19:21
Tamino
ground is connected to only two places: one of the three leads of both transistors (i'm guessing the emitter). power is connected to both LEDs. so i guess the power as seen by the rest of the circuit is whatever the voltage is minus the forward voltage drop of the LEDs. then it goes through one of the resistors and into both (1) the capacitor and (2) the second lead of "your" transistor (i'm guessing this is the collector). meanwhile the other end of the capacitor is connected to two places: (1) the power, through another resistor, and (2) the third lead of the *other* guy's transistor (I'm guessing this is the base).
so it seems to me that the "natural" state of the capacitor as long as your transistor isn't passing current, would be for both of its sides to be high, cuz whatever difference in voltage there is between the two sides would have a tendency to be equalized through R3/R4. and the whole thing is being tugged high through R1/R2.
and i guess if both sides of the capacitor are high, then no current flows through the LED and that side is off. and, the signal you're giving to the base of the other guy's transistor is also high.
so i guess the interesting stuff happens because "your" transistor is also getting *its* base pulled high by the other guy, so that it turns on... now suddenly current flows and your LED turns on. BUT, now suddenly current is also flowing through your capacitor, too. at least until the capacitor fills up, you're now pulling the base of the other guy's transistor low. so the other guy isn't turned on. but you are.
but when your capacitor fills up, then even though it wants to keep pulling the other guy's base low, it just can't anymore. so the other guy's base goes high again, and he turns on, and you turn off. and then while he's turned on and his capacitor is filling, your capacitor is draining again because both sides are connected through some quantity of resistors, to power.
