Understanding thyristors (?) (was:AW: My new VCO)

[Haible Juergen]

	>  is this chinese or what?   :)

Japanese, actually. Designed by the clever people of Korg. (;->)

But seriously, I'm not sure if I have understood the circuit myself.

I'm not that familiar with thyristors at all.
What I *think* to know is this:
If you connect a npn and pnp to a typical scr circuit (base to collector,
and vice versa), it would normally not conduct, even if there is a voltage
applied between the two emitters (what you'd call anode and cathode
on a scr). Only if you have a trigger pulse of the right polarity on either
base (what you'd call gate then), one transistor would conduct, and 
make the other one conduct as well, even if the gate pulse is not there
anymore. The circuit would lock up in a conductive state. Only when the
current thru the scr would get close to zero because of external reasons
(e. g. the external voltage breaking down because a capacitor is
discharged).
This would end the locked state, and even when the external voltage is
growing again, another gate pulse is needed to start the action again.

Now there would be another way to trigger the scr - I think it's called
"über Kopf zünden" in German (is that "overhead trigger" (?) ): When
an external voltage would exceed the BC reverse breakdown voltage of one
of the transistors, the current would start to flow even without a gate
pulse.
(Please note that this is more a *question* than an explanation - I really
don't know for sure.)

Anyway, if you look at the VCO circuit, neither of these mechanisms would
happen, would it ? No external gate pulses, and no excess voltage.
So I suspect that the leakage currents of the transistors would be *used*
for the desired operation. If you remove Q3, wouldn't the tinyest current
from the collector of Q2 be amplified in Q1, and again amplified in Q2,
and the circuit would lock up at once ? 
This would be prevented by Q3. I'd say there is a *current* divider built
from Q1 and Q3, the current distribution being determined by the voltages
on the emitters (E) of Q1 and Q3. As long as E(Q1) is more positive
than E(Q3), the vast majority of current would flow thru Q3 and prevent
the positive feedback described above. Only when E(Q1) - the capacitor
voltage ! - comes down near +5V, the current amplification circle would
begin.
Now the cap becomes discharged, E(Q1) rising again. But as we now have
*high* currents to distribute between Q1 and Q3, the resistor R6 would
prevent
the most of the current to flow thru Q3, and the Q1/Q2 circle is not broken.
You'll get a voltage pulse across R6, however, which is used to trigger
the CMOS divider.

Does this make sense ? And is it different from "normal" thyristor action ?
(What prevents the leakage currents in a normal tyristor from forming
a positive feedback without external gate voltage, btw. ?)
And what is the purpose of the diode ?

I don't know if this shed some light. If not, I hope it will "trigger"
someone
else to put it right or explain it further. (I admit when I did my first
experiments
with these circuits, I was looking for a reverse breakdown to trigger it
(;->) )

JH.