bosoz output level

Sounds like it must be a mistake in the construction. What is the amplitude of the wave applied at the input? It must be awefully small to get a 1Vpp wave at the output, since the BoSoZ without attenuation has 20dB voltage gain.

I would start by checking all of the connections. It is a pretty simple circuit, so first I would make sure that the gate, drain, and source on each FET is wired up properly. Then I would measure all of the voltages with both inputs grounded. Should be 4V below ground at each source pin when idle, if you used +/-60V power rails. Drains should both be about 30V above ground. The idle voltage across 124Ω R15 should be around zero.

If all of that checks out fine, I would remove the output capacitors and measure the peak-to-peak amplitude at each output.

If you still haven't found the problem I would investigate the FETs. Perhaps one of them was subject to ESD.

Nelson Pass

The one and only
Paid Member
2001-03-29 12:38 am
When converting unbalanced to balanced, there is some
loss on the "undriven" side, which can by corrected by
having greater resistance biasing the diff pair, and gets
practically perfect when you bias the diff pair with a
constant current source. Your other alternative is to
reduce the Source - Source resistance of the diff pair,
which increases the gain also.
Got your post Nelson, so I have to use constant current source.. Can you or someone else post schematic of modofied bosoz with current sources then. or just schematic of current source..
I use 1V input which falls if i connect it directly to the amp and i like balanced out too, connected to boz it falls to 0.6v, so there is almost no need for gain.
But if i set the value of r15 to 124ohms i get 3.3V for +phase and 2.6V for - phase, I have a feeling that it should be much more (in the Nelsons article I read that the o/p voltage swing can be 120v at +-60v rails, is that meant for 1v input?)..
And what will be the o/p voltage at, say 20db, of gain for 1V input voltage, how it is calculated?
Danger! I mean only to measure the voltage with the output cap removed, to check for a defect there. DO NOT use it without the output caps to drive an amp. Everything will be kaput.

Anyway, it sounds like yours is working in light of the advice from Nelson. My incarnation of BoSoZ uses a contant current source, so I never noticed this assymetry. Apparently it is normal, which makes some sense now that it has been pointed out.
CCS or not?


As Mr. Pass said: "When converting unbalanced to balanced, there is some loss on the "undriven" side".
You have a loss at the output at around 30%, it seems too much to me.

I am using my BOSOZ as an unbal. to bal. converter, I do have some loss at "the undriven side" but not more than around 10%, and until now i have not considred this lack of mismatch to be a problem, but i could be wrong. I would like to other oppinions on this question!

The question is, do we get better performance by adding a Constan Current Source, or will the small mismatch from the passive circuit be les significant than an eventually degrade in sonic performance caused by the change from passive to active currentsource.
Grey has "bin there" in a long thread, but that was in relation to SOZ, and since BOSOZ an SOZ do have the same topology, one could get the idear, that his experince could be usefull here.

I my self like the passive topology very much.

You should read Nelsons artickel "DIY OP Amps", you can get from .
I have simulated a simple Constant Current Source.
The upper left diagram shows the original BOSOZ, and the one under BOSOZ with CCS, simulated from the circuit at the right.


  • balanced zen linestage ccs.jpg
    balanced zen linestage ccs.jpg
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Regarding your first post.
I have done some sims on this.
If R15=124E the loss at undriven vill be around 8%.
If R15=375E the loss at undriven vill be around 20%.
The loss at the undriven increases with lower value for R15.
So if you have chosen the lowest gain and distortion with R15 >= 400E tis could be an explanation or a part of it.