I'll give it a go

Posted by Bob Janova at 4:32pm Sep 28 '13
You must sign in to send Bob Janova a message

I wouldn't stand a chance of guessing the behaviour here if you hadn't described it, but since you have ...

Without the capacitors the LEDs are just directly transistor-controlled, initially the current goes through the resistors and opens the transistor and then after that the current can flow through directly. If you put a big enough resistor in one side at that point you'd probably make it not elevate the base voltage enough and you could get one side on and one off.

I think the magic is happening because the capacitance 'steals' some of the gate voltage while that side is active, so the other transistor doesn't get its gate voltage. I.e. when side 1 is 'on', the point below D3 is at a higher voltage than the point below R3, so the capacitor will charge and nothing will go through the base of T2. Once it's charged, it's like the no capacitor situation so current will flow through T2 and open the other side. At that point C2 starts charging and takes T1's base current away.

The resistance makes a difference because the PD across C1 is a little smaller initially than that across C2, so it ... hmm actually never mind that should make it finish first and open side 2 first.

So yeah I don't fully understand either, heh.