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Old 28th December 2005, 06:18 AM   #1
boxedin is offline boxedin  Canada
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Default Another question about class A.

where is the power actually lost in CLASS A?. Power side? or Signal side?.

All i see is you can't transfer power without dissipating heat. How do you transfer power ? Across a current source or put a resistor in parallel to speaker 8 ohm?. which is better way ?

What is best used efficient BJT current source circuit ?

i appreciate it.

Thanks
Alexk
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Old 28th December 2005, 06:35 AM   #2
AKSA is offline AKSA  Australia
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Hi Alex,

Power side.

In a Class A single end, the output stage is biased on all the time. To accommodate positive and negative waveforms, the output stage is biased at around half the maximum current, so that at full excursion positive it will be on hard but not saturated, and at full excursion negative it will be on but not switched off.

Thus this output stage is always on about 50% capacity, passing appreciable current. The product of this current and the voltage across the output device fixes the dissipation, and it is always high.

In a push pull, the output stage is biased so that all devices remain on for the entire waveform. So, at maximum positive excursion, the upper device will be close to saturation, while the lower device will be almost off. Vice versa for maximum negative excursion. Once again, you must select a heavy bias current to flow at all times, and the product of this current and the voltage across all devices (normally twice the rail voltage with a dual rail supply) dissipates a lot of power.

Note that all this power is dissipated in the output stage, which is required to handle the heavy currents required by the speaker.

Cheers,

Hugh
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Old 28th December 2005, 10:13 AM   #3
boxedin is offline boxedin  Canada
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Hi hugh,


http://www.rudistor.com/sound%2Dlab/ar3eng.htm

Whats wrong with above circuit ?

I see 330 ohms in parallel wasting signal AC voltage unnecessarily. Unnecessary heat is dissipated on those resistor. its a bad design i think.

Can we replace those resistor with current source?. I don't mind transistor dissipate the heat not the resistor. I don't like any resitor and capacitor on my Output Stage and Input Stage. As it distort signal.


let me know.


Thanks
Alexk
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Old 28th December 2005, 10:29 AM   #4
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This circuit was designed by i....
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Old 28th December 2005, 10:32 AM   #5
AKSA is offline AKSA  Australia
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Alex,

Yes, you can replace the three 330R resistors with a CC sink. No problem; this will give superior performance, no question, as the output transistor will then operate in constant current, rather than vary with signal.

HOWEVER, this is a headphone amp, and signals will not be all that high in amplitude, perhaps 5Vpp at the very most.

Thus the current variation will be low anyway, and the worsening of distortion minimal.

I'm not overly impressed with the circuit either; it's simple, but there is cap coupling there which I'd try to avoid with direct coupling and bipolar rails. And while there is no global feedback I feel that the distortion of the first stage would be excessive. Further, the use of two 3055 transistors, very slow devices, as a darlington configuration is a bit sloppy. More modern devices with appropriate driver, such as 2SC4793 and BC550, might be more suitable, and there should be a resistor from the emitter of the driver to ground to increase its quiescent draw for greater linearity would be a good idea.

Hope this is helpful,

Cheers,

Hugh
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Old 28th December 2005, 10:41 AM   #6
ilimzn is offline ilimzn  Croatia
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Default Re: Another question about class A.

Quote:
Originally posted by boxedin
where is the power actually lost in CLASS A...
...
What is best used efficient BJT current source circuit ?
It seems to me you are not clear about some very basic concepts in electricity, so your questions do not seem to make much sense.
Power in class A/AB/B amps is lost on output devices because they have to waste 'excess' power from the power supply which is not at that moment in time required on the output. For instance, in class A amplifier, using 30W power from the power supply, all 30W are lost as heat whenever there is silence in the input signal (and also in the output). Also, power is ALWAYS lost with these type amplifiers because the power supply id DC, and the only AC signal which you can create from DC with minimum losses is a square wave - which the output from the amplifier is not )or rather, never is under normal operating conditions, unless you are trying to listen to full volume square waves - instant death for your tweeters... and not good for your ears either ). If just has to do with the basics of AC, ohm's and joule's law.

Quote:
Originally posted by boxedin
http://www.rudistor.com/sound%2Dlab/ar3eng.htm
Whats wrong with above circuit ?
I see 330 ohms in parallel wasting signal AC voltage unnecessarily. Unnecessary heat is dissipated on those resistor. its a bad design i think.
Can we replace those resistor with current source?. I don't mind transistor dissipate the heat not the resistor. I don't like any resitor and capacitor on my Output Stage and Input Stage. As it distort signal.
Again, you seem to be missing some basics. Resistors and capacitors distort, but transistors do not? What mechanisms do you think are responsible for distortion in amplifiers in general?
Without these basics I don't see how you could feel qualified to judge how good or bad the design is. Subjective evaluation of how something sounds is one thing, but there is very little subjectivism about the vary basic laws of physics - unless you are capable of creating your own universe

The amplifier you linked has several problems, one of which is trying to drive a 32 ohm load with a 110 ohm output impedance. Fortunately, this is not a major issue except for the fact that it is very inefficient and produces a hump in the frequency response at the resonance frequency of the headphones.
Second, the power devices used are not well chosen for the application, but again, it will work - just not as good as it could.

Changeing the resistors for a current source is not a good idea unless some form of NFB is added. As far as efficiency, it will be the same with the resistor or the current source - either has to dissipate (standing current times half supply voltage) Watts of heat. Wether it's a resistor or a current source makes no difference. What would make a difference, would be an inductor, but then again you need a different circuit.

The heat dissipated on those resistors is necessary because otherwise there would be no way to get an AC component out of the amplifier. The transistor in the output pulls the output in the positive sense, the resistors in the negative - to have AC you need both ways, a positive and negative going signal. No resistors, no negative going signal, no AC. It's a different matter altogether what alternatives we have to resistors.
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Old 28th December 2005, 11:49 AM   #7
boxedin is offline boxedin  Canada
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Actually i am capable of making my universe. Its called illusion.

Lets do the critiqing part with only circuits not the person's intellectual capacity or his electronics fundamentals.

There are obvious things that when you amplify things you also amplify noise. All components dissipate to various degree. What i said is, i like to limit the output signal dissipation only to transistors not resistors. Thats what wrong with circuit. You amplify and why waste the amplified signal in disspation on a resistor?. There are many other parts wrong in that ciruit. But my main concern is why waste amplified sifgnal?.

Resistor can dissipate dc all it want.

CLASS A dissipating DC is a one thing , dissipating well amplified signal is bad thing!.

I have n't found a solution yet, when i find it i will share with you all.

This is all part of my reseach for Decent Class A amplifier. (I am not trying to make perfect no distortion amplifier as of yet). Taking baby steps to understanding theory and purpose behind every component in a CLASS A amplifier.

Thanks anyway for responding though.

Thanks
Alexk
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Old 28th December 2005, 01:06 PM   #8
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"All i see is you can't transfer power without dissipating heat. How do you transfer power ? Across a current source or put a resistor in parallel to speaker 8 ohm?. which is better way"

Power is not transfered, it is dissipated, ideally in the load, in audio output stages this is normally taken as the speaker.

Power is (simplified - ignoring phase for this discussion) the product of the voltage across the load and the current through the load.
Thus the idea for an output stage is to generate a voltage source which will drive the load from as low a resistance as possible and faithfully follow th input signal. In fact a speaker is a current operated device - the current in the coil reacts with the field of the permanent magnet to produce movement. Thus in reality it is the current which must faithfully follow th input waveform.

As described by AKSA, in order that current is available to drive the speaker, it is necessary to have the drive device (FET, MOSFET, BJT, Valve) conducting at a value which permits the current to be 'diverted' to the speaker when the signal requires.
Thus in a single ended output the output device needs to be conducting the maximum instantaneous value of current that the load will be driven with so that when the device turns off (say on the +ve o.p. cycle) the full current can flow in the load, similarly the output device must be able to conduct the full current of the load (speaker) plus the current into the internal load (e.g. a resistor) (say on the -ve. o.p. cycle). In practice complete turn off and on of the output devices is avoided as, amongst other effects, this leads to serious distortion.

The use of a current souce would not assist as it would need to dissipate the same average power as the equivalent resistor.

Please note. This description has ignored all the issues relating to the reactance of the speaker etc., and the changes in its impedance with frequency. Also issues of damping, feedback and speaker filters to seperate the signal to the various speakers.

Hope this helps.

Richard
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Old 28th December 2005, 01:06 PM   #9
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Quote:
Originally posted by boxedin
Actually i am capable of making my universe. Its called illusion.

Lets do the critiqing part with only circuits not the person's intellectual capacity or his electronics fundamentals.

There are obvious things that when you amplify things you also amplify noise. All components dissipate to various degree. What i said is, i like to limit the output signal dissipation only to transistors not resistors. Thats what wrong with circuit. You amplify and why waste the amplified signal in disspation on a resistor?. There are many other parts wrong in that ciruit. But my main concern is why waste amplified sifgnal?.

Resistor can dissipate dc all it want.

CLASS A dissipating DC is a one thing , dissipating well amplified signal is bad thing!.

I have n't found a solution yet, when i find it i will share with you all.

This is all part of my reseach for Decent Class A amplifier. (I am not trying to make perfect no distortion amplifier as of yet). Taking baby steps to understanding theory and purpose behind every component in a CLASS A amplifier.

Thanks anyway for responding though.

Thanks
Alexk


Greetings from the Uk

I realy think that you ought to go out and buy a good book on electronics, and read the section on amplification, this will explain the conceps of amplifiers, and how they operate.
Without a knowledge of these fundamentals your understanding of the various components making up an amplifier will be limited.
The function of any given component often is different in the dc and the ac analysis of the circuit.


With regard to your comment ref. noise, you should consider the fact that a good resistor generates less noise than a transistor or other active device. This is not normally a major consideration in a power amplifier, only in a pre-amp for a low level signal e.g. microphone or disc pick up.

Richard
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Old 28th December 2005, 02:32 PM   #10
ilimzn is offline ilimzn  Croatia
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Quote:
Originally posted by boxedin
Lets do the critiqing part with only circuits not the person's intellectual capacity or his electronics fundamentals.
I used to have a maths teacher, who would, when students would insist to a wrong answer, say that he will accept it as long as they prove it. This would. of course. put the student into ever deeper trouble. The moral of this story is, you cannot critique a circuit without knowledge of the funadamentals (note: FUNDAMENTALS, I am not talking about finesse here). It was not my intention, and indeed, it will never be my intention to critique anyone's intellectual capacity. The latter becomes evident soon enough, for all of us

Quote:
There are obvious things that when you amplify things you also amplify noise. All components dissipate to various degree. What i said is, i like to limit the output signal dissipation only to transistors not resistors. Thats what wrong with circuit. You amplify and why waste the amplified signal in disspation on a resistor?. There are many other parts wrong in that ciruit. But my main concern is why waste amplified sifgnal?.
Resistor noise voltage equals square root of (4 x Boltznam Konstant x resistance x temperature x bandwidth). For your 3x330 ohm paralleled resistors (=1x110 ohm resistor), working at an all too high 150 deg. C this comes out as approximately 227 nanoVolts - your signal is on the order of Volts, so you have at least 120dB of distance between it and noise generated by this resistor. Assuming you replace the resistors with transistor(s), do you want to try adding up all the noise mechanisms that work inside it, and do not forget that the transostor itself will amplify this noise, whereas the resistor will not. And then there are all the other components ahead of the output stage...

As far as wasting amplified signal, I have given you an answer to that problem, I believe it has been known for some 75 years, probably more, in this sort of application, and it is called an inductor or choke. Otherwise, how does X Watts wasted on a resistor, differ from X Watts wasted on a transistor? I would say about as much as the difference in the weight of one ton of iron and one ton of feathers.

There are other ways of doing this that do indeed waste less power - such as a current steering topology, which is normally done with complementary transistors (and is not single ended, therefore). In this sort of topology, there is no resistor in parallel to the load, and power from the power supply is constant and equal to maximum output power on the actual load - what is not used by the load, is dissipated by the transistors.

Quote:
Resistor can dissipate dc all it want.
CLASS A dissipating DC is a one thing , dissipating well amplified signal is bad thing!.
In this sort of topology, if it dissipates one, it dissipates the other.

Quote:
This is all part of my reseach for Decent Class A amplifier. (I am not trying to make perfect no distortion amplifier as of yet). Taking baby steps to understanding theory and purpose behind every component in a CLASS A amplifier.
In that case, it may be prudent to look at a basic design, one transostor, such as Nelson Pass' Zen.
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