> But i am curious as to what you see is the
> problem with it? That it won't oscillate or that
> Q1 (top) won't turn off?

I think it will start up as you intended.  The cap on the base of Q2 will
charge up and Q2 will turn on, which will turn off Q1, which forces the
inductor to suck current from somewhere else.  So far so good.  Obviously
you meant the inductor current to then come from the diode and contribute a
little negative charge to the -15V supply.  However, when the inductor
voltage gets negative, so does the emitter of Q1.  Even with Q2 off, the
base of Q1 will be driven when it gets below the Q2 collector on voltage,
around 200mV.  Q1 will therefore act like a -500mV voltage source on
attempts to pull its emitter lower, which will prevent the -15V from ever
being generated.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, olin@embedinc.com, http://www.embedinc.com

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