1/4 vs 1/2 wavelength antenna

There is noth**************************************************************************************** " ..the load, in first approximation, is not related to the device, except for VCE(sat). The load value is primarily dictated by the required output power and the peak voltage; it is not matched to the output impedance of the device. "

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When device people talk about "matching", they mean matching the load to what the transistor wants to see, which is not the conjugate of the output impedance. The way this is done is to build an amplifier, and vary the load until maximum output power is reached. The transistor is then removed, and the impedance looking into the coupling network is measured. The conjugate of this is sometimes listed as "output impedance" on data sheets. Newer data sheets will have an asterisk * next to that, and a note explaining what it means. If you look at Philips literature, you will see exactly the same explanation.

This whole thing has been hashed out here about 5 times during the past year.

Tam

Reply to
Tam/WB2TT
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Ken, Motorola 721 specifically says the device IS NOT MATCHED.

Tam

Reply to
Tam/WB2TT

SWR foldback is part of impedance matching?

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73, Cecil
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Reply to
Cecil Moore

I'm sorry, but they are not. Nor are any power amps that I know of. Efficiency (and thus necessarily output swing) is what matters for power amps. To maximize swing requires load line matching, not impedance matching.

If you want to study RF PA's I suppose Cripps is one of the best I know of for a modern text:

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Reply to
gwhite

Exactly Cecil. That's exactly why I said the concept of impedance is dubious for large signal devices and thus provides a second argument, albeit weaker, against the "impedance matching" idea for power amps.

"The concept of 'output impedance' breaks down for large signal devices. For example, what is the output impedance of a class C or D amp taken when the transistor is on or off?"

The strongest argument for dropping the impedance matching concept is PA efficiency, and therefore maximum signal swing. Obtaining maximum swing is a load line issue.

Reply to
gwhite

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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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

I answered your other copy of the posting.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

This needs a 12V transistor that switches in about 25pS. Would that be likely on a current chip?

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith
[...]

Motorola seems to disagree with you. Perhaps more importantly, I disagree with you. The large signal impedance of a transistor working into a tuned load still applies quite well.

When dealing with an RF output section you don't deal with just one part of the cycle. You can tell if the real part of the impedance is matched if:

(1) If you increase the resistance of the load does the power decrease?

AND

(2) If you decrease the resistance of the load does the power decrease?

What do you mean by "maximum signal swing" in this context. I can get a bigger swing by leaving the output completely unloaded and hence causing the actual efficiency to be zero.

The reactive component issue is still there too. Reactive loads cause increased currents in the output stage without delivering any power to the load so they still need to be reduced as much as practical.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

I still say they are. Motorola AN-721 takes on the theory. AN-758 does a practical example matching 12.5 Ohms into 50 Ohms

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

Tam

An audio amplifier can have an improved capability of minimizing a speaker's deviation from its desired displacement if it has a low output impedance. That is, the speaker's cone tends to swing past its desired displacement when its terminals are loaded to a high impedance.

Jerry

Reply to
Jerry Martes

Arguments about whether the power ammplifier is matched or not matched to 50 ohms arise due to misunderstandings about the meaning of "matched".

Meaning 1.

------------- The PA has been designed for maximum, linear, undistorted power output when loaded with Ro ohms and the load reistance has actually been adjusted to equal Ro. Ro is usually 50 ohms. (There may be additional criteria to define what constitutes an optimum match.)

Meaning 2.

------------- The load impedance Z = R+jX has been adjusted to equal the conjugate of the internal impedance resistance of the PA. (The internal impedance of the PA is usually unknown but the circuit is assumed to behave as if a conjugate match exists.)

The two meanings are entirely different from each other. If there is danger of confusion then the meaning should be stated. Some people already use the descriptions "Zo match" and "Conjugate match".

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Reg, G4FGQ
Reply to
Reg Edwards

Meaning 3: The PA has been designed to deliver the maximum power at that load impedance and the distortion is not an issue.

Actually meanings 2 and 3 are effectively equal in the case of the tuned system, if you define the Zo based on the change in output power vs connected impedance for small changes. Since a lot of such systems aren't linear, this is the way you end up having to define the impedance. You can't use open circuit voltage and short circuit current.

Remember that this all started with the OP having a "transmitter". This would include any needed filtering. He was just connecting a 1/4 and 1/2 wave lengths of bent up wire. His output filter, I assume is just a bunch of LC sections.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

In article , Tam/WB2TT wrote: [...]

It say " Strictly speaking..." is that the section you are refering to?

Yes they do say that but then if you follow the design through, I think you will find that they try to cancel the reactive component. If you then put in the output device protection they didn't include, you end up with the matching as I explained elsewhere. As the impedance moves away from what the designer intended, the radiated power decreases.

The point is that the OP was taking about hooking a "transmitter" to a length of wire. I was talking about the matching from a complete transmitter which I assume contains such protection to this length of wire.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

In article , Tam/WB2TT wrote: [...]

Actually, in this case, I was speaking of matching the transmitter's output to the load. The transmitter already contains gawd knows what L and C components etc. The OP has a completed transmitter and a hunk of wire. If he matches the wire to what the transmitter wants to see, the transmitter will be happy. If he causes a reactive current to flow that the designer did not design for he will cause added heating in the output device. If the designer did a good job, the transmitter will protect its output devices and thus end up producing less power.

Also if he makes the real component of the impedance vary from what the designer intended, the output power will decrease. Which direction gets limited by the Vcc and which by the protection circuit depends on the collection of Ls and Cs inside the transmitter.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

The phrase "output impedance" in connection with amplifiers is ambiguous and likely to result in arguments.

The correct description is "internal impedance" or "internal resistance" and should always be used.

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Reg.
Reply to
Reg Edwards

I read in sci.electronics.design that gwhite wrote (in ) about '1/4 vs 1/2 wavelength antenna', on Fri, 25 Feb 2005:

What is a 'load line'? A straight line on an I/V graph? What does the gradient of that line represent?

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Regards, John Woodgate, OOO - Own Opinions Only. 
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
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Reply to
John Woodgate

I read in sci.electronics.design that Larry Brasfield wrote (in ) about 'Say what you mean.', on Fri, 25 Feb 2005:

The problem is that people say 'output impedance' when they mean 'load impedance'. To prevent misunderstanding I use the term 'output source impedance'.

There are also some people who use 'input impedance' when they mean 'source impedance'. I don't talk to them. (;-)

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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
Reply to
John Woodgate

My "Bailey" amps (Wireless World c. 1970) have lots of internal resistances, all soldered in neatly by hand.

In your comment below, I think that you have "output" much "impudance".

and

and

Reply to
Airy R.Bean

What's the S22 of an IC-756PRO?

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73, Cecil
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Reply to
Cecil Moore

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