Aleph J balanced input transformer

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ZM. Any input/interstage transformer. What I am trying to understand is how negative feedback works with transformer balanced inputs.

I understand on unbalanced circuit the two resistors at the negative input, 22k and 220k, form a voltage divider that sets the amount of negative feedback and determines the gain. But with the transformer I lift the 22k resister from ground and attach it to one end of a coil. I take the 22k resistor from the positive input and attach it to the other end of the coil to form a isolated balanced input.

This is where I get confused. How is the gain set and how does the feedback interact with the out of phase signal coming from the transformer?
 
simple - when running balanced - signal side(source) is in same time virtual gnd for each input ..... if you're going to look "from input side"

so - input doesn't know and doesn't care are you driving it with battery charger or photoflash circ or signal xformer , as long signal level is polite

😉
 
Ok so you are saying the coil between R1 and R3 is virtual ground?

How does negative feedback work in this situation? On the attached schematic there is an 'in phase' signal going to the + input and an 'out of phase' going to the - input. The negative feedback is an 'in phase' signal being fed into the - input. When the 'out of phase' signal coming from the transformer reaches the gate of the jfet the 'in phase' feedback merges and cancels part of the 'out of phase'. The 'out of phase' signal at the - input is now smaller in amplitude then the 'in phase' signal at the + input.

Does this upset the balance of current through the two jfets?
 

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  • ALEPH-J_input_trans.png
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replace that xformer on input with floating voltage source symbol , and everything will be clear ;

or - if still isn't clear -draw center tap on secondary and ground it ; even if that is not necessary for proper work of A J ;
 
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