Long awaited explanation that is really sticking down to earth rather than bumping here and there. Thank you Mr Zahi for making my life in understanding simple op amp subject simple ... rather than the simple op amp subject made complex!
After I read the comments, I realized he used improper word. When I heard, I couldn't think he was really using that kinds of word. I gave him a thumb up for his explanation.
i have a question: i am using opamp in open loop configaration vcc=12v, vee=0v, v- =2v, v+=(0 to 5v) ac square wave output is=(3 to 12volts)ac sqr???? i am expecting output=(0 to 12 volts) ac sqr. what happened here please explain.i want 0 to 12 instead of 3 to 12
@@pspmaster2071 because he explains his stuff very well, I have watched almost every video on circuit 1 playlist, and he is 100% better than my prof at University hahahha
The non inverting input has the more positive voltage 10v compared to the inverting input 9v. Gain in calculated using super position of both inputs. Gain will be noon inverted
Form the diagram containing a voltage dependent current source,i can tell that the op amp is an inverting amplifier but your answer is positive.clarification please.
The circuit connected to the op-amp is the small signal model of a common emitter amplifier, which inverts the input. The op-amp again inverts this output, so the final output has the same sign.
this was very helpful, except that i think answering the phone during the video & having another guy guess the answers was time consuming. greatly helpful though
the potentials are not zero. it starts from Vout potential and ends in the ground, so clearly there is a potential difference for the current to flow. As of why he writes 0 and 0, it probably would be 0 and some negative small number @AAMIR
Open circuit voltage is a potential difference thing. You only have one end of it the other end is no where there, if there would have been ground attached to it, Iz wouldn’t have to be zero necessarily
Why the current through the 1k resistor is 0? If you convert the current source into the voltage source then the current has to go through the 1k resistor, there is no any other way left.
Current will always choose the path of least resistance. Why go to the 1k resistor when you can go to a path with no resistance. It is also connected to ground
