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In Reply to: RE: What tubes to use with this Interstage Transformers? posted by megasat16 on September 04, 2016 at 21:22:58
Putting a series resistor outside the FG puts a series or parallel resistance to the load and I always look it as part of the load.
it is part of the load to whatever is driving it. (whatever is upstream) and it becomes part of the source impedance for whatever is downstream.
Let's say it uses 550ohm resistor for 600ohm impedance, do you can think you can substitute a 550ohm resistor inside that FG with 9950ohm resistor to make your FG look like 10K source impedance?
yes... that is how it works.
I would love to have something like that.
this whole discussion has stemmed form your refusal to accept this simple fact so you can either accept that you already can do it or deny it and continue to redefine electrical theory.
dave
Follow Ups:
Firstly, there are a lot of Dave Slagle on the world so I wanted to make sure you are whom I think you are.
Are you from Intact Audio who makes Transformers?
Secondly, I don't redefine anything. I followed what I learned in school many decades ago and do either AC Analysis or DC Analysis completely (not both together as I see fit).
If there is anyone redefining it, it is you who insists upon it as the convenience of simplifying circuits to your theory and advantages instead of the analysis and theory that is taught and learned by millions of E.E or E.T around the world.
Thirdly (or) lastly, I now understand where your misunderstanding and confusion arises from.
I attached a couple of pics taken straight from the schematic of your WaveTek 288 user manual downloaded from the BAMA site.
Please tell me exactly how these attenuators are used in the circuit?
Are these series resistors part of the circuit or not part of the circuit (outside of the circuit even though they are put inside the box) in your WaveTek?
Regards,
James
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
you have the correct dave slagle....
Please tell me exactly how these attenuators are used in the circuit?
Are these series resistors part of the circuit or not part of the circuit (outside of the circuit even though they are put inside the box) in your WaveTek?
for attenuation closing K1 shorts out the π attenuator and for output impedance it appears to me that K2 and K3 either add or remove 25 ohms or 550 ohms in series with the output which coincides nicely with the selectable 50, 75 and 600 ohm output impedance. Now are you insisting that since these added series resistors are on the upstream side of J19 keeping them inside the box they behave any differently than if they were added on the downstream side? Or are the designers at wavetek now the ones taking liberties?
dave
The WaveTek Engineers are in the right and did exactly what I thought before too. They put these resistors inside the circuit with a closed feedback loop (R79) to the compensator circuit and microprocessor circuit which corrects any phase and voltage inaccuracies and send the compensated signal through the power amplifier circuit...Even then, they know the compensator circuit can correctly predict and correct the frequency below 1MHz so they put a lockout function for the higher impedance setting above 1MHz on the unbalanced output J79. It seems to me like a pretty good implementation to me.Read the user manual yet?
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
Edits: 09/05/16 09/05/16
From my understanding R79 simply allows for the sampling of the output voltage for the self calibration feature and is only used when the calibration cycle is running.
Indeed they do lock out frequencies above 1mhz when the larger resistors are used but we are talking about sub 100hz signals in this case and only looking for a rough estimation of actual in circuit behavior to decide if there is merit in actually lashing it up. If the transformer doesn't meet the desired low frequency results with a 5K series resistor there isn't a chance it will do better when it is loading a 10.
Applying rationale that may be extremely important in the megahertz+ range to this situation is probably not the best use of anyones time here.
dave
I presented you two pics together so you can connect the dots and at least come up with knowing the resistors are part of closed circuit with the feedback resistor of some sort going back into the more intelligent circuit compromised of the calibration circuit, microprocessor and the driving circuits.Instead, I got rebuttal and you come up with another explanation of what you think it is.
Anyway, observe the attached pic carefully including the V levels on the scope and explain to me what's happening at the transformer.
Have you seen anything like that before?
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
Edits: 09/05/16
I presented you two pics together so you can connect the dots and at least come up with knowing the resistors are part of closed circuit with the feedback resistor of some sort going back into the more intelligent circuit compromised of the calibration circuit, microprocessor and the driving circuits.
you want me to connect the dots you see and at no point do I see any mention or reference that that 100K resistor is a feedback resistor. I do see it connects to a "black box" called the "internal calibration network circuit" and know that the 288 has a calibration button but that is about as far as I get.
as for your measurements.... I have no idea what I am looking at. first the pictures are not very clear and the numbers are hard to read but you state that you used a 20V signal (didn't mention whether it was rms peal or p-p) and then show what appears to be a 40V p-p sine so that makes some sense. the other trace voltage label is not clear but it does appear to be 40mv which is a 1000:1 ratio or 60dB of attenuation and the only way I can even conceive of that happening is if the primary of the DUT is shorted.
Since you have gone as far as hooking up the circuit, why haven't you done the actual sweep with and without the 5K series resistor and shown the results?
dave
Don't worry about whether you see it or not. But now you know that the series resistor you see inside your FG is not quite the same as "Series Resistor" that you conveniently add to change impedance of the FG as you see fit.
If you still have in doubt, you can ask any Engineer about it.
If you add an external resistor to the source (FG, Amp, Preamp, Battery; etc.), it's just become part of the load circuit and does not do any impedance transformation to the source circuit. I guess it only took about 60 posts to get to this point.
Without further ado, I'll just go back and thinker what tubes to use with my IT. I think it'll do a lot of benefits to all of us following this thread.
I know you and Tre wanted to help and I appreciate it a lot. But I simply can't accept the wrong information presented for the sake of Engineering.
No hurt feelings meant to anyone!
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
"If you add an external resistor to the source (FG, Amp, Preamp, Battery; etc.), it's just become part of the load circuit and does not do any impedance transformation to the source circuit. I guess it only took about 60 posts to get to this point."
It seems t me you are misunderstanding a basic point about equivalent circuits. What Dave Slagle was saying seems to be correct. A source of 50 ohm impedance, in series with a 5K resistor, can be viewed as a source of 5050 ohms impedance.
It doesn't matter, from the point of view of analysing the total circuit, whether you view it as a source of 50 ohm impedance, driving into a load of 5K resistance in series with the transformer, or whether, on the other hand, you view it as a source of 5050 ohms impedance driving directly into the transformer. It is the same circuit either way you view it.
But it is perfectly valid, and useful, to view the 50 ohm output impedance generator plus the 5K series resistor as a 5050 ohm output impedance generator. The circuit as a whole doesn't care whether the 5K resistor sits inside the box labelled "signal generator," or if it sits outside the box.
Chris
Nope! You are just as confused as Dave and mixing up AC circuit with the DC ESR.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
"Nope! You are just as confused as Dave and mixing up AC circuit with the DC ESR."No, actually not. It is really just an application of the Thevenin theorem.
Chris
Edits: 09/08/16
Chris,
I have said Tre and Dave before and I will tell you the same.
Even though it is not the same for the engineering point of view, the calculation would come out the same for the DC circuit with purely resistive load. I have no argument about it.
The problem with the (FG Z + external series resistor = new FG Z) when the load circuit has reactive component and it will change depends on the AC source. The results aren't the same everytime a signal from the FG increases or the frequency increases or decrease.
Regards,
James
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
In principle, the 50 ohm output impedance of the signal generator might have some reactive component (frequency dependent) to it. But when connected in series with the 5K resistor, the non-reactive 5K will be completely swamping the 50 ohms, and so any errors arising from treating the generator plus 5K resistor as simply being a "new generator" with a 5050 ohm output impedance (essentially non-reactive, to good approximation) will be very tiny.
Chris
Yes, it is true when the FG+5K resistor is connected to purely resistive LOAD.But we can't do that when the reactive load is connected on the other end of FG+5K series resistor.
I don't think I can explain you any better than I did to Dave.
You need to see a FG as an amplifier with a feedback loop so the output impedance remains near constant as possible under different loads. Any drop in voltage and current and phase coherence is being compensated by the FG internal circuitry. That is something an external series 5K resistor can't do. And that series resistor is not the origin of the actual DC/AC current or source.
That's not even remotely the same or close to a true 5K output impedance AC source.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
Edits: 09/08/16
"Any drop in voltage and current and phase coherence is being compensated by the FG internal circuitry. That is something an external series 5K resistor can't do."
But the whole point is that a signal generator with 5K output impedance will NOT compensate for voltage drop in the way you are saying. If it did, then its output impedance would be lower. The output voltage of a signal generator with 5K output impedance will drop to one half of its unloaded output voltage if it drives into a 5K load. That is what it means to have a 5K output impedance.
If instead the signal generator compensated by means of internal feedback circuitry, so that connecting the 5K external load caused less than 50% drop in output voltage, it would mean that the output impedance of the signal generator was actually, by definition, less than 5K.
A signal generator with unloaded AC output voltage V and output impedance R can be modelled by an ideal zero-output-impedance voltage source with AC voltage V, in series with a resistor of resistance R.
Yes, there are certain approximation issues concerning some possible reactive component to the output impedance of the generator. But I would think that any modern half way decent signal generator will have an essentially resistive (non-reactive) output impedance, at least at low (like audio) frequencies.
And as I said before, if one is making a 5K output impedance (or 5050 to be more precise) signal generator by putting a 5K resistor in series with a 50 ohm output impedance signal generator, the issue of whether the 50 ohm impedance is reactive or not will pale into insignificance when put in series with 5K of purely resistive impedance.
Chris
Chris,
You are contradicting some partial truth with some false information. The 5K impedance of FG has some (even if limited) capability to supply the voltage without sagging when a load with 5K or more load is connected (regardless of it's resistive or reactive).
The 5K series resistor external to the FG has 0 capability of supplying any load current to the 5K LOAD unless the 50ohm FG supply some current to it first. So, based on the impedance of the load, it is just dropping voltage (stealing from the source) rather than working in union with the Source.
Very different, yeah?
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
I think it wiuld be helppful if you were to give *your* definition of what a signal generator with 5K output impedance means. I mean, be quantitative and say exactly what measurement would you perform that would produce 5K as the answer?I believe I have given a conventional definition of output impedance, namely it equals the resistance of the load you have to drive in order to cause the no-load ourput voltage to drop by 50%
So what is your quantitative measurement that would allow you to determine the output impedance?
Chris
Edits: 09/08/16
It is a voltage source with an active components as the current / voltage provider (namely tubes or transistors with some passive devices such as resistors and Xc and Xl in a closed loop circuit).
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
No, I'm not asking about a narrative discussion about what is "inside the box." I'm asking what measurement you would perform, entirely "outside the signal generator box," to determine the quantitative answer for what its output impedance is.
For example, what external measurement would you make in order to derermine whether the output impedance of 5K claimed by the maker was an accurate claim or not? I.e. How exactly are you defining output impedance? How would you measure it?
Until you give your operational and quantitative definition of output impedance, it is not really possible to have a sensible discussion.
Chris
Chris,I don't need this discussion. You can read to the beginning of this thread if you want to know why. I am not the guy started it.
I think you'll also find the answers to your questions at the beginning of this thread.
Thanks,
James
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
Edits: 09/08/16
The problem with the (FG Z + external series resistor = new FG Z) when the load circuit has reactive component and it will change depends on the AC source. The results aren't the same everytime a signal from the FG increases or the frequency increases or decrease.
this is case with any source with a greater than 0 output impedance. For this particular application it is the fact that the source Z is 5K that we are using to our advantage in order to find the effect of the reactive component (primary L). The variation you speak of is exactly what we are looking for so we can work it against the other knowns to find the unknown (primary L)
dave
Dave,
From your theory to work, you will need to the DCR, and either the inductance and the capacitance at that frequency. You also need to consider which is predominant factor (i.e. primarily resistive, or capacitive or inductive). You will also need to consider the phase angle based on the predominant factor. And then, you will need to figure out the amount of reflection to the source. And once you know that, that's the result for the one fixed setting (such as 2Vp-p, 20Hz). Once you change a little value, you are required to recalculate once over again.
Again, the source is the TRUE origin of the current / voltage.
I know what you have been saying all along on this topics. I wish you are right and things are that easy too. But it's not and I learned that very long ago. In fact, all of us (my classmates and I) got it wrong at the very beginning and puts a big smile on the Instructor face. Some good memory there.
Regards,
James
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
You seem to be completely missing the application here. The ultimate function generator in this particular situation is a tube with an Rp of 5K amplifying the random signals we call music. This is what needs to replicate for a precursory test.
You mention your ideal for a function generator being an amplifier with a feedback loop yet this actual application doesn't meet that criteria so I guess we now need to wrap feedback around our tubes to make them work properly.
Back to the original task, the 5K series resistor will work perfectly for this application. It may show issued in different applications however this is not about designing a universal device it is about testing a very specific application. The main thing of interest is to determine the primary inductance of the DUT to see what tubes it is suitable for loading across the audio band.
dave
Dave,
Please don't start on the topics of amplifiers and feedback? It took me a good part of a year on that subject learning all types. It definitely doesn't serve the purpose of this thread.
The tube with Rp of 5K-10K is what I want to use as a source to test the transformer for an actual bandwidth and FR curve.
I have very little interest in the actual L value or the Z value but a nice HP Impedance Analyzer would be an addition to my toys in the future.
Regards,
James
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
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