A resistor that distorts can be represented by a resistor that dynamically changes its value with the signal, allright for the discussion?
A grid leak resistor can distort (modulate its value) all it wants, but since there's no current trough it there's no signal modulation so no distortion.
Exact , and not : You forget the EXCESS noise produced by current flowing through, even very low ...
This is not trivial, as the usual NFB network is a voltage divider with significant voltage across the resistor connected to the output.
This would be thermally modulated in value, and unlikely to track the associated resistor to ground, because of the considerable power difference in the two (other than at a very low gain).
I've always used considerable overkill for the power rating on the resistor connected to the output for this reason.
This isn't a feedback network. Do you know what RC coupling is and the function of a grid leak resistor?
At the risk of being shot, here's how I look at it.
A resistor that distorts can be represented by a resistor that dynamically changes its value with the signal, allright for the discussion?
A grid leak resistor can distort (modulate its value) all it wants, but since there's no current trough it there's no signal modulation so no distortion.
Enter the input volume control.
NOW the grid sits at the node of a voltage divider (vol. stuff and grid leakage resistor) and NOW the resistor distorting is distorting the signal because it is modulating the divider ratio.
But only because of the vol. control stuff - the voltage division.
So if the resistor distorts say 0.01% and the ratio between vol control stuff and grid leak is 10k/400k = 1/40, the resultant distortion is 0.01/40 = 0.00025%.
Can I go play outside now?
Jan
Not quite, if the resistor distorts the voltage across it, that is the voltage into the amplifier. If there is no voltage across the resistor then there is no signal.
The only effect the volume control has is to increase the source resistance. So at full volume the source resistance to the grid is lower say 1000 ohms. When you turn the volume all the way down the source resistance goes to zero (almost.) At any setting in between the source resistance goes up.
Now if we do our Norton circuit equivalents we can consider the input to be a current feeding two parallel resistors. One the input volume control and the other the grid leak resistor. So the signal is reduced as is the distortion generated but the ratio stays the same. It is not reduced by the volume control or any voltage divider.
The resistor by itself will produce .01% THD or a bit more at 1000 hertz. The distortion changes with frequency. (Rising as the frequency goes down & the energy driving it goes up as the frequency goes down until around 100 hertz in most recorded material.) This adds to the amplifier's own distortion. If the distortion products are in phase they add.
As we have harmonics besides the second they also add and raise the Total Harmonic Distortion. Now when the new distortions are added to what already was being produced by the amplifier you certainly can get past the detection threshold. Keep in mind the following stages and even the loudspeaker all distort and so what started out as second harmonic distortion may end up as much higher order (at a lower level) and many more harmonics.
This isn't a feedback network. Do you know what RC coupling is and the function of a grid leak resistor?
Yes, it just a little off topic, but I was agreeing that a voltage divider is not as trivial as it seems, especially when significant power or higher impedance levels are involved.
Yes Rayma, ther is nothing trivial with resistors : Another example to feed this mistery topic 😀 Did someone try a HIGHER wattage resistors in a passive crossover ? i did it on my VOTTs A5... 100watts resistors sound A LOT BETTER - more airy , less stressed - than the 10watts stock ones , why ???? 😕
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Yes Rayma, ther is nothing trivial with resistors : Another example to feed this mistery topic 😀 Did someone try a HIGHER wattage resistors in a passive crossover ? i did it on my VOTTs A5... 100watts resistors sound A LOT BETTER - more airy , less stressed - than the 10watts stock ones , why ???? 😕
Were the two types of resistors of the same brand, materials, and construction?
If not, try it with resistors that differ only in power level to narrow down the variables.
a voltage divider is not as trivial as it seems
In this case, it is. Ignore the dust cloud. The transfer function is load resistance divided by the sum of source (Rs) and load resistance. If the theoretical load resistance (call it Ro) has a variation with voltage (call it delta), then the actual load resistance can be written as Ro(1 + delta). Write the voltage divider equation. It's Ro(1 + delta)/{Ro(1 + delta) + Rs}.
Now in the case of RC coupling, the load is much higher than the source, so you reduce to Jan's basic calculation.
Were the two types of resistors of the same brand, materials, and construction?
If not, try it with resistors that differ only in power level to narrow down the variables.
Not the same brand : CAL-R black silicon instead enameled Sfernice. Strange .
Yes Rayma, ther is nothing trivial with resistors : Another example to feed this mistery topic 😀 Did someone try a HIGHER wattage resistors in a passive crossover ? i did it on my VOTTs A5... 100watts resistors sound A LOT BETTER - more airy , less stressed - than the 10watts stock ones , why ???? 😕
Let me guess: your wife/girlfriend/steady buddy came running from the attic asking what you had changed as it now sounds sooo much better.
You crack me up! Which is good. Not too much humor here.
In this case, it is. Ignore the dust cloud. The transfer function is load resistance divided by the sum of source (Rs) and load resistance. If the theoretical load resistance (call it Ro) has a variation with voltage (call it delta), then the actual load resistance can be written as Ro(1 + delta). Write the voltage divider equation. It's Ro(1 + delta)/{Ro(1 + delta) + Rs}.
Now in the case of RC coupling, the load is much higher than the source, so you reduce to Jan's basic calculation.
That was a little hard for me to follow, so using regular calculus, differentiating the transfer function T = Ro / (Ro + Rs) with respect to Ro,
I got : Rs / (Ro + Rs)^2
It's evident, on acoustic classical guitar, she heard the difference 😉 i used vitrous enamel before. (307A SE trioded amps)
It's pretty simple algebra, just load divided by sum of load and source.
If load >> source, then dividing top and bottom by load gets you 1/(1 + source/load), which you can expand into a Taylor series and drop the terms smaller than the square.
If load >> source, then dividing top and bottom by load gets you 1/(1 + source/load), which you can expand into a Taylor series and drop the terms smaller than the square.
With what signal voltage across it? THD in CC resistors is proportional to level, remember.The resistor by itself will produce .01% THD or a bit more at 1000 hertz.
With what signal voltage across it? THD in CC resistors is proportional to level, remember.
No 3rd order THD is proportional to voltage squared and the inverse of the square root of frequency. Other terms may or may not have a voltage coefficient. Only Carbon Composition resistors have so much distortion among resistors intended to be ohmic.
The resistor in question goes across the input signal to the amplifier. The total distortion (Not the ratio!) goes down as the volume is changed but so does the input signal. So the third order harmonic distortion will decrease significantly, but not the other orders. Look at the distortion as shown in post 37. The second picture is of the distortion of a good carbon composition resistor, with about 1 volt across it at 1000 hertz. The carrier is suppressed by the measurement technique.
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It's pretty simple algebra, just load divided by sum of load and source.
If load >> source, then dividing top and bottom by load gets you 1/(1 + source/load), which you can expand into a Taylor series and drop the terms smaller than the square.
SY,
Can you write the circuit equation for a voltage divider? Can you do a Norton equivalent?
You explanation reminds me of the saying "Plurality must never be posited without necessity."
ES
As Cliffforrest noted, those "measurements" were meaningless. What I really don't understand is how someone who apparently has enough money and interest to own an Audio Precision could post something so utterly meaningless, and also be unable to comprehend how meaningless it is 😕No 3rd order THD is proportional to voltage squared and the inverse of the square root of frequency. Look at the distortion as shown in post 37.
As Cliffforrest noted, those "measurements" were meaningless. What I really don't understand is how someone who apparently has enough money and interest to own an Audio Precision could post something so utterly meaningless, and also be unable to comprehend how meaningless it is 😕
Did you look at the article these were taken from in Linear Audio? Did you search for the long string of posts on how these measurements were done?
Did you ask a question about it?
WTF should we have to look at articles in publications that we do not subscribe to understand what you are saying?
There is a big disconnect between you and others in this forum. You seem to be carrying on in your own private world. I think you are just taking the p***, but that's just me.
There is a big disconnect between you and others in this forum. You seem to be carrying on in your own private world. I think you are just taking the p***, but that's just me.
WTF should we have to look at articles in publications that we do not subscribe to understand what you are saying?
There is a big disconnect between you and others in this forum. You seem to be carrying on in your own private world. I think you are just taking the p***, but that's just me.
Do you expect me to recount an entire years worth of research and the final article in a few posts? If you haven't read the article or any of AP's follow ups or even the posts in the Blowtorch thread showing the start to finish progress on the measurements, then just complain.
FFS, all we want is a diagram or at least a half decent explanation of the TEST SETUP. Otherwise all we see is just a couple of FFTs you could have cribbed from anywhere. A year's research? Don't make me laugh. It's a couple of distortion measurements on a couple of resistors (supposedly). 30 mins work, tops.Do you expect me to recount an entire years worth of research and the final article in a few posts?
You didn't even say what volume of Linear Audio, or what thread. You think that's how academic 'research' works? Post pretty pictures and then (200 posts later) expect people to solve some kind of riddle about references? I'm guessing you didn't do science in high school.
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