There's still a contradiction in there. The first sentence says "With Qbar high, the discharge transistor turns on..." A couple lines below it says "With Qbar low the transistor turns on..."
My understanding is that whenever Qbar is high the transistor is on and the cap discharges. Whenever Qbar is low the transistor is off and the cap charges.
I'm glad I posted that before creating new pages!
Of course you are right, again!
Very easy to slip up even when you know what you want to say. I think the edit below may be OK:
"With
Q high, the discharge transistor turns on and shorts pin 7 to ground, discharging
the capacitor through
R2. When the capacitor's voltage drops below 1/3
VCC, comparator 2's
output jumps to a high level, setting the flip-flop and making
Q low and the output high.
With
Q low, the transistor turns off, allowing the capacitor to start charging toward
VCC through
R1 and
R2. When the capacitor voltage exceeds 1/3
VCC, comparator 2 goes low,
which has no effect on the
SR flip-flop. However, when the capacitor voltage exceeds
2/3
VCC, comparator 1 goes high, resetting the flip-flop and forcing
Q high and the output
low. At this point, the discharge transistor turns on again and shorts pin 7 to ground,
discharging the capacitor through R2. The cycle repeats over and over again. The net
result is a squarewave output pattern whose voltage level is approximately
VCC - 1.5V and
whose on/off periods are determined by the
C,
R1, and
R2."