At 15:18: The current ID1 should be: ID1 = (Iss/2) * sqrt(Un * Cox * W/L * Iss) * Delta_V He forget to write down Iss into the equation. "" please correct me if I am wrong""
Does anyone know how in the last example R1/2 was connected to ground? I don't fully understand how differential changes cause the resistor to be connected to ground. Why don't we consider the output impedance offered by the other transistor?
Since we proved at around 20:00 that the source voltage (Vp) does not change for small signal inputs, professor has deduced that the points of symmetry in a symmetrical circuit are points which are not affected by small signals and are thus AC grounded. Alternatively, if you want to consider the effect of the other MOS device while finding the gain, what you could do is find Vx and Vy for Vin1 considering Vin2 to be grounded and vice versa. Hence you could find (Vx-Vy) wrt Vin1 and wrt to Vin2 from which you could find out Av. It is just a little bit more complicated process.
In Lec12 it was said that Vp changes when Vcm changed. at 18:44 it was said that Vp doesnt not change when Vin increases. is this Vin signal source? How these two scenarios are different
That's correct, but in this case, with Vin1 & Vin2 changing differentialy, Vp remains constant (it's a virtual GND); observe that ideal current sources don't behave in this way, only current remains constant for any load connected; then, in small-signal analysis this node is ac GND. I suggest to you review lecture 10 of this course, for a better explanation.
In this equation Vgs is changing as well. You are keeping Id constant and changing W, so Vgs has to change, so you cannot use this equation. I am commenting after one year, you must have figured it already :)
Long Kong is a student of Professor Razavi, he himself is a PhD from UCLA, working in Apple as an senior RF IC design Engineer and has also published various papers along with proff Razavi.
Thank you sir!! I have one doubt. Why are we placing current source in the bottom in NMOS Diff pair and in the top for PMOS diff pair. What will happen if I do the reverse?
The *biasing current* in NMOS devices flows from the drain to source; whereas in PMOS devices it's the opposite. So, when 2 NMOS( or 2 PMOS) devices are interconnected as a differential pair, the tail current source (= _Iss_ ) should be placed to ensure that the biasing current flows in the appropriate direction through the respective devices.
that current source in real design is a biased transistor who plays role of current source, you can not do the reverse because NMOS is steering-current device and PMOS is a pushing-current device
1:40 - Review of lecture 13
Examples
07:57 - Small-Signal Behavior of a MOS Diff Pair
22:40- Example withe Source Degeneration
38:00 - P-type differential pair examples
At 15:18: The current ID1 should be:
ID1 = (Iss/2) * sqrt(Un * Cox * W/L * Iss) * Delta_V
He forget to write down Iss into the equation.
"" please correct me if I am wrong""
yes，he missed Iss in the root.but it begins with "(Iss/2) + "also xd
yes you're correct just the plus sign between Iss/2 and the square root
ID1 = (Iss/2) + sqrt(Un * Cox * W/L * Iss) * Delta_V
awesome ..... sir sent me a mail n its here thank you so much sir. u r an angel
Cool
He's God
@@mrpossible5696 No He is God Gifted. I think as a Muslim he personally will not like being called god.
@@mutiurrehman4418 I agree
@@mutiurrehman4418you don't actually know though if he is muslim, he is just from iran, ask him first
Does anyone know how in the last example R1/2 was connected to ground?
I don't fully understand how differential changes cause the resistor to be connected to ground. Why don't we consider the output impedance offered by the other transistor?
Since we proved at around 20:00 that the source voltage (Vp) does not change for small signal inputs, professor has deduced that the points of symmetry in a symmetrical circuit are points which are not affected by small signals and are thus AC grounded. Alternatively, if you want to consider the effect of the other MOS device while finding the gain, what you could do is find Vx and Vy for Vin1 considering Vin2 to be grounded and vice versa. Hence you could find (Vx-Vy) wrt Vin1 and wrt to Vin2 from which you could find out Av. It is just a little bit more complicated process.
rewatch at 18:10
In Lec12 it was said that Vp changes when Vcm changed. at 18:44 it was said that Vp doesnt not change when Vin increases. is this Vin signal source? How these two scenarios are different
VCM in lecture 12 is DC common mode voltage. whereas change in vin in this lecture he is talking in small signal change.
38:20 P type Diff pair
So satisfying 😌😌😌
is it the official yt channel?
great lecture
So good
DC current devices should be open circuited...why did u put GND while considering small signal analysis?
That's correct, but in this case, with Vin1 & Vin2 changing differentialy, Vp remains constant (it's a virtual GND); observe that ideal current sources don't behave in this way, only current remains constant for any load connected; then, in small-signal analysis this node is ac GND.
I suggest to you review lecture 10 of this course, for a better explanation.
32:24 if i use the equation: gm= μn cox w/l (Vgs-Vth) , then gm => 2 gm, not to 2^0.5. What is going on ?
In this equation Vgs is changing as well. You are keeping Id constant and changing W, so Vgs has to change, so you cannot use this equation. I am commenting after one year, you must have figured it already :)
@@venkatasaketramgoteti8726 hello is this happens because of iss??
Is channel kaa naam Long Kong kyon hai? 🤔
Long Kong is a student of Professor Razavi, he himself is a PhD from UCLA, working in Apple as an senior RF IC design Engineer and has also published various papers along with proff Razavi.
thank you!! greatest atitude!
Thank you sir!!
I have one doubt. Why are we placing current source in the bottom in NMOS Diff pair and in the top for PMOS diff pair. What will happen if I do the reverse?
The *biasing current* in NMOS devices flows from the drain to source; whereas in PMOS devices it's the opposite. So, when 2 NMOS( or 2 PMOS) devices are interconnected as a differential pair, the tail current source (= _Iss_ ) should be placed to ensure that the biasing current flows in the appropriate direction through the respective devices.
that current source in real design is a biased transistor who plays role of current source, you can not do the reverse because NMOS is steering-current device and PMOS is a pushing-current device
Does anyone else wonder why there aren't any other good lecture videos for electronics?
yes
There is bro like Ali Hajimiri
Thank You
Waiting eagerly for circuit theory I & II and the rest of the series.
I am hoping my son will be able to watch the circuit theory I & II and the rest of the series in Year 2050
@@ayushmeena5708 I believe it's out there but not leaked to youtube yet
@@mnada72 Ohh can you please share it, God will surely bless you for that
19:00
8:15
43:01
18:00