Dual Biased Emitter Follower

I don't see how that can work. The floating battery, and D1, mean that the upper-circuit power rails (namely the B1 terminals [1] ) will flail all over the place as a function of the input and output currents. Looks like you'll zener the base of Q1 on negative input swings, adding to the fun.

What are you trying to do?

If I understand correctly, John, if the diodes were removed the circuit would work.

The question is, why are they there?

As I attempted to indicate on the diagram, the two outputs are meant to be floating (dual single wire system), one above and one below the grounded battery common by 0.7V.

IOW one swings 0.7V positive to 9V positive. The other is 9V negative to 0.7V negative.

Putting aside convention, for the moment, the unique point is NOT to have them referenced to ground. Perhaps we are going about it the wrong way, but we seem to have tried everything else.

I would like to go ahead and build the circuit if anyone can tell me how to give it half a chance.

How about replacing the diodes with +/- 5V voltage regulators? We could use +/- 12VDC rails to compensate for the voltage loss.

David

It looks to me like you are trying to make a class B or class AB output driver. Why not look at those circuits first before rolling your own?

With diodes removed, I assume one output would swing between +9 and 0 Volts, and the other -9 and 0 volts. Can we say that much will work?

The intended circuit is a learning exercise, and, as yet, has no practical application.

The only object is to output two identical signals with no direct ground reference to each other or the supply; one being offset positively from ground and the other negatively.

Not being a conventional approach, it appears difficult to get one's head around. But I assume it is possible ... somehow.

David

"David White"

You show the inputs with a common ground too - contradicting your own words.

... Phil