What would happen if you flipped this transistor...?

[earthtonesaudio]

So take your Darlington pair:

...and rotate the second transistor 90 degrees clockwise, so that the emitters are coupled, and the collector of the first feeds the base of the second, and take the output from the second tranny's colector.
Assuming it's biased correctly, it looks like this would provide non-inverting gain with negative feedback through the junction of the two emitters.
Yet, I've never seen this anywhere before.  Any ideas why?

[Gus]

post a drawing

then there is this
http://en.wikipedia.org/wiki/Sziklai_pair

[earthtonesaudio]

I'll see about making a drawing... but to put it another way,

Take the Sziklai pair you mentioned:


...flip the second transistor upside down, and make it NPN.  Then take the output from the collector of the second.

[earthtonesaudio]

drawing:

[GibsonGM]

Something out of phase would happen, I bet!  ;o)  Try it, I want to see what that will do! 

[earthtonesaudio]

Hm... I think there might need to be a resistor between the emitters, and/or between C of Q1 and B of Q2.

[slacker]

I just simmed it in a rangemaster type circuit and you don't seem get anything 
I guess this is because the Q1 collector Q2 base junction has no voltage on it. If you add a resistor from there to the supply it looks a bit like this.

http://www.eskimo.plus.com/fxstuff/squarepie.jpg which is essentially a discrete schmitt trigger.

[earthtonesaudio]

Cool.  Seems obvious now...
http://www.national.com/an/AN/AN-32.pdf
There's a Schmitt trigger in there that looks pretty similar.

Ian, how does that "freq" control work in the SquarePie?

[edit]Ah yes, the 'search function is my friend' strikes again!
http://www.diystompboxes.com/smfforum/index.php?topic=58718.msg458230#msg458230

[frank_p]

All the voltage availible  to Q1  would be Vbe Q2 (constant).  The potential difference between c of Q1 and E common is stuck at Vbe fo Q2.

So the current going through Q2 would always be constant also since the base of Q2 is biased at a constant voltage.

It's a guess.

[calpolyengineer]

From what I can tell, the first transistor would turn on, but it would be trying to pull current out of the base of the second transistor which would would turn that one off. Since the second transistor is the one connected to any kind of power, the only thing going on in that circuit is a very small current in the B-E loop of the first transistor.

-Joe

[earthtonesaudio]

I didn't put it in the drawing, but there should obviously be a resistor from the first transistor's collector to the positive supply.  That's sorta what I meant by "assuming it's biased correctly."  But shame on me for forgetting what happens when you assume... 

[frank_p]

I didn't put it in the drawing, but there should obviously be a resistor from the first transistor's collector to the positive supply.  That's sorta what I meant by "assuming it's biased correctly."  But shame on me for forgetting what happens when you assume... 

A connection from the collector of Q1 that goes also to the base of Q2 ?

[earthtonesaudio]

Biasing details: