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Author Topic: Amplifier Efficiency >100% What do you think?  (Read 2548 times)
ZS6BIM
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« on: February 08, 2012, 11:43:45 PM »

Below is an image of test instruments measuring the DC input power and RF output power of an HF solid state power amplifier.

http://i620.photobucket.com/albums/tt283/mwmikep/FWD_RFL_PWR_1.jpg

As you can see the amplifier is transmitting into an open circuit – i.e. nothing is connected to the LHS of the Bird watt-meter.

The Rhode & Schwarz inline power meter records the forward power as 30W and the load SWR as 2300 - as one would expect with an open circuit load.

The Bird watt-meter records the reflected power as 30W – also as expected.

However look at the DC input to the RF amplifier shown on the two Fluke DMM’s.

The supply voltage is 28V and the current 0,66A – this works out as 18,48W!

How do we then get 30W out with 18,48W in?

73
Mike
 Wink
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WX7G
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« Reply #1 on: February 09, 2012, 12:02:28 AM »

Unloaded the amplifier output voltage is doubled (assuming it has a 50 ohm output Z). Double the voltage and the wattmeter will read 4X the forward power and 4X the reflected power. Loaded the amplifier in question would output 7.5 watts.

BTW, I have achieved 1000 watts forward and reflected measured at the center of a 90 degree t-line with 100 watts applied thru a matching network. I designed this for 1 kW power burn-in of a through-line RF device using a 100 W RF source.
« Last Edit: February 09, 2012, 06:09:36 AM by WX7G » Logged
TANAKASAN
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« Reply #2 on: February 09, 2012, 12:04:26 AM »

1) The instruments you're using are calibrated for a fifty ohm load, give them an open circuit load and that calibration is no longer valid.

2) At an SWR of 2306:1 some very strange things are going to happen and you cannot trust your instruments.

3) The picture shows that the R&S instrument has nothing connected to the sensor input.

Tanakasan

Two edits already, I need my coffee  Grin
« Last Edit: February 09, 2012, 12:11:34 AM by TANAKASAN » Logged
ZS6BIM
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« Reply #3 on: February 09, 2012, 12:18:19 AM »

The instruments are calibrated and there is no significant measurement error.

Sorry over-cropped the image - the transmitter output connects to the RHS of the R&S measuring head.

73
Mike
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TANAKASAN
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« Reply #4 on: February 09, 2012, 12:38:01 AM »

Right, but the instruments are calibrated to work with a fifty ohm load, not an open circuit.

Tanakasan
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ZS6BIM
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« Reply #5 on: February 09, 2012, 01:02:52 AM »

The instruments are in a 50 ohm transmission line - it just happens to have an open circuit at the end.

 Smiley
Mike
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VK4TUX
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« Reply #6 on: February 09, 2012, 02:16:56 AM »

Below is an image of test instruments measuring the DC input power and RF output power of an HF solid state power amplifier.

http://i620.photobucket.com/albums/tt283/mwmikep/FWD_RFL_PWR_1.jpg

As you can see the amplifier is transmitting into an open circuit – i.e. nothing is connected to the LHS of the Bird watt-meter.

The Rhode & Schwarz inline power meter records the forward power as 30W and the load SWR as 2300 - as one would expect with an open circuit load.

The Bird watt-meter records the reflected power as 30W – also as expected.

However look at the DC input to the RF amplifier shown on the two Fluke DMM’s.

The supply voltage is 28V and the current 0,66A – this works out as 18,48W!

How do we then get 30W out with 18,48W in?

73
Mike
 Wink

You didn't get 30w out, it was reflected. Fluke 87 mk III meter(s) for true rms values would be better also. Why not do a comparison with a 50 ohm pure resistance load?


Adrian ... vk4tux
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ZS5WC
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« Reply #7 on: February 09, 2012, 02:34:05 AM »

Hi Mike,

As one poster pointed out --a RF wattmeter is normally designed for a specific impedance (Z).
Otherwise how would you calibrate the meter?..
Normally this is 50ohms.
The reading you are getting is false.
The old and trusted way would be to use an RF milliampmeter for any (Z) impedance.
My take on it anyway--but interesting question.

73 de ZS4L
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W5DXP
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« Reply #8 on: February 09, 2012, 05:43:09 AM »

How do we then get 30W out with 18,48W in?

Power delivered by the source to the load equals forward power minus reflected power at the source output.

Power delivered by the source to the load = 30w - 30w = 0w

The efficiency of your source is zero. Some (or all) of your forward power is reflected power that is being re-distributed back toward the open-circuit. That re-distribution of energy is caused by some combination of re-reflection and/or interference associated with the superposition of the forward and reflected waves.

-------------------------------------------------------------------------------------------

Using a 100w source that is delivering 100 watts to a 2000 ohm load through some low-loss 50 ohm coax, it is possible to obtain a forward power of ~1000 watts and a reflected power of ~900 watts. The SWR is 40:1. Some tuners will match a 2000 ohm load to a 50 ohm source so the above conditions are entirely possible.

Power delivered by the source to the load = 1000w - 900w = 100w
« Last Edit: February 09, 2012, 05:44:44 AM by W5DXP » Logged

73, Cecil, www.w5dxp.com
Random length "tuned feeders" usually de-tune an antenna system!
AD4U
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« Reply #9 on: February 09, 2012, 08:43:30 AM »

Was this a Dave Made amp?  He is the only person I know of who can get more output than input.

Dick  AD4U
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ZS6BIM
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« Reply #10 on: February 10, 2012, 06:23:26 AM »

This simple experiment shows is that even though the R&S watt-meter shows a forward power of 30W and the Bird a reflected power of 30W no power is actually passing through the instruments. (assuming a lossless transmission line)

Does this then mean that there is no such thing as reflected power?

I think so.

73
Mike
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WX7G
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« Reply #11 on: February 10, 2012, 08:51:23 AM »

There certainly is reflected power!

This is easily seen and imagined by analyzing a transmission line in the time domain. SPICE is great for this exercise. Or in the real world use a TDR as is done everyday in industry.
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W5DXP
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« Reply #12 on: February 10, 2012, 09:06:28 AM »

This simple experiment shows is that even though the R&S watt-meter shows a forward power of 30W and the Bird a reflected power of 30W no power is actually passing through the instruments. (assuming a lossless transmission line)

Power doesn't actually flow - it is EM energy that is being transferred. All EM energy must necessarily move at the speed of light in the medium because nothing else is possible. Power is the energy per unit time that is passing a fixed measurement point. The power in an EM wave, measured at a measurement point, is ExH.

Quote
Does this then mean that there is no such thing as reflected power? I think so.

Consider a one second long lossless transmission line with an SWR of 5.83:1 being driven by a 100w source. The forward power is 200w and the reflected power is 100w. 100w is being delivered to the load.

A one second long lossless transmission line is interesting because if one puts 100 watts into it for one second, the line contains 100 joules of energy.

In the above example, it can be shown that during steady state, the transmission line contains 300 joules, 200 of which are in the forward wave and 100 of which are in the reflected wave. It can also be shown that the 300 joules existing in the transmission line have been sourced by the source but have not yet made it to the load. It is physically impossible for EM energy to stand still or to move at any speed other than the speed of light in the medium. We know that to be a technical fact of quantum physics.

Standing "waves" are an illusion and do not even meet the technical definition of "wave".

Let's take that one second long lossless transmission line and remove the load. For one second, we source a kilowatt RF signal. During second number two, we disconnect the source and you grab one conductor in one hand and the other conductor in the other hand. What happens during second number three? I'll bet it would make you believe in reflected power.
« Last Edit: February 10, 2012, 09:13:05 AM by W5DXP » Logged

73, Cecil, www.w5dxp.com
Random length "tuned feeders" usually de-tune an antenna system!
ZS6BIM
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« Reply #13 on: February 10, 2012, 11:17:54 AM »

I don’t have a problem with reflected power during the time it takes for the standing wave to become established on the transmission line.

However consider your example of an open ended one second long lossless 50 ohm line connected to my amplifier (Amplifier Zs = 50 Ohms and the transmission line must be a multiple of a half a wavelength)

During the first two seconds of transmission the amplifier will see a load of 50, j0 and therefore deliver 30W. At the input to the line the R&S power meter will indicate a forward power of 30 watts and the Bird 0W of reflected power.

The amplifier has an efficiency of about 33% so at 30W output it will draw about 90W from the DC supply.

After 2 seconds has passed the Bird will suddenly indicate a continuous 30W of reflected power while the R&S will continue to indicate 30W of forward power – however the power drawn from the DC supply falls to 18,45W!

What has happened to the reflected power?

73
Mike
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N4RSS
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« Reply #14 on: February 10, 2012, 11:36:42 AM »

The reflected power is out of phase with the load and will be absorbed as heat in the line to the extent it's not absorbed by the load.

The output of the amp drops to the extent that the input to the line is no longer a match to the amp, which is a consequence of the out of phase voltage/current now developed at the input due the mismatch at the load.

That's my understanding.

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