confused about opamp circuit from practical electronics for inventors

hello, OK, I've been breaking my head over this one for a while and just can't figure it out. It's an example from a book called "Practical Electronics For Inventors" It's basically a "Simple Triangle-wave/Square-wave Generator" It has a integrator opamp circuit feeding the non-inverting input on a positive feedback.... (schmitt trigger or comparator not about the terminology here) opamp circuit, and then that feeds the inverting input back on the integrator...

The book then shows this equation for the threshold voltage (Vt): Vt = Vsat / (R3 - R2) , but i can't for the life of me see how they got this..... the circuit is as follows, anyone have any idea of how Vt is found? The rails are at +/- 15V for each opamp

Thanks Joshua

(view in courier font)

--------------------------------------- | | | | | C R3 | | |---| |--| _/\/\_ | | | | | | | | R1 | |\ | R2 | |\ | | ----/\/\---|-\____|__/\/\_|_|+\_ |____| ___|+/ ___|-/ | |/ | |/ | | --- --- /// ///

You might like to think about this in a different way:

At some moment during the changeover the output voltage must be zero - and that can only be true when there is no differential input voltage. Thus the non-inverting input must be at 0v at that time.

The input is a summing point for two currents - one from the output of the second op amp, determined by the output voltage divided by R3. The other the output voltage from the first op amp divided by R2.

Most of the time, the output of the second op amp is at Vsat (positive, or negative). However, the output of the first op amp is changing with time.

When those two currents are exactly equal and opposite, the voltage at the non-inverting input will be at 0v and the op amp will be in its linear region. The output from the second op amp will go to 0v, as there is no differential input voltage. However, this will cause the input current at the summing point, produced by the output voltage, to change, due to the change in output voltage. The result of this positive feedback is to drive the output of the op amp to its (opposite sign) Vsat.

Thus the only factors that determine the changeover point are the voltage from the first op amp, divided by its resistor and the voltage from the second op amp, divided by its resistor. At the instant that they are equal (and opposite) change-over takes place.