Switching current source for Class A?

[darkfenriz]

The CCS shifts between shortcircuit, and double the wanted pull down impedance with a 50% duty cycle smoothed out by the coil = result: continous current in the follower mosfet = class A sound!
Yeah?

But where is the oscilator?

The whole circuit is an oscillator, as every peak-current mode SMPS. The duty cycle will vary with output voltage ("hot" mosfet's source).

QED047
I meant a buck regulator of audio bandwidth tracking output signal and thus keeping the Vds at constant low value. Yes, that's nearly a class D amp, but why are you afraid of doing something that you seem to be more then competent to do?

[QED047]

Originally posted by darkfenriz


QED047
I meant a buck regulator of audio bandwidth tracking output signal and thus keeping the Vds at constant low value. Yes, that's nearly a class D amp, but why are you afraid of doing something that you seem to be more then competent to do?

Good question! I have had a little bit of experience with Class D and I'm never completely comfortable with it. I'm almost afraid to say that I'm just addicted to single ended Class A sound. Much of this game is about sentiment (IMHO) but there are such objective things as "tube sound" and I'm convinced that there's an objective sound to Class D. For all I know it's what a perfectly linear recreation of a given recording should sound like, but I don't get the same listening pleasure.

Also, as I said before, all the Class D designs I have encountered depend on negative feedback to account for non-linearities in conversion and variations in load. I seem to be particularly sensitive to IM distortion so I try to avoid feedback although I won't pretend to understand what's going on in this department.

[Mr Evil]

Originally posted by darkfenriz
P.S. You can reduce the dissipation further by supplying the output mosfet follower from a tracking step-down-regulator diven by input signal. That would become some incredible ultra-efficient single-ended class A design!!!

A continuously variable supply voltage is Class H. It's a good idea, but I haven't seen many implementations.

[Nrik]

QED047 : thanks for the extended explanation, wich also clarifies the oscillator question, and thanks to Darkfenriz for supporting that.

But what I still don't get is: Where do the power go?
It pulls 2A, but only dissipate 1W so the DC point at the followers source is ½volt or ...?

The coil must take some of the heat also.

[stinius]

Originally posted by Nrik


The coil must take some of the heat also.

Yes of course.

[QED047]

Originally posted by Nrik
QED047 : thanks for the extended explanation, wich also clarifies the oscillator question, and thanks to Darkfenriz for supporting that.

But what I still don't get is: Where do the power go?
It pulls 2A, but only dissipate 1W so the DC point at the followers source is �volt or ...?

The coil must take some of the heat also.

When stinius replies "Yes of course" it is of course true that a real inductor (as opposed to a theoretical, resistance free device) will dissipate some power as heat - but this is a small percentage (part of the 10% losses) of that which is usually lost in a convenional current sink.

This is the point I think you may be missing:- while the follower "sees" what appears to be a resistive element "inside" the Constant Current circuit - the power is actually going into charging up a magnetic field. When the transistor connecting the inductor to ground opens, the magnetic field collapses - generating a potential (at the switch/inductor junction) that is conducted back into the supply reservoir via the diode. Hence the unwanted power that would have otherwise gone out as heat in a linear CC circuit is restored back to the supply and reused (well, around 90% of it anyway).

I think you may be unused to considering power "stored" in a magnetic field. If you look at the junction between the inductor and the transistor to ground with a scope you will see the node alternating between ground and the supply rail in a rigid square wave. If the diode was disconnected the voltage at this node would go way up (until it reached the breakdown voltage of the transistor!) - but the diode dumps it all into the supply - which keeps the lid on the voltage raise.

This last point presents a slight design challenge (easily overcome) as any inductance between where the diode dumps its potential and the resevoir caps may "pump up" the supply rail locally - but by returning the diode cathode to the reservoir directly, and having plenty of low ESR caps there (and maybe a little resistive/inductive filtering to allow the volts out into the amplifier in a more orderly fashion) we can easily get around this.

If you play around with a buck voltage regulator instead of a linear regulator, you get used to watching the current from your bench supply go down as you wind up the output voltage - hence the power (V x I) stays put. With a linear regulator the current stays put as you wind up the voltage (you hope!) so the power dissipated goes up. I guess you need to watch these things first-hand to get the feel for it. I remember scratching my head a bit when I first met switching topologies

[darkfenriz]

Hi QED047

Sorry if you find me pestering, but I have a question, that came to mind:
How does it behave at negative side overdrive?
To my understanding the switching frequency should go down with output voltage because of less steep current ramp. Isn't there a point at which the switching frequency falls into audio range?

[Omicron]

Originally posted by darkfenriz
P.S. You can reduce the dissipation further by supplying the output mosfet follower from a tracking step-down-regulator diven by input signal. That would become some incredible ultra-efficient single-ended class A design!!!

Wouldn't it be much easier in this design to vary the bias current instead of the supply voltage? I mean you could simply let it track the setting of the volume potentiometer. That way you would have reduced power dissipation at lower listening levels. All the machinery needed seems to be in place already...

[stinius]

Originally posted by QED047



I think you may be unused to considering power "stored" in a magnetic field.

No

Originally posted by darkfenriz
Hi QED047

Sorry if you find me pestering, but I have a question, that came to mind:
How does it behave at negative side overdrive?
To my understanding the switching frequency should go down with output voltage because of less steep current ramp. Isn't there a point at which the switching frequency falls into audio range?

Good point.

[Bigun]

Originally posted by QED047
I often wonder if anyone has thought about using a digital delay between two DACs to "prepare" in advance a suitable supply voltage for the output stage. Without such "premonition" a significant loss of response to low to high signal level seems inevitable.

Variable operating biass output ?