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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