In article , gwhite wrote:
Richard Clark wrote:

On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote:

RF transmitters are not ....

Sorry OM,

This was all nonsense.

Nice articulation. I don't know who OM is, but RF transmitter power amps are
not "impedance matched." Neither are audio power amps for that matter.

"OM" is an amateur radio term. It is short for "Old Man". It is a
respectful term for all other males that is quick to transmit via Morse
code. Richard Clark appears to be an amateur radio operator or the like.

RF transmitter power amps are certainly "impedance matched" to the
intended load. Take a look in the ARRL "The radio amateur's handbook".
If you have the 1944 addition, you will need to start reading at page 96
in the lower right column. If you don't have that, try Motorola's AN-721.

As for audio amp, you are 1 for 3 my friend.

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Ken Smith wrote:
RF transmitter power amps are certainly "impedance matched" to the
intended load. Take a look in the ARRL "The radio amateur's handbook".
If you have the 1944 addition, you will need to start reading at page 96
in the lower right column. If you don't have that, try Motorola's AN-721.

A CMOS Class-E amp is in full saturation (0.5v at 2a)
for 10% of a cycle and off (12v at 0a) for the other
90% of a cycle. The tank circuit changes the digital
energy to analog energy by filtering out everything
except the fundamental frequency component. How
in the world does one determine the steady-state
impedance of the CMOS source? Isn't the best one
can do with a digital switch is to keep it within
specified parameters? The CMOS device dissipates 2
watts for 10% of the time - therefore 0.2 watts
steady-state.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:
The CMOS device dissipates 2
watts for 10% of the time - therefore 0.2 watts
steady-state.

Sorry, should have been: The CMOS device dissipates
one watt for 10% of the time - therefore 0.1 watts.


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In article , Cecil Moore wrote:
Ken Smith wrote:
RF transmitter power amps are certainly "impedance matched" to the
intended load. Take a look in the ARRL "The radio amateur's handbook".
If you have the 1944 addition, you will need to start reading at page 96
in the lower right column. If you don't have that, try Motorola's AN-721.

A CMOS Class-E amp is in full saturation (0.5v at 2a)
for 10% of a cycle and off (12v at 0a) for the other
90% of a cycle. The tank circuit changes the digital
energy to analog energy by filtering out everything
except the fundamental frequency component. How
in the world does one determine the steady-state
impedance of the CMOS source? Isn't the best one
can do with a digital switch is to keep it within
specified parameters? The CMOS device dissipates 2
watts for 10% of the time - therefore 0.2 watts
steady-state.

For what you say here really to be true the transistors must switch very
fast. About 25pS switching speed is needed at about 400KHz. If we take
that to be the case however, I think you will see why matching still
applies.

Lets take the reactive component first. If there is a reactive component
to the loading, the current in the switch will have a higher RMS value
without that increase in RMS increasing the radiated power of the system.
So the reactive component of the matching is fairly obvious.

Imagine that you have a well designed Class-E circuit loaded with the load
the designer optimized it for.

Now imagine that you slightly increase the resistance slightly. When you
do so, the current into the load will decrease but the voltage will not
increase enough to compensate for this.

Now lets assume that you slightly decrease the resistance. Since we are
assuming that this is a well designed case, we can assume that the
designer took steps to ensure that the output devices would be protected
from excess currents. This could be done by reducing the operating
voltage of the output section, for example. In any case, the voltage on
the load will decrease by a larger factor than the current will increase.

So it is obvious that the reactive part is matched and the resistive part
is matched just as it would be in a non-class-E output section.


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Ken Smith wrote:
Now lets assume that you slightly decrease the resistance. Since we are
assuming that this is a well designed case, we can assume that the
designer took steps to ensure that the output devices would be protected
from excess currents.

Let's assume the designer is an amateur who didn't provide
any protection for his tube's output. The lower the resistive
load, the more current the output device draws until it fails.
What is the output impedance of the device?
--
73, Cecil http://www.qsl.net/w5dxp


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In article , Cecil Moore wrote:
Ken Smith wrote:
Now lets assume that you slightly decrease the resistance. Since we are
assuming that this is a well designed case, we can assume that the
designer took steps to ensure that the output devices would be protected
from excess currents.

Let's assume the designer is an amateur who didn't provide
any protection for his tube's output. The lower the resistive
load, the more current the output device draws until it fails.
What is the output impedance of the device?

At the point where it fails, the output goes to zero, I assume. If so,
wouldn't that be the impedance as I've been defining it.


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Ken Smith wrote:
At the point where it fails, the output goes to zero, I assume. If so,
wouldn't that be the impedance as I've been defining it.

Is an amp that fails at one minute with 100w FM
better matched than an amp that fails at two minutes
with 100w CW?
--
73, Cecil http://www.qsl.net/w5dxp


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I read in sci.electronics.design that Cecil Moore
wrote (in ) about '1/4 vs 1/2 wavelength
antenna', on Sat, 26 Feb 2005:
Is an amp that fails at one minute with 100w FM
better matched than an amp that fails at two minutes
with 100w CW?

If the FM is what passes for music these days, it's MUCH better IMHO.
(;-)
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

In article , Cecil Moore wrote:
Ken Smith wrote:
At the point where it fails, the output goes to zero, I assume. If so,
wouldn't that be the impedance as I've been defining it.

Is an amp that fails at one minute with 100w FM
better matched than an amp that fails at two minutes
with 100w CW?

No, you've got the concept backwards. Obviously the worst matched is the
1 o=minute case, next would be the 2 minute case and so on up to one which
runs for about its MTBF at the connected load. This last case would
likely be the one the designer was targeting.

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Ken Smith wrote:
RF transmitter power amps are certainly "impedance matched" to the
intended load. Take a look in the ARRL "The radio amateur's handbook".
If you have the 1944 addition, you will need to start reading at page 96
in the lower right column. If you don't have that, try Motorola's AN-721.

It appears that I may have canceled an earlier posting
by accident so will repeat it.

A certain Class-E CMOS amp is in full saturation for
10% of a cycle, 0.5v at 2a. For the rest of the time
it is off. The supply voltage is 12v. What is the
steady-state impedance of the source at the fundamental
frequency?
--
73, Cecil http://www.qsl.net/w5dxp


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