The latest QEX has an article by k9la: "The Impact of Load SWR
on the Efficiency of Power Amplifiers". The amplifier is designed
to drive a 50 ohm load and simulated measurements were made for
8 reflection coefficients of 0.333 with phase angles in 45 degree
increments.

One thing I don't understand. When driving a 100 ohm load, the
drain current is 33 amps and the efficiency is 62%. When driving
a 25 ohm load, the drain current is 12 amps and the efficiency
is 83%.

The amp is a Class E/F design from Jan/Feb 2004 QEX. Why would
the drain current fall when the load is decreased from 100 ohms
to 25 ohms?
--
73, Cecil, http://www.qsl.net/w5dxp


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Cecil Moore wrote:
The latest QEX has an article by k9la: "The Impact of Load SWR
on the Efficiency of Power Amplifiers". The amplifier is designed
to drive a 50 ohm load and simulated measurements were made for
8 reflection coefficients of 0.333 with phase angles in 45 degree
increments.

One thing I don't understand. When driving a 100 ohm load, the
drain current is 33 amps and the efficiency is 62%. When driving
a 25 ohm load, the drain current is 12 amps and the efficiency
is 83%.

The amp is a Class E/F design from Jan/Feb 2004 QEX. Why would
the drain current fall when the load is decreased from 100 ohms
to 25 ohms?

The output filter of an amplifier can act like a 1/4 wave (or 3/4 wave)
transmission line, so a high impedance at the antenna terminal would
look like a low impedance at the final, and visa-versa. This extends to
the time-delay properties: older oscilloscopes with time delay used long
L/C filters operated way below cutoff to simulate even longer stretches
of transmission lines.

In fact, a filter that looks like this

___
--------o---UUU---o--------
| |
--- ---
--- ---
| |
| |
=== ===
GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

with all reactances equal to the design impedance (say 50 ohms) at the
design frequency is _called_ a 1/4-wave filter. Put two of them
together and you get a 1/2 wave filter:

___ ___
--------o---UUU---o---UUU---o------
| | |
--- --- ---
C --- 2C --- C ---
| | |
| | |
=== === ===
GND GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Stick on another section and it'll act like a 3/4 wave line, etc.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Tim Wescott wrote:

Cecil Moore wrote:
The amp is a Class E/F design from Jan/Feb 2004 QEX. Why would
the drain current fall when the load is decreased from 100 ohms
to 25 ohms?

The output filter of an amplifier can act like a 1/4 wave (or 3/4 wave)
transmission line, ...

I should have described the output circuitry. It's a push-pull
LC tank circuit with DC supplied by the coil center-tap. The
output is just a link coupler. The low-pass filter is a single
series L and single C to ground.
--
73, Cecil, http://www.qsl.net/w5dxp


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I shouldn't, but will. You don't say, but I think you are puzzled by a
decreasing load impedance causing a decrease in collector current. I'd have
to think about it, but I'd first Try modeling that LC filter at resonance to
see if it gives an inverse function like the 1/4 wave line.
A "gut" reaction tells me that a decreasing collector current for *either*
an increased or decreased load Z would not be a surprise...

Steve K9DCI


"Cecil Moore" wrote in message
...
Tim Wescott wrote:

Cecil Moore wrote:
The amp is a Class E/F design from Jan/Feb 2004 QEX. Why would
the drain current fall when the load is decreased from 100 ohms
to 25 ohms?

The output filter of an amplifier can act like a 1/4 wave (or 3/4 wave)
transmission line, ...

I should have described the output circuitry. It's a push-pull
LC tank circuit with DC supplied by the coil center-tap. The
output is just a link coupler. The low-pass filter is a single
series L and single C to ground.
--
73, Cecil, http://www.qsl.net/w5dxp


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Steve Nosko wrote:
I shouldn't, but will. You don't say, but I think you are puzzled by a
decreasing load impedance causing a decrease in collector current. I'd have
to think about it, but I'd first Try modeling that LC filter at resonance to
see if it gives an inverse function like the 1/4 wave line.
A "gut" reaction tells me that a decreasing collector current for *either*
an increased or decreased load Z would not be a surprise...

The output is link coupled through a series 216 nH coil and a 2.1 nF cap
to ground. The load is connected from the coil/cap junction to ground.
It's a Class E/F 40m amp.

Load Drain Current Drain Efficiency Load Power
25 ohms 12.2 amps 82.6% 129 watts
50 ohms 20.1 amps 77.7% 200 watts
100 ohms 33.1 amps 62.3% 264 watts

Can these be thought of as load-pulling data points? Maximum load
power appears to occur when the load is greater than 100 ohms.
--
73, Cecil http://www.qsl.net/w5dxp

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Can I drive a class E amp with an SSB signal?

Thanks,

The Eternal Squire

"Cecil Moore" wrote in message
...

Load Drain Current Drain Efficiency Load Power
25 ohms 12.2 amps 82.6% 129 watts
50 ohms 20.1 amps 77.7% 200 watts
100 ohms 33.1 amps 62.3% 264 watts

Can these be thought of as load-pulling data points? Maximum load
power appears to occur when the load is greater than 100 ohms.
--
73, Cecil http://www.qsl.net/w5dxp

Have you considered the possibility that he might have the data for 25 Ohms
and 100 Ohms interchanged? The amplifier is voltage limited at, it appears,
12.8 V. I would believe 25 Ohms, 33.1A, and 264W.

Tam/WB2TT

Cecil Moore wrote:

The latest QEX has an article by k9la: "The Impact of Load SWR
on the Efficiency of Power Amplifiers". The amplifier is designed
to drive a 50 ohm load and simulated measurements were made for
8 reflection coefficients of 0.333 with phase angles in 45 degree
increments.

One thing I don't understand. When driving a 100 ohm load, the
drain current is 33 amps and the efficiency is 62%. When driving
a 25 ohm load, the drain current is 12 amps and the efficiency
is 83%.

The amp is a Class E/F design from Jan/Feb 2004 QEX. Why would
the drain current fall when the load is decreased from 100 ohms
to 25 ohms?

How has the QL changed as a function of load? How has the power output
changed from 110 to 25 ohms load? Have you calculated the effective load
lines for the amplifiers?

We need a lot more information.

I don't subscribe to QEX so I don't have the circuit available.