I’m in the same situation as Dennis Ruffing. I had to remember how to perform the derivative of Id, once I did that, it all made sense from start to finish. Thanks again for posting this on line for free. I think it’s time to down load the text book and follow along more closely
Yes, the text has more detail than I can hope to fit into a video of reasonable length. Besides, the price is right! And if you do want a hard copy, you can always follow the link to Amazon for an inexpensive print version.
14:33 my high school calculus slumbering in my brain wakes up and says, "take the derivative of the curve to find the slope? I used to know how to take the derivative of an equation. Let me see here..." Thank you for this explanation. The missing puzzle piece for me was my (mis)understanding of Vpinch. Quick question: assuming a properly biased AC signal Vgs, a Vds can be set anywhere in the range between Vpinch and BVdgs and the current output at the drain will perform the pretty much the same, right?
Pretty much, but don't forget that the larger the output signal, the greater the distortion. Also, the AC drain current will depend greatly on the load impedance.
@@ElectronicswithProfessorFiore Thank you. I headed on over to Khan Academy for a review of derivative concepts and algebraic mechanics and boom, all the "prime" notation terminology came flooding back. High school (mid 80s) was a long while ago. I have taken to studying your videos by drawing right along and referring to your textbook. Slow and steady, but I have the time.
Thx for the content. What I don't understand fully yet, is how the current increases with increasing source voltage although the channel gets narrower. Is the electric field much stronger than this effect until a certain point where the pinch-off/saturation happens?
Prior to pinch-off, the device is operating in the ohmic region. IOW, think of the channel as just a resistor. Thus, as the voltage across it increases, current does as well. We exploit the ohmic region in certain applications (for example, the FET can be used as a voltage controlled resistance where R is a function of Vgs).
What a great Professor! Thank you for the great presentation!
I’m in the same situation as Dennis Ruffing. I had to remember how to perform the derivative of Id, once I did that, it all made sense from start to finish. Thanks again for posting this on line for free. I think it’s time to down load the text book and follow along more closely
Yes, the text has more detail than I can hope to fit into a video of reasonable length. Besides, the price is right! And if you do want a hard copy, you can always follow the link to Amazon for an inexpensive print version.
@@ElectronicswithProfessorFiore On order! Thanks
Excellent, thank you very much and Happy Christmas...
14:33 my high school calculus slumbering in my brain wakes up and says, "take the derivative of the curve to find the slope? I used to know how to take the derivative of an equation. Let me see here..."
Thank you for this explanation. The missing puzzle piece for me was my (mis)understanding of Vpinch.
Quick question: assuming a properly biased AC signal Vgs, a Vds can be set anywhere in the range between Vpinch and BVdgs and the current output at the drain will perform the pretty much the same, right?
Pretty much, but don't forget that the larger the output signal, the greater the distortion. Also, the AC drain current will depend greatly on the load impedance.
@@ElectronicswithProfessorFiore Thank you.
I headed on over to Khan Academy for a review of derivative concepts and algebraic mechanics and boom, all the "prime" notation terminology came flooding back. High school (mid 80s) was a long while ago.
I have taken to studying your videos by drawing right along and referring to your textbook. Slow and steady, but I have the time.
Thx for the content. What I don't understand fully yet, is how the current increases with increasing source voltage although the channel gets narrower. Is the electric field much stronger than this effect until a certain point where the pinch-off/saturation happens?
Prior to pinch-off, the device is operating in the ohmic region. IOW, think of the channel as just a resistor. Thus, as the voltage across it increases, current does as well. We exploit the ohmic region in certain applications (for example, the FET can be used as a voltage controlled resistance where R is a function of Vgs).
thanks for video :)