Re: Androcles' logic (was Bilge's logic)
From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 07/21/04
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Date: Wed, 21 Jul 2004 23:38:55 GMT
"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
news:HNDFWjAtjs$AFwi8@kennaugh2435hex.freeserve.co.uk...
| Androcles writes
| >| Paul:
| >| | Do you know what is meant by "a 50 ohm coax",
| >| | Androcles? :-)
| >| | No?
| >| | We mean that the characteristic impedance of
| >| | the coaxial cable in question is 50 ohm.
| >| | Not 50 ohm capacitive impedance.
| >| | Not 50 ohm inductive impedance.
| >| | 50 ohm RESISTIVE impedance.
| >|
| >| Androcles replied:
| >| | I wonder how hot it gets when you apply 250 volts at 50 Hz
| >| | across that 50 ohm resistance?
| >| | Let's see... thats err... 5 amps? I^2R = 25*50 = 1250 watts?
| >| | Ought to be enough to melt the insulator, don't you think?
| >| | ROFLMAO!
| >|
| >| Beyond blunder! He has gone crazy! :-)
| >
| >Try it, then. Plug your coax into a house socket.
| >I'll stand well back.
|
| Oh dear you will keep digging a deeper hole won't you?
|
| I am not interested in your petty squabbles with Paul but I am
| interested in trying to get things right.
He continues to raise the point. Why, I have no idea.
|
| In electronics a resistor is something which makes electrical power
| disappear. It becomes some other sort of power e.g. heat, so is no
| longer electrical power. Thus an aerial for example, if properly
| designed, will look resistive in that electrical power goes into it and
| disappears from the electrical circuit to become radio energy. A massive
| amount of power may be going into an aerial, i.e. into a resistive load,
| and the aerial doesn't even get warm. If you charge a battery it looks
| resistive. I recharge AA cells at 1.5A. They don't get hot unless you
| overcharge them because the power which is disappearing is going into
| chemical energy not heat. So the first thing you have to get straight is
| that something other than a resistor can look resistive and power into a
| resistive load does not always result in heat only as a loss of energy
| from the electrical circuit.
That's what I'm trying to tell Andersen.
|
| If you have a long piece of coax (ideal, loss-less) and you start to
| feed power into it (dc, 50Hz or 1000GHz makes no difference if it is
| ideal) then initially it will look resistive because the power is
| disappearing into it. It travels along it at about 0.6c. If it is open
| circuit or short circuit at the other end then when that power gets
| there it will be reflected and travel back towards the source. It is
| only when that power reaches the source that it ceases to look
| resistive. Power is no longer disappearing.
|
| However if instead of a short circuit or an open circuit it is
| terminated at the other end with a resistor equal to the characteristic
| impedance of the coax (matched) the power will not be reflected but be
| dissipated in that resistor.
The point you are missing is that part about "characteristic impedance".
It should be obvious that if I put 250V DC at one end of a coax and a
50 ohm resistor at the other, 5 Amps DC will be drawn. That's 1.25 kW.
Any real coax will have SOME resistance and so there will be heat dissipated
in it.
I can attach a simple reversing switch and toggle it by hand at 1 Hz, so now
there is AC in the coax, square wave. If I'm quick enough, I might be able
to toggle at 7 Hz.
I have a long way to go before I get to VHF, and 50 Hz isn't that high
either.
Therefore the coax will cook. However, it will not cook at VHF, it's
characteristic impedance at that frequency.
Now, if you or Andersen want to argue with that, do the experiment.
|
| Matched as in perfect power factor. Matched as in maximum power transfer
| theorem, equals no power reflected.
|
| As far as the send end is concerned power is disappearing down the coax
| and not coming back. That is what a resistor does - makes power
| disappear. So as far as the send end is concerned it has a resistor
| across it. There is no way it can tell the difference between a 50 ohm
| resistor and a mile of 50 ohm coax terminated in a 50 ohm resistor.
Try it with DC and 50 Hz.
|
| OK There is no such thing as ideal loss-less coax but the thing which
| gives it its characteristic impedance is the ratio of its inductance per
| unit length to its capacitance per unit length. You could in theory make
| 50 ohm coax 2 ft in diameter with a 6 inch diameter copper core with as
| near zero resistance as you want to eliminate resistive loss. In
| practice you get what you pay for. The inner core has some resistance so
| a bit of power will not make it to the load - As far as the source is
| concerned it is still disappearing so it makes no difference it will
| still see 50 ohms. Loss will increase with frequency. This is for two
| reasons. Firstly because as frequency increases more of the current
| flows in the surface of the metal rather than being evenly distributed
| through it so any conductor has a higher resistance at high frequency.
| (that is why VHF coils are silver plated to give the surface a lower
| resistance) and secondly because it is difficult to keep the central
| conductor in the centre using vacuum as an insulator so you have to use
| some sort of plastic (or ceramic) and that has dielectric loss which
| increases with frequency. If you want low loss at 1000 GHz you have to
| pay rather a lot for your coax but it will still work quite happily down
| to dc.
| --
| John Kennaugh
| to email convert the number from hex to decimal
In other words the more expensive coax has a characteristic impedance at
1000 GHz
and the ordinary stuff has a characteristic impedance way down in the VHF in
UHF range, and neither is suitable for operation at the other's frequency.
Androcles
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