I'm reading this page about a BJT astable multivibrator:

http://www.learnabout-electronics.org/Oscillators/osc41.php

I'm confused by the following two passages, can anyone help explain?

"it is the nature of a capacitor that when the voltage on one plate changes rapidly, the other plate also undergoes a similar rapid change, therefore as the right hand plate of C2 falls rapidly from supply voltage to almost zero, the left hand plate must fall in voltage by a similar amount."
*my confusion: Why does the cap discharge, and not keep charging? It still has a 0v potential at one side, and Vcc through a resistor on the other.

and

"With TR1 conducting, its base would have been about 0.6V, so as TR2 conducts TR1 base falls to 0.6 −9V = −8.4V, a negative voltage almost equal and opposite to that of the +9V supply voltage."
*My confusion: why does it fall -9v?

Thank you for any help

 

The circuit is two one-shots, with the output of each triggering the other and the result is oscillation.

Your question: Why does the capacitor discharge and not keep charging?
At the beginning of a half cycle TR1 is on and C1 has voltage on it from the previous half-cycle. Ideally the voltage across C1 would be the power supply voltage less 0.7 volts base-emitter voltage of TR2. The voltage across C1 is a negative voltage and it holds TR2 off until current flowing from R2 charges C1 to the point that TR2 turns on, which in turn dumps the voltage across C2 into the base of TR1 turning it off, which begins another half cycle.

(Some text removed for clarity)
"With TR1 conducting, its base would have been about 0.6V, so as TR2 conducts TR1 base falls to 0.6 −9V = −8.4V, a negative voltage almost equal and opposite to that of the +9V supply voltage."
*My confusion: why does it fall -9v?

When TR2 begins to conduct C2 is charged up to about 8.4 volts as you said. At this point you have a capacitor chared to +8.4 volts on the collector of Q2 which is nearly at +9V, and then when you take the positive plate of the capacitor to ground as TR2 turns on the negative plate goes to -8.4V.

Note: With a 2N3904 and most other small signal transistors the power supply should be limited to 5 volts in order to eliminate timing errors resulting from breaking down the base-emitter junctions with excess reverse bias.

 

See

 

Thank you... very helpful!
So basically, When TR1 turns on and thus Out 1 = 0v, the voltage across the capacitor is now referenced opposite it was, and so it has ~ -8.4v across it.

 

You understand