Re: Androcles' logic (was Bilge's logic)
From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 07/23/04
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Date: Fri, 23 Jul 2004 16:06:40 GMT
"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
news:5yi6JbLVjRABFwLS@kennaugh2435hex.freeserve.co.uk...
| Androcles writes
| >
| >"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
| >|
| >| 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.
|
| yes.
|
| >That's 1.25 kW.
|
| yes
|
| >Any real coax will have SOME resistance and so there will be heat
dissipated
| >in it.
|
| yes
|
| >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.
|
| I don't see your point. The characteristic impedance is the same from DC
| to UHF and beyond. There will be some heat dissipated in the cable
| because as you say any conductor has some resistance but it will be
| small and will be distributed along its length so actual temperature
| rise will be minimal. The majority of the 1.25KW will be dissipated in
| the terminating, 50ohm resistor (1kW electric fire :o).
Let us first concentrate on the word "impedance" rather than the combined
term
"characteristic impedance".
A series LR circuit will behave as a potential divider, with the maximum
voltage
across the resistor ranging from V at DC to 0 at infinite frequency, agreed?
A------(L)-------B-------(R)-----D
D-----------------------------------D
(All D's are the same connection).
If we then place a capacitor in parallel with the resistor;
A------(L)-------B-------(R)-----D
(C) |
D---------------- D----------------D
it will have no effect at DC, and effectively short circuit the resistor at
infinite frequency.
So from DC to infinite frequency, we have a range from V to zero across R.
At the RESONANT frequency of the coil and capacitor, the maximum power will
be dissipated in the resistor.
So...from DC to resonance, the circuit is lossless.
That is your point, I believe.
However, when the frequency is infinite, L will restrict the current and
C will short-circuit the resistor. No power will be dissipated in R.
Now, a coax cable is more like a series of coils and capacitors,
A------(L)---B---(L)--B---(L)---B---(L)---B----(R)---D
(C) (C) (C) (C) |
D-------------D---------D---------D----------D-----------D
but that doesn't change anything.
So for frequencies greater than resonant, the impedance falls to zero.
So I had it backwards. My mistake.
|
| >Therefore the coax will cook.
|
| No. The electric fire will warm the room.
|
| >However, it will not cook at VHF,
|
| Ignoring complications such as an electric fire would look like an
| inductance at VHF and 1KW is a lot of power - Exactly the same applies
| except the coax will dissipate very slightly more power (not less)
| because of dielectric loss and surface effect.
|
| > it's
| >characteristic impedance at that frequency.
|
| You really haven't got your head around 'characteristic impedance'. A
| piece of coax is a tube of copper with a wire down the middle. There is
| capacitance because you have two conductors and a gap between. A certain
| capacitance per unit length. Does a capacitor change its value with
| frequency? No!. The wire down the middle has inductance. If you have
| ever worked at UHF it is not uncommon to have a straight wire inductor
| but if you are an LF man expecting loads of turns it may seem strange
| but take my word for it a piece of wire has inductance. Does inductance
| change with frequency? No. The characteristic impedance is the ratio of
| inductance/capacitance. Does that change with frequency? No!
|
Ok.
| >Now, if you or Andersen want to argue with that, do the experiment.
|
| I don't need to. It is connecting an 1KW electric fire to the mains with
| two bits of copper insulated from each other.
Ok.
|
| >|
| >| 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.
|
| Look if you apply a 1Hz square wave with fast edge speeds (say 2ns) it
| will contain frequency components up to 250MHz at least. At the other
| end of the coax you get an undistorted square wave. Not only do all the
| frequencies from dc to 250MHz get through but do so without phase
| distortion either. Wideband from dc to 250MHz without phase distortion -
| that alone should tell you it is resistive.
|
| Why distributed capacitance/inductance should result in a resistive
| characteristic impedance is not easy to understand, but it does and we
| would be stuffed if it didn't. I did the maths more than 40 years ago
| and can't reproduce it for you. If you ask Paul nicely he might :^)
|
| >| 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.
|
| >In other words the more expensive coax has a characteristic impedance at
| >1000 GHz
|
| You didn't read it did you! Repeat:
|
| >|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.
| Paul is correct 1000GHz is too high for coax but there is one in my
| catalogue for use up to 12.4GHz and it would I assure you be quite happy
| at all frequencies below that. Happier in fact.
Yes, but not above it.
|
| >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.
|
| Taking one from the catalogue I have in front of me
|
| Attenuation per 10m length.
|
| 1000MHz 1.7dB
| 200MHz 0.75dB
| 100MHz 0.5dB
| 50MHz 0.35dB
|
| The attenuation carries on going down to DC where the loss is purely the
| resistance of the copper. Always!
Of course, and therefore goes up as the freqnecy increases.
|
| This cable has a good return loss figure (a measure of its
| characteristic impedance accuracy) up to 2150MHz again it gets even
| better the lower the frequency. Price 0.74ukp per m so we are not
| talking exotic.
|
| The loss at DC/50Hz will be around 0.1dB/10m or less. (about 2% loss /m)
| So it is quite safe to connect your electric fire on the end and it will
| still be ideal to connect your UHF TV aerial to your TV.
|
| Trust me. I do know what I am talking about.
| --
| John Kennaugh
| to email convert the number from hex to decimal
Thanks.
Now, what is the "characteristic impedance" at 1000 GHz?
Androcles
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