# Transistor feedback loop

In an attempt to create a single transistor flipflop I made this :

The Idea : When the base is triggered by a pulse ,current starts flowing from collector to emitter . From the emitter it divides and goes to the base and ground keeping the circuit open .

The Reality : Current stop flowing when trigger is removed .

Adding a resistor to the base made no difference .Adding a capacitor from the base to the emitter got me the expected results but when the capacitor was fully charged the current stopped flowing from collector to emitter .

I have two question ,
1) Why will this not work ?
2) What did the capacitor do to make it work ?

• The wire from base to emitter is rendering Vbe=0 always. A capacitor is making it equal to V_capacitor. Commented Aug 26, 2015 at 16:51
• You effectively made a diode (base to collector) and a poor one by shorting the base to emitter diode. Are you familiar with the equivalent circuit of a BJT? Commented Aug 26, 2015 at 16:53
• Plus there isn't any feedback going on here.... Commented Aug 26, 2015 at 16:55
• Even ignoring the Vbe drop, the emitter does not go up by as much as the base when the transistor turns on (and the collector goes down or stays the same, at best) so there is no (practical) way to make a flip-flop with a single transistor. You need positive feedback. Commented Aug 26, 2015 at 17:09

1) To current flow , $V_{BE}$ have to be around .6~.7V based on the transistor type. When you shorten them out, it become zero and so no current flow.
2) When you add a resistor , as there is no voltage difference between base and emitter , it has no effect.But if you add a capacitor , capacitor create a voltage difference $V_C$ . But after some times , when the capacitor is fully charged , it just act as an open circuit as it's dc current . So base became float and circuit stops working.