Dual Biased Emitter Follower

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.

formatting link

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

Reply to
David White
Loading thread data ...

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?

[1] A battery is a "B", not a "BT"
--

John Larkin, President
Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro   acquisition and simulation
Reply to
John Larkin

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

Reply to
David White

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?

Reply to
miso

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

Reply to
David White

"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 f****it approach.

... Phil

Reply to
Phil Allison

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.