Class B w/o crossover distortion (1975)

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There is a quite great difference between Hans Hartsuiker's and Peter Blomley's circuits : the common base transistors are driven on their emitter by a low impedance (ouput of the 5534 + 10 Ohm) in the first circuit and by a high impedance (about 20 kOhm) in the second circuit, which then works more like a current source. To me, this last one seems the good way to implement the idea.
 
Hahfran,



I'd be interested in your attempt.... this has defied some of the best brains on the planet for fifty years.... you may be moving towards error correction feedforward, of course.

However, how about we strike a compromise? A conventional, voltage driven output stage, BUT, with no switching off at the output devices at any time? Would this suffice? It's been done, of course, a gasoline engine doing diesel tricks, and this could avoid a large Class A clunker with its grossly inefficient quiescent current.


Hugh

Zero bias diodes based on zero point energy don't have a pn junction these are not Schottky diodes in fact, theoretically, they should conduct at bias levels in the range of nanoVolts.
However the idea with CMOS audio switch controlled by zero cross detector
to split the signal in the frontend or input doesn't work that will produce more
crossover distortion than a class B....
So what about the compromise ? I think however after reading a translated excerpt of Baxandall's comment on the output triple cascade the idea of current driving has merits over voltage drive since there are now power transistors with a large signal current amplification versus frequency that suffice the requirements. These did not exist in 1969.
 
Coming back to the original Visch circuit, it does suffer from pretty nasty dynamic artefacts:

The sim shows what happens when a significant level of signal (15V peak ie. 25% of nominal power) is applied after an idling period: the .ic statement defines the voltage across the hold capacitor, 1.69V when idle.

During the first few cycles, the initial Iq of ~50mA is insufficient, and results in clipping of the negative part of the waveform.
After some tens of ms, the Iq rises to 650mA, and the clipping disappears.
 

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This is not a bootstrap: boostrapped amplifier have no problems with sudden transients, quite the contrary. They are unable to pass DC, but the capacitor can be made as large as practical to increase the LF response.
Here, if the capacitor is made larger, it will only increase the response time. And if it's too small, the amplifier will distort. There is no acceptable trade off.
 
Tief, To answer these excellent questions you must build the circuit, subject it to measurement and listening tests over a large sample. And once you have the answers, you may be reluctant to share your findings publicly! These questions precipitate forum wars, with everyone having an opinion, but no one willing to build the prototype.....
Have a great week, Hugh

................ but no one willing to build the prototype. I hope, some guy's will do this in the next time
 
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This is not a bootstrap: boostrapped amplifier have no problems with sudden transients, quite the contrary. They are unable to pass DC, but the capacitor can be made as large as practical to increase the LF response.
Here, if the capacitor is made larger, it will only increase the response time. And if it's too small, the amplifier will distort. There is no acceptable trade off.

A bootstrap is always a compromise of two contradicting requirements.
The cap has initially a charge according to C*V. If the transient is negative the cap must be charged and the time constant is proportional to C. Thus the contradicting requirements are according to the polarity of the transient signal relative to idling. Exactly a bootstrap functions perfectly if the amp always idles.
 
A bootstrap is always a compromise of two contradicting requirements.
It isn't:
In the (crude) example BS1, the amplifier is operated at its maximal power, with a bootstrap capacitor of normal size: the voltage across the cap is centered about 5.4V.
BS1idle shows the no-signal condition: C2 now has a stable 5.4V voltage.
There is no difference, because the circuit is linear: the average potential on <**> and the average potential on out> remain identical, independent of the signal level.
The cap has initially a charge according to C*V.
This means that the CV product remains constant, provided the capacitor is large enough.
Thus the contradicting requirements are according to the polarity of the transient signal relative to idling. Exactly a bootstrap functions perfectly if the amp always idles.
The only contradicting requirements are that of size vs. performance:
Hicap shows an extreme, with 100 farad capacitance, and locap shows the other extreme, no cap at all.
Ideally, C2 should be replaced by a battery, there would be no LF response issue.
 

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It isn't:
In the (crude) example BS1, the amplifier is operated at its maximal power, with a bootstrap capacitor of normal size: the voltage across the cap is centered about 5.4V.
BS1idle shows the no-signal condition: C2 now has a stable 5.4V voltage.
There is no difference, because the circuit is linear: the average potential on <**> and the average potential on out> remain identical, independent of the signal level.

This means that the CV product remains constant, provided the capacitor is large enough.

The only contradicting requirements are that of size vs. performance:
Hicap shows an extreme, with 100 farad capacitance, and locap shows the other extreme, no cap at all.
Ideally, C2 should be replaced by a battery, there would be no LF response issue.

ic=dU/dt*C however the current ic is limited by R3
I think however this is academical because if U(t)=a*sin(2pift) then
ic=a*cos(2pift)*C if R3=0 ( then it isn't bootstrapping) BUT no acoustical
instrument starts up with full amplitude a! Rather the sonic character of an acoustical instrument is determined by the attack -sustain- release envelope
and the dynamically changing harmonics spectrum in this time.
The attack time of for ex. a violin is about 10 ms same for a flute for large organ pipes its already seconds. So the case you brought up happens only
in very rare cases, especially in electronic instruments.
In practice this kind of distortion does not apply except if you prefer to listen to sine bursts 5 on 50 off.
 
ic=dU/dt*C however the current ic is limited by R3
I think however this is academical because if U(t)=a*sin(2pift) then
ic=a*cos(2pift)*C if R3=0 ( then it isn't bootstrapping)
I don't see what you mean. Which schematic are you referring to?
BUT no acoustical
instrument starts up with full amplitude a!
I am no expert on the subject, and if a member with good expertise on the subject could give his opinion, it could be helpful.

However, I see at least two examples of such instruments:
-Percussion instruments, drums, cymbals, bells...
-Pinched cords instruments, typically guitar
Anyway my opinion is, that an amplifier has to be technically flawless (blameless?) before proceeding any further
Rather the sonic character of an acoustical instrument is determined by the attack -sustain- release envelope
and the dynamically changing harmonics spectrum in this time.
The attack time of for ex. a violin is about 10 ms same for a flute for large organ pipes its already seconds. So the case you brought up happens only
in very rare cases, especially in electronic instruments.
In practice this kind of distortion does not apply except if you prefer to listen to sine bursts 5 on 50 off.
Even it were the case, you (I anyway) do not build amplifiers to suit specific cases: you build one amplifier, capable of handling any situation, any type of program, however rare.
In my opinion, cutting corners from the start of a design is not a good idea.
If the task really seems impossible, you can begin to think about "simplifications", but it is only a last resort option (for me anyway).
 
I refer to the Vish schematic. Every bootstrap cap has to deliver current ...
The bootstrap cap must supply the 15 mA when the current source is off at negative swings and is recharged by the current source. With pencil-and-paper methods I cannot confirm the effect shown in the simulation. I will attempt to find out "who is wrong".
 
The capacitor in the Visch circuit is not a bootstrap. It is an analog memory element, holding the actual Iq setting.
The charge and discharge paths are entirely non-linear, and the reasoning pertaining to a bootstrap capacitor cannot be applied.

The maximum negative swing is ~30.5V, and this shows that the presence of the capacitor does nothing to improve that aspect: see the clipping example.

The circuit needs about 500ms to reach its full amplitude, and at that time, the Iq amounts to ~2.2A.
 

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Well a bootstrap capacitor is supposed an analog memory because it is assumed that
q=CV= const and therefore the voltage swing is transferred in full.
However every bootstrap cap must supply current and must be recharged. Therefore at low enough frequency the assumption does not hold. With discharge q decreases and V=q/C drops. This is why there must be something wrong with the sim. Initially the C is charged and if dq/dt=i=15mA leads to a V drop of dV that makes the clipping then this must be always so there is no reason for a "recovery". The "time consant"
of 500 ms cannot follow from iconst*C
You may do a sim of i(R3)
 
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I think I(R3) is only moderately interesting, because the voltage across R3 is constrained by the Vbe of Q1.
More interesting is the current in R7.
Q7, D1 and Q2 act as a sort of PWM modulator, sharing the 15mA between Q7 and Q2. The process operates in a closed loop and takes time to reach an equilibrium, and only part of this current is available to charge the capacitor.
 

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I think I(R3) is only moderately interesting, because the voltage across R3 is constrained by the Vbe of Q1.
More interesting is the current in R7.
Q7, D1 and Q2 act as a sort of PWM modulator, sharing the 15mA between Q7 and Q2. The process operates in a closed loop and takes time to reach an equilibrium, and only part of this current is available to charge the capacitor.

Ok. in the schematic you provided "my R3" is your R7. Is D1 a Schottky?
In the amp I made back in 1976 or so D1 is actually a big power germanium transistor C/B shortened ( as it seems I don't know the configuration of the 3 pins ).
I still think the problem is in the sim right now its way too cold down in the basement ( for me not for the amp) to measure the amp but as far as I recall
no effect like that has been observed.
 
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