View Full Version : One-question quiz #4

It's been too quiet here lately. Time for another one-question quiz.

Many a ham uses an antenna tuner between a transceiver (or an amplifier) and the feedline. Some of us know that this device doesn’t really tune the antenna. Few hams know what it actually does.

What does an antenna tuner actually do?

24-hour rule, please -- don't give the answer for 24 hours, so others can have a chance to ponder it.

Radio

04-04-2015, 08:45 PM

I have a built in antenna tuner in my Kenwood TS570.

It goes clicky chatter chatter click clicky BEEP. And then I can talk.

Oh, I was supposed to wait on that, wasn't I?

Sorry guys. :bag:

electricflyer

04-05-2015, 02:31 PM

Pick me, Pick me, I know.

coupevillefish

04-05-2015, 05:29 PM

I Googled it. Found a great ARRL article. It basically prevents the return voltage creating high SWR from a mismatched antenna from reaching the radio.

Now another question :

coupevillefish

04-05-2015, 05:59 PM

I Googled it. Found a great ARRL article. It basically prevents the return voltage creating high SWR from a mismatched antenna from reaching the radio.

Now another question :

Another try at posting the follow on question. I understand that the antenna tuner can be in the radio, immediately behind the radio or at the base of the antenna. I one or the other better? More cost effective?

Another try at posting the follow on question. I understand that the antenna tuner can be in the radio, immediately behind the radio or at the base of the antenna. I one or the other better? More cost effective?

At the antenna end of the feedline is always best, because then the SWR will be low for the whole length of the feedline, which means minimum loss of power in the feedline.

Now back to the original question --

I Googled it. Found a great ARRL article. It basically prevents the return voltage creating high SWR from a mismatched antenna from reaching the radio.

And just how does it do that? ;)

Pick me, Pick me, I know.

Lay it on us.

Answer, with discussion:

Briefly, an antenna tuner intercepts power reflected back down the feedline by a mismatched antenna, and reflects it back up the feedline, returning it to the antenna for another try.

Consider a 100-watt 50Ω transmitter, connected with a 50Ω feedline to a 50Ω antenna. Everything is perfectly matched. One hundred watts come out of the transmitter, go into the feedline, and are radiated by the antenna. This system will have an SWR of 1.0 everywhere. Let’s assume the feedline is perfect and has no losses.

Replace that antenna by a folded dipole whose impedance is about 300Ω. The SWR is now about 6. The antenna radiates just half the power, or 50 watts, and reflects the other 50 watts back down the feedline. These 50 watts will re-enter the transmitter.

So what? Well, let’s assume that that transmitter is 80% efficient. It takes 125 watts from its power supply, puts out 100 watts of RF signal, and dissipates the other 25 watts as heat. Its designer will have provided enough cooling (heat sinks and fans) to handle those 25 wasted watts. Now, add 50 watts of reflected power that the amplifier must also get rid of, and the cooling load on the amplifier is trebled. The final stages may burn up, a common failure in transmitters. That’s why many modern rigs are designed to reduce power, or even shut down, with high SWR.

What to do? We could improve the antenna, but it’s easier to use an antenna tuner. The tuner will intercept and re-reflect that reflected power, sending it back up toward the antenna. Now there’s 150 watts going up the feedline even though the transmitter puts out only 100 watts. The antenna is radiating half the original power, and half the re-reflected power.

Of the re-reflected 50 watts, the antenna will radiate 25 and reflect the other 25 back down the feedline. The tuner will send it back up. Now there’s 175 watts going up the feedline.

After enough round trips, there will be 200 watts going up the feedline; and the mismatched antenna will be radiating 100 watts, which is what the transmitter puts out. Put a wattmeter in the line (between the tuner and the feedline) and – sure enough – it will show 200 watts going up and 100 watts coming back. This may have been surprising in the past, but now you understand it. The “extra” 100 watts has been making multiple round trips up and down the feedline.

Notice that some of the power is delayed by its extra travels. If the feedline is really long, this may actually be noticeable in the transmitted signal and is an example of ringing.

This also explains why, with a lossy feedline that loses some power as heat, the losses increase disastrously with high SWR. Much of the power goes up and down multiple times, with losses each way. That’s why an antenna tuner, though it saves the transmitter, isn’t as good as an antenna that’s properly matched to the feedline. It also explains why a feedline that’s perfectly adequate with a well-matched antenna may fail with a mismatched one. Imagine: If the mismatch were so bad (SWR=100) that the antenna radiated only 4% of the power that reached it, there would be 2500 watts going up and 2400 watts coming back. Common amateur feedlines can’t handle 4900 watts. Fortunately, amateur tuners can’t handle SWR=100 either.

After you have adjusted the tuner so that no reflected power re-enters the transmitter from the feedline, the transmitter will “see” a load that matches the impedance for which it is designed (probably 50+j0 Ω). Thus, if you answered that the tuner “matched” or “transformed” the feedline (or antenna system, really) impedance to the transmitter, your answer is strictly correct but unenlightening.

There are other ways to keep reflected power out of the transmitter. One of these is by the use of a terminated circulator (also called an isolator). That’s a device that has a third port on which you place a dummy load, so that direct power goes up the feedline, while returned power goes out the third port and into the dummy load. This would waste the reflected power; better to use a tuner to send it back up to the antenna.

wa8yxm

04-06-2015, 07:49 AM

The tuner adjusts either the ractance or capatance of the antenna (or both) so as to match whatever impedance the antenna is to the impedance of the transmitter so as to maximize power transfer. NOTE that there is loss in the tuner and the greater the adjustment it needs to make the greater the loss so resonate is best for the antenna.

Note also that though an invrted V can approach 50-52 ohms... Neither a dipole or a folded dipole nor a verticle are 50 ohm Dipole 75, folded 300 verticle 32.5

W7JZE

04-08-2015, 07:13 AM

Answer, with discussion:

Briefly, an antenna tuner intercepts power reflected back down the feedline by a mismatched antenna, and reflects it back up the feedline, returning it to the antenna for another try.

Consider a 100-watt 50Ω transmitter, connected with a 50Ω feedline to a 50Ω antenna. Everything is perfectly matched. One hundred watts come out of the transmitter, go into the feedline, and are radiated by the antenna. This system will have an SWR of 1.0 everywhere. Let’s assume the feedline is perfect and has no losses.

Replace that antenna by a folded dipole whose impedance is about 300Ω. The SWR is now about 6. The antenna radiates just half the power, or 50 watts, and reflects the other 50 watts back down the feedline. These 50 watts will re-enter the transmitter.

So what? Well, let’s assume that that transmitter is 80% efficient. It takes 125 watts from its power supply, puts out 100 watts of RF signal, and dissipates the other 25 watts as heat. Its designer will have provided enough cooling (heat sinks and fans) to handle those 25 wasted watts. Now, add 50 watts of reflected power that the amplifier must also get rid of, and the cooling load on the amplifier is trebled. The final stages may burn up, a common failure in transmitters. That’s why many modern rigs are designed to reduce power, or even shut down, with high SWR.

What to do? We could improve the antenna, but it’s easier to use an antenna tuner. The tuner will intercept and re-reflect that reflected power, sending it back up toward the antenna. Now there’s 150 watts going up the feedline even though the transmitter puts out only 100 watts. The antenna is radiating half the original power, and half the re-reflected power.

Of the re-reflected 50 watts, the antenna will radiate 25 and reflect the other 25 back down the feedline. The tuner will send it back up. Now there’s 175 watts going up the feedline.

After enough round trips, there will be 200 watts going up the feedline; and the mismatched antenna will be radiating 100 watts, which is what the transmitter puts out. Put a wattmeter in the line (between the tuner and the feedline) and – sure enough – it will show 200 watts going up and 100 watts coming back. This may have been surprising in the past, but now you understand it. The “extra” 100 watts has been making multiple round trips up and down the feedline.

Notice that some of the power is delayed by its extra travels. If the feedline is really long, this may actually be noticeable in the transmitted signal and is an example of ringing.

This also explains why, with a lossy feedline that loses some power as heat, the losses increase disastrously with high SWR. Much of the power goes up and down multiple times, with losses each way. That’s why an antenna tuner, though it saves the transmitter, isn’t as good as an antenna that’s properly matched to the feedline. It also explains why a feedline that’s perfectly adequate with a well-matched antenna may fail with a mismatched one. Imagine: If the mismatch were so bad (SWR=100) that the antenna radiated only 4% of the power that reached it, there would be 2500 watts going up and 2400 watts coming back. Common amateur feedlines can’t handle 4900 watts. Fortunately, amateur tuners can’t handle SWR=100 either.

After you have adjusted the tuner so that no reflected power re-enters the transmitter from the feedline, the transmitter will “see” a load that matches the impedance for which it is designed (probably 50+j0 Ω). Thus, if you answered that the tuner “matched” or “transformed” the feedline (or antenna system, really) impedance to the transmitter, your answer is strictly correct but unenlightening.

There are other ways to keep reflected power out of the transmitter. One of these is by the use of a terminated circulator (also called an isolator). That’s a device that has a third port on which you place a dummy load, so that direct power goes up the feedline, while returned power goes out the third port and into the dummy load. This would waste the reflected power; better to use a tuner to send it back up to the antenna.

Another amateur myth busted! WoW, Carl. Great explanation and surprising to me. I've heard for years that a tuner only makes the transmitter happy and does nothing for power levels at the antenna.

I've heard various explanations about what happens to the reflected power, usually along the lines of "it is dissipated in the tuner", but that never seemed totally believable to me since my tuners (when I used to use them) never seemed to get hot, or even warm.

So... back up the line it goes for a second, third, fourth, etc... "try" at re-radiating.

Hard to wrap my head around "... it will show 200 watts going up the transmission line..." even though the xmtr is only 100 watts. Intuitively that seems close to getting something for nothing. That old "energy can neither be created nor destroyed..." thing. But, I bow to the master on this one. RF isn't always intuitive. DC dead shorts can disappear at RF freqs with some matching circuits (I am told).

Neat-o. Thanks for a great explanation.

Hard to wrap my head around "... it will show 200 watts going up the transmission line..." even though the xmtr is only 100 watts. Intuitively that seems close to getting something for nothing. That old "energy can neither be created nor destroyed..." thing.

Think about the tuner-feedline-antenna system. There's 100 watts going out (radiated by the antenna) and 100 watts coming in (from the transmitter). That's the same rate of energy-per-second (100 watts = 100 joules per second), so the same amount of energy is going out as coming in.

There's additional energy running back and forth. Where did it come from? When the key was first pressed and the transmitter started pouring 100 watts in, it took a while for the power radiated by the antenna to build up. At first, with 100 watts coming up, the antenna was radiating only 50 watts. Then, after the first bounce-bounce-round-trip, it was radiating 75 watts. As energy bounced back and forth, eventually the antenna reached a full 100 watts radiated. So, at first, the transmitter was pouring in 100 joules per second and the antenna was radiating less than 100 joules per second. The extra energy got stored and is running back and forth. Energy that's traveling is power, and that's how we get 200 watts going up and 100 watts coming down.

When you release the key, that gradual buildup is undone by a gradual slowdown, and the stored energy finally goes to zero.

Simple stuff, no?

N3LYT

04-09-2015, 05:37 PM

I worked as the tech for a Sheriff dept my office was in the country jail needless to say we had a lot of cameras and they were feed with 75 ohm coax you could actually see the results of a miss match generally when there was an open end some where like an un terminated through port.

Radio

04-09-2015, 06:54 PM

Did anyone ever fool around with a Smith Chart? :whistle:

It's been so long since I've even seen a Smith chart that I'd have to learn it all over again from scratch. But there was a time when I could use Smith charts.

W9WLS

04-10-2015, 06:58 AM

I'm like Carl, It;s been quite a while since I actually USED one.

There are several programs out there that replaced it, and a few that emulate and do a graphic output of a Smith Chart.

wa8yxm

04-10-2015, 09:15 AM

Let us assume a 2/1 SWR and a line loss of 10% (Makes math easy)

Transmitter out 100watts

Power delivrd to antenna 90 watts

Power returned 81 watts

Power re-reflcteback to theantenna (Very good tuner) 80 watts

power delivered to antenna 2nd time72 watts

Power reflectd 36 We have now radiated 86 watts

As you can see every time we make the trip we loose line loss. TWICE.

My tuner is o.00 inches or mM from the antenna.. Thus the line loss between tuner and antenna is 0.00 watts.. All my losss are in the wire (Antenna) and tuner... This is because the tuner is the antenna feed point. Long wire antenna. End fed.

Let us assume a 2/1 SWR and a line loss of 10% (Makes math easy)

Transmitter out 100watts

Power delivrd to antenna 90 watts

Power returned 81 watts

Power re-reflcteback to theantenna (Very good tuner) 80 watts

power delivered to antenna 2nd time72 watts

Power reflectd 36 We have now radiated 86 watts

As you can see every time we make the trip we loose line loss. TWICE.

My tuner is o.00 inches or mM from the antenna.. Thus the line loss between tuner and antenna is 0.00 watts.. All my losss are in the wire (Antenna) and tuner... This is because the tuner is the antenna feed point. Long wire antenna. End fed.

Correct in principle, but not in detail. With an SWR of 2.0 at the antenna, the antenna reflects 1/9, not 9/10, of the power coming up the feedline. So (corrected) we have:

Transmitter out 100 watts

Power lost in feedline 10 watts

Power delivered to antenna 90 watts

Power radiated by antenna 80 watts

Power reflected by antenna 10 watts

Power lost in feedline coming down from antenna 1 watt

Power returned to (perfect) tuner and reflected back toward antenna 9 watts

Power delivered to antenna 2nd time 8.1 watts

Power radiated by antenna 2nd time 7.2 watts

Thus, after the first round trip, the antenna is radiating 87.2% of our original 100 watts. An SWR of 2.0 isn't so terribly bad.

Repeating with SWR about 5.8, the antenna reflecting half the power, we have:

*** Note: my math was wrong below, too. I've corrected it now ***

Transmitter out 100 watts

Power lost in feedline 10 watts

Power delivered to antenna 90 watts

Power radiated by antenna 45 watts

Power reflected by antenna 45 watts

Power lost in feedline coming down from antenna 4.5 watts

Power returned to (perfect) tuner and reflected back toward antenna 40.5 watts

Power making its second trip up the feedline and reaching antenna 36.45 watts

Power radiated by antenna 2nd time 18.225 watts

Thus, after the first round trip with SWR=5.8, the antenna is radiating less than 64% of our original 100 watts. Remember that all of the loss is in the feedline. It wasn't so bad at SWR=2, but at SWR=5.8 it's sickening.

W7JZE

04-11-2015, 09:16 AM

Think about the tuner-feedline-antenna system. There's 100 watts going out (radiated by the antenna) and 100 watts coming in (from the transmitter). That's the same rate of energy-per-second (100 watts = 100 joules per second), so the same amount of energy is going out as coming in.

There's additional energy running back and forth. Where did it come from? When the key was first pressed and the transmitter started pouring 100 watts in, it took a while for the power radiated by the antenna to build up. At first, with 100 watts coming up, the antenna was radiating only 50 watts. Then, after the first bounce-bounce-round-trip, it was radiating 75 watts. As energy bounced back and forth, eventually the antenna reached a full 100 watts radiated. So, at first, the transmitter was pouring in 100 joules per second and the antenna was radiating less than 100 joules per second. The extra energy got stored and is running back and forth. Energy that's traveling is power, and that's how we get 200 watts going up and 100 watts coming down.

When you release the key, that gradual buildup is undone by a gradual slowdown, and the stored energy finally goes to zero.

Simple stuff, no?

Thank you, Carl. I am learning lots from this thread. You are a gifted teacher.

I especially enjoyed the watts : joules/sec ratio. I knew the power * time = energy relationship. Comfortable using that for a long time to the point of it being intuitive. Hardly (if ever) thought about rearranging it as energy/time = power. I see the math but I had to think about it for a minute to "feel" it (If that makes any sense).

Also I was unaware of the exact relationship of watts (100) to joules/sec (100). That whole "joules thing" I always recognized as energy but never knew "How much" it was. J = W * S. Simple. I should have looked that up decades ago. Hearing that a megaton atom bomb releases ~4.18 x 10^15 joules never meant too much to me. I had no scale in my head to relate to it.

Now I know that it gets as "hot" as 4 Trillion toasters running for a full second.

Toasty, indeed :bounce:.

(PS: I also liked: "Energy that's traveling is power". I never thought of it like that. Makes sense. I just hadn't looked at it that way.)

Any homework tonight, Teach?

wa8yxm

04-11-2015, 10:19 AM

Correct in principle, but not in detail. With an SWR of 2.0 at the antenna, the antenna reflects 1/9, not 9/10, of the power coming up the feedline. So (corrected) we have:

(*!^@#(*&#@(!@B B (unprintable comment)

Thanks for the correction... That was what I meant to type but I messed up.. So thanks.

Now.. Why did I mess up like that (Must not have finished my first cup of coffee).

(*!^@#(*&#@(!@B B (unprintable comment)

Thanks for the correction... That was what I meant to type but I messed up.

I blew it too. My arithmetic was wrong when I was fixing your arithmetic. I've edited the correction, and now I think I've got it right.

Any homework tonight, Teach?

Yup. Memorize the following:

A watt is a joule per second.

A volt is a joule per coulomb.

A volt is a coulomb per farad. Thus a farad is a coulomb per volt.

An ampere is a coulomb per second.

A henry is a volt per (ampere per second), i.e. a volt-second per ampere.

An ohm is a volt per ampere. Thus a mho is an ampere per volt.

Manipulate these to show that:

An ampere per second is a volt per henry. Understand this to mean that, to change the current through an inductor by one ampere per second (increasing or decreasing), you must impose one volt across the inductor for each henry of its inductance.

A watt is a volt-ampere.

A joule is a farad-volt-squared.

... and so forth.

Now then, how many joules are stored in a two-farad capacitor charged to three volts?

W7JZE

04-11-2015, 10:41 PM

... <snip> ...

A watt is a volt-ampere.

A joule is a farad-volt-squared.

... and so forth.

Now then, how many joules are stored in a two-farad capacitor charged to three volts?

Ouch! Ouch! Just thinking about that makes my...

Head hurt... Ouch!

(I will try to work it out, though)

HMmmm... "A joule is a farad-volt squared"

1 farad * 1 volt = 1 farad-volt and 1^2 = 1 Joule. OK. Then, maybe...

2 farads * 3 volts = 6 farad-volts and 6^2 = 36 Joules.

36 joules? Maybe?

Actually I threw you a curve there, and also erred when I wrote, A joule is a farad-volt squared. It ain't. A farad (volt squared) is half a joule. Only the volts get squared, and I omitted the 1/2. To derive that from the simple equations requires the integral calculus.

Now then, how many joules are stored in a two-farad capacitor charged to three volts?

Most people would use the calculus to solve this. Let's see if we can do it another way.

Let's charge those two farads [C] with a constant two-ampere current. The voltage e will go up at one v/s.

At start, the capacitor is at zero volts, and the current is two amperes, so the power going in is zero. But the power starts rising immediately, because the voltage [e] starts rising immediately and the current is constant.

After one second, it's charged to one volt. Our two amperes are now flowing at one volt, so the power going in at that instant is two watts.

After two seconds, it's at two volts, and still two amperes, so the power going in is four watts.

After three seconds, it's at three volts, and still two amperes, so the power going in is six watts.

Since the voltage went up linearly and the current was constant, the power went up linearly too, from zero to six watts. So the average power was three watts over the three seconds. The energy, then, must be nine joules.

Does this consist with the answer from calculus? Yes. The result from integral calculus would be that the energy = (1/2)Ce^2, using your excellent notation. (I wish I could use superscripts here.) C = 2. e = 3. e^2 = 9. Thus (1/2)Ce^2 is nine joules.

We're good to go. We did it at constant current, but actually the energy in that capacitor doesn't depend on how it got there. It depends only on the capacitance and on the voltage across it. At two farads and three volts, it's nine joules, regardless. The capacitor doesn't "remember" the history, it only remembers the end result.

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