Thanks Hannes!
In doing a "cost-benefit analysis" I have to try and weigh the "hash" you refer to. At the moment It seems to be an entirely subjective cost as I'm finding that the ripple present at the sense resistor is immeasurable at the loudspeaker. On the other hand, the objective reduction in dissipation is huge.
With Class D the signal has to be accurately converted into a PWM train, and requires signal feedback to make it work in practice (I've not seen any designs that don't use signal feedback). For this reason alone I find Class D to be undesirable - no feedback is one of my key objectives.
In doing a "cost-benefit analysis" I have to try and weigh the "hash" you refer to. At the moment It seems to be an entirely subjective cost as I'm finding that the ripple present at the sense resistor is immeasurable at the loudspeaker. On the other hand, the objective reduction in dissipation is huge.
With Class D the signal has to be accurately converted into a PWM train, and requires signal feedback to make it work in practice (I've not seen any designs that don't use signal feedback). For this reason alone I find Class D to be undesirable - no feedback is one of my key objectives.
It's become obvious to me at least that I don't understand this.
In the ultimate implementation we could have a single ended follower that dissipates Vce * Ic and the other dissipations approach zero.
That would be twice as efficient as a CCS loaded follower and 4times as efficient as a resistor loaded follower.
In the ultimate implementation we could have a single ended follower that dissipates Vce * Ic and the other dissipations approach zero.
That would be twice as efficient as a CCS loaded follower and 4times as efficient as a resistor loaded follower.
darkfenriz said:
Maybe, but a "tracking step-down-regulator driven by input signal" sounds a bit like a complete Class D amp
Although I suppose you might mean for the supply to change outside the audio band (very slowly?) so as not to interfere with the signal. Perhaps simply slave the supply voltage to the preamp volume control?But this would necessitate a switch-mode power supply design - which is a reasonably proposition but not one that is altogether appealing. Having said that, a buck regulator topology fed from a conventional transformer defining the maximum DC supply might be more palatable than a complete off-line design using HF transformers.
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.
Nrik said:
I really haven't invented anything here. Besides, I think patents suck big-time. They're only worth the money you have to defend them with - and I deplore the practice of companies patenting stuff that's not inventive but merely common sense to everyone (e.g. Color Kinetics patenting the use of RGB LED lighting when the properties of additive light have been exploited ever since Sir Isaac Newton played around with a prism).
But you're right - half an amp based in this would be cool (to the touch)
EDIT: If not already established, this thread may serve as evidence of PRIOR ART dating from July 16 2009 (just in case some idiot does try to patent it)
AndrewT said:
I think - correct me if I'm wrong - that it goes down like this:
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?
Nrik said:
I don't think you have uinderstood it completely. I'll have another go at explaining how it works: The inductor prevents current flowing into a dead-short when the bottom transistor switches on. Instead, due to the reactive nature of the magnetic field opposing the current, the current rises linearly in time - in proportion to the value of inductor and the potential across it. The differential amplifier senses the current build-up and trips the comparator (switching off the transistor) when it reaches a set point. The current then decays (linearly) and when the hysteresis of the comparator circuit is exceeded, the transistor switches back on. Hence it is self-governing and oscillates at a variable rate depending on the potential of the follower stage. The frequency is determined by a number of factors, but they are all controlled (do not depend on load).
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?
darkfenriz said:
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.
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.
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.
Nrik said:
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
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?
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?
darkfenriz said:
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...
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