Sorry I can't resist.
Quote"Maybe the fingertip of God." (pmillet, "you called").LOL
Hello Mr P.
Regards
M. Gregg
Quote"Maybe the fingertip of God." (pmillet, "you called").LOL
Hello Mr P.
Regards
M. Gregg
You didn't say that the delays were small. You said they didn't exist. Perhaps I am nitpicking but I think it is important to help the original poster have the same level of understanding that we do, don't you?
My Japanese is a little rusty but it looks like the 20KHz shift is about 15 degrees from 180... Is that at max load?
I read 20° at 20Khz.
This is quite decent for a SE OPT and yes, if you don't apply feed back this will have unoticeable effect on the "sound".
But what if your loudspeakers need to be highly damped ?
There are two ways:
- Use other loudspeakers.
- Go PP because you can have a better OPT and use more feedback w/o killing the "sound"
Or even better: use transformerless transistorized amplifier.
But that's another story 😉
Yves.
A few years back..maybe 20 now, how time flies, scientists first announced they had actually measured the time it takes for one electron to bump into another to transmit power. Is there anything in the universe that is instantaneous? My first two responses I think were pretty clear and non elite.
Well,
I think it's been an interesting topic so far. I hope Mr P is going to give his views?
Regards
M. Gregg
I think it's been an interesting topic so far. I hope Mr P is going to give his views?
Regards
M. Gregg
The original question appeared to be about signal propagation delay through the amplifier. For audio this is negligibly small and so can be ignored, despite what some people say. The amplitude and phase response of the amplifier is extremely important, as it directly affects the sound. Feedback can make things better or worse, depending on how it is done and the quality of the original amplifier.
Amplifiers with variable feedback almost always have low levels of feedback. Low feedback is usually a bad idea, unless the amplifier is extremely linear without it (so not SE - I mean 'measures linear', not just 'sounds nice'). The result is that most tests of 'some NFB' vs 'no NFB' will make 'no NFB' sound better. This does not mean NFB is bad, just that 'not enough NFB' is bad.
Amplifiers with variable feedback almost always have low levels of feedback. Low feedback is usually a bad idea, unless the amplifier is extremely linear without it (so not SE - I mean 'measures linear', not just 'sounds nice'). The result is that most tests of 'some NFB' vs 'no NFB' will make 'no NFB' sound better. This does not mean NFB is bad, just that 'not enough NFB' is bad.
DF96 Question,
Do you think that NFB creates distortion problems when applied to a "perfect linear no NFB" amp. I know this is a difficult question to answer. Just interested in your thoughts.
Regards
M. Gregg
Do you think that NFB creates distortion problems when applied to a "perfect linear no NFB" amp. I know this is a difficult question to answer. Just interested in your thoughts.
Regards
M. Gregg
The way I look at this small shift at high freqs. is that (1) it is a small loss of feedback efficiency since the instaneous voltage level is about 20% smaller or 20% higher depending on the cycle position. (2) It's a small price to pay for a part of the spectrum we can barely hear, anyway.
We have a winner!
There are two competing problems:
For feedback to reduce distortion over a given frequency range (let's say 20Hz - 20kHz), the amplifier (before the application of feedback) needs to have a significantly higher bandwidth than 20Hz - 20kHz. Think about it - if there is no gain in the amplifier at 20kHz, then any amount of feedback cannot correct the nonlinearities at 20kHz.
The result is that, if the amplifier bandwidth is not much greater than the desired signal bandwidth, at low frequencies, distortion is effectively cancelled by the feedback, and at high frequencies, it is not. So if you apply feedback to the typical SE amp, the effect is that the low-order distortion products (2nd harmonic, 3rd harmonic...) are greatly reduced. But the high-order products are not. High-order products are much more audible, and even if they were present before feedback is applied, they would have been masked by the relatively large lower-order products.
(Ever wonder why good audio opamps have an open-loop bandwidth of several MHz? It's to push the HD products out so far in frequency that they don't matter.)
So there can be no doubt - feedback will change the sound. Better or worse, depends on the person and the speakers, phase of moon, whatever. But different.
OK, so let's make our SE amplifier bandwidth 200kHz, so we can apply feedback without messing up the sound too much. Take a look at that transformer datasheet again (a very good OPT, by the way). The phase shift at 200kHz is about 120 degrees. So even if there is no other phase shifts in the amp, you are only 60 degrees away from positive feedback. The result will most likely be a big peak in the frequency response at HF, and some ringing on fast transitions (like a square wave). It might still sound good...
But there are other phase shifts going on. The miller capacitance, for example, will form a pole with whatever source impedance is driving it. So it is entirely possible that the total phase shifts will add up to make an oscillator.
Anyway... as for opinion - I don't think there's anything wrong with applying overall loop feedback. But it is not a trivial exersize. It will affect the way things sound. In general, I seem to prefer amps without feedback, or with as little as possible, though I cannot really say why. Psychoacoustic, probably...
Pete
It depends if you are of the "school" that think the combination of frequencies above or below hearing range create harmonics in the audible range.
Sorry off topic.
A perfect amp, with perfect feedback, would provide perfect output. The gain would be lower, the output impedance would be lower (assuming voltage sensing feedback) and the bandwidth would be higher. You would simply have the ideal situation described in textbooks. Feedback does not introduce distortion in itself. It does mix up any distortion already present, so higher orders appear.
Not true. All that is required is that the amplifier has some gain left, otherwise almost all solid-state amps and op-amps would not work at all. These have open-loop bandwidths of maybe 1-10Hz, yet can amplify 20kHz with feedback because they still have enough spare gain. There is an argument that this increases the relative importance of higher frequency errors, which I think is what you are trying to say. The 'integrator' in the amp open-loop forward path gets turned into a 'differentiator' in the closed loop handling of distortion, which can make things like crossover distortion look (and possibly sound) worse.
Not at all, I think it's a very common ''wonderment''. It's tough to visualize all the factors at once, for me anyway.
The feedback is applied as a much smaller signal than the original, and it is also negative, so at worst it works to lower the original gain. That is more of a FR deviation than a distortion source. As far as mixing frequencies creating distortion, the music I listen to is FULL of frequency mixing! YEEEEEHAAA!
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It's what makes this so interesting.
Got to sign off now. Have fun catch you later🙂
Regards
M. Gregg
Feedback always reduces distortion. Let's play with an awful transfer function: Vo = G*Vin^2. Pure second harmonic, if biased around zero, or a mixture otherwise.
Notice the gain (as Vo/Vin) is NOT G, though it's proportional to G. It varies. Which is one possible definition of distortion. In particular, dVo/dVin = 2*G*Vin, so the small-signal gain is proportional to bias. This is sort of like a remote-cutoff tube, where bias can be used to vary gain.
If we put a biased sine wave through this thing with bias A and peak amplitude B, we have (note w == omega, angular frequency):
Vin = A + B*sin(w*t)
Vo = G*[A + B*sin(w*t)]^2
= G*[A^2 + 2*A*B*sin(w*t) + B^2*sin^2(w*t)]
= G*[A^2 + 2*A*B*sin(w*t) + B^2*(1/2 + 1/2 sin(2*w*t)]
In other words (omitting overall gain G for now):
- DC bias of A^2 + B^2 / 2,
- Undistorted signal of 2*A*B*sin(w*t),
- Distortion of (B^2 / 2)*sin(2*w*t), second harmonic,
- and no other harmonics.
Since we're only interested in AC, we can ignore the DC component. (It's noteworthy that the DC component depends on the AC amplitude. In fact, experience tells us that overdriven amplifiers experience a bias shift!)
We can look at the relative amplitudes of signal and distortion. The signal amplitude is 2*A*B, so the gain is linear with bias A (= variable gain, as noted before), and also linear with the input amplitude B. The distortion is B^2 / 2, independent of bias instead, and proportional to the square of the input amplitude. If we look at the ratio, we see the relative amounts: distortion per signal is B / (4*A), so the distortion goes up linearly with amplitude, and goes down linearly with increasing bias.
This is interesting, because tubes have a roughly x^3/2 transfer curve, and are known to be more linear at higher bias. Also, distortion goes down at small signal levels, i.e., if B = 1mV, then B^2 = 1uV, which is just about negligible. But it goes up significantly for signals which are very large relative to the bias!
If we now put this into a feedback loop, first we have to talk about how we do it. The transfer function is not a two-input system (like an op-amp), nor is it differential (like the grid and cathode of a tube), so we have to come up with some way to implement feedback.
If we make the substitution Vin <= Vin - H*Vo, where H is the feedback ratio and Vo is the output voltage, then we can implement the loop:
Vo = G*(Vin - H*Vo)^2
= G*(Vin^2 - 2*H*Vin*Vo + H^2*Vo^2)
Yuck, we have Vo on both sides, one of which is squared. We'll need to use the quadratic equation to solve for Vo:
G*H^2*Vo^2 + Vo*(-2*G*H*Vin - 1) + G*Vin^2 = 0
With the solution,
Vo = (Vin / H) + 1 / (2*G*H^2) + sqrt(4*G*H*Vin + 1) / (2*G*H^2)
which is pretty ugly, but we can figure a few things about it:
- It's linear in Vin, and has first-order gain 1/H, which is exactly what we expect from a feedback circuit. (A voltage divider has H = R2 / (R1 + R2), which will give useful voltage gain for typical values.)
- It has an offset of 1 / (2*G*H^2), independent of applied signal, which is kind of funky, but okay. It goes down as G --> infty, so we can think of this as input offset. For this particular case, the amp itself only has useful gain for nonzero signals, since it's "cut off" at 0V, so this is a reasonable interpretation.
- It has this TERRIBLE nonlinear term of sqrt(4*G*H*Vin + 1) / (2*G*H^2). How do you even take the sort-of-square root of a sine wave signal?
Maybe you can't take it directly, but you can find the Taylor series expansion of the function. Then it's easier to see what effect it has.
In particular, the expansion is:
[1 / (2*G*H^2)] * [1 + u / 2 + u^2 / 8 + u^3 / 16 + 5*u^4 / 128 + ...]
where u = 4*G*H*Vin. We already have the first two terms in Vo, so let's group them:
Vo = (2*Vin / H) + 1 / (G*H^2) + [1 / (2*G*H^2)] * [u^2 / 8 + u^3 / 16 + 5*u^4 / 128 + ...]
- DC component is now 1 / (G*H^2).
- Signal component is now 2*Vin / H, which is kind of weird...
- Distortion is now in polynomial components, so we can estimate the harmonic content. However, we can't find the exact harmonics just by looking at it, because (sin(x))^n is not equal to sin(n*x). In general, the harmonic terms will also produce lower order terms, so the DC offset and fundamental amplitude will change. If we evaluate the first couple terms (x^2 and x^3 are easy to do), we'll probably get a more realistic gain and offset for this system without getting too tedious, but I won't go that far right now.
If u << 1 (a condition of small input signal, low gain, or low feedback), we can approximate this equation by simply truncating the series at u^2 or so. The assumption being, if u = 1mV/V, then u^2 = 1uV/V, and u^3 and further are completely negligible.
For large parameters, we're better off using the original form,
sqrt(4*G*H*Vin + 1) / (2*G*H^2)
which has the asymptotic form (4*G*H*Vin >> 1) of
2*sqrt(G*H*Vin)/(2*G*H^2)
= sqrt(Vin) * G^(-1/2) * H^(-3/2)
Which goes to zero as G and H go to infinity (i.e., increasing loop gain).
So it's always true that:
1. Nonlinear systems are a pain to analyze,
2. Feedback linearizes a system, and
3. The asymptotic result, as loop gain approaches infinity, is zero distortion.
One final note: this analysis is for a static system only. Real systems are rarely stable at infinite gain, and when they are, the loop gain at signal frequencies is nowhere near infinite. Real systems are time dependent and the arithmetic shown above turns into nonlinear differential equations. Even so, we can still take confidence in this result as a first-order approximation to real systems.
Tim
Notice the gain (as Vo/Vin) is NOT G, though it's proportional to G. It varies. Which is one possible definition of distortion. In particular, dVo/dVin = 2*G*Vin, so the small-signal gain is proportional to bias. This is sort of like a remote-cutoff tube, where bias can be used to vary gain.
If we put a biased sine wave through this thing with bias A and peak amplitude B, we have (note w == omega, angular frequency):
Vin = A + B*sin(w*t)
Vo = G*[A + B*sin(w*t)]^2
= G*[A^2 + 2*A*B*sin(w*t) + B^2*sin^2(w*t)]
= G*[A^2 + 2*A*B*sin(w*t) + B^2*(1/2 + 1/2 sin(2*w*t)]
In other words (omitting overall gain G for now):
- DC bias of A^2 + B^2 / 2,
- Undistorted signal of 2*A*B*sin(w*t),
- Distortion of (B^2 / 2)*sin(2*w*t), second harmonic,
- and no other harmonics.
Since we're only interested in AC, we can ignore the DC component. (It's noteworthy that the DC component depends on the AC amplitude. In fact, experience tells us that overdriven amplifiers experience a bias shift!)
We can look at the relative amplitudes of signal and distortion. The signal amplitude is 2*A*B, so the gain is linear with bias A (= variable gain, as noted before), and also linear with the input amplitude B. The distortion is B^2 / 2, independent of bias instead, and proportional to the square of the input amplitude. If we look at the ratio, we see the relative amounts: distortion per signal is B / (4*A), so the distortion goes up linearly with amplitude, and goes down linearly with increasing bias.
This is interesting, because tubes have a roughly x^3/2 transfer curve, and are known to be more linear at higher bias. Also, distortion goes down at small signal levels, i.e., if B = 1mV, then B^2 = 1uV, which is just about negligible. But it goes up significantly for signals which are very large relative to the bias!
If we now put this into a feedback loop, first we have to talk about how we do it. The transfer function is not a two-input system (like an op-amp), nor is it differential (like the grid and cathode of a tube), so we have to come up with some way to implement feedback.
If we make the substitution Vin <= Vin - H*Vo, where H is the feedback ratio and Vo is the output voltage, then we can implement the loop:
Vo = G*(Vin - H*Vo)^2
= G*(Vin^2 - 2*H*Vin*Vo + H^2*Vo^2)
Yuck, we have Vo on both sides, one of which is squared. We'll need to use the quadratic equation to solve for Vo:
G*H^2*Vo^2 + Vo*(-2*G*H*Vin - 1) + G*Vin^2 = 0
With the solution,
Vo = (Vin / H) + 1 / (2*G*H^2) + sqrt(4*G*H*Vin + 1) / (2*G*H^2)
which is pretty ugly, but we can figure a few things about it:
- It's linear in Vin, and has first-order gain 1/H, which is exactly what we expect from a feedback circuit. (A voltage divider has H = R2 / (R1 + R2), which will give useful voltage gain for typical values.)
- It has an offset of 1 / (2*G*H^2), independent of applied signal, which is kind of funky, but okay. It goes down as G --> infty, so we can think of this as input offset. For this particular case, the amp itself only has useful gain for nonzero signals, since it's "cut off" at 0V, so this is a reasonable interpretation.
- It has this TERRIBLE nonlinear term of sqrt(4*G*H*Vin + 1) / (2*G*H^2). How do you even take the sort-of-square root of a sine wave signal?
Maybe you can't take it directly, but you can find the Taylor series expansion of the function. Then it's easier to see what effect it has.
In particular, the expansion is:
[1 / (2*G*H^2)] * [1 + u / 2 + u^2 / 8 + u^3 / 16 + 5*u^4 / 128 + ...]
where u = 4*G*H*Vin. We already have the first two terms in Vo, so let's group them:
Vo = (2*Vin / H) + 1 / (G*H^2) + [1 / (2*G*H^2)] * [u^2 / 8 + u^3 / 16 + 5*u^4 / 128 + ...]
- DC component is now 1 / (G*H^2).
- Signal component is now 2*Vin / H, which is kind of weird...
- Distortion is now in polynomial components, so we can estimate the harmonic content. However, we can't find the exact harmonics just by looking at it, because (sin(x))^n is not equal to sin(n*x). In general, the harmonic terms will also produce lower order terms, so the DC offset and fundamental amplitude will change. If we evaluate the first couple terms (x^2 and x^3 are easy to do), we'll probably get a more realistic gain and offset for this system without getting too tedious, but I won't go that far right now.
If u << 1 (a condition of small input signal, low gain, or low feedback), we can approximate this equation by simply truncating the series at u^2 or so. The assumption being, if u = 1mV/V, then u^2 = 1uV/V, and u^3 and further are completely negligible.
For large parameters, we're better off using the original form,
sqrt(4*G*H*Vin + 1) / (2*G*H^2)
which has the asymptotic form (4*G*H*Vin >> 1) of
2*sqrt(G*H*Vin)/(2*G*H^2)
= sqrt(Vin) * G^(-1/2) * H^(-3/2)
Which goes to zero as G and H go to infinity (i.e., increasing loop gain).
So it's always true that:
1. Nonlinear systems are a pain to analyze,
2. Feedback linearizes a system, and
3. The asymptotic result, as loop gain approaches infinity, is zero distortion.
One final note: this analysis is for a static system only. Real systems are rarely stable at infinite gain, and when they are, the loop gain at signal frequencies is nowhere near infinite. Real systems are time dependent and the arithmetic shown above turns into nonlinear differential equations. Even so, we can still take confidence in this result as a first-order approximation to real systems.
Tim
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