I watch your videos to learn teaching techniques and I have learned and used some of them in my lectures... amazingly I saw yesterday in the news that you received an award in teaching ... well deserved and unique. I can only compare you to Richard Feynman.
One thing I like the most about professor Hajimiri apart from his great enthusiasm for teaching that he has never said "wrong" to a student's answer even if it is.........he always pointed out something great out of a student's answer.........Need more profs like you........Thank you so much sir for this great content.......greetings from your virtual student from India🤎🤎🤎🤎🤎
I have read quite a few of your IEEE papers on PLLs. Accidentally came across these lecture videos and it was a joy to listen with the way you explain the thought processes involved. Great work, Professor!
I love the way you elicit the differential pair. It is absolutely fabulous comparing to someone just show a differential pair and feed me the knowledge. Great teaching! You are right, the thought process/ understanding of the ideas behind it is so important. The knowledge/technology will update, but the thoughts behind it are essential. I am so lucky that I find your videos.
I find your lectures an excellent and important resource for the humanity. Thanks. P.S. Why do you not indicate the polarity of the voltages for every "two points" of the circuits you draw?
17:39, I have never seen such an explanation of differential amplifier. Thanks, sir. Where can I get more related to stability and Miller's compensation. Why capacitor decreases stability (in terms of voltage and current).
I understand adding a common current source at the emitter fixes the sum. Hence, If both the Vbe's are increased, mathematically the sum restricts a change in emitter currents.. but we have literally increased the Vbe's right, what exactly restricts or decouples the transistors' from the exp'l relation of Ic to the vbes. By raising vbe arent we opening up the transistor to pass more current. After some thought, I think ( I'm not sure ) my question boils down to; How does the current source compensate for the increase in Ie (due to increase in Vbe) and maintain a constant Ie
What happens is that when you increase the base voltages by A, the emitter voltages also go up by the same amount (exact if you ignore Early effect) and hence Vbe do not change at all. That is why we use a current source and not a voltage source.
@@AliHajimiriChannel However, I still don't understand how the current source will ensure equal increase at the emitter if there's an increase at the base. Yes, it makes that we don't have a voltage source tied at the emitter. Because a voltage source will keep the emitter node voltage constant. So from ΔVBE = ΔVB - ΔVE, if there's an increase in VB there will be a significant change in VBE because ΔVE ~ 0 ( VE remained constant). But I don't see how the current source would enable equal change in VE (i.e ΔVB = ΔVE) so that ΔVBE = 0. Please help me out, sir. Thanks in advance.
Edward Norton exactly looks like you. But of course not as intelligent as you are. I like two three professors the most one is ofcourse you sir other two are prof. Niknejad and prof. Behzad Razavi. And Yes you are Richard feynman of Analog IC Design.
![[MV] A leap of faith...ความเชื่อใจครั้งสุดท้าย (From "Petrichor the series")](http://i.ytimg.com/vi/U1piZH2CNXA/mqdefault.jpg)
I watch your videos to learn teaching techniques and I have learned and used some of them in my lectures... amazingly I saw yesterday in the news that you received an award in teaching ... well deserved and unique. I can only compare you to Richard Feynman.
Thank you for your remarks. You are very kind.
One thing I like the most about professor Hajimiri apart from his great enthusiasm for teaching that he has never said "wrong" to a student's answer even if it is.........he always pointed out something great out of a student's answer.........Need more profs like you........Thank you so much sir for this great content.......greetings from your virtual student from India🤎🤎🤎🤎🤎
He also answer questions from non caltech students. In Hong Kong, some professors even ignore their own students
Nothing is more inspiring than seeing a competent man sharing his passion. Many thanks Sir!
I have read quite a few of your IEEE papers on PLLs. Accidentally came across these lecture videos and it was a joy to listen with the way you explain the thought processes involved. Great work, Professor!
I love the way you elicit the differential pair. It is absolutely fabulous comparing to someone just show a differential pair and feed me the knowledge. Great teaching! You are right, the thought process/ understanding of the ideas behind it is so important. The knowledge/technology will update, but the thoughts behind it are essential. I am so lucky that I find your videos.
You are a brilliant teacher.. 😍😍
i would show this to my "professor" to teach him how to teach.
there are good scientist around, but not so many teachers.
Thankyou so much Sir ,for making such hard concept easy to understand for us , Respect from India
Thank you for this clear and fluid explanation
Great and clear teaching. Thanks a lot
I find your lectures an excellent and important resource for the humanity. Thanks.
P.S. Why do you not indicate the polarity of the voltages for every "two points" of the circuits you draw?
17:39, I have never seen such an explanation of differential amplifier. Thanks, sir. Where can I get more related to stability and Miller's compensation. Why capacitor decreases stability (in terms of voltage and current).
Watch videos 169N, 170N, and 171N
@@AliHajimiriChannel Thanks sir
differential pair as if i ve never seen before
You are just superb :)
Legend
I understand adding a common current source at the emitter fixes the sum. Hence, If both the Vbe's are increased, mathematically the sum restricts a change in emitter currents.. but we have literally increased the Vbe's right, what exactly restricts or decouples the transistors' from the exp'l relation of Ic to the vbes. By raising vbe arent we opening up the transistor to pass more current.
After some thought, I think ( I'm not sure ) my question boils down to; How does the current source compensate for the increase in Ie (due to increase in Vbe) and maintain a constant Ie
What happens is that when you increase the base voltages by A, the emitter voltages also go up by the same amount (exact if you ignore Early effect) and hence Vbe do not change at all. That is why we use a current source and not a voltage source.
@@AliHajimiriChannel Oh the current source drives the emitter voltage up, is that it? Thank you very much for responding, you are the best
Thank you very much.
You asked the same question I needed to ask. That's why I checked the comments to see if the question as been answered.
@@AliHajimiriChannel
However, I still don't understand how the current source will ensure equal increase at the emitter if there's an increase at the base.
Yes, it makes that we don't have a voltage source tied at the emitter. Because a voltage source will keep the emitter node voltage constant. So from ΔVBE = ΔVB - ΔVE,
if there's an increase in VB there will be a significant change in VBE because ΔVE ~ 0 ( VE remained constant).
But I don't see how the current source would enable equal change in VE (i.e ΔVB = ΔVE) so that ΔVBE = 0.
Please help me out, sir. Thanks in advance.
why is it Vcc-R*I?
Edward Norton exactly looks like you. But of course not as intelligent as you are. I like two three professors the most one is ofcourse you sir other two are prof. Niknejad and prof. Behzad Razavi. And Yes you are Richard feynman of Analog IC Design.
He lost me when he started doing the thing with the pens.