G.Kleinschmidt said:
But all those non-linear mechanisms are still there, and the VAS will still need to drive them. The only difference adding a resistor will make is that the VAS will then have to drive the resistor as well.
Also, unless you are running a really low current VAS, 67k would be negligible loading anyway. 750uA peak will drive a 67k load to over 50V. In fact, if your amplifier has any frequency compensation done at the VAS stage (either miller or feedback or shunt), the loading of the VAS from that at high frequencies will be a couple of orders of magnitude greater than the loading provided by a 67k resistor.
Well, you have to make up your mind. You cannot say that the 67k is insignificant and at the same time that the non-linear parallel impedance of several megohms with the non-linearities is significant.
I think what the 67k does is to reduce significantly the modulation of the Vas load and this may well reduce the distortion.
Jan Didden
Jan:
"I think what the 67k does is to reduce significantly the modulation of the Vas load and this may well reduce the distortion."
Jan,
thats true...but also it reduces the gain one can use for FB,
so at least I believe thats a drawback.
Things become clearer with some exaggeration :
the modulation get better with 1K...100E....!?
Regards
Heinz!
"I think what the 67k does is to reduce significantly the modulation of the Vas load and this may well reduce the distortion."
Jan,
thats true...but also it reduces the gain one can use for FB,
so at least I believe thats a drawback.
Things become clearer with some exaggeration :
the modulation get better with 1K...100E....!?
Regards
Heinz!
There exists an optimum of course.
Well, you would get more gain ommiting this resistor, but all in all you must get rid of this gain at 20kHz to keep your amplifier stable.
Things look slightly different if you use double pole compensation with one or more zeros, but this technique relies on accurate cancellation of natural poles by employing zeros and is not trivial to keep the whole unconditionally stable.
Well, you would get more gain ommiting this resistor, but all in all you must get rid of this gain at 20kHz to keep your amplifier stable.
Things look slightly different if you use double pole compensation with one or more zeros, but this technique relies on accurate cancellation of natural poles by employing zeros and is not trivial to keep the whole unconditionally stable.
powerbecker said:Jan:
"I think what the 67k does is to reduce significantly the modulation of the Vas load and this may well reduce the distortion."
Jan,
thats true...but also it reduces the gain one can use for FB,
so at least I believe thats a drawback.
Things become clearer with some exaggeration :
the modulation get better with 1K...100E....!?
Regards
Heinz!
Indeed, but it means that you cannot put up a blanket statement that because a parallel resistor decreases available loop gain for feed back that therefore distortion rises. As most things in life, It Depends.
Suppose that the two effects cancel exactly so the RMS THD stays constant. Wouldn't you then expect that because the Vas is linearized that the THD products would be more benign?
Jan Didden
darkfenriz said:There exists an optimum of course.
Well, you would get more gain ommiting this resistor, but all in all you must get rid of this gain at 20kHz to keep your amplifier stable.
Things look slightly different if you use double pole compensation with one or more zeros, but this technique relies on accurate cancellation of natural poles by employing zeros and is not trivial to keep the whole unconditionally stable.
Hmmm..a optimum?
When my destination is linearity at a given bandwidth I need lot´s of gain for FB.
Recently I simulate (yes...only) a Fet-amp with 800kHz bandwidth and get a *THD20* 0.000 001%.
The amp has about 140dB FB @ 20kHz.
I use no double pole comp. ,but nested FB.
Heinz!
powerbecker said:
...
*THD20* 0.000 001%.
The amp has about 140dB FB @ 20kHz.
This clearly translates to 10% THD in forward path according to classical feedback theory.... interesting
Still final number looks nice, even for simulation result.
janneman said:
Indeed, but it means that you cannot put up a blanket statement that because a parallel resistor decreases available loop gain for feed back that therefore distortion rises. As most things in life, It Depends.
Suppose that the two effects cancel exactly so the RMS THD stays constant. Wouldn't you then expect that because the Vas is linearized that the THD products would be more benign?
Jan Didden
I think, I don´t make a statement, because I say "I believe.."!
I write so, because I have never experimented with that, and so I don´t have "numbers" or examples.
May be you have some?
powerbecker said:
I think, I don´t make a statement, because I say "I believe.."!
I write so, because I have never experimented with that, and so I don´t have "numbers" or examples.
May be you have some?
No numbers, just reasoning. I say "I think" because I am not sure. The reasoning is as follows. The increase in high order harmonics with feedback happens because of the non-linearities in the forward path. The stronger the non-linearities, the more complex and higher-order the harmonics. Also, the stronger the feedback, the more complex and higher order the harmonics. You lose both ways.
So if we have less non-linearity to begin with, and less feedback also, even if the total RMS sum of the harmonics is the same, I 'think' 😉 that the thd spectrum will be more benign. You win both ways. What do you think?
Jan Didden
darkfenriz said:
This clearly translates to 10% THD in forward path according to classical feedback theory.... interesting
Still final number looks nice, even for simulation result.
I don´t test it before, but now I do it:
Bingo, you are right @ 20kHz there are 5.16% THD!
janneman said:
No numbers, just reasoning. I say "I think" because I am not sure. The reasoning is as follows. The increase in high order harmonics with feedback happens because of the non-linearities in the forward path. The stronger the non-linearities, the more complex and higher-order the harmonics. Also, the stronger the feedback, the more complex and higher order the harmonics. You lose both ways.
So if we have less non-linearity to begin with, and less feedback also, even if the total RMS sum of the harmonics is the same, I 'think' 😉 that the thd spectrum will be more benign. You win both ways. What do you think?
Jan Didden
*No numbers, just reasoning. I say "I think" because I am not sure.*
🙂 nice....exact the same with me!
Also your arguments be clear to me!
But then we came to a difficult region, we had to weigh up what SOUNDs better :
less THD with higher harmonics or vice versa!
Well, the answer seems to be clear.
But it is also clear that that depends from the level of the harmonics!
Also my last hearing comparison (25 years back) was not a illumination, because I fail to distinguish between my transistoramp against a well done tubeamp!
So, as you can see also from my further post I advocate the smallest possible levels of thd for shure.
On the other hand I can imagine that, when I will (may be) start again to design audioamps first it has to be one with "zero" distortion.
But also I will add a switched "distorter" for the users who like the "warm" sounds!
Heinz!
PB2 said:
Bob, just to make sure that I understand you. Are you saying that given the open loop case, and say for example a two tone IM test of 100 and 110 Hz that the measurement will be the same with and without the resistor, and the same output level?
I realize that this is not specifically a PIM test but it should serve the purpose. I've deliberately chosen low frequencies.
Pete B.
Hi Pete, I'm sorry I took so long to get back to answer your question. The assumption is that we are applying your two-tone test to an open-loop amplifier that has been compensated for a given amount of open-loop gain at 20 kHz. The question I think you are positing, is, if at a given output voltage level, the presence or absence of the VAS load resistor makes a difference in the two-tone LF IM. Is that right? The presumption is that the net shunt resistance we apply causes the open loop bandwidth to increase. This is the same as saying that the shunt resistor causes the LF gain to be reduced to approximately the same level as the gain will be at the new open loop corner frequency.
If the output stage is a triple, with a high input impedance of about 500k, the gain even at 100 Hz will still largely be controlled by the necessary Miller compensation capacitor when there is no load resistor (especially if the input stage is loaded with a current mirror). When we add the load resistor, the p-p current of the VAS will have to increase from a very small amount necessary to drive the 500K load, to a larger amount necessary to drive the shunt resistor. If the shunt resistor is 50 k, this current will have to increase by a factor of 10. This still may not cause much distortion, but it is, nevertheless, going to be an increase in the distortion caused by nonlinearities in the VAS. Suppose the VAS is biased at 10 mA, and we want 100v p-p. The p-p current to drive the 50 k resistor will be about 2 mA, so the VAS collector current is changing by 20% to achieve this swing.
So, yes, the 100 Hz & 110 HZ twin tone IM will tend to increase when the load resistor is applied.
Cheers,
Bob
janneman said:
Well, you have to make up your mind. You cannot say that the 67k is insignificant and at the same time that the non-linear parallel impedance of several megohms with the non-linearities is significant.
Ummm, I think that you need to read my post again. I didn't say one thing or another about the significance of "non-linear parallel impedance of several megohms". I was simply pointing out the fact the 67k is hardly any loading at all. At 20kHz, the loading of even a small miller compensation capacitor will be much, much greater, and it won't make the non-linearities, regardless of their significance, go away.
I think what the 67k does is to reduce significantly the modulation of the Vas load and this may well reduce the distortion.]
Well I think that it would not. 67k is totally insignificant, especially when any frequency compensation begins to load the VAS.
The only thing that light resistive loading of a VAS such as this may do is to prevent the open-loop gain from reaching astronomical levels at very low frequencies. That can, in some situations, despite actually worsening low frequency THD, improve the amplifiers overall low frequency THD+N performance.
But not for reasons that anyone here has cited thus far.
darkfenriz said:Compensation I use is 68pF cap to ground, which has impedance of j120kohm at 20kHz.
OK, I should have expressed my point a bit better. At 20kHz the VAS will be working much harder to drive the compensation capacitor at the harmonic frequencies of 20kHz, than into a 67k load resistor.
As far as high frequency THD is concerned, you have to consider the VAS performance several octaves above the audio frequency range.
At the fifth harmonic of 20kHz, the 68pF cap has an impedence of about 23k.
Bob Cordell said:The presumption is that the net shunt resistance we apply causes the open loop bandwidth to increase.
That's a common misconception. A typical frequency compensated power amplifier may have an open loop crossover frequency in the vicinity of 1 MHz or more. When you consider just how much the miller compensation capacitor will be loading that VAS at those frequencies (not to mention the decreasing input impedence of the output driver stage), a 50k load resistor hardly makes a difference at all. As far as distortion is concerned, the only significant thing such a resistor will do is to increase the low frequency THD due to limiting the open loop gain, making not only the VAS work harder, but the preceding stage(s) as well.
powerbecker said:Recently I simulate (yes...only) a Fet-amp with 800kHz bandwidth and get a *THD20* 0.000 001%.
The amp has about 140dB FB @ 20kHz.
I use no double pole comp. ,but nested FB.
That amp would need a seriously low noise input stage! 😀
You're talking professional communications reciever like gain there!
powerbecker said:
*No numbers, just reasoning. I say "I think" because I am not sure.*
🙂 nice....exact the same with me!
Also your arguments be clear to me!
But then we came to a difficult region, we had to weigh up what SOUNDs better :
less THD with higher harmonics or vice versa!
Well, the answer seems to be clear.
But it is also clear that that depends from the level of the harmonics!
Also my last hearing comparison (25 years back) was not a illumination, because I fail to distinguish between my transistoramp against a well done tubeamp!
So, as you can see also from my further post I advocate the smallest possible levels of thd for shure.
On the other hand I can imagine that, when I will (may be) start again to design audioamps first it has to be one with "zero" distortion.
But also I will add a switched "distorter" for the users who like the "warm" sounds!
Heinz!
Heinz!,
Agree with you. In my opinion, lowest possible distortion is a worthy goal in itself. I try to built amps that change the signal as little as possible. I know that very often, in discussions, when someone runs out of arguments, they say 'but it sounds better'. The best way to kill any intelligent exchange. But what can you do?
Jan Didden
I hear this from a friend. Age 50's, max hearable frequency about 13khz. Age 60's, max hearable frequency about 9khz (or 5khz?)
But how come the elders knows a good sounding amp, bad sounding amp, while they cannot hear the trebles?
This means that high order harmonics are also obvious in low-mid frequencies, not only obvious in treble frequencies?
But how come the elders knows a good sounding amp, bad sounding amp, while they cannot hear the trebles?
This means that high order harmonics are also obvious in low-mid frequencies, not only obvious in treble frequencies?
David,
I think you know that 10th harmonic of 200Hz is only 2kHz ....
Ear is most sensitive in the midband.
I think you know that 10th harmonic of 200Hz is only 2kHz ....
Ear is most sensitive in the midband.
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