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In Reply to: RE: What tubes to use with this Interstage Transformers? posted by megasat16 on September 08, 2016 at 09:23:34
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
Follow Ups:
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
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