In message <bk87dc$ijf$1@news.onet.pl>, Mathew Orman <orman@nospam.com>
writes
"John Kennaugh" <john@kennaugh.yourfingerdemon.co.uk> wrote in message
news:OiV39+SuN3Z$EwfB@kennaugh.demon.co.uk...
In message <bim8q8$1r6$1@news.onet.pl>, Mathew Orman <orman@nospam.com>
writes
..........Single turn loop 1 m in diameter of 3 mm gage copper wire is
energized with source of 1 Ohm impedance at 1 MHz sin.
OK so you have a signal generator with a load connected across it
consisting of a loop of wire which will be an inductor of approx 5 micro
Henry . At 1MHz that presents an impedance of about 30 ohms.
The total length of wire is aprox. 3.14 m.
Well it would be wouldn't it if it approximates to a circle.
The current waveform at one end is identical with the current waveform at
the other end.
Here you seem to be a bit confused. You are not applying a current at
one end and waiting to see how long it takes to reach the other. All you
are doing is measuring the current flowing when a voltage is applied to
a load. There is only one current. There isn't one terminal that the
signal comes out of and another which it goes back into. Think of the
signal generator as having a transformer output with neither side
connected to earth and a winding resistance of 1 ohm. You wouldn't
expect different currents at different ends of the transformer winding
would you. Same thing applies even if it does not have a transformer O/P
stage.
Instead of considering it in the frequency domain think of it in the
time domain. Let us apply a 1V step in voltage. Initially the current
will be small because the load is inductive and energy is being stored
in the inductance. Current will increase in accordance with i = Ldv/dt
until it is limited by the source resistance of 1ohm. to 1A. Try the
same thing again except this time reverse the voltage every 0.5us. You
now have a square wave at 1MHz. Because it takes time for the current to
build up after each transition the current waveform will lag the voltage
waveform. This is what happens with a sin wave.
Now suppose you and the signal generator are behind a screen and you
cannot tell what is being connected. You can tells the load is inductive
because of the current phase lag. You could, by very accurate
measurement of phase determine the resistance but you have no way of
knowing whether it is a loop of wire 1m diameter or a few inches of wire
wound around a high permeability former to give the same inductance.
OK now suppose you bent your piece of wire into a long hairpin shape.
Two parallel wires with a short circuit at the other end. You have now
increased the capacitance between the wires. The wires themselves have
inductance. When you apply your step function it propagates along the
pair of wires. Each bit of wire has inductance which slows up the build
up of current and each bit of wire has capacitance which has to be
charged up sequentially. It will propagate along the pair of wires at
approximately 0.5c until it reaches the end where the short circuit is,
where it will be reflected. Energy is traveling down the line and there
is no resistance to turn that energy into heat so it is reflected. Back
at the source some energy will be dissipated in the source resistance
but unless it matches the characteristic impedance of the line some will
be reflected back. A steady state voltage will be eventually be reached
when the series of reflections subsides. If you drive it with a sin wave
energy still travels back and forth again at around 0.5c and this can be
shown with the use of a directional coupler which samples the energy
traveling in each direction. A VSWR meter works on a similar principle.
As far as the generator is concerned and the engineer behind the screen
all he sees with a sin wave is a fixed relationship between the phase of
the voltage and current and the amplitude from which he can determine
the equivalent load. As the relationship varies depending on how long it
takes the signal to go there and back it can end up looking inductive or
capacitive. This phenomena is used in RF circuits for matching. If a
circuit looks like a capacitor + a resistor then at one particular
frequency it can be made to look purely resistive by connecting it by
just the right length of transmission line to get the current and
voltage waveforms in phase = resistive.
If you go back to a step function there is a piece of kit called a 'time
domain reflectometer' this uses a step with an extremely fast edge and
by looking at the reflections coming back you can see each impedance
discontinuity. The engineer behind the screen can now get a lot more
information. For example if in bending your circle into a hairpin shape
you didn't get the two wires exactly parallel then a time domain
reflectometer would show reflections coming back from each kink. You
could tell the distance the kink is and whether it represented a greater
or smaller gap than the norm.
Whether or not you can consider your original circle as a transmission
line with an enormous kink in it or a transmission line with a
characteristic impedance which increases then decreases with distance I
don't known. It will certainly take time for the energy leaving the
signal generator to reach the far side of the loop and as there is no
appreciable resistance to turn it into heat it will be reflected. What
would happen in the time domain is an interesting question. Certainly as
you increased the frequency the delay would become an increasingly
significant proportion of the period and it would not look purely
inductive. If you increase the frequency enough it will become a loop
antenna and radiate the energy.
Thank you for the question. It gave my brain cells a work out if nothing
else.
Hope that un-confuses you.
Also there is no phase shift between the two waveforms.
Why is that?
According to Einstein-Lorentz it takes aprox. 9.5 nanosecond for the
electric field
to propagate such distance.
--
John Kennaugh
to email take 'yourfinger' out
Thank you for taking time and explaining what is happening in the loop.
The point was not the current and voltage distribution across the loop
but the current waveform relation at the source terminal points
and you've stated that they are identical in all aspects.
The same is true for a pierce of coax.
The current waveform of coax core insertion
is identical with coax shield return current
providing that the source terminals are floating (example using the
transformer).
Even if not
You see the electric force field that is applied from the source
acts instantaneously across entire length of the loop.
What 'force field'? If you apply a voltage to the end of a piece of coax
the initial current simply charges the capacitance in the line
immediately adjacent to the generator terminals via the line inductance,
it has absolutely no immediate effect anywhere else and has no effect at
the far end of the coax until it has had time to propagate down the line
and reach it. If you have 10km of coax and apply a voltage to one end it
won't reach the other end until about 50us later.
Consider two lines a fixed distance apart. The lines have inductance per
unit length L1,L2,L3 etc. and capacitance per unit length C1,C2,C3 etc.
A - L1 - L2 - L3 - L4 - ...................
C1 C2 C3 C4
B - L1 - L2 - L3 - L4 - ...................
Put a voltage across A,B Current flows in the loop A -L1-C1-L1-B. An
inductor slows down the rate at which current will build and will limit
the current into C1. As C1 charges up, the current into C1 reduces as
the voltage increases but as the voltage has now increased this has
applied a driving voltage to L2 C2 L2. The current flowing in C1 is
falling and at some time t eventually -> zero. C2 is a bit behind C1
etc. Meanwhile at time t current is just reaching its maximum in Cn
further down the line. The current is now flowing A - L1 - L2 - L3
...........Ln some will still be flowing in Cn-1, Cn will be coming to
its maximum and current will be starting to flow in Cn+1. All this
current flows back to B via L1 - L2 - L3 ..........Ln. Because at any
one time the current in Cn-1 is falling, the current in Cn is a maximum
and that in Cn+1 is rising the current flowing is a constant. The route
it takes is progressing along the line.
I don't know what you mean by electric force field and nothing acts
instantaneously at a distance. I think you have been watching too much
Star Trek and not paying enough attention to basic electric theory. You
seem very confused.
--
John Kennaugh
to email take 'yourfinger' out
.
