Texas RC Question, Analog Interview

[@analogeasy-peasy7559]

@ 8:30 I have said the black wavefoem to be the input which clearly is by mistake. As stated in the video, the input is varying from Vp and 0 as a pulse.

[@anirbanghosh2577]

you can also discuss what if the input goes to -Vp rather than 0V

[@pratikbharbat9451]

U people are doing some next level things n making engineers life easy . Thnx for sharing experiences

[@analogeasy-peasy7559]

So nice of you

[@tomcat1184]

wow, wonderful explanatin, thanks !!

[@praveenkumarmaddeti6668]

At the end the input signal was taken as symmetric square wave instead of pulse signal. Apart from this.. It was nice and brief explanation..

[@saurabhtrivedi6181]

Pulse is a symmetric square wave right. You can confirm this from the Cadence Virtuoso tool bro.

[@praveenkumarmaddeti6668]

My point is he analysed the circuit for pulse input but in the diagram he has written symmetrical square wave..

[@saurabhtrivedi6181]

@@praveenkumarmaddeti6668 yeah, but the pulse input one time period is a square wave right!.. That's is what being analysed over here. The same output will be repeated for all the cycles.

[@praveenkumarmaddeti6668]

He analysed it for input that changes between 0 and Vp. But in the output waveform he has shown the input waveform that changes between -Vp and Vp.

[@saurabhtrivedi6181]

@@praveenkumarmaddeti6668 please see the last part where the two capacitor comes in parallel. They will come parallel only if input voltage is zero right!

[@shashankmourya4366]

p.s...my voice got disconnected due to bad network😐

[@jashanjain8405]

Your presence is enough 😌

[@mai-vv3yo]

Appreciate your work and effort. Its really helpfull 👍🙏I request you to Please do solve some questions on rc circuit with diodes

[@analogeasy-peasy7559]

Please mail some if possible.

[@jashanjain8405]

Well explained bro!! Thanks for these kind of series✌🏻😍

[@analogeasy-peasy7559]

Glad you liked it

[@shashankmourya4366]

We try🙂

[@Gyan.250]

Why in the negative cycle of input the out is not starting from - V. They all are parallel voltage should be equal. Why are sharing it between two cap.

[@Zerjditywnskg455]

thanks for sharing...there is no ground?

[@amitjana8172]

very nice video Dada :)

[@MIKL7701]

Thank you very much sir for the great video, at 4:49 you state that vo is v/2, I would appreciate an explanation why?

[@prashanthr3776]

Apply voltage divider rule for parallel capacitor. When input is =1. Solve nodal equation You’ll arrive with vout= (c.c)vin/((c+c)c) = vin/2

[@mirzaamaan6937]

Bro if this same circuit is solved by laplace then the whole scenario is different at infinity output will be vin/2

[@JasonXavier-rk6kl]

Nice one 👍

[@analogeasy-peasy7559]

Thank you! Cheers!

[@roswaldfrancis463]

A thevenin equivalent: vin/2-2C-R,hpf

[@mrinalkantidebnath3046]

Can we solve this using laplace transform.

[@hakuna3132]

I was shortlisted for fit challenge texas instrument 2022,interview went good but still I didn't get through the technical round😕

[@analogeasy-peasy7559]

Happens man! Don't worry if you are doing your job honestly then something good must be waiting for you :) BTW what is FIT? written test I guess

[@jashanjain8405]

Shashank 🤜🏼always rocking 😎

[@shashankmourya4366]

Learned from the best 😜

[@nidhigupta8970]

in last part of video; it says when input is zero initially it will got -v/2 ; bt according to me it should go -V becasue both cap connected in parallel ; correct me if i am missing something?

[@biprobhadra1102]

No it should be -V/2 as the capacitor will instantly divide the charge among themselves as the charge should be conserved after that then it will discharge through R.

[@anuragmishra3227]

@@biprobhadra1102 Shouldn't the current across resistor be infinite for it to instantly change its voltage?

[@BALA-jw7nd]

Bro im having a small doubt. At t=0 the moment when applying sudden change in voltage 0-V volts the capacitors will be short circuited and the voltage at the second cap i.e,.will be zero right(in waveform at t=0,Vout=0)..But why V/2..can u clear it.

[@analogeasy-peasy7559]

Can you please see the video named " Most Important RC Circuit Concept" in this channel itself.

[@amitjana8172]

For the voltage of a capacitor to jump we need infinite current. At t=0+ infinite amount of current flowing through the capacitors. Here the value of both the capacitor is same. If we apply voltage division rule of the capacitor then the voltage across each capacitor will be V/2 at t=0+.

[@arslanbhatti9771]

Hi, Thanks for the review. I had a question about the Vout(t=INF).. Will it be equal to 0 because capacitor discharging into C1 (the first capacitor) or will it be a capacitive divider like in your previous video .. in this case since C's are equal Vout(t=INF) = Vin/2 ? Appreciate your help. You are a great teacher.

[@analogeasy-peasy7559]

It (Vout(t=INF)) should be zero right as i have mentioned in the video. Thanks for the kind words ☺️

[@arslanbhatti9771]

@@analogeasy-peasy7559 appreciate the reply. Can you explain again in words why it's 0 at V(t=INF)?

[@tanmayvadhera4250]

@@arslanbhatti9771 at steady state the capacitor connected to source will block the current. This means that the drop across the resistor , 'R' will drop to 0 i.e. Vout will drop to 0

[@biprobhadra1102]

@@arslanbhatti9771 at t=infinity means at steady state omega=0 for a step voltage Capacitor will behave as a Open Circuit C1=C itself and in C2 || R the Resistor will dominant here since capacitor is at open ckt so o/p terminal equal to 0. As C2 has discharged all its voltage across R terminal and C1 has hold its voltage.

[@ShubhamJaiswal-v9q]

great job , but what is this sound in every ten sec 🤖🗣🗣

[@TRUELiGHTERS]

the man in the lower side died with his eyes open

[@saurabhtrivedi6181]