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Old 7th Aug 2018, 11:07 pm   #21
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Default Re: Back terminated cable?

I can't recall having been taught about this subject in my transmission line theory lectures in the late 1960's but did come across it in the book "Semiconductor Circuit Design Volume II", Texas Instruments, 1973, where it is called "Reverse Termination" in connection with TTL and Schottky TTL transmission lines, scans attached. Interesting that terminating with a value different from the characteristic impedance of the line can be use to optimise the balance between the respective waveforms at the sending and receiving ends.

When I was designing with TTL in the early 1970's, it was not an issue: our connection paths were short, and our design philosophy was to use clocked logic to ensure that conditions had reached a steady state before inputs were enabled.
Attached Files
File Type: pdf rev term 29_34 Schottky TTL.pdf (280.2 KB, 33 views)
File Type: pdf rev term 12_15 TTL.pdf (204.5 KB, 28 views)

Last edited by emeritus; 7th Aug 2018 at 11:22 pm.
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Old 8th Aug 2018, 5:12 am   #22
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Default Re: Back terminated cable?

It's also, I believe, the reason for the resistors (often 33 Ohm) in series with the dynamic RAM address signals on many (older) computers.

It generally wasn't necessary to design the interconnections between TTL ICs as transmission lines, certainly not on small designs. ECL is an entirely different matter. They you design everything as a transmission line, and the circuit works properly as a result!
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Old 8th Aug 2018, 6:10 am   #23
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Default Re: Back terminated cable?

I think there could be an issue if the output impedance of a transmitter or power RF amplifier doesn't match the transmission line(or applied load impedance) even though as pointed out if the correct impedance transmission line is terminated in the correct load impedance, that no reflections are sent back to the transmitter or amplifier's output.

Usually with the amplifier's own Zo matched to its load, it is possible to transfer the maximum power to the load within the limitations of the power supply and device ratings and heat sinking and thermal considerations which determine the maximum possible power output.

In many cases with modest power outputs it really wouldn't matter most of the time if the amplifier's true Zo didn't match the load, especially if the amplifier's Zo was a lot lower than the load presented to it. After all, this has been the custom in transistor amplifiers for audio work where the amp's output Zo is usually much lower than the the speaker's Z. It is harder to get the Zo as low from RF power amplifier outputs though.

But isn't it correct that for optimised power transfer and efficiency, at full possible power output, while keeping the output devices inside safe limits, it's still better to have a match between the amplifier's Zo and the presented or applied load ?

So the question might not relate to impedance bumps and reflections at the junction of the amplifier's output connection and the impedance presented there, but rather maximum power transfer and safe operating conditions for the output devices.

For example: If the impedance presented to the amplifier's output is too high, then the output devices (plate, collector or drain terminals) clip to the supply rail long before full power output is delivered. On the other hand, if the applied load impedance is too low, more energy is dissipated in the output devices and less in the load, so at full power in the load, the output devices might be pushed past their limits. Then if the drive levels are wound back to correct this problem, the output power delivered to the load is too low.

One advantage also I found with transmitters having a Zo that matched the load (in this case 50R) it is easily to see that the amplifier is working normally with the dummy load applied at full power with the scope across the load or the output devices and that at lower output levels the amplitude drops by 6dB when the load is applied helping to confirm all is normal with the RF amplifier . When the RF amplifier's Zo has not been designed for a specific target value and is not known, it's not as helpful.
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Old 8th Aug 2018, 6:39 am   #24
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Default Re: Back terminated cable?

We got away with it in TTL logic circuitry, but where transmission line approaches came to the fore was with ECL logic. Outputs were emitter followers and inputs were open bases with just a high value pulldown and the speeds were much higher than Schottky TTL. I did one design which took in serial data at 139.264 MB/s, converted it to 16bit parallel form. shoved it into an elastic store and pulled it out with an asynchronous clock and squirted it out as serial under the new clock. It was used to apply controlled jitter to passing data. Everything was synchronous to one clock or the other and the clock periods varied around 7.18ns. There were around a hundred terminated transmission line connections and the two clock distribution systems had to be dead right. Got it working and shipping to paying customers!

There is some very good material on transmission lines in the Motorola 'MECL system design handbooks - worth reading even if you're not using that logic family.


Some transmitters do act like source impedances close to 50 Ohms, but it's coincidence rather than design. Output stages have networks designed to transform the (usually 50 Ohm) impedance that they should be presented with at their output connector into whatever impedance resonates with the output device and extracts the planned amount of power. What this network transforms the anode/collector/drain impedance to when seen looking in the output connector could be anything and will vary across the band of operation.

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Old 8th Aug 2018, 7:03 am   #25
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Default Re: Back terminated cable?

Crossposted with Hugo.

The VHF transmitters I've done are required to produce at least 50% of their rated power into a 2:1 VSWR and 40% into 3:1. This is due to American and European requirements being specced differently. So the transmitters have to handle strong reflections, but the length of coax to a bad antenna could be anything, so I have to test the designs with all phase angles of reflection using trombone lines extended by a one or other from a set of different length cables.

What genuinely throws the spanner in the works for the idea of really matching output devices is that the output impedance of the device in class B, C, and onwards operation changes dramatically around the cycle.

There is a chap over in the states writing for the amateur radio market who takes the view that all transmitter output devices must be used in a full both-ways match. He never mentions cyclic impedance variation! This caused a very large and somewhat heated debate on the internet in the nineties. It may still be going. One guy calmly putting the view of single direction loading was Warren Bruene of Collins radio fame. A very calm and thoughtful man with an awful lot of professional RF design under his belt. That doesn't guarantee someone is right, but it does suggest that what they say ought to be listened to and thought about. Once you do the detailed maths rather than just talking averaged values, you soon see what's going on.

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Old 8th Aug 2018, 10:58 am   #26
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Quote:
Originally Posted by Radio Wrangler View Post
I've usually used the phrases "source terminated" and 'load terminated" for single-terminated lines because their meaning is clearer.
Excellent! For me, 'source terminated' is a much clearer and descriptive phrase to use than 'back-terminated'.

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Old 8th Aug 2018, 11:13 am   #27
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Quote:
Originally Posted by Thyristor View Post
I had a similar incident that you describe: happened to me at work recently. I said to a (new) engineer "....it won't work, because t'other end isn't terminated...." which was met with a blank look.
Ditto. Recently, after a long period of internet crashes, I had a visit from an OR engineer. He quickly identified the problem by examining the line from the house to the pole in the street using a TDR: it showed a 'hump' on the display. So I commented on that, using the term 'TDR'.
"Huh? 'TDR': what's that?" came the reply.
"Time domain reflectometer" I said.
"Oh" - and I received a blank, puzzled look.
A replacement cable solved the problem.

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Old 8th Aug 2018, 11:38 am   #28
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Default Re: Back terminated cable?

Back on topic please.
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Old 8th Aug 2018, 2:09 pm   #29
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Quote:
Originally Posted by Argus25
But isn't it correct that for optimised power transfer and efficiency, at full possible power output, while keeping the output devices inside safe limits, it's still better to have a match between the amplifier's Zo and the presented or applied load ?
The maximum power transfer theorem is taught to all new engineers, but in real life it rarely applies. There are various reasons for this:
1. it applies only to the situation where you are varying the load impedance to maximise the power transfer - lots of people wrongly try to apply it to varying the source impedance
2. it applies only when there is no constraint on maximum terminal voltage/current for a current/voltage source - this is rare in real life where finite voltage supply rails etc. impose limits
3. it does not tell you how much power will be dissipated in the source impedance, so it tells you absolutely nothing about efficiency - many people fail to grasp this

So transmitters (and audio amplifiers, and power generators) rarely have an output impedance which matches the load and this will continue to cause no end of confusion and arguments.
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Old 8th Aug 2018, 2:19 pm   #30
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Default Re: Back terminated cable?

David,

Regarding your remarks in post 25. And a very simplistic look at it.

I would have guessed that when an amplifier is considered as a whole, regardless of the class of operation of the devices and then three scenarios are considered:

Open circuit (load = infinity) and short circuit; neither of those scenarios have any functional ability to deliver any external power.

So the load must be in some range in between to extract any power.

Also, the load should be considered as purely resistive or Ohmic (for test purposes) even though, in reality it won't be and there could be a range of VSWR's that might have to be tolerated and still have amplifier stability and reliability. For example if the antenna is unplugged,preferably the output devices don't fail.

In any case it should be found that as the load resistance drops below infinity there will be a value of it which corresponds to maximum power delivery to the load, and that resistance value, in theory, should correspond to the amplifier's global output impedance?

That means, that the target value of the amplifier's output impedance should be set to match that of a dummy load resistance equal to the target load impedance of the transmission line and its load (still knowing that it won't be a perfectly Ohmic load and that reflections will be there to one extent or another).

Obviously though, from what you have said for RF amplifiers , some designers want to alter that match, in the interest of other factors such as stability or tolerating high VSWR loads. But isn't that just fudging an ideal power transfer match because the load is imperfect and those imperfections, in practice, have to be tolerated , or is there another reason why a deliberate mismatch would confer some advantage that outweighs the loss of maximum power transfer ?
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Old 8th Aug 2018, 4:56 pm   #31
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Quote:
Originally Posted by Argus25
In any case it should be found that as the load resistance drops below infinity there will be a value of it which corresponds to maximum power delivery to the load, and that resistance value, in theory, should correspond to the amplifier's global output impedance?
No, that is your fundamental misconception. Anything you derive from this false statement is likely to be untrue too. It is extremely likely that the load impedance giving maximum power out (subject, perhaps, to other issues such as acceptable distortion) will not be equal to the output impedance and possibly not even the same order of magnitude.

Consider a load driven either from the collector of a BJT (grounded-emitter) or the emitter of a BJT (emitter follower). Given the same supply rail voltage and same heatsinking etc. it is likely that the optimum load impedance will be the same because you have the same voltage swing and possible current. The output impedance will be vastly different, being much higher than the load impedance for collector drive and much lower than the load impedance for emitter drive.
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Old 8th Aug 2018, 5:12 pm   #32
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Default Re: Back terminated cable?

If we take a step back.

The maximum power transfer theorem is quite right. Knowing the impedance of a source, it tells you what load impedance will draw the greatest amount of power from it that is theoretically possible. What it fails to do is to tell you whether you'd want to do this. Many supplies of power have voltages and output impedances that the max power transfer theorem would extract far mopre power from than they would survive. Reckon on 0.3 Ohms at the light bulb socket above your head and 240v. A 100W bulb takes about 416 mA, the wiring dissipates about 52mW. Your room is bright, your house is safe, your bills are affordable. But that maximum power transfer theorem niggles at you things would be so much better if...

So you get a new light bulb made that runs at 0.3 Ohms.

Total circuit 0.6 Ohms, so current is 400A. The bulb gets 50% of the 240v = 120v @400A = 48kW and so does your wiring!

Your room is very bright, Your house is on fire... no, the fuses save you. But notice that you lost control and choice over the power. You got it all, all that could theoretically obtained. Notice also that the efficiency was 50%. The wiring that made up that 0.3 Ohm got as much power as the bulb.

So if you want a controlled amount of power, if you want high efficiency and if the source impedance is real honest resistance in physical objects, matching may not be good.

Another step back. Source impedance comes out of the black box theory, a Thevenin source. This has a disclaimer that while the thevenin source may fully represent a real world source in terms of what goes on outside the black box, there is no guarantee of representation of what goes on inside the box. So any thoughts about efficiency are on a stick wicket. They may be OK, but they may not.

The output Z of an active device is the slope of a voltage/current characteristic. Dimensions of volts versus amps for a gradient are LIKE resistance, but they may be created by a mechanism other than by a resistor of that value. Considerations of efficiency are risky. You really have to take each case down to details to be sure.

But good old sig gens, at least good ones have an attempt at a feedback amp as a voltage source and create their Zout with a series resistor. So they really do do it the Thevenin way. Rum one into a Zo load and the source terminating resistor gets as much heat as the load, and the amplifier has to provide twice as much power as the load gets. OK for milliwatts from a sig gen, but you go looking for ways to keep your bills down if you are broadcasking kilowatts 24/7

I can make a current source set to 1mA and I can make it have a source impedance of 1 megohm. Mr Thevenin would model this as a 1kV supply and a 1 Megohm resistor. My circuit might contain neither of these elements. It'll not go above a few volts and it won't be doing the power Thevenin's PSU and resistor would imply.

A lot of people use models and forget that they are just that, models.

It's important to keep an eye on the areas of validity of models.

David
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Old 8th Aug 2018, 6:14 pm   #33
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Quote:
Originally Posted by G8HQP Dave View Post
The maximum power transfer theorem is taught to all new engineers, but in real life it rarely applies. There are various reasons for this:
1. it applies only to the situation where you are varying the load impedance to maximise the power transfer - lots of people wrongly try to apply it to varying the source impedance
2. it applies only when there is no constraint on maximum terminal voltage/current for a current/voltage source - this is rare in real life where finite voltage supply rails etc. impose limits
3. it does not tell you how much power will be dissipated in the source impedance, so it tells you absolutely nothing about efficiency - many people fail to grasp this

So transmitters (and audio amplifiers, and power generators) rarely have an output impedance which matches the load and this will continue to cause no end of confusion and arguments.
I'd add a '4' to the above:

4. it applies only when the voltage (or current) source is fixed. In practice this is rarely so, because you have full freedom to alter drive levels In an audio amplifier, making load impedance equal source impedance will only maximise power in the very rare case where you just can't add another stage of low-level amplification; and the signal levels are low enough such that maximum power is below the onset of distortion.
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Old 8th Aug 2018, 6:55 pm   #34
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Default Re: Back terminated cable?

It is obvious that you could have an amplifier output stage that acts as a pure voltage source and has a theoretical zero Z output impedance, or an output stage that acts as a pure current source and has a theoretical infinitely large output impedance.

Therefore the apparent impedance of the output stage(as seen looking into its input) should not be used as a value to calculate what the ideal load impedance might be for maximum power transfer to the load. So I think everybody is in agreement on that.

If the output impedance is very low though, as it is from a near perfect voltage source and it is a low power application, a series resistor can then set the amplifier's output resistance, which then can determine the ideal load resistance. Often this is the way video and RF amplifier's are designed with 75 and 50R resistors respectively to suit the cables and resistive terminations.

For transferring any power however, whatever the impedance of the output stage is (which is never zero or infinite because there is no such thing as a perfect current source or voltage source) the important thing is that the load restricts the voltage swing to avoid clipping, but not to the extent that the power output is excessively limited. Would that be a better statement about ideal load impedances applied to RF amplifier outputs ?
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Old 9th Aug 2018, 11:28 am   #35
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Default Re: Back terminated cable?

I think - and I'm not an RF engineer - that an amplifier with an internal 75 ohm (or 50 ohm) series resistor would have that resistor because:

1) it provides back-termination to stop reflections, in case of a bad match at the remote end of a transmission line;

2) it stops the amplifier blowing up its output stage if said transmission line goes short-circuit (quite important with dodgy cables and connectors).

75 ohm is fairly standard for RF simply because it is the impedance of a resonant dipole. 50 ohms is commonplace for video and most other stuff, because (why anyone??!)

If there is such a resistor, then the load on the output stage is crudely 100ohms (or 150 ohms) in normal operation.

Whatever the impedance of the output stage, the load and output power are determined by lots of things! For a simple, single-ended class A stage, you can load it and wind up the drive till it starts clipping. Too low a load and it will clip in one direction. Too high a load and it will clip in the other. Ideal load is when it just starts to clip symmetrically - and this will give maximum power at the onset of clipping. For a Class B push-pull stage, you can reduce the load resistance more and more - increasing the drive if necessary - and power will go up and up, until either a current limit is reached or a device goes pop.
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Old 9th Aug 2018, 11:58 am   #36
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Quote:
Originally Posted by kalee20 View Post
75 ohm is fairly standard for RF simply because it is the impedance of a resonant dipole. 50 ohms is commonplace for video and most other stuff, because (why anyone??!)
Many moons ago someone did some research into coax cable: For a given screen diameter, if the conductors are copper and the dielectric is polythene, how does the insertion loss per distance vary with the characteristic impedance the cable is made for?

The answer was unexpected. There isn't a slope, there is a dip and it's very close to 75 Ohms.

It's 50 Ohms that seems to have been more arbitrary.

So, for a given distance to span and a set lump of copper to do it with 75 Ohm wins. The world's telcos latched onto this very quickly for their FDM systems. High-Z open wire feeder would have been better, but they wanted cables that could be bunched as they reached network nodes, bunched in ducts etc, so it had to be self-screening like coax.

With their amounts of cable, a slight shift in the cost of copper versus insertion loss was significant. TV studio practice picked up coax practice from the telcos, after all their lines would take signals from studios to transmitter sites.

Domestic TV went 75 Ohm because of dipoles, or 300Ohm ribbon because of folded dipoles.

David
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Old 9th Aug 2018, 11:59 am   #37
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Default Re: Back terminated cable?

Quote:
Originally Posted by kalee20
I'd add a '4' to the above:

4. it applies only when the voltage (or current) source is fixed.
No, it applies to a variable source too.

Quote:
Originally Posted by Argus25
It is obvious that you could have an amplifier output stage that acts as a pure voltage source and has a theoretical zero Z output impedance, or an output stage that acts as a pure current source and has a theoretical infinitely large output impedance.

Therefore the apparent impedance of the output stage(as seen looking into its input) should not be used as a value to calculate what the ideal load impedance might be for maximum power transfer to the load. So I think everybody is in agreement on that.
Yes. Yesterday you appeared to be saying the very opposite; maybe we misunderstood
you?

Quote:
For transferring any power however, whatever the impedance of the output stage is (which is never zero or infinite because there is no such thing as a perfect current source or voltage source) the important thing is that the load restricts the voltage swing to avoid clipping, but not to the extent that the power output is excessively limited. Would that be a better statement about ideal load impedances applied to RF amplifier outputs ?
The load has to be whatever the load has to be for optimum operation of the output stage. Whether this is a matter of voltage or current or some linear combination depends on the output stage topology.

75 and 50 ohms were chosen as coaxial cable standards because of optimising things like power handling and attenuation for a given cable size or weight. I forget which is which. It is a fortunate coincidence that a halfwave dipole in free space has an impedance near 75 ohms.

Last edited by G8HQP Dave; 9th Aug 2018 at 12:01 pm. Reason: add citation
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Old 9th Aug 2018, 12:08 pm   #38
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Default Re: Back terminated cable?

Kalee20,

I agree with all that you said.

In practice I have found with the class B push pull transistor RF amplifiers I have made for the 2MHz vicinity, the thing that stops you from lowering the load impedance seen by the output transistor's and winding up the drive, is the limitation imposed from the size of the RF transformers, not so much a transistor going pop. As the currents are increased trying to squeeze more power out of it, the output and or driver transformer cores add severe non linearity and it would require going to a bigger sized core. I guess though in the the transistors would be pushed too hard too.

So with the ones I have built the output power I have settled on has the transformation ratio set so that into a 50R load the collector waveform at full power is not quite clipping and the peak currents are not high enough to cause significant non linearity in the waveform.
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Old 9th Aug 2018, 12:18 pm   #39
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Dave regarding your post #37 (sorry I can't reply with a quote easily on my mobile device).

Still I wonder, would it not be an advantage (putting optimal operation of the output stage aside and power transfer issues) to strive for an RF amplifier output impedance that matched the coaxial line it was connected to, for the purpose of ensuring that reflected waves heading toward the amplifier do not see an impedance bump there and are absorbed rather than re-reflected into the line ? Or is that not considered too important ?
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Old 9th Aug 2018, 1:26 pm   #40
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Quote:
Originally Posted by G8HQP Dave View Post
Quote:
Originally Posted by kalee20
I'd add a '4' to the above:

4. it applies only when the voltage (or current) source is fixed.
No, it applies to a variable source too.
The max power transfer theorem applies if you consider that dP/dRl is zero at Rl = ro (where the derivative should really be the partial derivative, but the keyboard won't do that symbol - pedants might insist that Zl = zo*). What I meant was, after "mismatching" the load, you can restore power output, and maybe get even more, simply by turning up the wick!

A simple, single-ended triode operated with a fixed input level (which is small enough that clipping never occurs) gives its maximum power when Rl = ra. This is the Max Power Theorem at work. But if you allow yourself to always vary the input level up to the point at which clipping occurs - whether it is grid-current clipping, anode current cut-off clipping, or clipping due to the anode voltage 'bottoming' then max power is reached at quite a different value of load.

Quote:
Originally Posted by Argus25 View Post
So with the ones I have built the output power I have settled on has the transformation ratio set so that into a 50R load the collector waveform at full power is not quite clipping and the peak currents are not high enough to cause significant non linearity in the waveform.
That sounds like you've got the thing just about as good as possible, Argus!
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