4 hours of lectured explained in like 14 minutes thanks dude :)
@leonard-riccardowecke27736 жыл бұрын
I believe that he forgot to mention why he left the bar over the nmos logic away. If the logical equation for the NMOS part gives TRUE then the output on the CMOS is LOW. He implemented that by leaving out the bar over the Nmos part(double negation of cmos logic). Therefor the equation for the Cmos part is implemented. In Mnemonics: Task: Cmos("given logic") != High Rule(that he didnt mentioned): NmosLogic=True => CmosOutPut=Low => IF Nmos = false THEN Cmos=True
@siliconforestry45084 жыл бұрын
Thank you
@JoshTronicsECE9 ай бұрын
This is more understandable and clear than what I've been taught. Thank you for your lecture. Keep it up!
@ashishgautam59896 жыл бұрын
This is one of the best ways to learn something!
@robertwitt12763 жыл бұрын
seriously such an awesome two videos dude. I watched 3 lectures and was lost, now I am about to watch them again and understand what is going on. Thank you so so so so much keep up the good work
@Adam-gp3ij5 жыл бұрын
Because of you I understood the entire chapter ! I appreciate your explanation
@thetruthabr98794 жыл бұрын
how do u know whether or not the inverted D in the PUN goes right above the PDN under the circuit you drew or above it right under +Vdd?
@ajmaln23872 жыл бұрын
litterly a week of being lost in lectures. All done in 2 vids... THank you!
@nkululekostefaans15793 жыл бұрын
This just summed the whole topic in less than 8 mins.. thank you.. SUBSCRIBED!
@achrafelkhandouli5 жыл бұрын
thnx
@phamhuutri19965 жыл бұрын
what if the function is not inverted? let say just like this [(A+B*C)*C+D]. how do i slove this ?
@tamannachowdhury23067 жыл бұрын
This really helped me alot. Thanks..
@mrohitsingha16155 жыл бұрын
daam!!!!!! this guy makes everything easy
@eunicechiu87988 жыл бұрын
in my understanding, did u just ignore the inv(the bar) over the logic equation? what is the difference between the layout of [Inv(A+B*C)*C+D] and (A+B*C)*C+D ? Anyway, steps are clear. Thanks!
@tarcisiojosedearaujopereir6647 жыл бұрын
Eunice Chiu same question, but as i remember the cmos always inv normally, u can think in the basics inv nand and nor
@moustafahammoud41296 жыл бұрын
check morgan theorem
@leonard-riccardowecke27736 жыл бұрын
I believe that he just forgot to mention why he left the bar over the nmos logic away. If the logical equation for the NMOS part gives TRUE then the output on the CMOS is LOW. He implemented that by leaving out the bar over the Nmos part(double negation of cmos logic). Therefor the equation for the Cmos part is implemented. In Mnemonics: Task: Cmos("given logic") != High Rule(that he didnt mentioned): NmosLogic=True => CmosOutPut=Low => IF Nmos = false THEN Cmos=True
@usrakhan9756 жыл бұрын
HE ADDED AN INVERTER WHILE DESIGNING XOR and XNOR BUT Y DID HE NOT ADD AN INVERTER HERE . THIS EQUATION HAS ALSO GOT WHOLE BAR
@mindylee5023 Жыл бұрын
Thank you so much for posting this video. You save my life !!!
@mengxili37315 жыл бұрын
First thankyou so much for making this video. It helps a lot. Then I have a question in when you drawing the D-PMOS into the pull up network. Does it matter if I draw D bellow where it will not be connected Vdd later but with output. Actually those two different position where D can be placed later will influenced the stick diagram. So if the position of D does a matter here and also in stick diagram, then do you have any tip for deciding the position of D_
@bommenavishwateja74685 жыл бұрын
its a series configuration doenst matter where you place it
@joelrcc4 жыл бұрын
@@bommenavishwateja7468 no, error. when is PULL, for example AB, in down position is A, when is DOWN, in down position is B
@bommenavishwateja74684 жыл бұрын
@@joelrcc I don't think so buddy
@joelrcc4 жыл бұрын
@@bommenavishwateja7468 Well. At end, is not neccesary but by standar is in this way
@bloe26034 жыл бұрын
Hello sir, is it okay if i contact you personally regardings the designs of the CMOS circuits? Need help
@zerap81583 жыл бұрын
Your video really helped me! 🧡 (It's much better than what my teacher taught 😂
@caesarxi13033 жыл бұрын
Thank you so much. I finally got it. Actually, PMOS performs the operation. NMOS configuration is only for activating GROUND. PMOS does the job by negating the expression. BUT when the expression is true, its negation is false, and then PMOS configuration needs to activate GROUND, so it wont deliver VDD to the output. If the expression is false, PMOS output is true, and it needs to deactivate the GROUND. To activate/deactivate the ground, we need the OPPOSITE configuration on NMOS, which is responsible for delivering the GROUND signal. To activate GROUND, NMOS must evaluate the expression as true. To deactivate GROUND, NMOS must evaluate the expression as false. I did this with the NAND port, so I could visualize it better, since it has an easier configuration. 2 PMOS in paralel = NAND 2 PMOS in series = NOR 2 NMOS in paralel = OR 2 NMOS in series = AND NAND needs an AND to activate/deactivate GROUND.
@caesarxi13033 жыл бұрын
The input signal for PMOS comes from above. The input signal for NMOS comes from below.
@vaishnavipalyam86286 жыл бұрын
Thanks for the detailed explanation. It helped me a lot
@mervatarafat25784 жыл бұрын
You are the best keep omving on♥️♥️♥️
@loki-qd4zp4 жыл бұрын
U save my whole interview session
@absolute___zero3 жыл бұрын
so.... the pull up and pull down networks are just joined together by wires and that's it ???? Are you sure this joined circuit can't be simplified any further?
@javariababar36163 жыл бұрын
Best video great 🙌
@teov34203 ай бұрын
You are a life saver.
@rohantyagi52797 жыл бұрын
Totally confused here isn’t n-mos transistor a pull up network as it pulls the signal from ground?
@TheFullcharge135 жыл бұрын
Really good video, easy to follow. Great job
@Parth.Deshpande4 жыл бұрын
Awesome bro , thanks
@rohittanwar19464 жыл бұрын
thanks you so much 😍😍😍😍 for explaing this topic very easly 😊😊😊
@Play4fuNNNNNnn7 жыл бұрын
Can you make video on how to design a 3-times NOR-Gatter and a 2 Times AND-Gatter by using NMOS technology? i d be really thankfully.
@cubeyd273 жыл бұрын
Thank you bro 3 hours like 7 minutes 👏👏
@manasbohat68834 жыл бұрын
can we simplify the function by boolean algebra and then implement?
@jamesblooom8670 Жыл бұрын
amazing job mate
@zorkan3 жыл бұрын
Excellent
@shaikrehman32365 жыл бұрын
Your video is very helpful thank you sir
@javismc7099 Жыл бұрын
Thanks for this helpful video
@hemanthar49745 жыл бұрын
Very well explained 🔥
@happilyevernever42894 жыл бұрын
Is this the 2 step implementation of cmos or single step?
@samiya8774 жыл бұрын
its really helpful.thank u
@domulko444 жыл бұрын
thanks a lot :) really useful video
@koray_kara3 жыл бұрын
Thank you very much sir.
@asr_ast3 ай бұрын
for PDN take inverted of given boolean for PUN take dual of given boolean
@zombiedude3474 жыл бұрын
The example logic should be simplified as (A | B & C) & C is equivalent to (A | B) & C.
@YoO1616 жыл бұрын
how isnt your pmos wrong? shouldnt it be (a*(b+c)+c)*d? so the same you got, without the negated variables? i cant see how your pmos is right, as you create the pmos by taking your function f(a,b,c,d) and turn it to f(!a,!b,!c,!d).
@zombiedude3474 жыл бұрын
When Negating, AND/OR also switch, in addition to the inputs being negated.
@lakshyarana675 жыл бұрын
what about that bar !!!!
@markharrisllb3 жыл бұрын
Brilliant, thank you.
@blazin7299Ай бұрын
You saved my life!
@josealmanza68403 жыл бұрын
Could you do Y=((A'+B)*C')+D'
@SANJAY-ve9ub Жыл бұрын
Can anyone explain how this entire process happens in real time?
@govinddeshmukh45715 жыл бұрын
Why do is taken upper
@BrutalGames20135 жыл бұрын
In my opinion the title should be [Inv((A+B*C)*C+D)] rather [Inv(A+B*C)*C+D], since the complete term is inverted