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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?
 

neon

Oct 21, 2006
1,325
Joined
Oct 21, 2006
Messages
1,325
coaxial cables are specifid as 50,75 300 ohms impedance not resistance so to get maximumn power transfer the load is xx. because of the frequency involved if THE CABLE is not cut at the right lenght guess what 0 inpedance.the scope have 50 ohm inpedance but in actuality the inpedance is 11megaohms which is allmost standard for practically all good meters and T/E
 
M

Michael A. Terrell

Jan 1, 1970
0
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?


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.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
A

AJ

Jan 1, 1970
0
Right, however it is recommended you read and understand the difference
between resistance, capacitance, and inductance. Each of these variables
will affect the overall impedence of a circuit and the circuits overall
effeciency. Having a missmatch in overall impedence can, and has, rendered
many circuit designs inoperable.
Find a good tutorial on electronics and have at it. Don't be afraid of
asking questions, others will respond, some in a helpful manner, some with
raw datum, others with theory, and yet others with left field opinions. 73's
 
J

John Fields

Jan 1, 1970
0
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?

---
A 1/4 wave whip exhibits about a 50 ohm resistive impedance at
resonance so, in order to keep from having reflections in the
transmission line, the transmission line's impedance is made equal
to the load (antenna) impedance and the source impedance is made
equal to the load impedance as well.

That also guarantees that with the system "matched", maximum power
will be transferred to the antenna from the source.
 
M

Michael A. Terrell

Jan 1, 1970
0
AJ said:
Right, however it is recommended you read and understand the difference
between resistance, capacitance, and inductance. Each of these variables
will affect the overall impedence of a circuit and the circuits overall
effeciency. Having a missmatch in overall impedence can, and has, rendered
many circuit designs inoperable.
Find a good tutorial on electronics and have at it. Don't be afraid of
asking questions, others will respond, some in a helpful manner, some with
raw datum, others with theory, and yet others with left field opinions. 73's


Who are you replying to?


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
B

Bob Pownall

Jan 1, 1970
0
Michael said:
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.

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
 
B

Bob Pownall

Jan 1, 1970
0
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
 
M

Michael A. Terrell

Jan 1, 1970
0
Bob said:
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


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. ;-)


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
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?

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:-
http://www.setileague.org/askdr/imped.htm

Robin
 
C

Charlie Siegrist

Jan 1, 1970
0
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?

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.
 
C

Chuck

Jan 1, 1970
0
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:-
http://www.setileague.org/askdr/imped.htm

Robin

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
90. 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
 
J

John Fields

Jan 1, 1970
0
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.

---
Not true with output transformerless amplifiers, since the amplifier
looks like a voltage source with an output impedance on the order of
milliohms.

What's important is to not try to take more than the rated current
from the amp by having it try to drive an impedance lower than it's
rated to drive.

For example, take a 100 watt amp driving an 8 ohm load to 100 watts.

Since:


P = I²R,


the amp will have to deliver:


P 100W
I = sqrt --- = sqrt ------ ~ 3.5 amperes
R 8R


Similarly, trying to drive a 4 ohm load to 100 watts will require:


P 100W
I = sqrt --- = sqrt ------ = 5 amperes
R 4R


which will more than likely overload the amp.
 

neon

Oct 21, 2006
1,325
Joined
Oct 21, 2006
Messages
1,325
in actuality a 50 ohms coax is 47 ohms and in tv cableing is 59 ohms. but wecall it 50 ohms.
 
B

Bob Myers

Jan 1, 1970
0
Chuck said:
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
90. 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.

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.
 
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.

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
 
P

Phil Allison

Jan 1, 1970
0
<robin.pain in the [email protected]
"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..."



** 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
 
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.

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


Robin
 
P

Phil Allison

Jan 1, 1970
0
Apart from, for instance, the permittivity of free space.


** So context snipping and false quoting is your puerile game.

Go **** yourself - imbecile.




....... Phil
 
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