I rarely comment on TH-cam, but this video deserves appreciation. The way you explain, without using any complex terms, and reinforcing the basic concepts, is how every teacher should teach. Thank you so much for the quality content!
Thank was wonderful! What you taught in 15 minutes made the whole semester make sense! The reasoning behind things that you add makes it so much more understandable, and thus easier to remember and figure out intuitively! Thank you so much!
Sir for the first time in my life of engineering about analog circuit analysis I understood stuff. I'll re view your video when I am off tension cause I got my lab exam tomorrow for VLSI Lab. I love the way u explain the stuff do simple it just clicks right with a guy who got into electronics engineering by mistake😶 having a way of an electrical engineer
Thanks! We don't necessarily need Vgs to be zero, we just need the *input* voltage vin to equal zero. This is because output resistance measures how a change in the *output* affects the system, while completely ignoring the input (setting it to be zero).
I've got extremely confused here. Is this in saturation region? Why are we not mentioning the bias if so? Why do we make the Vdd as zero. Why do we direclty connect the current source to ground?
Yes, it's in saturation. Expected for Common Source but could have been explicit. Bias resistors aren't necessary for MOSFETs if we assume saturation, meaning we are certain that Vds > (Vgs - threshold) and Vgs > Vthreshold. We aren't calculating the Q-point either where a bias voltage may help. Vdd is needed for Id, which is needed for gm and ro but he doesn't calculate gm and ignores ro. Current source to ground is because there is no Rsource resistor. It isn't as necessary as in the BJT equivalent. In Thevenin/Norton + small signal analysis, we short circuit constant voltage sources, including Vdd, and open circuit constant current sources. Since the current source is voltage-dependent and Vgs = 0V from finding Vthevenin aka Vtest, it must have no current and be an open circuit. He's deliberately choosing the most simple example. That's what's causing confusion.
Very nice videos! What year were you in when they began going over these concepts? Im teaching myself these things and I'm trying to learn things in a coherent order. I would say at this point the most difficult thing I know how to do is taking thevenin equivalents of different componentns in series parallel AC RLC circuits. Im pretty good at analyzing circuits with only passive components. But adding in active things like Fets Tubes and BJTs is intimidating. Ive made some simple Hartley and colpitts oscillators on my breadboard, and experimenting with those has left me with so many questions. So I figured Id start here in answering those questions.
just thought I'd let you know, it may depend on a lot on the school, but my degree started active components (op amps) late second year and moved in BJTS and MOSFETS in third
Mostly. MOSFETs are characterized by a thin oxide layer that galvanically isolates the gate from the source and drain - thus the infinite input impedance. JFETs physically connect the bits of semiconductor without an oxide dielectric, so the input impedance is less than infinite.
@@jacobfaseler5311 Thanks for the reply. Finally, I found a website where the whole process of biasing a jfet comes explained, and that's what I did, and it works fine.
This is excellent for the most basic explanation. I noticed though that you did not talk about the resistance from channel length modulation, the body effect, the capacitances between each node, and the frequency response. Is there a reason for this and do you have any video recommendations for them?
Yup, I have a few videos on the body effect (none on channel length modulation I don't think). In general, I find it's good practice to simplify things like my life depends on it, and then introduce complexity as needed. One thing I could do better and I always liked is sprinkle in little hints here and there about that complexity without talking about it too much.
For this you need to know where all the capacitors are - then you can figure out what the -3dB frequency is where the transfer function falls to 1/sqrt(2).
Because the source is grounded we call it common source? So the drain is grounded for common drain? I don't think so. I would say it's called common source because it is connected to AC ground and likewise for common drain.
It's because we are using the Thevenin equivalent circuit to get Rout. Voltage sources are considered short circuit and current sources are considered open circuit.
You do in 15 minutes what my professors were trying for a whole term... thank you so much!
your professor only teach common source circuit?
@@myerwerl lol
I rarely comment on TH-cam, but this video deserves appreciation. The way you explain, without using any complex terms, and reinforcing the basic concepts, is how every teacher should teach. Thank you so much for the quality content!
Thank was wonderful! What you taught in 15 minutes made the whole semester make sense! The reasoning behind things that you add makes it so much more understandable, and thus easier to remember and figure out intuitively! Thank you so much!
Your 15 min explanation equals to 3 hours of our professor.
Thank you.
I AM FROM TAIWAN your explaination is fabulous
Sir for the first time in my life of engineering about analog circuit analysis I understood stuff. I'll re view your video when I am off tension cause I got my lab exam tomorrow for VLSI Lab. I love the way u explain the stuff do simple it just clicks right with a guy who got into electronics engineering by mistake😶 having a way of an electrical engineer
i have my test tmr too lol
Please contact me I need some explanation
Great video. Was looking this up as a refresher for a problem in the work force and this was very well explained.
you explained this in a simplest way and that is so helpful! thank you, may god bless you!
thank u so much , as someone else said , u did in 15 minutes what my professor failed to do in 2 houres
Very crisply explained! Well done
Wow. That was really great ! Love you my man, thank you for making stuff clear.
Thank you:)
wow you made my life so easier then my prof did.
thank you so much
THIS IS BY FAR THE BEST IN THIS TOPIC
Thanks for making this video -- it helps me a lot! I love the way you explain these concepts :)))
It made it easy for me to understand things I didn't understand. Thank you.
Thanks! What was the piece of understanding you were missing that this filled in?
@@JordanEdmundsEECS It's embarrassing, but it was the part that used KVL. Thank you once again.
Very good explanation sir.... It was very clear...
👉 What is gm and how it is Calculated sir...? That you missed in this lecture
damn, love the way you shared that knowledge, respect
Thank you so much for this video, finally I understood a lot : )
Glad to hear :) Anything specific you think helped you understand?
Why dont you connect R1 and R2 resistors at gate terminal? Is it necessary to connect plz give me clarification sir
Appreciable 🤞
Wonderful explanation! :)
Thanks for the video!Intuitive explanations!! It would be great if you explain why we take Vgs=0 when we find output resistance
Thanks! We don't necessarily need Vgs to be zero, we just need the *input* voltage vin to equal zero. This is because output resistance measures how a change in the *output* affects the system, while completely ignoring the input (setting it to be zero).
@@JordanEdmundsEECS Got it, thanks!!!
It was just great..Clear explanation.Thank you!
thank you. it was really helpful
Really well explained, thanks!
thank you very much for this great explanation.
You are welcome! What specifically did you enjoy about it?
Thanks for the video, very useful
Thanks!
Very good explained.
Glad you liked the explanation! Anything in particular you found helpful about it?
Can you please explain to me what exactly the "gm" is?
I've got extremely confused here. Is this in saturation region? Why are we not mentioning the bias if so?
Why do we make the Vdd as zero. Why do we direclty connect the current source to ground?
Yes, it's in saturation. Expected for Common Source but could have been explicit. Bias resistors aren't necessary for MOSFETs if we assume saturation, meaning we are certain that Vds > (Vgs - threshold) and Vgs > Vthreshold. We aren't calculating the Q-point either where a bias voltage may help. Vdd is needed for Id, which is needed for gm and ro but he doesn't calculate gm and ignores ro. Current source to ground is because there is no Rsource resistor. It isn't as necessary as in the BJT equivalent.
In Thevenin/Norton + small signal analysis, we short circuit constant voltage sources, including Vdd, and open circuit constant current sources. Since the current source is voltage-dependent and Vgs = 0V from finding Vthevenin aka Vtest, it must have no current and be an open circuit.
He's deliberately choosing the most simple example. That's what's causing confusion.
Very nice videos! What year were you in when they began going over these concepts? Im teaching myself these things and I'm trying to learn things in a coherent order.
I would say at this point the most difficult thing I know how to do is taking thevenin equivalents of different componentns in series parallel AC RLC circuits. Im pretty good at analyzing circuits with only passive components. But adding in active things like Fets Tubes and BJTs is intimidating.
Ive made some simple Hartley and colpitts oscillators on my breadboard, and experimenting with those has left me with so many questions. So I figured Id start here in answering those questions.
just thought I'd let you know, it may depend on a lot on the school, but my degree started active components (op amps) late second year and moved in BJTS and MOSFETS in third
Thanks for sharing. Really. I am here because I am having troubles with a Jfet preamp for an electric guitar... Is this valid for JFETs as well?
Mostly. MOSFETs are characterized by a thin oxide layer that galvanically isolates the gate from the source and drain - thus the infinite input impedance. JFETs physically connect the bits of semiconductor without an oxide dielectric, so the input impedance is less than infinite.
@@jacobfaseler5311 Thanks for the reply. Finally, I found a website where the whole process of biasing a jfet comes explained, and that's what I did, and it works fine.
can you sir please explain this with Rs RD and R1 R2 configuration and Vdd ,-
Vdd
he is a genius period.
Thank you♥️ i love this video
THANK YOU FOR THIS VID
wagbayi joor 💪🏽 idea! 👍🏽
Excellent
This is excellent for the most basic explanation. I noticed though that you did not talk about the resistance from channel length modulation, the body effect, the capacitances between each node, and the frequency response. Is there a reason for this and do you have any video recommendations for them?
Yup, I have a few videos on the body effect (none on channel length modulation I don't think). In general, I find it's good practice to simplify things like my life depends on it, and then introduce complexity as needed. One thing I could do better and I always liked is sprinkle in little hints here and there about that complexity without talking about it too much.
why you are so amazing!
how to Derive the high corner frequencies of the amplifier of the common source
For this you need to know where all the capacitors are - then you can figure out what the -3dB frequency is where the transfer function falls to 1/sqrt(2).
thank you so much
Got it! 🙂
hello can you help me in something very important in electronics please using common source amplifier pmos based
please very important
My teacher said there was no way to teach this in an easier way, his last day of teaching is in 2 weeks🎉
Because the source is grounded we call it common source? So the drain is grounded for common drain? I don't think so. I would say it's called common source because it is connected to AC ground and likewise for common drain.
Yup!
I discovered some similarities in BJT aplifiers with the inspiration of this vedio.
good voice
tq sir
QUESTION. Why are we shorting Vin to find out Vout/Iout(output resistance)?
what i understood is to get rid of the influence of input voltage
Because that's the definition of Rout. To find Rout, we always need to short Vin.
It's because we are using the Thevenin equivalent circuit to get Rout. Voltage sources are considered short circuit and current sources are considered open circuit.
because that's how two-port analysis is performed
:3 thx
Sexy explanation bro, thanks
Yeah as an ME I'll stick to Navier-Stokes equations thank you next lol
LOL
so why is this useful? other than that it's used in a lot of analog circuits lol is it that we can amplify the voltage but not the current?
It’s used in essentially *all* analog circuits xD. It’s also the simplest case to analyze and so a good place to start.
Then someone asks you to calculate the cascaded noise figure for multiple MOSFET devices--this is huge!
You divided by zero. RIP.
Very crisply explained! Well done
Thank you so much