I've read several articles on polarization in class a of the op amp's final stage, such as inserting a resistor between the power supply (negative or positive) as a current generator or btw with fets. the problem that I believe is at the basis of this solution is that in any case the dynamic parameters of the op amp change, because the current sources are not ideal. In your opinion, wouldn't it be much simpler to power the op amp with unbalanced voltages like +18 -12 ?... wouldn't the final stage work anyway in class A?
Note that the bias does not always work as expected: I was doing a comparison between various "good" opamps, like the NE5534, LM4562, etc., and for some I was astonished to find that the THD was in fact increased, not reduced!
Obvious. Bias has an optimal value above and below it simply goes wrong. Take an example any triode and write down some loadlines and watch it by yourself.
It is probably a bit more complicated than that: true, an AB output stage has a sweet spot for bias current, but when a sufficient static bias is added, it is not AB anymore, it becomes pure class A, which tends to eliminate a number of distortion mechanisms.
In fact most of the samples benefited from class A, but not all. I have posted the results somewhere on the forum, but it was a side discovery when studying something else, and I didn't make systematic measurements for that. With lots of time and patience, the post(s) in question should be retrievable, but I am not going to attempt it as the new search function sucks.
The possible explanations are a): distortion cancellation occurs between the two halves of the OP and is upset when one is disabled
b): I pulled the bias from the "slave" side of the PP in circuits that have no symmetrical drive
c): Another unidentified reason....
Summary: it can be beneficial, even in top end opamps, but it needs to be carefully evaluated and tested in reality and for each case, otherwise it could have the opposite effect
The output stage symmetry was reduced since one device got more current, hence there is more even order THD.
OK. But most opamps I know operate in class AB in order to limit the power disipation inside the waffer and the current drain. Moreover in multiple devices as DIL 14 packages (TL084 for example). Class A needs more room and surely an external heatsink, and for most purpuses are useless. Let aside finest audio applications, opams used in voltage sensing, motor control, integration/differentiation of signals, analog computing in general, don't take advantages of pure class A.
If I remeber ok, most of the wafer area was occupied by the compensation capacitor.
If I remeber ok, most of the wafer area was occupied by the compensation capacitor.
That won't help, the trick is to let the output current always flow in the same direction.
As I said, this was a side discovery, and I didn't investigate further: I simply looked at the global THD figure, and saw that it went up (in some instances only: most of the times, it went down as expected from common wisdom theories).
It could be that pulling the current from the opposite polarity could be beneficial, but I have no idea: I was busy testing something else. Could be worth investigating though
To answer the OP's question, operating with an asymetrical supply will only limit the peak output voltage to the lower voltage supply. It will still run as Class AB unless you add the "forced Class A" resistor/current source.
I recall the "forced Class A" trick is originally from Walt Jung in an article and/or one of his earlier op-amp books several decades back. More recent op-amps have distortion so low it's hard to measure, and presuming forced Class A even makes for an improvement in measured distortion, it's questionable whether it would be audible.
I recall the "forced Class A" trick is originally from Walt Jung in an article and/or one of his earlier op-amp books several decades back. More recent op-amps have distortion so low it's hard to measure, and presuming forced Class A even makes for an improvement in measured distortion, it's questionable whether it would be audible.
but feeding it asymmetrically the 0p amp is like working with offset. if for example I needed a 4Vpp maximum excursion, it would work in the point indicated in the image with lower distortion. Am I doing something wrong in the reasoning?
That graph would not apply if there were a current source offset. Of course, the feedback could cover up much of the increased distortion.
You are misinterpreting the graph. It doesn't show harmonic distortion, but distortion plus noise, THD + N. To measure harmonic distortion accurately at low levels, they would have had to measure in narrow bandwidths around each harmonic, calculate the root of the sum of the squares and compare it to the fundamental. Instead they just measured everything but the fundamental and compared that to the fundamental.
In the part to the left of the minimum, each factor of ten signal increase causes a factor of ten decrease of the THD + N, because it is actually the noise floor rather than the distortion that dominates.
In the part to the left of the minimum, each factor of ten signal increase causes a factor of ten decrease of the THD + N, because it is actually the noise floor rather than the distortion that dominates.
Patrizio88 - You need a sophisticated analyzer to measure distortion below the THD+N floor. This is done via FFT or averaging of periodic signals.
The ear too can detect signals buried in noise, but not at such low levels. I would not worry about such levels of distortion.
Ed
The ear too can detect signals buried in noise, but not at such low levels. I would not worry about such levels of distortion.
Ed
"Zero-crossing" crossover distortion is zero current not zero voltage.
And modern opamps have almost no idea where their supply pins are, as long as they have several Volts headroom so the crystal does not starve.
And modern opamps have almost no idea where their supply pins are, as long as they have several Volts headroom so the crystal does not starve.