Please explain what is going on with this transistor, I'm stuck.

[Skruffyhound]

This transistor is preventing pin 5 from becoming more positive than ground when under a load. How?
How is this transistor set up.
I'm guessing I have about 60mA of load at maximum (2 dual opamps, 2 triode stages, 2 LED's)
Vin is +11.75V, Vout is -11.60V . How can I work out Ic and Ib to get Beta so I can calculate R1.
The datasheet says

RX should be chosen to provide enough base drive to the external transistor so that it is saturated under nominal output voltage and maximum output current conditions

I have a load of questions : is the transistor acting like a diode or is base turning current off if the emitter approaches ground. Which way are electrons flowing here in reality/convention. etc.,etc.

I've been sitting staring at this for two days now, but there are just too many things I don't understand yet.
Help put me out of my misery and tell me approximately how I should go about this.

[PRR]

> preventing pin 5 from becoming more positive than ground when under a load. How?

Transistor will not conduct until its E (and B) are seriously negative.

Transistor E to C path will only flow as much current as hFE times the E to B current flowing through R1.

If B is at zero, R1 current is zero, transisor E-C path current is zero.

If E (and B) are pulled to say -5V, now R1 has about 4.4V across it. If it is say 5K, then 4.4V/5K= 0.88mA will flow in R1 and through the E-B path. Transistor Beta multiplies that. Since a cheap transistor has Beta over 100, we can assume over 100*0.88= 88mA are available. (If the load pulls less, it will only flow the load current.)

If we are very chicken, we might assume hFE could be as low as 5. Then we need 60mA/5= 12mA in R1, at 4.4V is is 366 ohms. However the *total* load on the poor LT1054 is now 60mA+12ma= 72mA, of which 13% is silly waste in R1. That's not awful, but we do want to find higher transistor hFE for larger Rq and less waste. Any of the jellybeans sold around pedal builders will have hFE at 60mA over 100.

[Skruffyhound]

Thank you Paul,
                      that clears a few things up.
In the other thread I just got a value of 23K for R1, so I might need to just evaluate the evidence again 

(Vout(11.60)xBeta(100))÷Iout(0.05)=R=23.200Ω

(11.60V is just the negative voltage I measured on the chip)

[PRR]

> I just got a value of 23K for R1

That "366" value is for 5V and a real-unlikely hFE>5.

If we assume hFE>100 (good for nearly any jellybean) the 366 obviously becomes 7320.

If the 5 is now twice that, then almost 15K.

I think I had 60mA in mind, now you say 50mA.

So 23K is not wrong.

[Skruffyhound]

Brilliant. Thanks Paul.
Learnt a lot doing this and got a little bit more confident.
It's one transistor, one resistor and a load, very simple stuff but a big step from only being able to paint by numbers, now I know why it is how it is.