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-   -   BOSOZ with CCS, and a question (http://www.diyaudio.com/forums/pass-labs/30477-bosoz-ccs-question.html)

jh6you 18th March 2004 09:24 AM

BOSOZ with CCS, and a question
 
2 Attachment(s)
Having been listening for about 1.5 years, I have arrived to my final BOSOZ as shown below. On the way to this, I have had various tweaks.

Brief info:
The differences from the original BOSOZ are self-explanatory in the attached circuit. Anyhow, CCS¡¯ are applied. The bias current is set to 30mA. The rail voltages are +/-30. The gain is minimized to 6dB. I am fully satisfied with the sound.

A technical question:
Mine is for the unbalanced-balanced use, and the volume control is arranged at input, not at output. In principle, this could be not a good idea with respect to CMRR and the level of output noise. In this respect, do you recommend me to have the volume not at the input, but at the output? Of course, if I want to have the volume at the output, I have to use certain money. Still is the output volume worth a try? Your opinion would much appreciate. Thanks.

hugonot 18th March 2004 09:54 AM

Little changes
 
If you don't use Q2 input, you have to connect its gate to ground. R19 = 100k and this could limit bandwidth of your preamp. When you connect its to signal source then impedance gate to ground = output impedance of signal source ( usually small value < 200ohm) + R9.

jh6you 22nd March 2004 04:28 AM

hugonot

Afraid to say that I don’t clearly understand your point.

I compared the sound with and without R19, but heard no noticeable difference. Nevertheless, the reason why I have R19 is to think about the symmetric ciruit. In order for me to have a most probable symmetry, I am also considering to replace the input volume with the (balanced) output volume. My question came from this thinking, though no reply...

MBK 22nd March 2004 04:54 AM

For optimal CMRR in a differential amp, the (-) in and (+) in source impedances should be equal. So, you could use 0ne double ganged pot per channel, and have the "volume control" voltage divider apply to your (-) input in the exact same way as for your (+) input. Then you connect the signal line of your unbalanced source to (+) in and the ground line of your unbalanced source to your (-) in. If you know the source's impedance, you could use a small resistor of same value as signal source impedance, before the (-) input to make it a perfect balance with ideal CMRR.

Basically what's missing in your circuit is to put the exact same volume pot on (-) in *before* R19, same as before the (+)in.

MBK

jh6you 22nd March 2004 07:48 AM

MBK

Well understood. Thanks to your explanation, I can draw a clear picture about the ideal symmetric circuit. And, I presume that if I install the output voltage dividers at +out and –out instead of the one at the input, I could have the same result indicated by you. Mmm... thanks again.

JH

PS/
By the way, being compared with the present circuit, would the ideal symmetric circuit (i.e. with the enhanced CMRR) give me noticeably better dynamic sound? Is it worth a try? I wonder where in Shanghai I could buy a four-gang 12-step rotary switch. Always, a question bites another...

MBK 22nd March 2004 10:45 AM

Hi jh6you,

if you experience a lack of dynamics, from your circuit the reason could be that your input impedance is quite low (3.2k pot??). That's good for low noise, but it puts quite some load on the source (the source = CD player etc., must be capable of delivering high currents). If your source is a stock consumer product it may be worth using a higher impedance pot, say 20k, just use a cheap one to see if that was "it". Or you could buffer the output of your source to make it able to deliver more current.

Before you go and look for a 4 ganged pot, I would try the above variation of input impedance, and also, you could use a fixed resistor network at (-)in and set it to an average value similar to your "usual" pot position at the (+)input. That will give you an idea if it makes a huge difference.

MBK

jh6you 22nd March 2004 01:49 PM

Hi MBK

You kindly confirm that the low input impedance demands the source which is to be a proper combination of voltage and current. As far as I remember, Mr Nelson Pass also said the same in his Zen amplifier papers. Meanwhile, in fact I have no practical problem in dynamic range of the sound at the moment. Nevertheless, if there is a room for further improvement of it, I have no reason to hesitate to get it.

I appreciate your good and practical advice. First of all, I will try the fixed resistor network at (-)in. I will come back to the thread with the result.

:)

JH

jh6you 25th March 2004 03:02 AM

Have tried the fixed network at (-) input, which must renew CMRR on certain degree. Nevertheless, I could not feel it with my ears. Not at all. Maybe too early to say, but I might give up the thinking that I have wanted to remove the input voltage divider (2 ganged pot) and instead install them at the output (4 ganged pot) with respect to CMRR and attenuation of output noise. I would keep the input voltage divider as it is.

Meanwhile, as a trial, I have lowered the gain value down to 0dB by increasing R5 from 500 to 1k, and, YES, got a good result. How much? Enough to convince myself: “the lower gain, the better sound.” Soft details are much improved, further helping my easy listening.

It has been interesting trial.

MBK 25th March 2004 08:34 AM

Hi JH,

CMRR matters sometimes more, sometimes less. It is most important in systems where many devices are connected over long lines. This creates differences in the grounding and ground loops. That is why pro audio uses balanced lines and tries to get maximum CMRR. BUT, if no obvious grounding problems exist, the CMRR is actually not so important to the sound. Me I tried to make my system all balanced and with great attention to CMRR. But sometimes I wonder if it was really necessary. My sources are still unbalanced, and don't create noise problems either.

Interesting what you said about the gain. Me about 80% of my chain is op amps (only my mid amp is an Aleph), so I have high open loop gains by default. I am interested in trying discrete buffers with low gain now...

stefanobilliani 25th March 2004 11:22 AM

Quote:

Originally posted by jh6you
I have lowered the gain value down to 0dB by increasing R5 from 500 to 1k.

Don't you experience more noise in that condition?
Yes , it depends on your source and amplifier/speakers conditions and carateristics , but wow!

In many tries I stay happy with R5 at 150 ohm ,it is a lot smaller than 1000 , no input pots , 10k at the output balanced.

Just a though , anyway.

:)


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