Class B w/o crossover distortion (1975)

[hahfran]

Originally Posted by Elvee

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.

[Elvee]

Originally Posted by hahfran

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).

[hahfran]

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".

[Elvee]

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.

[hahfran]

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)

[Elvee]

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.

[hahfran]

Originally Posted by Elvee

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.

[Elvee]

Substituting a schottky for D1 doesnt change the behaviour.

Here is the voltage across the cap: goes from 1.69V when idling to 4.4V at full power.

[hahfran]

Originally Posted by Elvee

Substituting a schottky for D1 doesnt change the behaviour.

Here is the voltage across the cap: goes from 1.69V when idling to 4.4V at full power.

Hum...when biased properly ( DC offset 1.3 V SINE) the voltage across the cap should be 15 volts with no ac signal applied.

[Elvee]

If the cap had 15V, this would mean that all the bias current goes into the string of predriver transistors, with nothing left for Q2.
It would be a hard-switching class B.