01:25 - Intro 06:25 - Example with Capacitor at the Output 10:20 - Poles and Zeros - Bode's Rules 18:00 - Pole Frequency 21:10 - Finding Poles by Inspection 27:10 - Example with the CS Amplifier and Capacitor at the Inpiut 30:32 - Example with the CS Amplifier and Capacitor at the Output 34:40 - Combining Examples - Multiple Poles 40:28 - Plotting Frequency Response 44:35 - A Complication
learned more in the first 10 minutes of this video than I learned in the first 1.5 semesters of my MSEE degree. Thank you for making this content available for everyone.
I love this guy, every example on this lecture led us to the contruction of the complete small signal model of a MOSFET with the capacitances included. Brilliant!!
I read past comments about how you teach us by simple steps the final outcome, but how this helped me has never been more obvious than this lecture. This is truly amazing. It took me only one hour to go from simple RC circuits to MOSFET ones
The reasoning at 9:20 is not correct even though conclusion is Id = const I = Id + Ic ; Vout = Vdd - I.Rd At w=0, I = Id => Vo = Vdd - Id.Rd At w !=0, I = (Id + Ic) > Id => Vo = (Vdd - (Id+Ic) Rd ) < (Vdd-IdRd), So as w increases, I increases, drop I.Rd increases and Vo decreases
By using "Small-Signal Analysis": At w=0, Vout = Rd *(Id) At w!=0, Vout = Rd*(Id - Ic) I think you guys forget Vdd is "ac GND" at "Small-Signal Analysis",lol
he kept it in s land. He didnt convert to s=jw b/c its bode magnitude plot anyways, dont need the j , just w. Also H(jw) = 1/(jwRC) , you forgot the w if you are going to s=jw
I don't think there is a better way to teach these concepts. You are a special teacher. Thank You
01:25 - Intro
06:25 - Example with Capacitor at the Output
10:20 - Poles and Zeros - Bode's Rules
18:00 - Pole Frequency
21:10 - Finding Poles by Inspection
27:10 - Example with the CS Amplifier and Capacitor at the Inpiut
30:32 - Example with the CS Amplifier and Capacitor at the Output
34:40 - Combining Examples - Multiple Poles
40:28 - Plotting Frequency Response
44:35 - A Complication
its great to master analog design, but even more greater to make it understandable for the learners. really admire razavi
Wow! How easy you explain things. Can’t wait to learn about Miller effect
learned more in the first 10 minutes of this video than I learned in the first 1.5 semesters of my MSEE degree. Thank you for making this content available for everyone.
I love this guy, every example on this lecture led us to the contruction of the complete small signal model of a MOSFET with the capacitances included. Brilliant!!
I read past comments about how you teach us by simple steps the final outcome, but how this helped me has never been more obvious than this lecture. This is truly amazing. It took me only one hour to go from simple RC circuits to MOSFET ones
Always thank you sir. Your attitude towards teaching the world is unmatched.
Behzad Razavi lectures always inspires me.King of Electronics..Great Prof. These Lectures Are Assets..Thanks ❤
21:41 Finding poles by inspection
What a fantastic lecture, thank you so much!
Thank you very much, Professor!
The reasoning at 9:20 is not correct even though conclusion is
Id = const
I = Id + Ic ;
Vout = Vdd - I.Rd
At w=0, I = Id => Vo = Vdd - Id.Rd
At w !=0, I = (Id + Ic) > Id => Vo = (Vdd - (Id+Ic) Rd ) < (Vdd-IdRd),
So as w increases, I increases, drop I.Rd increases and Vo decreases
nice, was wondering about the same,
You said Id is constant. So Id= Ir+Ic. If there is no capacitor, Id=Ir. Doesn't that mean drop across resistor decreased
By using "Small-Signal Analysis":
At w=0, Vout = Rd *(Id)
At w!=0, Vout = Rd*(Id - Ic)
I think you guys forget Vdd is "ac GND" at "Small-Signal Analysis",lol
@@joshmeiyb I think what you said is completely correct.
@@joshmeiyb Exactly. We're talking about AC input voltages. This is small-signal analysis.
What an explanation!
Waiting eagerly for circuit theory I & II and the rest of the series.
This guy is special.
Its just awesome .. u the best prof.
This is a Godsend. Thank you so much, you saved me!
OMG! It’s Razavi!
I didn't get the reasoning at 9:40, shouldn't the gain be higher because the voltage drop is smaller?
Capacitor acts as a short circuit at infinite frequency ,so Vout=0 =>Av=0
@Sai Charan Marrivada's reply is the real reasoning. Your logic is correct and Dr. Razavi didn't mention that properly.
nice explenation
Detailed Electronics Lectures available in Urdu Language
Ty
where did the j at 19:10 go? shouldnt the frequency be "1/(jRC)"?
he kept it in s land. He didnt convert to s=jw b/c its bode magnitude plot anyways, dont need the j , just w. Also H(jw) = 1/(jwRC) , you forgot the w if you are going to s=jw
can anyone tell me before this in which lecture prof. has discussed poles & zeros as I came to specifically look at that
that is great
Love yousir
Will we include ro for the resistance?
Can we also identify right-hand side poles by inspection?
can anyone please give me the link of that basic circuit theory lecture in which the pole and zero concepts had been taught?
what's the first lecture which bode plot has been introduced ?
11:50 Bode's Rules
Can you provide the link for quiz?
45:30
27:40
There is a mistake 13:58 bode rule 20Db slope increases when approaches to poles.....
No , he is correct