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08-27-2013, 10:46 PM
 Member Join Date: Nov 2011 Posts: 239
Understanding ohm's law

I've been struggling to understand the difference between different impedance antennas and feedline. When you have, for example, a feedline with an impedance of 1000 ohms, what is different about the electricity flowing through that line than a line with 50 ohms? Is the 1000 ohm line resisting the electricity more?

Based on R=V/I, if we have a 50 ohm feedline, then that must mean that only 1 amp can flow through the line if the voltage is 50... or is that not right?

Also, why is it that feeding a 50 ohm antenna with 75 ohm feedline produces standing waves? What is happening at the feedpoint that causes the electricity to flow backwards?
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K9LTC - Landon
08-28-2013, 1:58 AM
 Member Join Date: Feb 2007 Location: Inland Empire Posts: 4,267

Quote:
 Originally Posted by AgentCOPP1 I've been struggling to understand the difference between different impedance antennas and feedline.
You'll continue to struggle if you keep trying to apply ohms law to ac circuits. When you start talking about impedance mismatches, among other things, the dc formulas go out the window. There is an ac version of ohms law, much the same, only more involved. Replace the term "resistance" with "impedance". And in written form, R is replaced with Z. Impedance includes the ohmic (dc) losses in the wires, but also includes capacitive and inductive reactance. A complex impedance would have a resistive component AND reactance. You then also become concerned with phase angles, and RMS values.

The ARRL Amateur's handbook has an excellent section on ac theory written in understandable language.

Quote:
 Originally Posted by AgentCOPP1 When you have, for example, a feedline with an impedance of 1000 ohms, what is different about the electricity flowing through that line than a line with 50 ohms? Is the 1000 ohm line resisting the electricity more?
Not in the sense of ohmic (resistive) losses. In ac circuits, transfer of power from a source to a load becomes inefficient unless the source and load impedance are the same. A weak but useful analogy might be water pipes. If you have a large water pipe, reduced to a smaller size, there will be a discontinuity of flow at that transition.

Quote:
 Originally Posted by AgentCOPP1 Based on R=V/I, if we have a 50 ohm feedline, then that must mean that only 1 amp can flow through the line if the voltage is 50... or is that not right?
In matched conditions, with a sinewave, you could almost get away with that. As soon as there's a mismatch, or complex waveforms, the relationship between voltage, current, and power becomes too complex for a simple formula to deal with.

Quote:
 Originally Posted by AgentCOPP1 Also, why is it that feeding a 50 ohm antenna with 75 ohm feedline produces standing waves? What is happening at the feedpoint that causes the electricity to flow backwards?
Going back to the second paragraph of my reply, in a mismatch condition, power transfer from a source to a load is less than 100% (minus ohmic losses) if there is an impedance mismatch. The power that is not absorbed by the load has to go somewhere, so it is, in effect, reflected back toward the source.

When dealing with things like 50 ohm antennas, and 75 ohm feedlines, the term "ohm" has a more complex meaning than, say, a 50 ohm resistor. A resistor has resistance, and in the case of a 50 ohm resistor, it has (big surprise here) 50 ohms of resistance and, theoretically, zero reactance.

An antenna with a 50 ohm "impedance" is different. You could measure it with an ohm meter and find it's either a dead short, or wide open. But put a radio signal into it of the right frequency, and the transmitter would not be able to tell the difference between the antenna, and a 50 ohm dummy load. What this means is that the inductive and capacitive reactance (resistance to an ac signal) cancel each other out, and leave an overall 50 ohm resistive effect at that frequency. That is what is meant when an antenna is at resonance, btw - the capacitive and inductive reactance are cancelled, leaving only a resistive component. Please note, that resistive component may be something other than 50 ohms, so your resonant antenna could STILL have a high vswr. Trying another frequency, and the inductive and capacitive reactance will no longer cancel each other out, and the antenna will have a resistive component and a reactive component... and standing waves on it's feedlines.

Speaking of feedlines, the 50 ohms of a 50 ohm piece of coax won't be measured with an ohm meter. That is the equivalent impedance an ac signal would see if you applied it to an infinately long piece of that coax. If the coax is less tahn infinity long, and terminated in a resistor of the same 50 ohms, then the signal source can't tell if the coax is short with a perfect load on it, or infinitely long with no load - no energy gets reflected back.

Chomp on that a bit, and get an ARRL handbook.
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Last edited by zz0468; 08-28-2013 at 2:49 AM..
08-28-2013, 8:35 AM
 Member Premium Subscriber Join Date: Dec 2004 Location: Tulsa Posts: 1,070

You've got him really confused, now break out the Smith Chart!
08-28-2013, 8:44 AM
 Member Amateur Radio Join Date: Dec 2002 Location: Wichita Falls, TX Posts: 3,713

This video explains standing waves at points in a transmission line where the impedance changes.
AT&T Archives : Similarities of Wave Behavior (Bonus Edition)
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Tom
08-28-2013, 9:05 AM
 Member Join Date: Dec 2006 Location: Oklahoma Posts: 1,941

At one time the school supply list for electronics majors included aspirin in very large count containers. Usually more than one container was recommended, usually taken with your preferred alcoholic beverage. No idea why that's still not on the list, maybe to promote student initiative?
Probably the most important 'truth' in this thread so far is that AC and DC ain't the same at all. When something starts to 'alternate', whatever affects that 'simple' non-alternating thingy get's transformed into a more complex thingy in more than one direction. (Howz that for a mystical, odd, weird way of stating how this stuff makes you feel?) Confused yet? Don't feel too bad, all of us go through the same thing. Stick with it, it really does make sense after a while... so I'm told...
- 'Doc
08-28-2013, 11:51 PM
 Member Join Date: Jul 2010 Location: Victoria B.C. Canada Posts: 400

Quote:
 Originally Posted by LtDoc Probably the most important 'truth' in this thread so far is that AC and DC ain't the same at all.
For example a piece of wire won't carry a normal DC current unless each end is connected to a terminal of opposite polarity. But it will carry an AC current if the frequency the current alternates at is sufficiently high even if there is only one terminal. And, that AC current will cause the wire to radiate electromagnetic waves.

Or take a capacitor and stick it in a DC circuit. No current will flow (though if you take the capacitor out of the circuit and grasp both ends then if the capacitance is high enough you will get a shock). But put a capacitor in an alternating current circuit and it will conduct electricity. The higher the frequency of the AC the more current will flow.

AC ain't DC. Radio waves are produced by AC.

This is an oversimplification and I've left a few things out but I hope it makes the main point.

Last edited by Ed_Seedhouse; 08-28-2013 at 11:53 PM..
08-29-2013, 12:15 AM
 Member Join Date: Feb 2007 Location: Inland Empire Posts: 4,267

Quote:
 Originally Posted by rfradioconsult You've got him really confused, now break out the Smith Chart!
Excellent idea!
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08-29-2013, 7:55 PM
 Member Join Date: Nov 2011 Posts: 239

Quote:
 Originally Posted by Ed_Seedhouse For example a piece of wire won't carry a normal DC current unless each end is connected to a terminal of opposite polarity. But it will carry an AC current if the frequency the current alternates at is sufficiently high even if there is only one terminal. And, that AC current will cause the wire to radiate electromagnetic waves. Or take a capacitor and stick it in a DC circuit. No current will flow (though if you take the capacitor out of the circuit and grasp both ends then if the capacitance is high enough you will get a shock). But put a capacitor in an alternating current circuit and it will conduct electricity. The higher the frequency of the AC the more current will flow. AC ain't DC. Radio waves are produced by AC. This is an oversimplification and I've left a few things out but I hope it makes the main point.
For some reason, your explanation really clicked with me. ZZ's as well.

So ZZ, just to clarify, a perfectly resonant antenna will look the exact same to a radio that an infinitely long piece of coax will? Also, if you put a 50 ohm resistor at the end of a 50 ohm piece of coax, that resistor will act as a dummy load?

I forgot to ask this as well. What physically makes one piece of coax have a higher impedance than another? I'm guessing that something can be done to the coax to make it just about any impedance that you want.
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K9LTC - Landon
08-29-2013, 9:00 PM
 Member Amateur Radio Join Date: Dec 2005 Location: MS Gulf Coast Posts: 1,360

Quote:
 Originally Posted by AgentCOPP1 I forgot to ask this as well. What physically makes one piece of coax have a higher impedance than another? I'm guessing that something can be done to the coax to make it just about any impedance that you want.
The dielectric constant of the inner insulator and the distance between the inner and outer conductors.
08-29-2013, 9:42 PM
 Member Join Date: Feb 2007 Location: Inland Empire Posts: 4,267

Quote:
 Originally Posted by AgentCOPP1 For some reason, your explanation really clicked with me. ZZ's as well.
I'm glad to hear that. I read, and re-read my reply a dozen times, and I'm still not entirely happy with it. But if it made sense, I'll stop dissecting it.

Quote:
 Originally Posted by AgentCOPP1 So ZZ, just to clarify, a perfectly resonant antenna will look the exact same to a radio that an infinitely long piece of coax will?
Yes. Now, to be clear, that perfectly resonant antenna might not be a perfect match at 50 ohms, but it will not have any capacitive or inductive reactance, and will appear only as a purely resistive load.

Quote:
 Originally Posted by AgentCOPP1 Also, if you put a 50 ohm resistor at the end of a 50 ohm piece of coax, that resistor will act as a dummy load?
Yes. The ideal antenna presents a resistive load with no reactance, and a dummy load does the same thing without radiating. If that resistive load is of the same impedance as the characteristic impedance of the coax and the source, the vswr is 1:1, with no standing waves, and no mismatch.

Quote:
 Originally Posted by AgentCOPP1 I forgot to ask this as well. What physically makes one piece of coax have a higher impedance than another? I'm guessing that something can be done to the coax to make it just about any impedance that you want.
As KE5TLF stated, it is the physical dimensions of the inner and outer conductors that determines the characteristic impedance. The composition of the dielectric has some impact on that as well, and also weighs very heavily on the cable's velocity factor, or the percentage of the speed of light that a signal will pass through the cable.
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09-01-2013, 9:00 PM
 Member Join Date: Dec 2002 Location: Carlsbad, CA Posts: 701

ZZ0468,

Well, so far as I could determine your explanation was beautifully done - FWIW, I can't figure out how I could have said it any better, given the intended audience and necessary simplifications required, and most likely would have flailed badly in comparison had I tried before you did. In my mind, it had a sort of "sweet musical flow and feel" to it....I know, yeah, you'd have to know me well to make sense of that comment;-)!

Ed's info was also well stated, accurate, concise and clearly illustrated the concept for the OP.

The Smith Chart is my favorite mandala!

-Mike
09-02-2013, 4:16 PM
 Member Join Date: Feb 2007 Location: Inland Empire Posts: 4,267

Thank you, Mike. I appreciate hearing that.

And yes, I agree. The Smith chart is pretty neat.
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