Another member mailed me with a question about using the current mirror to set the DC offset in this amp.
Obviously people are getting confused - usuall, simple explanations don't use the mirror's degeneration or take it into account but just the load and that seems to be easy to mix up.
The point is that Vbe of both BJTs is (ideally) equal but the current here is not set only by Vbe because the voltage drop on the degeneration resistor also has to be added in the equation for calculation of current. We can manipulate that voltage drop (by changing the values of emiter resistors) on both sides of the mirror.
The current in first branch will be determined by JFETs (i.e. their Vgs) but change of the value of voltage drop in emitter circuit of that branch of mirror will be reflected in the other branch beacuse the Ohm's law won't have it any other way.
If it still isn't clear, you'll just have to make your own tests or trust me that it works...
Obviously people are getting confused - usuall, simple explanations don't use the mirror's degeneration or take it into account but just the load and that seems to be easy to mix up.
The point is that Vbe of both BJTs is (ideally) equal but the current here is not set only by Vbe because the voltage drop on the degeneration resistor also has to be added in the equation for calculation of current. We can manipulate that voltage drop (by changing the values of emiter resistors) on both sides of the mirror.
The current in first branch will be determined by JFETs (i.e. their Vgs) but change of the value of voltage drop in emitter circuit of that branch of mirror will be reflected in the other branch beacuse the Ohm's law won't have it any other way.
If it still isn't clear, you'll just have to make your own tests or trust me that it works...
