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In an attempt to create a single transistor flipflop I made this : enter image description here

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 ?

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  • \$\begingroup\$ The wire from base to emitter is rendering Vbe=0 always. A capacitor is making it equal to V_capacitor. \$\endgroup\$
    – Eugene Sh.
    Commented Aug 26, 2015 at 16:51
  • \$\begingroup\$ 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? \$\endgroup\$
    – jippie
    Commented Aug 26, 2015 at 16:53
  • \$\begingroup\$ Plus there isn't any feedback going on here.... \$\endgroup\$
    – MathieuL
    Commented Aug 26, 2015 at 16:55
  • \$\begingroup\$ 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. \$\endgroup\$ Commented Aug 26, 2015 at 17:09

3 Answers 3

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A flip-flop cannot be implemented by a single BJT or transistor,there is no feedback hence your output only depends on the current input,its not a flip-flop.

You have shorted the base and the emitter hence they are at same voltage,giving trigger will add same potential to both terminals,hence no current flows,when you add a capacitor between the base and emitter,it acts like a voltage source, hence current flows as long as the capacitor charges to the base supply or your trigger voltage,but when the capacitor voltage become equal to the trigger same problem arise the base Vbb and the emitter are at same potential Vee.hence Vbb- Vee = 0. (no current flows).

The base and emitter must have alteast have 0.6v difference for current to flow,also remember BJT is a current switching device.

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The reason this did not work is because the base needs to be at a higher potential then the emitter, by about 0.6v. the reason the capacitor worked is that you stored the pulse energy in the capacitor above the potential of the emitter and then the capacitor slowly discharged through the base. When the potential of the capacitor fell bellow the threshold of the base to emitter voltage the bjt turned off.

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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.

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