![]() ![]() ![]() When Q1 is on, it’s collector emitter junction has a voltage of about -0.2V and allows current to flow through LED1. When the unconnected end of the 100k resistor is connected to negative terminal of 7 (or 9V) battery, current will flow through the 100k to establish -0.7 across base emitter junction and this turn on Q1. Here though one transistor is connected directly to collector of first transistor so it will actually invert the operation of the first transistor. On being like a closed switch and off being open circuit between the collector and emitter terminals. This circuit is just two PNP transistors that are set to operate as either on or off. May I suggest Falstad, an online Javascript-based simulator that visualizes voltage and current. Want to understand it more fully? Try a sim. Yes, even with just a few tens of micro amps. One issue: the first LED won’t turn off totally when Q1 is off, as it will be sinking base current to the second PNP, so it will glow faintly. This action will make the LEDs light alternately. Q1 needs a base drive of at least -0.7V (the intention is probably that R5 be connected to battery (-) to turn Q1 on) and Q2 will invert that. It’s a pair of PNP common-emitter amplifiers set up to drive LEDs. Other than that, there’s nothing mysterious about this circuit. My opinion, that’s a crappy drawing and a mean trick, but perhaps that’s the point of the exercise. That’s the only clue to the power supply polarity. With a battery symbol, the ‘short’ side is ‘cathode’ (-) and ‘long’ side ‘anode’ (+). If you flip the drawing over it will make more sense. The (+) side is tied to ground, and the (-) side is supplying the LEDs and the rest of the circuit. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |