sake). The feedpoint impedance of a quarter-wave antenna (e.g., the typical "ground plane vertical") is 37 ohms, but this assumes an ideal ground plane, etc. - in practice, the feedpoint presents an impedance somewhat higher than this. 50-ohm line (or 52 ohm, which was the earlier version of this standard value) will drive most antennas reasonably well, and also is very close to the geometric mean of these two standard feedpoint values:

SQRT (37 x 73) = 52 ohms

...and you'll find that this value turns up a lot in transmission line work when creating "matching sections."

Once these standards had been established for RF/antenna work, there was little sense in coming up with different values for other uses (e.g., oscilloscope inputs, etc.) - might as well just leverage the existing standards.

Bob M.

Reply to
Bob Myers

You have just contradicted yourself. In line one you say "Free space had nothing to do with this" and then in lines 9~10 you say "...in free space is around 73 ohms..."

Robin

Reply to
robin.pain

** As the two comments are on different matters - there is no contradiction.

The impedance of " free space " ( calculated to be 377 ohms ) applies where there is no medium carrying the EM energy - ie out in space.

But a dipole antenna " in free space " is simply one with nothing near or attached to it that would alter its operation or characteristic impedance at it natural resonant frequency.

Do try to follow the *context* when reading posts from folk one hell of a lot smarter than you.

....... Phil

Reply to
Phil Allison

Apart from, for instance, the permittivity of free space.

Robin

Reply to
robin.pain
** So context snipping and false quoting is your puerile game.

Go f*ck yourself - imbecile.

...... Phil

Reply to
Phil Allison

Well, to begin with, I thought (possibly) the same as you i.e. Enull (or maybe it's Enought but Enull is shorter) is the constant defining (eventually) e.g. 73 ohms.

Then I found the above link and realised from Dr. Seti (demonstrably a better man than me) that the antenna must be matched to Zfs (Z of free space).

He continues with neat definition of Zfs that depends on Mu-null and Enull.

So it seemed to me that Zfs is the shortest (complete) way of expressing the constant that defines, eventually, e.g. 73 ohms (of centre loaded half-wave dipole).

Robin

Reply to
robin.pain

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