Does this look right?

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An externally hosted image should be here but it was not working when we last tested it.

I'm gonna run a BOSOZ unbalanced in/balanced out and hopefully someday feed a nice pair of Aleph 2 monoblocks. Anyways what I'm concerned with is how I terminated the negative input, I did a search and I believe this is what Nelson recommended.

The other main question is with regards to the portion of the circuit in the red box. I'd be using this pot as an input level trim with one of these (dual gang) on each input either before or right after the switching relays. Am I correct in thinking that this would work? I'd like to be able to use only dual ganged pots in my circuit for simplicity's sake (trim, gain, and volume). Any other suggestions for tweaks I'm getting real close to ordering up the parts.


My bad on the image. This time it should work.

Why is it that volume controls have to be after the relays? The pot on the input will be used to adjust for the different input signal levels depending on the sources used. To make the input signals all approximately the same level so that switching from say a CD player to a computer output doesn't dramatically change the volume.

I'm still gonna have my main volume control on the outputs. As discussed in this thread.

balanced volume control
And if you want unbalanced in to balanced out, there will be losses on undriven side and you will have unequal voltage on + output and - output so I suggest you that you install constant current source as discussed in this thread opened by me because I had the same problem with unbal -> bal.
You have 2 options, mosfet ccs or bipolar ccs, I chosed bipolar because it sounds better.. Its wery easy, just put enough heatsinking on transistors.
As I see it there is no other place than before the relay to put the pot (as mentioned by Peter). Since each pot is to attenuate any one input that input source will always be loaded by the pot, when in use, irrespective of the placing. BUT, if we put the pots after the relays then all pots will be in effect parallelled with their respective setting meaning that we will decrease the input impedance for ALL inputs this way (e.g. 6 inputs using 10 K pots set half way will have 6 times 5 K in parallell which is like 833 Ohms instead of the 5 K with one pot connected). Furthermore the change of one pot will affect all inputs as the parallelled value changes (and setting an input really low will almost short out the other inputs) kind of massive fake-law arrangement. If the pot is before the relay then only the used input pot is used and thus only one pot loads the input -> maximum input impedance and independent settings.

At least that is the way I understand the description earlier in the thread.

Is there really a problem with bosoz ubal in to bal out ? This the first time I hear about it. What are all those people out there with bosozs doing with their single ended sources ?
I'm confused now !

Secondly , can somebody please explain how to implement all those 5 pots in practice. I see on the pass site most people use only output vol pots which are 2x2ganged or 4ganged. So what do you do with the other ones do you use nominal r values or small pots or what?
Thanks Peter,
But I still can't understand what this loss means in practical terms. Is it a matter of loss of gain only or is somebody suggesting that there is a sonic degradation. I guess in one sense there will be less CMRR (?). Does this make a case for using the Aleph P over the bosoz although in another thread Nelson said that he couldn't detect any sonic difference between the two.
What about my question regarding use of pots ? Any comments ?
When converting unbal => bal @ 20db gain (R15=120Ohm) You looses 10% of the swing at the undriven side.

When converting unbal => bal @ 10db gain (R15=400Ohm) You looses 20% of the swing at the undriven side.

Wehter you uses constant current sources or not, you will loose half the gain when converting unbal=>bal. Acctually, you won´t loose anything, simce it has never bin there, but compared with bal.=>bal. You only get the half of that.

The losses at 10% or 20% will only degrade the CMRR, and proberly lowering CMRR with 10% or 20%.
But with ccs the output voltages of bosoz will be equal.. The only problm that i encountered with adding ccs to bosoz was that with then I can hear a little hum, which wasnt present before/it was dead silent:) I dont know why..
Is length of the wire to ccs transistors important from aspect of increasing hum (mine is around 4 cm, because I had to put transistors to proper heatsink)?
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