Best explanation, it should be on top search list on this topic. I've wasted multiple hours trying to understand from different videos, but none of the video explained everything like this video. Thank you for the video. Love from Nepal 🇳🇵
No brother. This theory wrong. Once capacitor charge 0.7 volt. And can not charge again 0.7. Because discharge not occur. How possible again capacitor charge ?
The way you have connected RL1 and R2 makes sense. I have seen other explanations that say the capacitor is charged through RL1 which doesn't make sense because the transistor won't turn on if the negative plate of the capacitor is facing Q2. Good job.
C2 charges through R2 because only that path can connect C2 with Vcc when Q2 is off. If C2 charges through RL1, there will be no direct connection between C2 and Vcc, so that cannot be the charging path
Awesome video! The only thing I am a little confused about is why int he very beginning, how come since one transistor is on the other is off? If both are trying to drive on at first, what makes one magically switch off? After this point though, I understand why the two switch states.
It does make sense so far I think, however when C2 charges, how comes C1 is not charging simultaneously also? Is it because the voltage is Vcc at the negative terminal of the capacitor?
At 7.28 min u said ,when switch us closed,only one transistor will be on ,why both the transistors are ON at a time when the vcc is same for both the transistors?sir..plz clarify my doubt sir
Nicely explained. But i have a doubt sir when we give Vcc in starting what is the path of current/ voltage so that one of the transistor will get 0.7 V and will get on.
+Sneha Murya, may be you know similat transistors do not have identical B-E and B-C junction voltage. the transistor with lower base-emitter junction voltage should turn on first.
