# Dual Biased Emitter Follower

Discussion in 'Electronic Design' started by David White, Feb 16, 2012.

1. ### David WhiteGuest

I posted this to sci.eelctronics.basic, where it probably belongs, but
things seem a bit quiet over there at the moment.

Based on previous advice from the groups, I have attempted
to revise our circuit utilizing a complimentary pair, and diodes to
establish a 1.4V DC offset between the two output signals.

Here is the schematic, if anyone would please be able to comment on
mistakes.

http://www.4shared.com/photo/-Uflsi2Z/Dual-Emitter-Follower.html

If this sounds familliar, there are several of us cooperating on the
same project, none obviously any more gifted than the other.

Thank you,

David White

2. ### mikeGuest

+1 on the "what are you trying to accomplish?

What are the characteristics of the two different inputs?
And what does it mean to have an offset between two different "audio
sine wave" inputs that are half-wave rectified?
And how does the base-emitter reverse breakdown voltage compare the the
input signals?
And what do the batteries do?
If you draw a box around the top half and apply norton's theorm,
the output current has to be less than the input current by the amount
of diode current.

3. ### David WhiteGuest

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

4. ### misoGuest

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?

5. ### David WhiteGuest

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

6. ### Phil AllisonGuest

"David White"

** You have no knowledge of electronics at all - do you ?

** Only way to do that involves isolation transformers.

** That implies they ARE referenced to each other.

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

** No, it is a fuckwit approach.

.... Phil