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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 

What load for maximum power transfer?
Poll ended at Wed Jan 06, 2010 11:21 am
50+j0: 13%  13%  [ 1 ]
0-j100: 0%  0%  [ 0 ]
50+j100: 13%  13%  [ 1 ]
50-j100: 50%  50%  [ 4 ]
100-j50: 25%  25%  [ 2 ]
Total votes : 8

 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
Silent Key

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Post Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
Folks,

I want to move on to something a bit "meatier" and look in detail at what a tuner actually does; but before doing that we need to make sure we understand something about matching. So here's the Puzzle:

Simon has been studying this simple diagram which shows an RF source connected to a load:

Image

He understands that the RF source is represented by the elements inside the dotted lines - a constant-voltage source in series with a source impedance Zsource. He's even been reading up on the "maximum power transfer theorem" and he understands that if Zsource were 50ohms, Zload would also have to be 50 ohms for the source to deliver the maximum power available to the load. But he now realises that Zsource might not be purely resistive - it might contain some reactance - and he's wondering what the load would then need to be for maximum power transfer. He asks the local club gurus what Zload would need to be for maximum power transfer if Zsource were a 50 ohm resistor in series with a 100 ohm inductive reactance - written in complex notation shorthand as 50+j100. As usual he gets a range of answers:


50+j0: A pure 50 ohm resistance, because the reactive component is irrelevant
0-j100: A pure 100 ohm capacitive reactance, because that cancels out the inductive reactance of the source and allows maximum current to flow in the load.
50+j100: A 50 ohm resistance in series with a 100 ohm inductive reactance, because for maximum power transfer the load impedance must be the same as the source impedance
50-j100: A 50 ohm resistance in series with a 100 ohm capacitive reactance, because the reactive components must cancel one another
100-j50: A 100 ohm resitance in series with a 50 ohm capacitive reactance, because that creates what is known as a "conjugate match"

Which of these should he believe?

Enjoy!
Steve G3TXQ

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Mon Jan 04, 2010 11:21 am
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
if he has 50+j100 , R50 with 100 ohms inductive reactance he only needs to cancel the inductive component out .

ill go with 2 answers

he wants 50+j0

but he needs 100ohms of capacitive reactance to cancel the inductive element so my other answer is........ he needs to "adjust" by

0-j100

i think this is what your asking and im not sure of my answers since iv not played with an analyser for a while and the old grey matter is rusting..im going with this before im tempted to go and have a quick look..

after editing 3 times to try and explain my choices i just cofused the issue so deleted the extra bit and ill stick with this choice..

billy

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Last edited by m0jha on Mon Jan 04, 2010 7:01 pm, edited 3 times in total.



Mon Jan 04, 2010 11:55 am
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
Gone with option 4 as to maintain max. radiation resistance you have to cancel out the inductive reactance with a equal and opposite force i.e 100ohm capacitance.

I am no expert on conjugate matching but isn't this a function of the combination of the source and the load i.e the complex resistance(s) to achieve max. rf power at the load i.e efficiency of the aerial system?

Suspect I am going learn something new here, so its all good.

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Mon Jan 04, 2010 12:52 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
Silent Key

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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
Folks,

I'm sure you appreciate that I deliberately avoid getting involved in the discussions so that I don't "give the game away" early and spoil it for others who may come late to the Puzzle

But I do enjoy reading the various postings; I'm encouraged to see them because healthy debate is just as likely as my answer to educate someone .... perhaps more so ;)

73,
Steve G3TXQ

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Mon Jan 04, 2010 4:49 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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I read this question, as asking what value in terms of reactance would be required to allow max power transfer i.e matching of the load and source. It seems to be the cancelling of the inductive reactance is key and option 4 has the correct value. No necessarily say that means the correct option\right answer.

Now my understanding of conjugating. would be if you could not achieve the cancellation of the inductive reactance by an equal and opposite force. Then by conjugating those know values i.e the 100 ohm inductive and 50 ohm resistive loss using opposite values to achieve max. transfer. Is that correctly defined in option 5? I dont know. Hence going for option 4.

I would not be surprised if option 5 is correct, but rather go on what little I know :roll: But as I said, I am sure going to a tad wiser once Steve has put us out of our misery :?


G7IGB wrote:
Zen Navigator wrote:
I am no expert on conjugate matching but isn't this a function of the combination of the source and the load i.e the complex resistance(s) to achieve max. rf power at the load

Now I'm confused Dave, is this not what Steve is looking for in his teaser? :?

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Mon Jan 04, 2010 5:04 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
G3TXQ wrote:
Folks,

I'm sure you appreciate that I deliberately avoid getting involved in the discussions so that I don't "give the game away" early and spoil it for others who may come late to the Puzzle

But I do enjoy reading the various postings; I'm encouraged to see them because healthy debate is just as likely as my answer to educate someone .... perhaps more so ;)

73,
Steve G3TXQ


How right you are Steve.

Rob's comments made me rethink and my further explanation, as placed a seed of doubt in my mind. If I am correct about conjugate matching then your definition in option 5 looks probably. But so does option 4!!

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Mon Jan 04, 2010 5:21 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
the question doesnt mention conjugate matching . its merely an option for an answer

unless theres more than 1 answer..

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Mon Jan 04, 2010 5:25 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
m0jha wrote:
the question doesnt mention conjugate matching . its merely an option for an answer

unless theres more than 1 answer..


Dont you bloody start :lol:

I know Billy, what I was trying to say in all the waffling was that if indeed my interpretation of conjugate matching is correct and the value in option 5 seems to match that citeria? Maybe I just disappearing up my own orifice :shock:

Option 4 rules ok :roll:

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Mon Jan 04, 2010 5:35 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17
G7IGB wrote:
Zen Navigator wrote:
Rob's comments made me rethink and my further explanation, as placed a seed of doubt in my mind.

This was not my intention, I'll keep my comments to myself in future. :oops:

Apologies Steve, I'll await your final answer without further interruption. :)[/quote

At times like these I like to employ the wisdom of Brain Clough... We talk about it for 20 minutes and then we decide I was right :lol:

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Mon Jan 04, 2010 8:25 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
Silent Key

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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
Folks,

I'll post my answer on this one tomorrow morning, so there's still this evening for anyone else to try the Poll. Only 6 have ventured an opinion so far; remember - it's anonymous, so there'll be no embarassment if you're wrong ;)

73,
Steve G3TXQ

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Tue Jan 05, 2010 4:31 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
I went with option 3 I did some reading on a site I found on google but It was a bit over my head.

It would be great if Steve or someone could recommend some basic reading on this subject as I have never came across all this 50+100J stuff before.


Thanks

Trevor

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Tue Jan 05, 2010 6:14 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
Silent Key

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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
ei2glb wrote:
It would be great if Steve or someone could recommend some basic reading on this subject as I have never came across all this 50+100J stuff before.


Trevor,

It's a bit difficult to recommend appropriate reading without knowing what level you're starting from, but if you can work your way through the 6 pages starting here:

http://www.allaboutcircuits.com/vol_2/chpt_2/1.html

you'll have an excellent grounding!

If you, or anyone else, wants to try those pages and has any difficulty understanding, why not post a question in this forum under a new thread: "Complex Notation" and I'll try to help out.

Once you understand complex notation, it greatly simplifies the analysis of AC circuits; and as you've probably noticed I tend to use it quite often in my Puzzles.

By the way, that web site is an excellent reference for a whole lot of circuit analysis info.

73,
Steve G3TXQ

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Tue Jan 05, 2010 6:54 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
Silent Key

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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
Folks,

The correct answer is #4: 50-j100

It's probably more obvious if I re-draw the schematic to separate out the resistive and reactive components and show them as an inductor and a capacitor:

Image

Now it should be clearer that if we have a -j100 reactance in the load, which cancels the +j100 reactance in the source, we have reduced the problem back to the purely resistive case; and we know that the load resistance must be equal to the 50 ohm source resistance for maximum power transfer.

So, in general, if Zsource is R+jX then Zload must be R-jX for maximum power transfer.

We refer to R-jX as the "Complex Conjugate" of R+jX; and of course R+jX is also the "Complex Conjugate" of R-jX:

* Complex because we are using complex notation - not because it's difficult to understand! The dictionary definition of "complex" is: "made up of different, connected, parts".
* Conjugate because they are a pair of numbers which are related in a complementary way - I'll leave you to picture why the word originates from the same Latin word as "Conjugal" ;)

I hope that doesn't all sound too difficult. The only thing you really need to remember is this: if we had a source with an impedance of 3-j5, we would need to provide a load of 3+j5 to ensure maximum power transfer; we then say that we have achieved a "Conjugate Match". In other words the resistive components are equal, and the reactive components are equal but of opposite sign. In part 2 we'll look at how this works out in our antenna systems.

Finally, just to comment on each of the possible answers:

1. 50+j0: Wrong - the reactance of the source does matter; if we do nothing about it, it will impede the current flow to the load.
2. 0-j100: Wrong - this load would cancel the reactance of the source, but because it has no resistance it can't dissipate any power. All the power dissipated would be in the source, and the power transfered to the load would be zero! The load would be a simple capacitor, and ideal capacitors don't dissipate power.
3. 50+j100: Wrong - we've now increased the total reactance in the "series loop" to +j200 which impedes the current flow even further.
4. 50-j100: Correct!
5. 100-j50: I slipped this in and mis-used the magic word "conjugate" to test that you knew what the term meant. 100-j50 is not the conjugate of 50+j100. Using a 100-j50 load we would neither fully cancel the source reactance, nor provide a match between the resistive components.

I know it's a "complex" topic, so if anyone wants further clarification I'm happy to give it, either by posting or by PM

73,
Steve G3TXQ

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Wed Jan 06, 2010 9:14 am
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
Silent Key

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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
In view of Dave's (Zen Navigator) posting on a new thread, let me just explain things a bit more.

Let's start by considering the case where the source impedance is a pure resistance. I'll change its value to 25 ohms so we don't get "hung up" thinking everything has to be 50 ohms:

Image

Let's think about what value of load would "extract" most power from the source. Here's a few possibilities:

1. Let's make R equal to zero - a short circuit. That will certainly make lots of current flow - in fact 100/25=4 Amps will flow. The problem is the voltage across the short-circuit load will be zero, so the Power=VxI=0x4=zero Watts.

2. Let's go to the other extreme and try to keep the voltage across the load high by making it an Open Circuit. Then the voltage across it will be 100v, but the current through it is V/R=100/Infinity=0 so now Power=VxI=100x0=zero Watts again.

3. Let's try something in between; say R=20 ohms. Then Vload=100x(20/(20+25))=44.44v and the current through the load will be 44.44/20=2.22Amps, so the Power=VxI=44.44x2.22=98.65W

4. Another possibility is R=30 ohms. Then Vload=100x(30/(30+25))=54.54v and the current through the load will be 54.54/30=1.82Amps, so the Power=VxI=54.54x1.82=99.26W. Compared to the 20 ohm load we've increased the voltage, but reduced the current.

5. Now try R=25 ohms. Then Vload=100x(25/(25+25))=50v and the current through the load will be 50/25=2Amps, so the Power=VxI=50x2=100W - better than either 20 ohms or 30 ohms.

I hope you can see intuitively that making the load resistance equal to the source resistance transfers the maximum power possible to the load. A complete proof requires the use of differential calculus, and I doubt you want me to go there ;)

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

Now let's add 200 ohms of reactance to the source impedance:
Image
If we keep the load as a 25 ohm resistor we'll find that, because the reactance is now impeding the current, less will flow and so less power will be transferred to the load. I wont trouble you with the maths, but the current will drop from 2A to 0.48A and the power dissipated in the load will be only 5.7W.

Pretty clearly we need something in the load that will "negate" the reactance of the source, allowing the original 2A to flow; and that of course is a capacitive reactance of -j200. Think of it as choosing a load reactance which will "tune out" the source reactance.

So we're back to our "conjugate match" theorem which tells us that if the source impedance is 25+j200 ohms we'll achieve maximum power transfer if the load is 25-j200

Hope that may be a little clearer.

73,
Steve G3TXQ

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Wed Jan 06, 2010 4:35 pm
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 Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1 
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Post Re: Technical Puzzle #17 - Matchmaker, matchmaker ...... Pt 1
m0jha wrote:
if he has 50+j100 , R50 with 100 ohms inductive reactance he only needs to cancel the inductive component out .

ill go with 2 answers

he wants 50+j0

but he needs 100ohms of capacitive reactance to cancel the inductive element so my other answer is........ he needs to "adjust" by

0-j100

i think this is what your asking and im not sure of my answers since iv not played with an analyser for a while and the old grey matter is rusting..im going with this before im tempted to go and have a quick look..

after editing 3 times to try and explain my choices i just cofused the issue so deleted the extra bit and ill stick with this choice..

billy



hi steve, the correct answer os #4.. 50-j100 for maximun power transfer. so isnt my 0-j100 the same thing?? needing to do nothing with the R as theres already R50 there but only adding 100 ohm of capacitance..

i understand the final value reading would be 50-j100 but my choice was based on waht would be needed to get to this figure from original reading of 50j+100

don't get me wrong im not trying to argue a point i just thought this may be the same thing adjustment wise..

billy

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Thu Jan 07, 2010 10:11 am
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