How did you think it should work?I built this circuit on my own thinking that it would work as blinking circuit.
You might see the LED flash on briefly when you connect the 9V. As soon as the capacitor charges the transistor will turn on making a short across the LED which will then be permanently off.But the LED is not glowing at all ! and obviously its not working as LED blinking circuit. I want complete explanation of this circuit.
Thanks Mr. David. Could you please tell me , Why this capacitor does not discharge again via 47K Ohm Resistor?How did you think it should work?
You might see the LED flash on briefly when you connect the 9V. As soon as the capacitor charges the transistor will turn on making a short across the LED which will then be permanently off.
Thanks! Could you please Recommend me basic led flasher/blinker circuits.. Links or Schematics will do.Welcome to the forum. Your capacitor is receiving a steady positive through the 47K. The voltage will rise and cause the transistor to conduct. There is nothing to discharge the capacitor. If you put your LED in series with the collector, the LED will turn on and stay on.
Also Kindly suggest Best Circuits Simulators.Welcome to the forum. Your capacitor is receiving a steady positive through the 47K. The voltage will rise and cause the transistor to conduct. There is nothing to discharge the capacitor. If you put your LED in series with the collector, the LED will turn on and stay on.
Great ! Explanation. Thanks .Hi adit,
The circuit is stable at some point, note the LTS sim
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Update;
This sim shows the effect more clearly , from 0V power up.
View attachment 247359
Simply Google it and select the images tab. That will give you quite a few schematics to choose from and links to their source for further information about them.Thanks! Could you please Recommend me basic led flasher/blinker circuits.. Links or Schematics will do.
There's also a trigger device that's meant for the job called 2N4991 or BS08D with a guaranteed breakover voltage.Yes, the transistor with the breaking down emitter-base junction is acting like half a diac.
View attachment 247394
Two transistors can simulate a whole diac, but usually at a much lower voltage.