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Hi sir! 😊 I have a question. Is there any explanation or principle behind the shorting of input source when solving for the output impedance, Z_o? I assume two reasons but I don't know which or if any of it is correct. I suspect it's either (1) because, in actual, we only measure resistances without supply; or (2) because these types amplifiers only operate small signals, hence, making the input signal negligible to 0-V and allowing for a convenient solving of output impedance. I apologize if these questions may seem basic. I'm just an undergraduate student and I just want to make sure that I really have a solid foundation when it comes to Electronics concepts. Hoping for your response, sir. God bless. 😊
Hi Raniel! Thx for asking! :) Impedance/resistance across a load, by definition, is dependent on the cummulative effect of all applied signals to the system you're analysing (think Superposition Theorem). Don't forget that the other assumption we make is that, at the operating frequency, all capacitors are at negligible impedance (i.e. 0 Ohm). So what's left of the system will be purely resistive. The effects of non-zero AC input sources will therefore not affect our output impedance analysis as both current and voltage have zero phase difference in a pure resistive circuit. In other words, for the circuits we're dealing with, independent sources have no effect on our output impedance analysis. We can therefore simplify our analysis by assuming that all independent sources are zero (i.e. for a voltage source, it's at 0V - short circuit, for a current source, it's at 0A - open circuit). Hope that helps. :)
@@cikgutaiff Given what you said sir, does this mean that, when it comes to DC circuits, a similar approach in determining output resistance can be carried out - that is, shorting the voltage sources? If so, then am I right in assuming that determining any circuit's effective resistance can be likened to simply determining its Thevenin's resistance? Thank you sir.
@@RanielDG Yes you are correct. We are indeed just determining Thevenin's resistance as these networks can indeed be modeled with either a Thevenin's or Norton's equivalent circuit.
Hi guys! Hope you've found this video useful in your learning journey. I'm planning to create more relevant content. Would be great if you could fill in this survey so that I know what direction I should take! Peace! ✌🏻 survey.qwary.com/f/AABlew?identifier=S_wSzSPnasH9Wc_FT15X0J1BuEcPl5gILyVTYFwsvWc=
EXCELLENT PRESENTATION!
Absolutely loved it... Soo detailed.. thank u so much💙
My pleasure 😊
Hi sir! 😊 I have a question. Is there any explanation or principle behind the shorting of input source when solving for the output impedance, Z_o?
I assume two reasons but I don't know which or if any of it is correct. I suspect it's either (1) because, in actual, we only measure resistances without supply; or (2) because these types amplifiers only operate small signals, hence, making the input signal negligible to 0-V and allowing for a convenient solving of output impedance.
I apologize if these questions may seem basic. I'm just an undergraduate student and I just want to make sure that I really have a solid foundation when it comes to Electronics concepts. Hoping for your response, sir. God bless. 😊
Hi Raniel! Thx for asking! :)
Impedance/resistance across a load, by definition, is dependent on the cummulative effect of all applied signals to the system you're analysing (think Superposition Theorem). Don't forget that the other assumption we make is that, at the operating frequency, all capacitors are at negligible impedance (i.e. 0 Ohm). So what's left of the system will be purely resistive. The effects of non-zero AC input sources will therefore not affect our output impedance analysis as both current and voltage have zero phase difference in a pure resistive circuit. In other words, for the circuits we're dealing with, independent sources have no effect on our output impedance analysis. We can therefore simplify our analysis by assuming that all independent sources are zero (i.e. for a voltage source, it's at 0V - short circuit, for a current source, it's at 0A - open circuit).
Hope that helps. :)
@@cikgutaiff Given what you said sir, does this mean that, when it comes to DC circuits, a similar approach in determining output resistance can be carried out - that is, shorting the voltage sources? If so, then am I right in assuming that determining any circuit's effective resistance can be likened to simply determining its Thevenin's resistance? Thank you sir.
@@RanielDG Yes you are correct. We are indeed just determining Thevenin's resistance as these networks can indeed be modeled with either a Thevenin's or Norton's equivalent circuit.
Sir i didnt understand all the "typically" part where you mentioned ro>>Rc, Bre
In analog electronics, when you see X