Dear Prof.Ali, first of all, I have to say thank you for sharing these videos, I watched all over the videos in 2 consecutive months and refreshed my memory about analog electronics in the best way, I have to say it was beyond helpful, (thanks again and best wishes from Hong Kong)
Thank you, Prof. A. H., for the video. I have a question I'm struggling to understand. At low frequencies, all the terms with 's' don't matter (since 's' approximates 0). However, as the frequency increases (far greater than 1), I think the higher-order 's' terms become more significant. For instance, if b1 is significant and the condition is 0 < S < 1, but the frequency Wh is usually in the kHz or MHz range, why can the denominator be approximated as 1 + b1s? Thank you 19:40
That happens under the assumption that there is a dominant pole. In addition, if you think about it, the b1 is the sum of the reciprocals of the poles while b2 is the sum of the reciprocal of products of two poles at a time. That means b2 is smaller than b1. That is why the s term effect is felt before the s^2 term.
@45:43 Why is the symmetrical relationship valid --> R_1^0 * R_2^1 = R_2^0 * R_1^2 R_1^0: Resistance looking into port 1 with all other ports at zero R_2^0: Resistance looking into port 2 with all other ports at zero R_2^1: Resistance looking into port 2 with port 1 at infinity and all other ports at zero R_1^2: Resistance looking into port 1 with port 2 at infinity and all other ports at zero To me it looks like the products are unequal.
Yes, this is correct and in someway a remarkable relationship. You can find the proof and discussion in the Analog book draft or the published paper, but in short in the derivation since it does not matter which element is indexed 1 and which one indexed 2, you will end up with this relationship. There is more general one with multiple elements too. Look at the paper.
Hello, I used this method to calculate the input impedance of an active inductor. But I get the wrong result when i zero the input voltage source for calculating the time constant. I get the right solution only for removing it. Do you know about why that is?
Professor if in your example in the common source stage R1 =0 that means that all the b2 coefficents are equal to zero? So the system has only one pole? Because if R1=0 then all time constants that include R1 are equal to zero
I assume you are talking about @23:00, which for a MOSFET has an r_pi of infinity. In that case, tau_ph^0=0 but not tau_mu^0, or tau_L^0 (as they have other terms in them that do not go to zero if R1=0. However, the second order time constants (tau_mu^L and tau_L^mu) are zero. This can be understood from the fact that you have a capacitive loop between C_mu and C_L (and the input voltage source which is also shorting C_ph), so you can define only one initial condition and the circuit as shown will become first order and will have one pole (b2=0).
@@AliHajimiriChannel Thank you professor for your quick reply . I tried to find Yin(s) of a common source LNA in order to see if the real part can be matched to 1/50 Siemens. The TTC is far more quick than KVL,KVC so i saw that when R1=0 ,Cgs plays no role in calculation. So in the case of Yin(s), b2=0, the system has one pole but a different numerator than the case of Vout(s)/Vin(s) as most importantly a2 is not equal to zero in the numerator. Thank you so much for uploading your work professor.
@Ali Hajimiri sir, please make the video on Analog design on FDSOI AND FINFET Technologies. that how to analog design efficeintly on nano tech( short channel devices). thanks
I really appreciate you share such a work with us. Don't leave us for years again, just keep uploading 💪 ❤️
Dear Prof.Ali, first of all, I have to say thank you for sharing these videos, I watched all over the videos in 2 consecutive months and refreshed my memory about analog electronics in the best way, I have to say it was beyond helpful, (thanks again and best wishes from Hong Kong)
After a year, Prof.Ali is back! Alhamdulilah 😎😎😎👍👍👍
Can’t wait for more new videos!
30:52 For a cascode with output buffer, if there is a complex pole, does this mean that the ZVT calculation is invalid?
Thank you Prof. A. H. for this one! Am sure you will upload many more, I would thank you for, over the following months, or better, years.
Thank you, Prof. A. H., for the video. I have a question I'm struggling to understand. At low frequencies, all the terms with 's' don't matter (since 's' approximates 0). However, as the frequency increases (far greater than 1), I think the higher-order 's' terms become more significant. For instance, if b1 is significant and the condition is 0 < S < 1, but the frequency Wh is usually in the kHz or MHz range, why can the denominator be approximated as 1 + b1s? Thank you 19:40
That happens under the assumption that there is a dominant pole. In addition, if you think about it, the b1 is the sum of the reciprocals of the poles while b2 is the sum of the reciprocal of products of two poles at a time. That means b2 is smaller than b1. That is why the s term effect is felt before the s^2 term.
Thank you very much for this TTC Summary and I am looking forward to getting your book "Analog: Inexact Science, Vibrant Art" when it is available!
After a long time .. good to see you prof .. 🙂 From India
Join to the greetings... 🙂From Russia
Amazing Presentation. Thanks, Dr. Hajimiri!
Please keep posting video sir!
Did I miss his other RF lectures at other channel?
Write back if u find something :D
54:00 Typo: tau⅓ = L3/R (not L2/R)
@45:43 Why is the symmetrical relationship valid --> R_1^0 * R_2^1 = R_2^0 * R_1^2
R_1^0: Resistance looking into port 1 with all other ports at zero
R_2^0: Resistance looking into port 2 with all other ports at zero
R_2^1: Resistance looking into port 2 with port 1 at infinity and all other ports at zero
R_1^2: Resistance looking into port 1 with port 2 at infinity and all other ports at zero
To me it looks like the products are unequal.
Yes, this is correct and in someway a remarkable relationship. You can find the proof and discussion in the Analog book draft or the published paper, but in short in the derivation since it does not matter which element is indexed 1 and which one indexed 2, you will end up with this relationship. There is more general one with multiple elements too. Look at the paper.
Awesome material, thanks for uploading
Thank you for your very important contact and reasoning that you bring into electronics!
Hello, I used this method to calculate the input impedance of an active inductor. But I get the wrong result when i zero the input voltage source for calculating the time constant. I get the right solution only for removing it. Do you know about why that is?
Is there a book or a printed version of this theory?
Please see the cert last slide of the presentation for where to find the book, the paper, and class lectures on TTC.
@@AliHajimiriChannel Thanks a lot!
Sir,I have a question while solving question on cascode can we use t model in one and pi model in other??
Professor if in your example in the common source stage R1 =0 that means that all the b2 coefficents are equal to zero? So the system has only one pole? Because if R1=0 then all time constants that include R1 are equal to zero
I assume you are talking about @23:00, which for a MOSFET has an r_pi of infinity. In that case, tau_ph^0=0 but not tau_mu^0, or tau_L^0 (as they have other terms in them that do not go to zero if R1=0. However, the second order time constants (tau_mu^L and tau_L^mu) are zero. This can be understood from the fact that you have a capacitive loop between C_mu and C_L (and the input voltage source which is also shorting C_ph), so you can define only one initial condition and the circuit as shown will become first order and will have one pole (b2=0).
@@AliHajimiriChannel Thank you professor for your quick reply . I tried to find Yin(s) of a common source LNA in order to see if the real part can be matched to 1/50 Siemens. The TTC is far more quick than KVL,KVC so i saw that when R1=0 ,Cgs plays no role in calculation. So in the case of Yin(s), b2=0, the system has one pole but a different numerator than the case of Vout(s)/Vin(s) as most importantly a2 is not equal to zero in the numerator. Thank you so much for uploading your work professor.
@Ali Hajimiri sir, please make the video on Analog design on FDSOI AND FINFET Technologies.
that how to analog design efficeintly on nano tech( short channel devices).
thanks
Thanks Dr. Hajimiri..
Thanks a lot sir
You might be going a tad bit too fast, professor.