Author Topic: I bet many hams will answer this question incorrectly ...  (Read 4504 times)

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Offline xrunnerTopic starter

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I bet many hams will answer this question incorrectly ...
« on: May 21, 2016, 10:42:51 pm »
"If you set your XYZ FM rig to a power setting of 20 Watts on it's menu display, how much RF power is the thing really generating to make that happen?"

I bet most of them will say "About 20W ... maybe a little more?"

What do you all think they will say?  :popcorn:

I might ask that question tonight on the local repeater.

I told my friends I could teach them to be funny, but they all just laughed at me.
 

Offline babysitter

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Re: I bet many hams will answer this question incorrectly ...
« Reply #1 on: May 22, 2016, 07:08:21 am »
The common understanding of "power generated" is "power delivered into a 50 ohm load", so if one is not careful listening he might go into your trap.

Most hams will still be aware that there is impedance matching, Zs=Zl with the implication that the source will convert about half of the generated power to heat just for that rule, and also know that there are things like driver level at the input of the PA which has to be generated by preceeding stages that will somehow also not make it to the socket, generally.

Also, clock signals for controllers...

Did I pass your test? :)

Best regards

Hendrik
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Offline xrunnerTopic starter

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Re: I bet many hams will answer this question incorrectly ...
« Reply #2 on: May 22, 2016, 12:19:22 pm »
The common understanding of "power generated" is "power delivered into a 50 ohm load", so if one is not careful listening he might go into your trap.

Most hams will still be aware that there is impedance matching, Zs=Zl with the implication that the source will convert about half of the generated power to heat just for that rule, ...

It's easy enough to tell, I just did it. Looking at my PSU for my FTM-100, I'm seeing little current draw on receive, maybe 0.5A.

Well, if it only generated 50W, it would draw about 50/13.8V = 3.6A when transmitting (Yes I know there is inefficiency and some of the RF circuits take a little more power when transmitting). But it draws nearly 7.5A @ 50W

100W / 13.8V = 7.2A, so it basically tells you right there.

What selecting output power on your rig really tells you is the same thing a high-end RF generator tells you when you select PD instead of EMF for the RF output display. PD is the power you get at the 50 ohm load, which is 1/2 the power the RF generator is developing. If the rig had a selection for EMF, it would show 100W on the display (for a 50W output into a 50 ohm antenna impedance).

Quote
Did I pass your test? :)

Yep.  :)
« Last Edit: May 22, 2016, 12:31:08 pm by xrunner »
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Offline KM4FER

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Re: I bet many hams will answer this question incorrectly ...
« Reply #3 on: May 23, 2016, 04:05:39 pm »
Oh but then the rules talk about Peak-Envelope-Power so which power are you asking about?

What's the relationship between PEP and dial setting?
 

Offline xrunnerTopic starter

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Re: I bet many hams will answer this question incorrectly ...
« Reply #4 on: May 23, 2016, 05:12:41 pm »
Oh but then the rules talk about Peak-Envelope-Power so which power are you asking about?

What's the relationship between PEP and dial setting?

Doesn't matter - PEP (SSB), CW, FM, the RF power amp makes about double the power that ends up in the load (the antenna).
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Offline rfeecs

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Re: I bet many hams will answer this question incorrectly ...
« Reply #5 on: May 23, 2016, 05:14:47 pm »
Most hams will still be aware that there is impedance matching, Zs=Zl with the implication that the source will convert about half of the generated power to heat just for that rule

This is not correct.

Maybe that idea comes from the image of an ideal voltage source generator with a resistor Zs in series with the output.  But that is not how a real amplifier works.

You can have a class C (or class F, G etc.) amplifier that has an efficiency much higher than 50%.  This can be followed by an isolator that provides excellent output match without losing much power.

When you say you lose half the power, you are just saying that your amplifier is about 50% efficient.  It fundamentally doesn't have to be that way.

It's easy enough to tell, I just did it. Looking at my PSU for my FTM-100, I'm seeing little current draw on receive, maybe 0.5A.

Well, if it only generated 50W, it would draw about 50/13.8V = 3.6A when transmitting (Yes I know there is inefficiency and some of the RF circuits take a little more power when transmitting). But it draws nearly 7.5A @ 50W

100W / 13.8V = 7.2A, so it basically tells you right there.

This is just telling you that your transmitter happens to be about 50% efficient.
 

Offline xrunnerTopic starter

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Re: I bet many hams will answer this question incorrectly ...
« Reply #6 on: May 23, 2016, 05:20:08 pm »
This is just telling you that your transmitter happens to be about 50% efficient.

The RF amp has an output impedance of 50 ohms, the antenna has a impedance of ~50 ohms Both are ideally matched impedances. Half the power is lost in the RF amp, just like any RF signal generator with an internal impedance of 50 ohms.
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Offline rfeecs

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Re: I bet many hams will answer this question incorrectly ...
« Reply #7 on: May 23, 2016, 05:27:47 pm »
This is just telling you that your transmitter happens to be about 50% efficient.

The RF amp has an output impedance of 50 ohms, the antenna has a impedance of ~50 ohms Both are ideally matched impedances. Half the power is lost in the RF amp, just like any RF signal generator with an internal impedance of 50 ohms.

It sounds like you are saying that an amplifier with greater than 50% efficiency is impossible.
 

Offline xrunnerTopic starter

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Re: I bet many hams will answer this question incorrectly ...
« Reply #8 on: May 23, 2016, 05:32:07 pm »
It sounds like you are saying that an amplifier with greater than 50% efficiency is impossible.

No, I'm saying what I'm saying, but even if we differ, my point is still the same - many hams don't realize that their radio generates a lot more RF power than ends up at the antenna to be radiated - because of the internal impedance of the amplifier section (~50 ohms).
I told my friends I could teach them to be funny, but they all just laughed at me.
 

Offline rfeecs

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Re: I bet many hams will answer this question incorrectly ...
« Reply #9 on: May 23, 2016, 05:50:55 pm »
No, I'm saying what I'm saying, but even if we differ, my point is still the same - many hams don't realize that their radio generates a lot more RF power than ends up at the antenna to be radiated - because of the internal impedance of the amplifier section (~50 ohms).

I think that's a fundamental misunderstanding.  But I'll leave it to others to explain further or agree or disagree.
 

Offline Richard Head

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Re: I bet many hams will answer this question incorrectly ...
« Reply #10 on: May 23, 2016, 06:21:21 pm »
Ok. Here goes.
An RF power amplifier has a certain impedance that it's designed to work into. It does not have an output impedance of 50 Ohms like test equipment. In fact, an RF PA has a very low output impedance, approaching zero Ohms. If it did have a 50 Ohm output impedance then as the previous poster mentioned you would not be able to achieve greater than 50% efficiency, which you can very easily, particularly with class C (85% or better).
Do this test. Connect the PA to an unmatched antenna (better still a 100 Ohm resistive load) and place an in-line power meter in between the two. Measure the reflected power, then measure the forward power. You will find that the forward power exceeds the output power of the PA. How is this possible? It's called "reflected gain". The power radiated by the antenna is actually only the difference between the forward wave(s) and the reflected wave(s). In fact, it's a little more complicated than that. The antenna mismatch reflects a portion of the forward wave back toward the source. When it reaches the PA it is re-reflected toward the antenna again (zero output impedance)where a portion is again reflected toward the source. This process continues until ALL the power is either radiated by the antenna or it is lost as heat in the feedline. The result is that even if your VSWR is not 1:1 virtually all your power is eventually radiated by the antenna. The bouncing back and forth of the signal does mean that the transmitted signal will have slight smearing of the information since some is delayed due to the bouncing (echos). In the old analogue TV transmitters there was a circulator in the feedline to dissipate the reflected signal so it wasn't transmitted as the feedline delay (2x) would cause ghosting.
 

Offline djacobow

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Re: I bet many hams will answer this question incorrectly ...
« Reply #11 on: May 23, 2016, 07:41:42 pm »
Wikipedia has a nice, succinct explanation of the Maximum Power Transfer Theorem, with graph

https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

As others have pointed out "amply" already, matching input and output impedances result in maximum power transferred, but not maximum efficiency, and so, no, an amplifier need not burn up as much power in its own output impedance as goes to the load.


 


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