วีดีโอ

Thank you for lecture. blessed that i can listen to one of the greatest lectures from the other country

Dude cramming all this 2 days before the exam is not it 🥲

EXjYu7_€🇨🇮🇦🇷🏢🇸🇭

Excellent Lecture Professor Razavi, Thank You.

These lectures save my midterm and they will save me again in final

젠장 라자비는 신이야

The Quiz music is painful.

30:24 will not the current also vary a bit I mean the current also will become Iee/2+del(I) and Iee/2-del(I) in this circuit. So the Vx and Vy also should change

12:00 "C" by itself is Uppercase; "c" in "coulomb" is Lowercase.

Bu dava sensiz olmaz

In cascode introduction we learnt that MOS transistor does not change the voltage with current like a resistor, but here we are making an opposite assumption. So are we treating MOS in triode region for active load?

Thank you so much for teaching such a clear and systematic approach.

ياريت كل الدكاتره زيك

Great lessons Mr. Razavi. One short question: Can we directly drive the load from the source of the transistor and avoid the Rs all together? I am thinking to build this circuit (and use as a buffer) to connect my HF radio and SDR radio to the same antenna. If the 1/gm would be close to 50 Ohm that would be a good match for the input of the SDR. Otherwise it can be lower, but maybe that is not a problem for the SDR input. Side note: I will build another small circuit to disconnect the SDR input from the HF radio output when transmitting.

When Delta V is small, the voltage at point P does not change. I totally agree with that. But, does that imply P is virtual ground? Yes, but why?

At 3:57, I am stucking on this: say, current is copied from the left side to the right side, left side current increases, right side current also increases, fine. But on the right side, there is another P type current source on the top, its SG bias voltage is fixed, so its current is also fixed. Since the copied current is in series with the P type current source, how could the copied current increase ?

This can be easily solved with Laplace

Iran❤❤

oshadharon sir

It is very interesting circuit, thank you Sir 👍

Thank you Sir At 9:40 Simple conpect~ That is a single pole circuit of Cs circuit and then Wp=BW=1/RdC(rds infinity) And then that have a dgeneration zero of Cs

I am lost. Assume M1 is the cascode device and M2 is the degenerating device, I can understand the output resistance is boosted to gm1 * r01 * r02. So far so good. But how could M2 help the narrower DS1 voltage range issue? To avoid M1 getting into the triode region, DS1 voltage needs to be narrower just like using a simple Rs resistance at M1 source... Correct? What's the advantage of M2 over Rs in this perspective?

For the logarithmic amplifiers, if I replace the NPN with a diode (anode connecting to Vout, cathode connecting to -in), will I also get a same logarithmic amplifier... Same question with the PNP, looks to me it can also be replaced by a diode with anode connecting to -in and cathode connecting to Vout.. am I correct? In my understanding, the logarithmic function comes from the i-v relationship of the PN junction in this context, so using a bipolar transistor or a diode in the feedback makes no difference.. am I correct?

50:35 Should be V0 - Vth not Vgs - Vth

43:19 Missing R_s in the circuit

greetings and special thanks from lebanon .. master 2 in microwave engineering physics

waiting for razavi electronics-3

very helpful content thankyou sir

23:00 laplace

30:25 The length of the channel in the denominator actually appears only after the integration of the left-hand side of the equation

46:00 ce review

47:00 bjt

It’s easy to forgive a typo wrt sign on PMOS drain terminal, and some arguable assumptions made wrt channel length modulation, because Prof. Razavi is an extraordinary gifted lecturer. I rarely see lecturers or teachers go to such painstaking lengths to drive home points or encourage thinking from many different perspectives. Many lecturers also avoid longhand writing on the white board, but not Prof. Razavi. The man is bursting with enthusiasm and energy it is as if he is teaching to an individual in a private tutorial. The fact that students readily spot typos and are ready to argue or debate one or two things is a very good thing. It means that they are learning more, questioning and supposing. Sometimes normal human error gives added challenge, and further opportunity for critical thought. It’s all good.

Thanks sir❤

Can any explain what happens to holes in p type semiconductor as he took only about electrons in consideration ,if we invert the voltage applied between semiconductor and capacitor as holes are more in p type it conducts more current. 15:03 min ?

40:00

26:50

razavi 才是真正的大师，读了五年大学从未见过这么好的老师

25:06 yes we did have Self driving cars by 2020.

32:00 vout/vx = gm2*Rc

الله يبارك في علمك، رائع جدا!

the goddddd!!!!!!!!!!!!

the goatttttt!!!!!!!!

God tier lecture 😭

Hallo Beautiful people, i dont know who is here in 2024 but i needed help with motors and generators i dont know who teaches those well like DR Behzad Razavi does an outstanding Job at teaching us Analog Electronics here.

Sir at the ending in the reverse breakdown you applied forward voltage it's a big mistake

my god never seen such an interesting and simple explaination of mos before

15:10

This Professor is too good. Biasing concept has been taught this way is too good, which I am able to explain others now after 20 years although I know what it was but was not able to put in words such nicely as he did...

Regarding drain and source terminals are inter-exchangable, I am confused. Source is connected to the substrate (internally, short in the die) but the drain does not.. how could they be symmetrical?