Te best astable multivibrator explanation on the internet. Love your thoroughness and repeating parts multiple times. Perfect explanation. Wish you had online courses for more circuits!
By far the greatest video I've seen on RU-vid for this circuit - and I've watched quite a few! Doing it from first principles was the key. Many thanks!
Thanks for including FET transistors towards the end. Also thanks for explaining that reversed biased BJT's Vbe acts like a Zener diode. Finally, how did you manage to keep your colored white-board pens fresh and writable throughout a 1:02:14 presentation?
I am in awe of this man. His ability to imagine his way into the thinking of a trainee, in order to lead the trainee from lack of understanding to full mastery of the subject, is evident at every step in the presentation. I have never seen a better technical training presentation on any complex technical subject. Bravo, bravo, bravo! I have three small suggestions for refinement in keeping with the underlying philosophy of the presentation. 1) When referring to the voltage across the capacitors, it might be helpful to refer to the voltage as a "differential voltage," to help a trainee further appreciate the fact that the voltage across the plates of the capacitor can be measured independently of any other reference point (such as ground or the positive side of the power supply). 2) When redrawing the circuit for the first time in the "traditional astable multivibrator view," it might be helpful to show the two resistors on each side of the circuit tied directly to each other and then to the power supply (rather than each tied independently to the power supply), to help the trainee fully grasp that the capacitor can be discharged through the two resistors, regardless of the power supply voltage. and 3) To help drive home the concept that a fixed differential voltage across a capacitor can appear to be above or below the power supply voltage, help the trainee imagine how the voltage reading for each side of the capacitor, relative to ground, would change if the voltage of the power supply were to be temporarily changed from, say, 10 to 20 volts, EVEN IF THE DIFFERENTIAL VOLTAGE ACROSS THE CAPACITOR REMAINS THE SAME. I bother to elaborate these possible refinements as a way of acknowledging that the presentation already incorporates dozens and dozens of such helpful aids to a trainee's understanding. I was not kidding: This presentation is the best I've ever seen, and I'm in awe of this man.
brilliant thank you. That idea that a discharged cap is effectively a short and a charged cap is effectively an open circuit ... never thought of that before .. such a powerful idea thank you.
50:44 Current can flow through a capacitor as long as the voltage is changing with time. An important factor in understanding how that circuit works is knowing that a capacitor works like a straight piece of wire when you consider short time scales. This means that if you change the voltage on one side of a capacitor, the voltage on the other side will immediately change by exactly the same amount. So when transistor turns on it immediatelly grounds one side of a capacitor lowering potential of one side by 10V, the other side follows, which means that it's new potential becomes + 0.7V - 10V = -9.3V. When this happens the -9.3V side of the capacitor which is connected via a resistor to Vcc +10V begins to charge to that level and as soon as it reaches +0.7V is the other transistor switches back on. I hope this helps in understanding.
In my understanding, that's because the field of the charge accumulating on the capacitor working against the source field depends on the amount of this charge, and in short time spans, the charge just doesn't have enough time to build up and work against the battery. That's just seen in the charging-time diagram where the initial current and other things are big initially.
This always has been the hardest one to wrap my head around, I get it much better now. I never thought of a cap in such a way with a voltage offset on one side or the other, thats how voltage multipliers work. They threw SO much stuff at us for my little AAS degree but I'm glad. Our instructor had a masters and he didn't hold back, if you wanted the material on that level he was happy to give it, some of the students complained, not me. Got my moneys worth then some. Digital was the best, your imagination is the only boundary.
10:15 ... ? Surely at t=0, the current will divide equally to the two resistors. As t increases, I would assume more current flows through the "vertical" resistor? Surely this is quite different from the simple switch (open) previously and must affect the charging time etc.?
I see that the capacitor seems to be put in series with the battery with the switch being opened, and we measure the voltage from one end of the series to the other. Now the restor connections somehow make it not 10V + 10V but 10V + 5V, but Idk how... Could you comment wether I'm understanding this rise in voltage correctly? And how it goes to 15V rather than 20V in series?
