can anybody help me to understand how this circuit works?

Hello, Can anybody please halp me to understand exactly how this circuit works?
This circuit is from third edition of IC OpAmp Cookbook from Walter G.Jung page382. I understand this much that when OpAmp goes to positive rail(+15V) 1/100 of voltage appears on base of Q1 which will be about 1.05V and as a result about 0.35V on emitter resistance of Q1 that gives about 1mA current on the other hand base of Q2 has a voltage about 1.4V and emitter resistance will see 0.7V that gives about 2.1mA.
If this circuit is going to work then collectors of Q1 & Q2 should have about -100V when OpAmp gives +15V and +100V when it gives -15V because Q5 & Q6 are emitter followers for Q1 &Q2 am I right ?, but how and why? What component and how forces this voltage on collectors of Q1 & Q2 what is the role of Q3 & Q4? What is the role of R5 & R6? I will greatly appreciate any help. Thanks.

Re: can anybody help me to understand how this circuit works

R1, R2, R3 and R4 in voltage divider configuration are used to control base voltages of Q1 and Q2 ..
For example, if the output of the opam is -15 (or close to) this should cause the Ube of Q1 to be ≈0.7V, so it should be fully open Q1, and the output voltage should be close to +125V, and the opposite way around, +15V should close Q1 and open Q2 ..
However, you have 1MΩ R5 (same with R6, but at the other end) which role is, as negative feedback, to lower the output voltage from +/-125 to possibly +/-100V ..

Q3+R11 and Q4+R12 limit the output current to ≈ +/-27mA ..

Other "funny" thing about this circuit is that you need an input voltage below -10V to change the output form -100 to +100, as there is quite strong positive feedback between the output and the opamp (+) input (R13) ..
In other words, for any input voltage above -10V the output will be always -100V (or so)

Regards,
IanP

thanks alot IanP. I remembered that you have helped me before too. thanks!