On Wed, 12 May 2004 10:13:10 +0200, Toni wrote:

This is something I have always thought: When driving loops, very
short monopoles (VLF) or other low-impedance antennas, why raise
transmitter Z to 50 ohms, send it through the line and lower it
again to whatever required? Couldn't the amplifier be built
directly at the antenna base so _no_ transmission line is
required and couple directly the low Z at the final transistor(s)
to the low Z at the antenna?

Hi Toni,

Exactly!

73's
Richard Clark, KB7QHC

Hi Cecil

En Cecil Moore va escriure en Wed, 12 May 2004 09:44:10 -0500:

40-50 years ago, we had to lower the transmitter Z usually with
a built-in adjustable pi-net tuner.

Do you mean before the existence of electronic simulators, when
people actually build electronic apparatus? :^) When I studied
electronics I was in the last course where students would
actually build things. From the next course on student's
practices would be made only by computer simulations :^(

As to your comment, I know valves exist but I have never learned
/ been taught how to use them. :^(

73s

Toni - EA3FYA

"Toni" wrote
En Richard Clark va escriure en Wed, 12 May 2004 00:32:35 GMT:

These designs also have the
benefit of low pass filters aiding a smooth transition of Z from the
device Z on the order of 0.5 to 1 Ohm to the System Z of 50 Ohms.

There is 0.5 Ohm when the transistor is on but not when is it off. So the
resulting impedance should probably be more than that. ??

This is something I have always thought: When driving loops, very
short monopoles (VLF) or other low-impedance antennas, why raise
transmitter Z to 50 ohms, send it through the line and lower it
again to whatever required? Couldn't the amplifier be built
directly at the antenna base so _no_ transmission line is
required and couple directly the low Z at the final transistor(s)
to the low Z at the antenna?

This is what I have. The amplifier is directly connected to the loop.

Here is a scope screen copy of the output of the amplifier:
http://www.fractalconcept.com/scope-screen.jpg

The C2 trace is the drain voltage and the C3 one is the 50 Ohm resistor
voltage.

The schematics is very simple:
http://www.fractalconcept.com/schema.pdf

The first one is the one that works on the scope screen.
The second one is a one that does not work at all: the torus of the 680nH
melted

I have a little more success by just connecting the drain to the loop
through a capacitor but it's still not good (25% efficiency)

Any idea ?

Marc

On Thu, 13 May 2004 10:58:27 +0200, "Marc Battyani"
wrote:
This is what I have. The amplifier is directly connected to the loop.

Here is a scope screen copy of the output of the amplifier:
http://www.fractalconcept.com/scope-screen.jpg

The C2 trace is the drain voltage and the C3 one is the 50 Ohm resistor
voltage.

The schematics is very simple:
http://www.fractalconcept.com/schema.pdf

The first one is the one that works on the scope screen.
The second one is a one that does not work at all: the torus of the 680nH
melted

I have a little more success by just connecting the drain to the loop
through a capacitor but it's still not good (25% efficiency)

Any idea ?

Hi Marc,

There are a number of questions.

First, how do you compute efficiency?

Where on the schematic are these scope connections?

What is your design source for this schematic, or if original, what
drove you to select the reactive components you did? The series load
on the drain does not resonate at 8MHz and is not particularly matched
to either the FET or the 50 Ohm load.

What is the inductance of the loop? Alternatively, what size is it?
This last is more important because it subsumes the radiation
resistance that must be known to perform any efficiency computation.

The two circuits are very, very different from each other to be
achieving the same purpose. The 50 Ohm resistor, as an equivalent
load appears to be grossly in error.

By all appearances of the magenta trace, you are running Class B or a
very long Class C. This is not a hallmark of high efficiency.
However, not knowing where this trace resides in the circuit, this is
simply a guess.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote
On Thu, 13 May 2004 10:58:27 +0200, "Marc Battyani"
wrote:
This is what I have. The amplifier is directly connected to the loop.

Here is a scope screen copy of the output of the amplifier:
http://www.fractalconcept.com/scope-screen.jpg

The C2 trace is the drain voltage and the C3 one is the 50 Ohm resistor
voltage.

The schematics is very simple:
http://www.fractalconcept.com/schema.pdf

The first one is the one that works on the scope screen.
The second one is a one that does not work at all: the torus of the 680nH
melted

I have a little more success by just connecting the drain to the loop
through a capacitor but it's still not good (25% efficiency)

Any idea ?

Hi Marc,

There are a number of questions.

First, how do you compute efficiency?

The power in the load divided by the power used by the amplifier.

Where on the schematic are these scope connections?

?? The C2 trace is the transistor drain voltage and the C3 one is the 50 Ohm
resistor voltage (not the ground ;-).

What is your design source for this schematic, or if original, what
drove you to select the reactive components you did? The series load
on the drain does not resonate at 8MHz and is not particularly matched
to either the FET or the 50 Ohm load.

I looked at a lot of class E app notes and picked one.
I started with approx values and changed them incrementally to try to have a
better result.

What is the inductance of the loop? Alternatively, what size is it?
This last is more important because it subsumes the radiation
resistance that must be known to perform any efficiency computation.

The loop is 65x25mm

The two circuits are very, very different from each other to be
achieving the same purpose. The 50 Ohm resistor, as an equivalent
load appears to be grossly in error.

Yes. This is the problem. The first one was to test if I was able to make a
working class-E amplifier. The second one is really bad. I have a third one
where the loop is connected in place of the resistor which is somewhat
better but still not good.

By all appearances of the magenta trace, you are running Class B or a
very long Class C. This is not a hallmark of high efficiency.
However, not knowing where this trace resides in the circuit, this is
simply a guess.

The C2 trace (the drain voltage) looks like the one expected for a class E
IMO.

Marc

On Thu, 13 May 2004 23:43:44 +0200, "Marc Battyani"
wrote:
This last is more important because it subsumes the radiation
resistance that must be known to perform any efficiency computation.

The loop is 65x25mm

Hi Marc,

Yes, this appears to be extremely problematic. The radiation
resistance of this loop is on the order of 30 nano Ohms. For you to
achieve a 50% efficiency in radiating the input power, no component in
the resonant loop must have an Ohmic resistance greater than this.

In other words, you don't stand a chance. What coupling that you are
getting is probably more capacitive or inductive than radiative.

The radiation resistance of a loop rises or falls by the 4th power of
its ratio to the wavelength of excitation. Double the radius and you
will multiply the radiation resistance 16 fold.n What this says, is
that the radiation resistance is overwhelmed by conductor loss (even
if only micro Ohms) that is turning your power into heat.

The two circuits are very, very different from each other to be
achieving the same purpose. The 50 Ohm resistor, as an equivalent
load appears to be grossly in error.

Yes. This is the problem.

An understatement with the loop dimension given.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote
On Thu, 13 May 2004 23:43:44 +0200, "Marc Battyani"
wrote:
This last is more important because it subsumes the radiation
resistance that must be known to perform any efficiency computation.

The loop is 65x25mm

Hi Marc,

Yes, this appears to be extremely problematic. The radiation
resistance of this loop is on the order of 30 nano Ohms. For you to
achieve a 50% efficiency in radiating the input power, no component in
the resonant loop must have an Ohmic resistance greater than this.

In other words, you don't stand a chance. What coupling that you are
getting is probably more capacitive or inductive than radiative.

Yes, this is exactly what I want to do : Inductive coupling.
Maybe you didn't see my first posts, what I want to do is transmit 3W of
power by induction over a 12mm distance.

The radiation resistance of a loop rises or falls by the 4th power of
its ratio to the wavelength of excitation. Double the radius and you
will multiply the radiation resistance 16 fold.n What this says, is
that the radiation resistance is overwhelmed by conductor loss (even
if only micro Ohms) that is turning your power into heat.

The two circuits are very, very different from each other to be
achieving the same purpose. The 50 Ohm resistor, as an equivalent
load appears to be grossly in error.

Yes. This is the problem.

An understatement with the loop dimension given.

Heh, this is why I'm posting here.

Marc

On Fri, 14 May 2004 13:57:58 +0200, "Marc Battyani"
wrote:

Yes, this appears to be extremely problematic. The radiation
resistance of this loop is on the order of 30 nano Ohms. For you to
achieve a 50% efficiency in radiating the input power, no component in
the resonant loop must have an Ohmic resistance greater than this.

In other words, you don't stand a chance. What coupling that you are
getting is probably more capacitive or inductive than radiative.

Yes, this is exactly what I want to do : Inductive coupling.
Maybe you didn't see my first posts, what I want to do is transmit 3W of
power by induction over a 12mm distance.

Hi Marc,

Then this returns us to the native Z of the source, the FET, and the
sink Z of the load AFTER the link. Your link shows a 1:1 coupling (as
best I can tell) and it follows that the load should be on the order
of half an Ohm. It also follows that the characteristic Z of the FET
load should also exhibit this value (it does not) as well as the
coupling circuitry to the link.

Move the primary loop back into the Drain lead to the positive rail
path, and connect the 2200pF cap (which may be too much) from the
Drain lead to ground path. The other circuitry is superfluous. The
characteristic Z of this load is roughly equal to the FET; and as the
FET on time is roughly 120°; and depending upon coupling, then you
might find 60 - 80% efficiency. It will require some tuning as the
Bandwidth will be 1 or 2 MHz around resonance. If you want some other
actual load resistance other than half an Ohm, then you need to boost
the transform ratio (it works by the square of the windings ratio).

Give this a try and report your findings.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote
On Fri, 14 May 2004 13:57:58 +0200, "Marc Battyani"
wrote:

Yes, this is exactly what I want to do : Inductive coupling.
Maybe you didn't see my first posts, what I want to do is transmit 3W of
power by induction over a 12mm distance.

Then this returns us to the native Z of the source, the FET, and the
sink Z of the load AFTER the link. Your link shows a 1:1 coupling (as
best I can tell) and it follows that the load should be on the order
of half an Ohm. It also follows that the characteristic Z of the FET
load should also exhibit this value (it does not) as well as the
coupling circuitry to the link.

Move the primary loop back into the Drain lead to the positive rail
path, and connect the 2200pF cap (which may be too much) from the
Drain lead to ground path. The other circuitry is superfluous. The
characteristic Z of this load is roughly equal to the FET; and as the
FET on time is roughly 120°; and depending upon coupling, then you
might find 60 - 80% efficiency. It will require some tuning as the
Bandwidth will be 1 or 2 MHz around resonance. If you want some other
actual load resistance other than half an Ohm, then you need to boost
the transform ratio (it works by the square of the windings ratio).

Give this a try and report your findings.

Much better!
A get 70% efficiency now
(with a 330pF cap)

Thanks

Marc

On Sat, 15 May 2004 19:02:11 +0200, "Marc Battyani"
wrote:
Much better!
A get 70% efficiency now
(with a 330pF cap)

Thanks

Marc

Hi Marc,

You are welcome.

73's
Richard Clark, KB7QHC