Simple Symetrical Amplifier

[Lazy Cat]

Hi Shaan

Wilson current mirrors as CCS nicely fit in SSA sch instead of current injection resistors. Signal distortions and AC performance in whole will be improved significantly. The only potential issue I see here is thermal stability of injected currents, so careful BJT matching and thermal coupling must be done. Two cascode BJT-s have to be located away from Wilson pairs, since cascodes dissipate most of the CCS power and having different thermal conditions as coupled pairs. I suggest you to assemble one Wilson current source/sink separately and test its current stability while heating its parts with suitable heat source (solder iron). This will give you an impression of how the circuit will fit/behave inside SSA sch and to pre-eliminate later surprises.

[Esperado]

Originally Posted by shaan

CSS with as low parts count as possible.

Fishing for compliments, Shaan ? ;-)
As far as i know, there is no better Accuracy/Simplicity ratio than Wilson Current mirror.
The only issue is it deteriorate with high frequency.
And temp variation effects, as pointed by L.C, but symmetrical on both rails.

Dado: Agree with all, Just a remark, with high voltages (>60V) 6v of drop just represent 10% of loss eq <1db.
Agree too with the keep-it-simple Zener noise : will try a low noise current source to see if any listening impact. If i worry about rail noise in preamps and converters, i use to neglect this aspect in main PSU for power amps. May-be i'm wrong ?

To LC, it would be interesting (i will do on my side if i find time for that) to study a perfect PSU for the SSA. I imagine a combination of serial and shunt regulation, ( with big loss of power expected ;-). Something like common serial and one shunt regulation on each amp board. To see if any noticeable sound improvement.
At this level of quality, no major improvement can be done amp side. PSU can bring the next big step. Don't you think ?

[Samuel Jayaraj]

Originally Posted by dadod

It will not detects small DC output offset and if one of the rails of the output fails it will shut down other one immediately.
This regulator powers whole amp, imput stage has additional c multipliers on the amp board.
dado

Thought as much that this regulated supply must be for the whole amp. Can we expect each rung to supply about 10Amps. Ideally how much capacitance should precede and succeed the regulator?

[shaan]

Hi LC.

Thanks for the points. I'll assemble the mirror tomorrow on the existing board replacing the resistors(seems it will be easy).

Hi Esperado.

Please understand this, you are over-estimating me; I learnt about the Wilson mirror just yesterday, from Wikipedia. I thought I would ask the kind guys here, diyaudio, as I have neither any experience with it nor do I have an electronics degree. To my naked eyes the Wilson seemed the best mirror with lowest parts count. I simulated it to verify. Do you not like all this? Then I'm sorry I disappointed you and wasted your time. But please either quote the full sentence when you reply or try avoiding use of poverbs in your posts. My mother language is Bengali, not English. Thanks.

[Esperado]

Originally Posted by shaan

But please either quote the full sentence when you reply or try avoiding use of poverbs in your posts.

Oh, please, Shaan, it was a smiley of sympathy !!!

You asked for a simple and good current source design and propose in the same message the best possible response.
Then, "the fishing for compliment" was just a second degree wink to say "indeed".

English is not my English language neither and i try my best.
Sorry for my bad sens of humor.

I don't know how experienced you are in electronic, and, if as you says you are an electronic beginner, you deserve hundred time more everybody's (and my) respect.
You work with passion, in a creative way, and i love people which experience by themselves, try various solutions instead of simply copy other's work, and are not afraid to take risk. You certainly not make loose time to nobody here. Please, keep on truckin, i'm sure you will be soon an amazing audio designer.

[dadod]

Originally Posted by Samuel Jayaraj

Thought as much that this regulated supply must be for the whole amp. Can we expect each rung to supply about 10Amps. Ideally how much capacitance should precede and succeed the regulator?

I use 4700uF + 10000uF per rail with 0.11R between and that was quite enough. This preceed the regulator, and succeed anly what is on the amp PCB, in may case 470uF per rail plus additional c multipliers for input stage.
dado

[shaan]

Originally Posted by Esperado

You asked for a simple and good current source design and propose in the same message the best possible response. Then, "the fishing for compliment" was just a second degree wink to say "indeed".

I see, got it. Sorry for the misunderstanding. Apologies.

[shaan]

I have simulated the temperature co-eff of the Wilson mirror. And must admit that it really has a big dependence on the device temp, drifting the offset as high as 400mV. This is when the +ve mirror is hot and the -ve is not.

But, when I heat both cascodes of the mirror to say 70 degrees(and hotter) then the offset only drifts about 25mV or so. It seems the offset is affected when the cascodes are at different temperatures, like the input pair.

Is it safe to thermally couple just the two + and - cascodes of the mirror, as well as the mirror pairs of each polarity?

Thanks.

[shaan]

About to connect them. Wish me good.

[Lazy Cat]

Originally Posted by shaan

But, when I heat both cascodes of the mirror to say 70 degrees(and hotter) then the offset only drifts about 25mV or so. It seems the offset is affected when the cascodes are at different temperatures, like the input pair.

As I said in #2931

Originally Posted by Lazy Cat

Two cascode BJT-s have to be located away from Wilson pairs, since cascodes dissipate most of the CCS power and having different thermal conditions as coupled pairs.

Originally Posted by shaan

Is it safe to thermally couple just the two + and - cascodes of the mirror, as well as the mirror pairs of each polarity?

Thanks.

You did it correctly, just right like it is in the picture.