Calculate Coaxial Cable Loss & Input Impedance at 1 MHz

[nelectrode]

Hey Guys,

How do I work out Transmission loses and input impedance for coaxial cable?

I’ve been given this website to use http://fermi.la.asu.edu/w9cf/tran/

I do have the characteristic impedance of 75Ω and frequency and I am asked to work the attenuation in dB/100ft and “input impedance for a 150Ω load resistance”
So I am not sure How to complete the input data.
Cable type: RG59
Frequency in Mhz: 1
Cable length in feet: 100
Load resistance: 150
Load reactance: How I find that? Do I have to use Z=R+JX?

Thanks

 

[davenn]

hi there

all cable manufacturers have datasheets for their cable
this always includes a table showing
losses usually per 100ft or 100 metres at a given frequency, sometimes it may state a shorter length
10 metres or 30 ft

for standard RG59 the loss at 1MHz is going to be less than 1dB for 100 ft ( 33 metres)

Load resistance: 150

resistance or impedance ??

you can't put a load with an impedance of 150 Ohms on a 75 Ohm cable
the mis-match and resulting SWR is going to be horrible

Your load impedance must equal your cable impedance for best power transfer, else you will end up with a large reflected power component

Cheers
Dave

 

[tfr000]

you can't put a load with an impedance of 150 Ohms on a 75 Ohm cable
the mis-match and resulting SWR is going to be horrible

I think that's the idea here. The impedance changes depending how many wavelengths you go down the transmission line.
Old-school, done on a "Smith Chart", but I guess that Java app is supposed to do the same thing.

 

[Averagesupernova]

you can't put a load with an impedance of 150 Ohms on a 75 Ohm cable
the mis-match and resulting SWR is going to be horrible

Your load impedance must equal your cable impedance for best power transfer, else you will end up with a large reflected power component

Cheers
Dave

This depends on the length of the transmission line in relation to the wavelength. An insignificant portion of a wavelength in transmission line length will not cause any problem at all. While we are on the subject of transmission line length what about the matching of the transmitter to the transmission line? Suppose the transmitter output impedance is 150 ohms and the load is 150 ohms but connected with 75 ohm line? If the line length is a small portion of a wavelength then no problem. If it is a half wavelength, then the line will appear as if it is not there at all minus some loss. I've only scratched the surface of this subject.

 

[sophiecentaur]

It's not usual to find that both transmitter and antenna are well matched to the transmission line. The amplifying device is often operated just for 'best' power and the antenna would be matched to make sure most of that power goes up the spout. (In the same way that a 'good' battery has a very low output resistance but you would never dream of putting a 0.05Ohm (max power) load on it.)
Another reason for matching the antenna is that a mismatch produces standing waves which can the maximum voltage on the amplifier output. This can harm the device.

To deal with your particular problem. The operating frequency is 1MHz, so cable loss will be negligible for 100ft of feeder. If you look it up on the cable spec (here, for instance) you can see just how low it is at 1MHz!

If you use the Applet they give you, you just need to put in the load impedance (which is 150Ω), cable impedance and length to find the impedance that the transmitter will 'see'. I think it will give you all the answers you want. (It does the difficult bits for you.)

 

[davenn]

This depends on the length of the transmission line in relation to the wavelength. An insignificant portion of a wavelength in transmission line length will not cause any problem at all. While we are on the subject of transmission line length what about the matching of the transmitter to the transmission line? Suppose the transmitter output impedance is 150 ohms and the load is 150 ohms but connected with 75 ohm line? If the line length is a small portion of a wavelength then no problem. If it is a half wavelength, then the line will appear as if it is not there at all minus some loss. I've only scratched the surface of this subject.

the problem is we don't cut coax lengths to a specific portion of a wavelength of the freq used in a particular system. We cut the coax to what is needed to get from the TX connector to the antenna connector ( ignoring any cavity filters, circulators that may be also inline). This means the coax is always just some random length depending on the height of the antenna up the tower.
Because of this we match the TX (RX) output (input) to the coax characteristic impedance and then that of the coax to the antenna.. If we don't, then we have the problems that Sophie and I have already stated concerning standing waves ( SWR) and the resulting high levels of reflected power back to the transmitter.
This can cause 2 common problems

1) as Sophie stated it can damage the transmitter output stage
in a lot of transmitter equip, protection is provided by measuring the reflected power level and it is used to lower the TX output power level till the problem is resolved

2) That reflected signal mixes in the TX output stages to produce intermodulation products that then get sent to the antenna and transmitted and can cause all sorts of hell with other comms services in the area

Matching TX, Xmission line and antenna is always the best method

Dave

 

[Baluncore]

In theory, a load mismatch is not line length dependent if the source impedance is matched to the line. That is because there will be no energy re-reflected again when returned to the source, since the source is correctly matched to the line.

Real sources such as transmitters will reflect the energy, because the output stage will be adjusted to a different impedance that reflects all the reverse energy. The transmatch will be adjusted to minimise absorption of energy by the transmitter by maximising forward power.

 

[sophiecentaur]

In theory, a load mismatch is not line length dependent if the source impedance is matched to the line. That is because there will be no energy re-reflected again when returned to the source, since the source is correctly matched to the line.

Real sources such as transmitters will reflect the energy, because the output stage will be adjusted to a different impedance that reflects all the reverse energy. The transmatch will be adjusted to minimise absorption of energy by the transmitter by maximising forward power.

I wondered what you were getting at there but I see what you mean; it's a strictly theoretical comment. A well terminated source will kill reflections but at what cost to the requirements for the transmitter design? It's hard enough to build a transmitter that will provide the right power at the right cost and efficiency, without also requiring it to look like 50Ω .

 

[sophiecentaur]

@davenn
To add to your list:
Tx line reflections will also affect the frequency response (short lines) and introduce echoes / ringing (long lines) which is particularly relevant to analogue TV and tall masts.

 

[Baluncore]

The OP was a transmission line question. The RG59 cable with Zo = 75Ω suggests a receive system with a probable maximum power of one mW. There is no need for a transmitter. Academic exercises like this are often based on theory, not practice.

It's hard enough to build a transmitter that will provide the right power at the right cost and efficiency, without also requiring it to look like 50Ω

That may be true for an inexperienced amateur, but it is not true for a competent professional.

[Generalisation] Amateurs place a tuner between the transmitter and the line, then call it an antenna tuner, but it is really a transmatch. Professionals tune the line to the antenna so as to avoid reflected energy on the line. [/Generalisation]

 

[sophiecentaur]

That may be true for an inexperienced amateur, but it is not true for a competent professional.

If a transmitter is to operate at high efficiency, it will need to have an output impedance that is not matched to the line. If it were matched, the highest efficiency (just in RF terms) would be 50%. That would hardly be a "professional" design.

 

[nelectrode]

Hey,

Thanks a lot, but as Baluncore stated this is purely a theoretical question,

so I basically have to use the Java app to work out my values.

I do know the following only:

Type: RG-59
Characteristic impedance: 75 Ω
frequency: 1Mhz

have to find:

Matched attenuation(dB/100ft): ?
Matched attenuation (dB/100m):?
input impedance for 150Ω load resistance: ?so on the app at http://fermi.la.asu.edu/w9cf/tran/

I have:

Cable type: RG-59
Frequency: 1Mhz
cable length in feet: 100
Load resistance: Do I have to put 150Ω here?
Load reactance: I thought it is going to be 75Ω but that's seems to be wrong.

Thanks

 

[sophiecentaur]

Load reactance will be zero if you're told 150 Ohms (i.e. no reactance in series)
That applet seems to do all you need.

 

[nelectrode]

That's what I thought but the values don't really match the one here:

http://www.embedded.com/print/4402915

So what is input Impedance for a 150Ω load resistance?

Because the app gives a Z and Z0?

 

[sophiecentaur]

Z0 should be 75 Ohms but the loss per foot in the cable (this is a real cable they're discussing) will introduce some effective reactance, I guess. It seems to be about 2%, which is quite a lot but I think that's what it's about. RG59 isn't high quality cable. Put 75 Ohms into the 'user defined' option cable parameters and it gives you the ideal lossless cable with Z0 of 75 Ohms
The Zin is what you will see, looking into the transmitter end of the cable when you put your 150 Ohms at the other end. The line acts as a transformer to change the load impedance, as seen at the transmitter.
Applets are sooo useful - as long as yo have an idea what they're doing for you.
You guys have it so easy these days. All we had was a hand calculator or, before that, a Smith chart and china graph pencil,

 

[Baluncore]

If a transmitter is to operate at high efficiency, it will need to have an output impedance that is not matched to the line. If it were matched, the highest efficiency (just in RF terms) would be 50%. That would hardly be a "professional" design.

I disagree. Under what restrictive conditions is your statement true. Please explain.
Maximum power transfer occurs when the transmitter is matched to the line.
Are you actually suggesting matching by using a resistive network?

 

[sophiecentaur]

I disagree. Under what restrictive conditions is your statement true. Please explain.
Maximum power transfer occurs when the transmitter is matched to the line.
Are you actually suggesting matching by using a resistive network?

This can be confusing but, if your source impedance is the complex conjugate of the load impedance then you do get maximum power. Butttttt - you dissipate half of the power in the source resistance. (The melting anode or collector of your output stage) That is the 'downside' consequence of the maximum power theorem. Doing it is not a good idea when you have a 500kW transmitter. (Class C with 90% efficiency is favourite when the modulation system is suitable)

You don't need to use a resistive network for the matching (that would be plain loopy). But any form of matching for maximum power will have the above effect. It's not done in any system that I can think of except for distribution amplifiers, where the power is low and the marked gain needs to be relied on.
That goes for audio power amps too, they are usually designed as a voltage source.

 

[Baluncore]

nelectrode.
Are you neglecting velocity factor?

Attenuation is simply specified by the physical length of the cable.
Phase shift is specified by electrical length = physical length / velocity factor.

Attenuation.
Your source will provide energy that is attenuated along the cable to the load.
The mismatch will reflect energy.
The reflected energy will be attenuated on it's way back to the source.

Input Impedance.
The line length must be corrected for velocity factor, maybe about 66% for a 75 ohm foam line.
The electrical line length, (phase shift), can then be calculated from the transit time.
The reflected wave and forward wave phase combine to form the apparent Zin.

 

[sophiecentaur]

The applet works all that out for you. (Effective length etc.)
If you accept that the Z0 is the value of termination to eliminate reflections then the effect of loss in the feeder is to introduce the requirement for a bit of reactance. Mr Smith and his chart doesn't allow for that to be included easily, although I'm sure it can be done by someone really familiar with using the chart.

 

[Baluncore]

This can be confusing but, ...

Yes, I think you are confused.

The reactive component does not waste real power. It is a “reactive” recycling of circulating energy.
RF energy efficiency of close to 100% is possible with a matched and non-reactive transmission line.

Playing reflective ping-pong with energy on a lossy line can be very expensive. By avoiding reflected energy on the line, the transmitter will see a non-reactive load and so does not have to dedicate supply current to the untwisting of output stage phase.

Resistive “Pi” or “T” pads are often used when coupling broadband microwave signal modules to reduce the reactive effect of other module's mismatch, so they stabilise the module and it's specifications. I agree resistive pads should never be used to match power transmitters.

I'm glad I don't need to worry about a 500kW transmitter. I only have a total of 75 kW here and I do my best not to use it. Unfortunately, if I run the anodes too cool, the “getter” on their surface does not function so well and the cathode emission begins to fall.

 

[Averagesupernova]

I think Baluncore and sophie are barking up different trees.
-
Sophie says that an output stage with a very low Zout relative to the load will dissipate little power in said output stage compared to the load. I would say this is correct. Notice that sophie did NOT say that a reactive component was involved. Sophie also said that this is most efficient. This is where it gets questionable. Maximum power transfer does not imply most efficient. It implies that given an output that has a Zout of R, then you can load said output with a load of R and no matter how you adjust the load you will not get more power to dissipate in the load. Raising the load resistance will cause less power to be dissipated in the load but will NOT cause more power to be dissipated in the source. Lowering the load resistance will cause less power to be dissipated in the load but will cause MORE power to be dissipated in the source.
-
As I have stated in a different thread, introducing a reactive component in the load is hard on the output stage since when the most voltage is dropped across the output transistor (zero crossing) there is more current passing through the device compared to if there were a resistive load. Voltage and current are not in phase, so this naturally occurs.

 

[sophiecentaur]

If you had a 50 Ohm load connected directly to a transmitter with a 50 Ohm internal resistance, you would agree that was a perfect match (I hope). It is possible to match any load to any source impedance (at least in principle), so you could achieve the equivalent of the above - if you wanted to. But, under those conditions, you would have equal power dissipated in the transmitter as the power dissipated in the load. That is 50% efficiency and not very kind to the output device.
If you have any doubt about the Maximum Power theorem then you can look it up. I don't need to find it for you.
Very few power distribution systems use a matched arrangement (name one, in fact). Car batteries would all melt, given a 0,05 Ohm load and all the Generating Company's equipment would fly off its beds with such a high current demand (the equivalent of several times the present National Load). I could ask you what the source resistance of your mains supply is? Quick calculation: The volts probably dip by, perhaps 5V when you take 50A - so that means 0.1 Ohms. Would you connect a 0.1 Ohm load and expect the equipment to survive (without those fuses in the way?)
Whatever it is that you know about the practicalities of transmitters, you need to realize that, when you 'tune for max power', the max power you are tuning for is set by the practical limits of the amplifying device you are using and not how well it's matched.
You have stated that you 'back off' your transmitter to stop it knackering itself. That is the case for all transmitter designs. I suggest that you don't in fact, have any way of measuring the output impedance of your transmitter so you aren't in a position to assert whether or not you are providing it with a conjugate match.

There's nothing special in the principles involved, whether it's a battery, generator of transmitter.

 

[sophiecentaur]

I think Baluncore and sophie are barking up different trees.
-
Sophie says that an output stage with a very low Zout relative to the load will dissipate little power in said output stage compared to the load. I would say this is correct. Notice that sophie did NOT say that a reactive component was involved. Sophie also said that this is most efficient. This is where it gets questionable. Maximum power transfer does not imply most efficient. It implies that given an output that has a Zout of R, then you can load said output with a load of R and no matter how you adjust the load you will not get more power to dissipate in the load. Raising the load resistance will cause less power to be dissipated in the load but will NOT cause more power to be dissipated in the source. Lowering the load resistance will cause less power to be dissipated in the load but will cause MORE power to be dissipated in the source.
-
As I have stated in a different thread, introducing a reactive component in the load is hard on the output stage since when the most voltage is dropped across the output transistor (zero crossing) there is more current passing through the device compared to if there were a resistive load. Voltage and current are not in phase, so this naturally occurs.

I agree (and have stated) that a perfect match is NOT efficient. It involves losing half the power from your source.
I took the case of a Class C transmitter, which can be 90% efficient and which cannot be matched - for the above reason. The actual efficiency of output stages is more to do with the physics of the devices themselves and the need to 'control' the current flowing through them whilst avoiding too much voltage drop. But that's another issue.
I agree that reactive components are rather a red herring.
My post ,above, states the case about not matching source to load in power systems.

 

[Averagesupernova]

Yeah sophie I was thinking about car batteries, 1000 watt subwoofer amplifiers, the source impedance of a typical electrical service, etc. also. None of them worry about loading to satisfy maximum power transfer. Nor should they. In car audio, preoutputs have a Zout of a couple hundred ohms maybe up to 1000 I believe but the Zin at the amplifier is typically 10K. Interesting point there also, they don't worry about transmission line impedance either. My comment in a previous post about insignificant portions of a wavelength would apply here.

 

[Baluncore]

nelectrode, first let me apologise for the sound of battle in the background.

If a transmitter is to operate at high efficiency, it will need to have an output impedance that is not matched to the line. If it were matched, the highest efficiency (just in RF terms) would be 50%.

If you had a 50 Ohm load connected directly to a transmitter with a 50 Ohm internal resistance...

Resistance and output impedance share the same dimensions, but they are certainly not the same thing.

The difference in the barking up trees appears to come from the fact that a 50 ohm internal resistance in a low frequency power supply, is quite different to the output impedance of an amplifier driving a 50 ohm transmission line. Zout is actually the slope of the RF amplifier's load line, the ratio of the voltage to current in the output stage.

A power distribution grid needs a very low resistance so it does not vary in voltage with variable load conditions. If you could match your electric light to the grid, you would get most of the power available at the time. That would not be good.

 

[Averagesupernova]

Resistance and output impedance share the same dimensions, but they are certainly not the same thing.

The difference in the barking up trees appears to come from the fact that a 50 ohm internal resistance in a low frequency power supply, is quite different to the output impedance of an amplifier driving a 50 ohm transmission line. Zout is actually the slope of the RF amplifier's load line, the ratio of the voltage to current in the output stage.

I don't think there is a difference at all. Source impedance is source impedance. And that is the Zout of the previous stage. You claim that Zout is simply the ratio of voltage to current in the output stage. That ratio depends on how the output is loaded and the load does not define the Zout.
-
Edit: Naturally a DC power supply isn't concerned with a reactive component in it's Zout. But for the sake of argument here, reactive components are being left out of the discussion.

 

[Baluncore]

I don't think there is a difference at all.

I beg to differ.

A 1 kW RF transmitter driving a matched line does not generate >= 1 kW internally.
Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

 

[Averagesupernova]

I beg to differ.

A 1 kW RF transmitter driving a matched line does not generate >= 1 kW internally.
Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

The reason it does not generate >= 1kW internall is because it has a Zout below 50 ohms which is exactly what sophie has been saying. The load at the end of the line may be 50 ohms and the line may be 50 ohms and we can call that matched but that does not mean the Zout of the transmitter is 50 ohms. It may well be spec'd to drive a 50 ohm load, but that is irrelevant.

 

[davenn]

The reason it does not generate >= 1kW internall is because it has a Zout below 50 ohms which is exactly what sophie has been saying. The load at the end of the line may be 50 ohms and the line may be 50 ohms and we can call that matched but that does not mean the Zout of the transmitter is 50 ohms. It may well be spec'd to drive a 50 ohm load, but that is irrelevant.

and right there is the same subject that others were arguing about on another place on the net several years ago.
I vaguely remember it coming down to that the Zout of the transmitter can actually be quite a bit lower a way down ~ 20 Ohms. I have to admit that I didn't fully understand the debate way back then and even again with the same debate over the last ~ dozen posts.
The 50 Ohms label on the back of the TX is only stating the transmission line impedance required
NOT the TX output impedance, as you stated in your last sentence, Averagesupernova.

I will see if I can dig up some of the emails that flew between various participants

cheers
Dave

 

[the_emi_guy]

Let me provide a quantitative example of what Baluncore is saying:

Let's say my RF source has an impedance of 10 ohms (10 + 0j) and I am running at 100MHz.

I have these choices:

1 - Go ahead and drive the 50 ohm transmission line directly and accept that anything that reflects back from the antenna will reflect again at source.

2. Add (40 + j0) ohms in series with RF source. Source is now 50 ohms so energy reflected from the antenna will not be reflected again at source. But I have introduced significant loss in the process.

3 - Install a network at output of RF source consisting of 32nH in series and 64pf in parallel. Assuming these reactive components are lossless, we have not impacted the efficiency or created any new losses. Yet energy reflected from the antenna returning to the RF source will see this as exactly (50 + j0) ohms and there will be no reflection at 100MHz.

Option 3 has built up source impedance to 50 ohm at 100MHz without loss of power, this is a narrowband match.

What Baluncore is saying is that Option 3 is generally chosen in favor of better VSWR.

Think of it this way, we can move any point on the Smith chart to the center using a couple of reactive components at a given frequency.

On the other hand, if we need the source impedance to be 50 ohms across a wide frequency range, such as within a spectrum analyzer, then you use resistors and accept the losses (broadband match).

 

[Averagesupernova]

the_emi_guy, you are causing the output of the transmitter to see a 10 ohm load correct? Looking into the LC network? If this is your aim, here is what happens which gets up back to what sophie and I are saying is something to avoid:
-
Take a transmitter with Zout of 10 ohms as you described. Feed it into a 50 ohm load. We will say there is a voltage across the load of 10 VAC. Thevenize and you will find the thevenin voltage is 12 VAC. The load is dissipating 2 watts, the Zout of the transmitter dissipates .4 watts. Use an impedance matching network and now the Zout of the transmitter dissipates 3.6 watts. From the thevenin voltage you find this to be the case, and assuming a lossless matching network the power dissipated in the load will also be 3.6 watts. This kind of power dissipated in the output stage is considerably higher than the .4 watts when driving the load 'mismatched' which is exactly what sophie was saying is to be avoided in the first place.

 

[the_emi_guy]

Sorry, my input was not well thought out.

The output stage of an RF transmitter is significantly more complex than what I proposed. It includes a tank circuit with a carefully selected Q (too low and we get distortion from class B or C amplifiers since it provides the freewheeling when active devices are off), too high and peak currents are excessive.

Then there is the matching network. This is required since the RF amp itself wants to see a specific load, usually lower than 50 ohms for transistor amplifiers (often higher than 50 ohms for tube amps).

I *believe* that this same matching network (and it is more than two components) is designed to make the RF source look like 50 ohms for a reverse flowing signal.

I will have to try to find (or work out) a better example...

 

[Baluncore]

The transmitter output stage has an impedance transformer called a tank circuit. To cancel the inductive component of the wound transformer it has a parallel capacitor, adjustable for different frequencies.

The load line or V:I ratio of the amplifier's active element is transformed to the impedance of the transmission line by the tank.

A reflected signal returning from the output t'line is reverse transformed by the tank from a V:I ratio of 50 ohm in the t'line to the V:I ratio of the amplifier's active element, and so is not reflected.

 

[sophiecentaur]

A few basic points would not go amiss here.
Reactive components do not dissipate any power. Their effect may be to change the maximum volts of current for a given power, which may, by implication, be problematic for the equipment and cause further internal dissipation (hence the Power Factor correction which is done in AC power systems). This is not relevant to the basic issue. Specific details of tank circuits and the like, just cloud things.

Anyone who cares to look at this link can see what my point has been. It's only one of many sources which back me up in this. A transmitter (any electrical / mechanical power source) that is designed with efficiency in mind will not be matched to its load. Yes, you will get maximum power into your load for a given emf but that will demand twice as much input power (DC power supply, in the end) and it will require you to dissipate half of this power in the amplifying device.

The confusion that other contributors are showing is that they are making assumption about what their practical results are telling them. What actual evidence do they have that the output impedance of their transmitter is '50 - or whatever'? How have they measured it? Efficient transmitters tend to be non-linear so it is hard to determine internal resistance. You can't just measure Open Circuit and then put 50Ω across the output to see how may volts are lost - as you can with the mains or a car battery. I guess you could see the differential effect of 50Ω and 51Ω loads on the output volts. Have any of you done this - or equivalent?

The idea of a conjugate match is not of much significance because the feeder is assumed to be providing the transmitter with a 'good enough '50Ω' at its output.

 

[sophiecentaur]

Here are a few links on the subject of matching. Read what they are actually saying about the significance of transmitter / load matching and efficiency. They all agree with what I have been saying.
http://www.jaycar.co.uk/images_uploaded/impmatch.pdf
http://users.tpg.com.au/users/ldbutler/OutputLoadZ.htm
http://urgentcomm.com/test-amp-measurement-mag/maximum-power-transfer

 

[Averagesupernova]

Thank you sophie. I will take a little more time later too look at the links you have provided. I skimmed through them very quickly.

 

[the_emi_guy]

Here are a few links on the subject of matching. Read what they are actually saying about the significance of transmitter / load matching and efficiency. They all agree with what I have been saying.
http://www.jaycar.co.uk/images_uploaded/impmatch.pdf
http://users.tpg.com.au/users/ldbutler/OutputLoadZ.htm
http://urgentcomm.com/test-amp-measurement-mag/maximum-power-transfer

Thank you sophie. I will take a little more time later too look at the links you have provided. I skimmed through them very quickly.

Agreed, I would like to digest these but it will take me some time. There does seem to be a divergence of opinion on this interesting topic (or maybe I am just not up to speed).

In the meanwhile here is what I have done:

I used Motorola AN267 which was created to help impedance match RF power amplifiers to transmission line. I designed two cases:

Case1: 1 ohm source impedance to 50 ohm line.
Case2: 5+5j source impedance (10 ohms parallel with 10j capacitive reactance) to 50 ohm line.

The results are attached (clipped from Mathcad).

Images 1,2,3 are case1.
Images 4,5,6 are case2.

Note that in both cases the network presents the desired impedance to the output of the amp while simultaneously presenting 50 ohms to any signal traveling in the reverse direction from far end reflection.

I would say that this is a "first order" method of matching. More elaborate schemes are used that take into account non-linearities and load pull characterization, but it seems that the basic scheme provides backmatch.

 

[sophiecentaur]

@the_emi_guy
The calculations are probably right; you can clearly drive the software correctly and your network will present the 50 Ohm load to the transmitter as a 1 Ohm load and vice versa. That would be a match for maximum power. It would result in an equal amount of power being dissipated in the 1 Ohm source resistance. No problem for a low power amplifier and it would be a well behaved piece of equipment - like a good Test Signal Generator, for instance but, with 50% efficiency, a high power transmitter would produce loads of embarrassing heat in its output stage and cost a lot of money to run. That was the point I have been making. I have never claimed that you can't match power amplifiers to deliver maximum power but I know that high power transmitters are built to be efficient (90% or so) and so they cannot be matched in that way - else, where do they get the 40% lost power back from and why do they not melt into the ground? Would you comment on that issue please and not on the nuts and bolts of matching networks?

 

[the_emi_guy]

@the_emi_guy
Would you comment on that issue please and not on the nuts and bolts of matching networks?

I hear you sophie, I understand your point and I need to think about this some more...

 

[sophiecentaur]

OK
No problem.
I think this is just to do with the directions we are coming from. I have no problem with your last post at all.

 

[the_emi_guy]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

 

[Averagesupernova]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

Any time we match with a transformer, Q-section, pi network, T-network, L-network, etc. we will match in both directions. If something is spec'd to drive a 1000 ohm load and you have a 50 ohm antenna system you will most certainly need to transform the impedance.

 

[sophiecentaur]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

If you think of any matching network as a transformer and if you ignore reactances then the ratio looking one way will be the inverse of the ratio looking the other way. If you transform the 50Ω feeder to look like 1kΩ from the transmitter's point of view (that's 1:√20 turns ratio) then the 1Ω output resistance of the transmitter will look like 1/20Ω (20:1 times 1), from the feeder side of the transformer. Afaics, you are stuck with that and it means that designers have to take that sort of thing into account and make allowance for reflected power. But, as long as the antenna match is good, there shouldn't be any signals arriving back at the Tx.
For UHF frequencies and above, it is possible to include an isolator or circulator to avoid the problems associated with mismatch at the transmitter. But then you need to accept a 1dB (iirc) or more extra loss.

 

[Baluncore]

My analysis shows that there are two quite distinct situations involved here.

Firstly; For the RF signal path through multiple stages to beyond the output transmission line, I believe that my assertion of “impedance matching is required for optimum efficiency” is correct for any RF signal path.

Secondly; for regional AC power distribution grids and local DC power supply rails, including 12V car batteries, I believe that sophicentaur's assertion of “zero impedance source is required for optimum efficiency” is correct.


RF signal paths.

Firstly, we should not dismiss the reactive coupling systems that transform the complex impedance between RF stages as being irrelevant. They are vitally important to the RF energy economy. It is the reactive components in switching power supplies and class C or D amplifiers that make those devices so highly efficient.

The second point I want to make regards a fundamental principle of RF design, one that has now been in use successfully for almost a century. When considering the design of a two stage RF amplifier, the load line of the first stage output is known and the input impedance of the second stage is known. A network that will efficiently match those quite different complex impedances is required in order to fully utilise the capabilities of the active elements in both stages. Likewise, we know the load line of the final output stage and we must match that to the transmission line impedance. I agree that without careful matching it will still work, but it will be wasteful of RF energy and equipment resources. The argument that the output impedance of signal path modules must be as low as possible for optimum efficiency is clearly false.

I therefore make the substantial claim that the RF impedances throughout the system need to be matched. Failure to efficiently match at RF represents an underutilisation of the components available and a reduction in the maximum RF energy that can be passed to the transmission line.

The characteristic impedance of a linear transmission line does not effect the energy it will dissipate as heat. It simply sets the relative phase and magnitudes of the voltage and current propagating in its two independent directions. Likewise, in free space, the ratio of the E to M field strengths represents an impedance of approximately 120*Pi ohms. That does not in any way represent a dissipation of power. An amplifier with a load line having a negative Gm represents power gain, not dissipation.


Power Distribution Systems.

A power supply can be derived from the regional AC distribution grid which has an extremely low impedance. The impedance of a power supply should not be matched to the grid, for obvious reasons. So let's buy a 90% efficient switching power supply that generates a DC supply voltage with an effectively zero resistance or impedance at the maximum current we specify, and pay for. We all should recognise that a switching power supply is a class D amplifier and as such it can have very high efficiency. But all this purchase has done is extended the “zero impedance source needed for efficiency” situation from the AC distribution grid to an internal DC supply distribution rail, and yes, as expected, it has done it efficiently without any need to waste 50% of the energy in the process.

Wherever modules share a common energy source, or distribute power to many users or stages with variable requirements, efficiency is gained by having a low supply resistance. Apart from thick wires and plenty of parallel capacitance, the low output impedance is often achieved by providing the source with some form of voltage control feedback. Two examples are the control of field current of an alternator, or the error voltage comparator that adjusts the duty cycle of a switching voltage regulator.


Summary.

The boundary between these two distinct regions occurs near to or within the RF amplifiers or modules. Just where the regions approach is actually determined by the operating class of the amplifiers employed. Close examination of the transition within an amplifier module reveals that for optimal efficiency the two regions are separated by a reactive network. That reactive isolation prevents the low impedance power distribution supply from short circuiting the signal path. It also prevents the RF signal from influencing other modules through the power supply rails.

Class A amplifiers will always be inefficient, Class B will be better while classes C and D can be very efficient. By mismatching the power distribution supplies, while matching the RF signal path, it is possible, (by using class C or D amplifiers), to make equipment that operates at significantly better than 50% efficiency.

To ascribe to the RF signal path the “zero impedance = high efficiency” concepts applicable to power distribution systems is a mistake.

 

[sophiecentaur]

You clearly have lots of experience and knowledge about transmitter practice but the notion that the Maximum Power theorem somehow only works under certain circumstances is a step too far. It involves Conjugate matching, of course, Introducing reactance is a red herring. Resistive components are the only ones that dissipate Power.
"Underuse of" components is not as heinous a crime as 'overusing' Power. Good Engineering aims at minimising the appropriate losses. Electrical supply costs soon outweigh the cost of a powerful broadcast transmitter - the electricity bills would make your eyes water. No one would ever chuck away 500kW of RF power into the cooling water.

Likewise, in free space, the ratio of the E to M field strengths represents an impedance of approximately 120*Pi ohms. That does not in any way represent a dissipation of power.

If you look at the Impedance, looking into a transmitting antenna, you will come across a resistive component (if not, then it is not radiating any power). That resistive component is referred to as Radiation Resistance. The Characteristic impedance of a 50Ω line is the ratio of Reactances, as is the impedance of free space, effectively. But, in both cases, the Energy transfer is due to where the energy goes and not to the characteristic impedance.

 

[Averagesupernova]

I have a question for you baluncore. If zero output impedance is not applying to RF transmitters, then why are we not dissipating the same amount of power in the class C amplifier that drives the antenna system that we are radiating in the antenna? You said it yourself in a previous post.

A 1 kW RF transmitter driving a matched line does not generate >= 1 kW internally.
Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

I call you on this by saying: The reason it does not generate >= 1kW internall is because it has a Zout below 50 ohms which is exactly what sophie has been saying. The load at the end of the line may be 50 ohms and the line may be 50 ohms and we can call that matched but that does not mean the Zout of the transmitter is 50 ohms. It may well be spec'd to drive a 50 ohm load, but that is irrelevant.
...and then you go on to say:

The transmitter output stage has an impedance transformer called a tank circuit. To cancel the inductive component of the wound transformer it has a parallel capacitor, adjustable for different frequencies.

The load line or V:I ratio of the amplifier's active element is transformed to the impedance of the transmission line by the tank.

A reflected signal returning from the output t'line is reverse transformed by the tank from a V:I ratio of 50 ohm in the t'line to the V:I ratio of the amplifier's active element, and so is not reflected.

The reflected signal is irrelevant since the assumption is that there is no reflected signal in this discussion. In the real world there will always be some reflected power but it is low enough to be disregarded with respect to the subject being discussed.
-
As for the tank circuit behaving as an impedance transformer. I suppose one can look at it in this manner. However, you cannot have it both ways. If the tank is acting as a transformer and matching the low impedance line to the transistor/tube, then you have to treat it as an actual transformer in that it is only changing the V:I ratio and not the power. So, by your logic the active element would once again be dissipating the same amount of power that the antenna is radiating and you clearly said this is not the case:

Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

 

[Baluncore]

As this subject is fraught by the many devils in the details, any general statement in a non-mathematical language must be expected to fail under some interpretation. So, full speed ahead and damn the torpedoes.

Unlike a linear supply regulator, a switching power supply can drop the voltage without the need to dissipate energy. It does it by employing a series inductor that acts as a reactive current limiter. The duty cycle of the switch, in conjunction with another switch or diode, effectively pulse-width-modulates the average value of the fixed inductive reactance at the switching frequency. That regulates the average current and so can be used to regulate the voltage efficiently.

The load line of an active device can also be modulated by oscillating between two points on the line rather than operating on the linear middle part of the line. That is a difference between the class A,B and class C,D amplifiers. The tank circuit will be different for different class amplifiers. That is because the effective output impedance of class A and B will appear resistive while the output impedance of classes C and D will be a very low resistance in series with a reactive network that functions as a flywheel.

Class A is inefficient because it operates in the linear mode. It forms a simple model of an externally matched amplifier in which a significant proportion of the energy is wasted in the output of the active device. Class D is efficient because the switching is between a point with high voltage with very low current and a point of high current with very low voltage, neither of which dissipates high power in the output of the active device. The output impedance is effectively the ratio of the pulsed current average to the pulsed average voltage. The lack of in-phase voltage and current is consistent with the use of a reactive element to limit the output power.

There is another way of viewing the design of a linear RF power amplifier. We can consider it to be two amplifiers in one. One is a voltage amplifier, the other is a current amplifier. The one active element performs both those tasks. The operating conditions of the active element can be chosen to maximise the output power, W_out, which is the product of V_out and I_out. The selection of active element operating conditions decides the input impedance. Matching between stages then becomes a case of juggling the voltage and current gain to maximise power output for hardware investment and/or energy expenditure. It should be clear that the ratio of V to I is the terminal impedance, and that it does not matter what V to I ratio is used so long as output power is optimised.

It should also be clear that when considering power output, neither voltage nor current is more important than the other. As such, arguing for a minimum ratio of voltage to current is unrealistic. What we actually need for economy is a minimum real loss resistance in series with the output. The output impedance Zo is then the ratio of output voltage to output current, which is quite different to the lossy real Rs component.

This demonstrates that a fixed voltage power distribution system should have a minimum series loss resistance. The output impedance as an output voltage to output current ratio is then unimportant.

On the other hand, the transfer of power between amplifier stages needs to transfer all available signal V and I for maximum power. That requires the V to I ratio, the impedance, be matched between sources and their loads for efficient power transfer. I agree that Rs << Zo to minimise the waste of the hard-won RF signal energy.

 

[AlephZero]

As this subject is fraught by the many devils in the details, any general statement in a non-mathematical language must be expected to fail under some interpretation.

Indeed - so why don't you back up your assertions with some math? (Give some references, if you don't want to duplicate standard theory)

Otherwise, you are imposing on the rest of the PF community to pick through your non-mathematical assertions and figure out which of them are correct and/or relevant.

For example, IMO much of your last post is confused through mixing different definitions of "efficiency", but I'm not interested in spending an hour of my time (literally) unpicking it in detail.

 

[sophiecentaur]

@Baluncore
You are totally missing the wood for the trees.
1. You can never ignore the maximum power theorem. It applies everywhere - with or without the detail.
2. Reactive components dissipate no power.
3. A low efficiency transmitter can be probably be well matched from the feeder as well as into the feeder
4. High efficiency transmitters (amplifiers) must have very low (or very high, aamof) source resistances. The way they are tuned to the feeder will (usually) assume the antenna is matched to the feeder and will be done to achieve efficiency. If the 1Ohm source is matched to the 50 Ohm feeder then they cannot be working at 90% efficiency because they will be seeing a 1 Ohm Load.
5. Introducing switched mode PSUs into your argument can only go against your ideas because they are designed to have a low internal resistance as possible (i.e. they don't get hot).


You are introducing so many irrelevant factors to back up the ideas you got from somewhere about matching. The only way you can prove this for yourself, and get it right, is to do the actual calculations with some real values. The analysis will be very hard. At any particular frequency, you can reduce all your networks to a simple emf and one source Z so you don't need to over-complicate things to find the ultimate truth in this.

I have very little argument with you except when you claim that matching both ways is possible to achieve with high efficiency. That just has to be wrong on basic, almost philosophical, grounds.

 

[Averagesupernova]

I just wish my questions would be answered.