Loop Antenna for Class-E amplifier
 

Hello,

I've made a Class-E amplifier which works rather well as I get 25VRMS into a
50 Ohm resistor.
But heating a resistor is not really want I do ;-)

I want to send this power into a loop antenna to transmit power by induction
to another aantenna 10 to 20 mm away.
The transmit loop antenna is in series with a capacitor and the receive
antenna is in parallel with a capacitor. Both antennas are tuned to the
correct frequency.

The problem is that though I have 125Vpkpk at the transistor drain, I only
have 15Vpkpk at my loop antenna and also 15Vpkpk in the receive antenna (not
loaded)
I guess I need to to some impedance matching but how do I compute the
impedance of those loops antennas (is it just the L//C ?) and how do I
compute the output impedance of the class-E amplifier ?

Any suggestion, URL etc ?

Thanks,

Marc Battyani

"Marc Battyani" wrote in message
...
Hello,

I've made a Class-E amplifier which works rather well as I get 25VRMS into
a
50 Ohm resistor.
But heating a resistor is not really want I do ;-)

I want to send this power into a loop antenna to transmit power by
induction
to another aantenna 10 to 20 mm away.
The transmit loop antenna is in series with a capacitor and the receive
antenna is in parallel with a capacitor. Both antennas are tuned to the
correct frequency.

The problem is that though I have 125Vpkpk at the transistor drain, I only
have 15Vpkpk at my loop antenna and also 15Vpkpk in the receive antenna
(not
loaded)
I guess I need to to some impedance matching but how do I compute the
impedance of those loops antennas (is it just the L//C ?) and how do I
compute the output impedance of the class-E amplifier ?

Any suggestion, URL etc ?

It sounds to me like you are talking about audio (or low) frequencies?

I think you are correct in that you need some impedance matching. The
impedance of your antenna is lower than 50 ohms so does not develop as much
voltage.

The best way to work out the impedance of the loop is to measure it. If you
are absolutely sure it is at resonance then it will be purely resistive and
you can measure the impedance of the loop with the equipment you have - if
it is not at resonance there will be phase errors in the following
measurement.

Connect your 50 ohm resister in series with your loop and then connect this
to the output of the amplifier. Measure the voltage across the output of
the amplifier. If it is reading over 100v pk2pk turn the drive down so it
does not saturate the output transistors.

Next disconnect the voltmeter and connect it across the 50 ohm resistor and
note the voltage (Vr). Disconnect the voltmeter and connect it across the
loop and note the voltage (Vl). You know one resistance Rr = 50 ohms.
Since the current is the same in both loads and they are in phase you can
use ohms law.

Vr/Rr = Vl/Rl

rearranging you get

R_loop = (E_loop * 50ohms) / E_resistor

it doesn't really matter if you measure RMS or pk2pk

good luck and let us know how you go.

Thanks,

Marc Battyani

"Michael" wrote
"Marc Battyani" wrote
Hello,

I've made a Class-E amplifier which works rather well as I get 25VRMS
into
a
50 Ohm resistor.
But heating a resistor is not really want I do ;-)

I want to send this power into a loop antenna to transmit power by
induction
to another antenna 10 to 20 mm away.
The transmit loop antenna is in series with a capacitor and the receive
antenna is in parallel with a capacitor. Both antennas are tuned to the
correct frequency.

The problem is that though I have 125Vpkpk at the transistor drain, I
only
have 15Vpkpk at my loop antenna and also 15Vpkpk in the receive antenna
(not
loaded)
I guess I need to to some impedance matching but how do I compute the
impedance of those loops antennas (is it just the L//C ?) and how do I
compute the output impedance of the class-E amplifier ?

Any suggestion, URL etc ?

It sounds to me like you are talking about audio (or low) frequencies?

Sorry I forgot to mention the frequency :(
It's at 8MHz.
The antennas are one turn loops.

I think you are correct in that you need some impedance matching. The
impedance of your antenna is lower than 50 ohms so does not develop as
much
voltage.

The best way to work out the impedance of the loop is to measure it. If
you
are absolutely sure it is at resonance then it will be purely resistive
and
you can measure the impedance of the loop with the equipment you have - if
it is not at resonance there will be phase errors in the following
measurement.

Connect your 50 ohm resister in series with your loop and then connect
this
to the output of the amplifier. Measure the voltage across the output of
the amplifier. If it is reading over 100v pk2pk turn the drive down so it
does not saturate the output transistors.

Next disconnect the voltmeter and connect it across the 50 ohm resistor
and
note the voltage (Vr). Disconnect the voltmeter and connect it across the
loop and note the voltage (Vl). You know one resistance Rr = 50 ohms.
Since the current is the same in both loads and they are in phase you can
use ohms law.

Vr/Rr = Vl/Rl

rearranging you get

R_loop = (E_loop * 50ohms) / E_resistor

it doesn't really matter if you measure RMS or pk2pk

good luck and let us know how you go.

I will try something like this to measure the impedance of the loop+cap.
But my problem for the impedance matching is that I don't know the amplifier
output impedance.
It's a basic class-E amplifier (A self + an IRL510 MOSFET)
Is there a simple way to measure it or to compute it ?

Thanks,

Marc

On Tue, 11 May 2004 16:41:22 +0200, "Marc Battyani"
wrote:

But my problem for the impedance matching is that I don't know the amplifier
output impedance.
It's a basic class-E amplifier (A self + an IRL510 MOSFET)
Is there a simple way to measure it or to compute it ?

Thanks,

Marc

Hi Marc,

For actual specifics of your device, go to:
http://www.irf.com/product-info/data...ata/irl510.pdf
The answer to your literal question requires more details of the
circuit, as the device shows a drain-source resistance that is common
of many finals transistors found in 100W amateur transmitters which
typically exhibit a 50 Ohm source Z at rated power.

This exhibited characteristic Z is a consequence of Z multiplication
through conventional transformer theory. 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.

If your amplifier lacks in these regards, it will suffer in efficiency
in spite of the claims of designs of its class. You may wish to
consider driving a small antenna directly to aid in matching. This is
because very small antennas (once the reactances are balanced out, an
important consideration) exhibit a much smaller load resistance that
corresponds to your device's natural output Z. However, to maintain
efficiency, you will need to be scrupulous about all Ohmic sources of
loss that come from construction issues.

73's
Richard Clark, KB7QHC

Amateur radio transmitters do NOT typically exhibit a source Z of 50 ohms at
rated power or any other power.

A conjugate match does NOT exist. It cannot play any part in transmitter
design.

Actual source Z, whatever its value, appears purely by chance AFTER the
design has been completed.

Even the designer doesn't know what value it might turn out to be. And the
amateur radio transmitter user couldn't care two hoots what it is.

I thought this had all been settled a few years back. But no! It seems the
Old Wives are still plagiarising each other.
---
Reg

On Wed, 12 May 2004 04:21:19 +0000 (UTC), "Reg Edwards"
wrote:

Amateur radio transmitters do NOT typically exhibit a source Z of 50 ohms at
rated power or any other power.

Creaky yarn spinning that went out with high button shoes.

A conjugate match does NOT exist. It cannot play any part in transmitter
design.

A tale told to unruly hams by curmudgeons to shut them up.

Actual source Z, whatever its value, appears purely by chance AFTER the
design has been completed.

Wishful thinking in its full technicolor and surround sound glory.

Even the designer doesn't know what value it might turn out to be. And the
amateur radio transmitter user couldn't care two hoots what it is.

Called projection of wish fulfillment.

I thought this had all been settled a few years back. But no! It seems the
Old Wives are still plagiarising each other.

Self fulfilling prophecy.
---
Reg

Who's Reg?

This was MUCH too tempting to pass up for another chuckle. The number
of logical inconsistencies within this short gasp begged further
illumination:

On Wed, 12 May 2004 04:21:19 +0000 (UTC), "Reg Edwards"
wrote:

Amateur radio transmitters do NOT typically exhibit a source Z of 50 ohms at
rated power or any other power.
....
Actual source Z, whatever its value, appears purely by chance AFTER the
design has been completed.

It thus follows there must be SOME Z exhibited, any value of which
occurs by happenstance and whose effect, easily determined, is a
probabilistic spin of the chambers in a capitalistic russian roulette
in the market place. Imagine trying to sell your rigs with this wild
inconsistency in quality and performance. 80% of your product would
be RMA'd to oblivion and your production line brought to its knees
with the loading docks running in reverse. Talk about a mismatch in
THAT line - I love it! We have a description of a conceptual death
wish embraced with a vengeance.

Even the designer doesn't know what value it might turn out to be. And the
amateur radio transmitter user couldn't care two hoots what it is.

Which, of course has been invalidated by the mere asking of the
question that lead to this thread. Boy, self snuffing predictions
come fast and furious. It would be so much more gothic if only there
were numbers behind this dire brooding.

This was better than any Dago Red you happen to be slugging down, Reg.
Do you write this stuff while looking through the bottom of the glass?
If you simply let yourself go, we could have a tale told by Heathcliff
pining for Catherine from the moors.

It seems the Old Wives are still plagiarising each other.
Yours was just syllables short of being lifted from Emily Brontë.

73's and thanx for the opportunity, I will of course allow you the
last comment ;-)
Richard Clark, KB7QHC

Hi Richard,

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.

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?

Aside, for VLF, audio is usually transferred at 8, 4 or even 2
ohms. VLF is not that far apart from audio, only 1 decade higher.

73s

Toni - EA3FYA

Sorry to say I've never read Emily Bronte. My style of writing has been
likened to Neville Shute but I've never read him either. Only seen films.

But I do believe fact is stranger than fiction.

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?

40-50 years ago, we had to lower the transmitter Z usually with
a built-in adjustable pi-net tuner.
--
73, Cecil http://www.qsl.net/w5dxp



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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