Analog Easy-Peasy

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@Shivshankar-c9f 5 күн бұрын

Please keep on posting videos like this there are students like us to support from whole heart❤

@Shivshankar-c9f 5 күн бұрын

Please keep on posting videos like this there are students like us to support from whole heart

@Shivshankar-c9f 5 күн бұрын

Please keep on posting videos like this there are students like us to support from whole heart

@Shivshankar-c9f 5 күн бұрын

Please keep on posting videos like this there are students like us to support from whole heart

@Shivshankar-c9f 5 күн бұрын

Please keep on posting videos like this there are students like us to support from whole heart

@HardikJain_YT 6 күн бұрын

in last directly output is input for G since G is 1 = GH/1-GH = 1/0.5 = 2

@user-kz8di7nn9g 27 күн бұрын

For what purpose this interview was ?

@landrover8135 Ай бұрын

When i try to solve the small signal gain around the switching threshold, i get the small siganl gain is {-(gmn+gmp)*{1-[1/(R*(gmn+gmp))]}}/(G+gmn+gmp)...If (gmn+gmp)*R < 1 , then this acts as a buffer otherwise as inverter...is my analysis correct ?

@Zerjditywnskg455 2 ай бұрын

thanks for sharing...there is no ground?

@AllenSA_airohaSun 2 ай бұрын

Thank you Sir

@Arjun_Choudahry 3 ай бұрын

sir please upload video regularly . It's really helpful for us🙏🙏🙏

@ROCKPWINCE 4 ай бұрын

Post the loop gain <1 the response will follow the open loop characteristics as if the feedback is broken hence the system does not blows up

@user-qc8pi4gt3o 4 ай бұрын

Hi, request you to kindly upload videos on a regular basis . It could be tremendously helpful for interview candidates. All the very best & thanks for helping a larger audience out there . #skyisthelimit

@analogeasy-peasy7559 4 ай бұрын

I will try my best

@srijandwivedi294 4 ай бұрын

After 9 months sir.... :/

@ebadurrahmankhan9033 4 ай бұрын

Please upload video regularly

@StandforAjdetray 4 ай бұрын

Much appreciated sir Very interesting and waiting next session

@bhavikparekh2044 4 ай бұрын

Its basically PWM modulation on a second order LPF, if the duty cycle is 50% then answer is half of vdd which is 2.5. If you change the duty cycle, then output will change accordingly

@HardikJain_YT 4 ай бұрын

gr888

@IcRDlayout 5 ай бұрын

Thanks 😊

@IcRDlayout 5 ай бұрын

Why not add two switches,one is above capacitor 1f and the other is above 10f capacitor??😊 Time is t1 and t2😊 First t1 closed and then or First t2 closed and then ....

@sumanbaur3021 5 ай бұрын

Along with this topics is there any necessity to have detailed knowledge of Digital electronics?

@analogeasy-peasy7559 5 ай бұрын

Basics is all you need like inverter understanding, set upholdtime and all

@bangaloriansvlog 5 ай бұрын

The same question was asked to me during my internship interview at Synopsys.

@TRUELiGHTERS 6 ай бұрын

the man in the lower side died with his eyes open

@saurabhtrivedi6181 5 ай бұрын

@anushatadamari3185 6 ай бұрын

Please do come up with more videos..

@user-mq1xc1rr2u 6 ай бұрын

IS Vx at -2 when t<0? It could be undefined. +2V is with respect to Vx node and not gnd. I think its inappropriate to refer Vx at -2V when t<0.

@Gyan.250 7 ай бұрын

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.

@Gyan.250 7 ай бұрын

If voltage supply is aplied then why capacitor is discharging ? The current should be supplied by the voltage source.

@zacharyscott8083 9 ай бұрын

8:28 this type of plot is called “Hysteresis” Basically, when an op amp has positive feedback, it exhibits hysteresis

@user-ld1bk2bk7l 10 ай бұрын

What about voltage across resistence can we neglect

@user-pm2rz7tv3o 11 ай бұрын

there is no gate current.what about that

@saadqayyum2148 11 ай бұрын

6:08 1/Ron is the derivative of Id wr.t. Vds and not simply the ratio of Id to Vds.

@KK-bj3mq Жыл бұрын

Very interesting, Waiting for next session

@kaustavguharoy4532 Жыл бұрын

for each of the two modes the time constant of the circuit remains same , Is it?

@analogeasy-peasy7559 Жыл бұрын

So as discussed in the video make an R and C model of the circuit. If the R charging and discharging is same then Yes. Clearly here the circuit gets changes so it should not be the case.

@mirzaamaan6937 Жыл бұрын

Please bring more such videos these are super useful

@maylok77 Жыл бұрын

please answer me why they add capacitor at the out?

@analogeasy-peasy7559 Жыл бұрын

Adding the capacitor makes the output voltage stable. The more the capacitor the more stable will be the voltage. In real designs as well we put dynamic capacitors inorder that our circuit just not start reacting to small noises and all. Would suggest you to see RC Network questions discussed here in the channel.

@IcRDlayout 5 ай бұрын

CL is placed to the Vo 😊

@amitbar7266 Жыл бұрын

A CMOS inverter with resistive feedback where transistors are self-biased in strong inversion is designed and optimized for low NF and high IIP3 in LNA

@aayushADE334 Жыл бұрын

what i can think is that since no current flows inside the mosfet so output is shorted to input and both the mos comes in sat as vds= vgs which results in vout = vdd/2 as per inverter characterstics .

@mohdkhairizulkalnain279 Жыл бұрын

Hi there thanks for the videos! Keep em coming! Just out of curiosity, may I know why the discharge and charging time constants are not equal initially?

@amitjana8172 Жыл бұрын

very nice video Dada :)

@amitjana8172 Жыл бұрын

Here if R tends to infinite then there will be no feedback, if R has some finite value then the circuit will acts as a negative feedback circuit. If we set the input to 0 and break the loop and increase the voltage by some +ve amount then NMOS will be on and PMOS will be off. So as it is a CS amplifier ckt then there will be 180 degree phase shift in the op, so for increase in input output decrease , so it is a negative feedback ckt, for finite value of R. Am i right bhaiya?

@amitjana8172 Жыл бұрын

Bhaiya could you please make a video of buffer related questions, like calculation of 3 dB frequency, or kindly refer any resources from where I can practice this kind of problems.

@amitbar7266 Жыл бұрын

Yes, - ve feedback,,and actually finite value of R will modify the overall Gm and overall Rout of the amplifier...

@analogeasy-peasy7559 Жыл бұрын

your analysis is right actually but just think about this- the feedback is the one which effects the input ( alter it change it) but in this case it is not possible since we have a forced input. This kind of feedback os called feed forward feedback.

@amitjana8172 Жыл бұрын

Thanks bhaiya

@AnalogABC Жыл бұрын

positive feed back ? , and ig here application of circuits might be other than inverter ig gain ? idk

@analogeasy-peasy7559 Жыл бұрын

it's actually a feed forward feedback

@AnalogABC Жыл бұрын

bro videos are really good pls dont stop making videos

@saurabhtrivedi6181 Жыл бұрын

Thanks man :)

@ntdglobal2510 Жыл бұрын

I have some questions. What is the effect of rise/false time of input on transient response? When can CGD be ignored? The resistor will be a feedforward path, not a feedback path if the voltage source is ideal.

@saurabhtrivedi6181 Жыл бұрын

Hii, so if your input is having a higher rise time or fall time then depending on the values of load capacitor, R and Ron( p or n) the output will get time to settle down to its final value ( which could be deduced by the formula mentioned in the video ). If the time constant is lower it will follow the input else will slow it down. If you have steep input with lower rise and fall time then also again depending on the loading it will follow or slow down the input. Now this question gets better if we have clock as an input to it then if the loading is higher and the input steep in nature than probably this will result in lower voltage swings at the output. Yep, you are right that this can’t be stated as a feedback until we have a forced supply at its input. Kudos on that🎉.

@mrx1167 Жыл бұрын

Can yu give little brief how do yu define it's not feedback, feedforward.

@saurabhtrivedi6181 7 ай бұрын

@@mrx1167 feedback is given directly to the input but since input is an ideal source so it won't be possible for it to be changed but this feedback is also connected to the output which is not tied to any ideal source hence known as feedforward not feedback ^_^

@thesailaja5941 Жыл бұрын

Everything is great guys why can't you start teaching❤

@mandlikprajwal466 Жыл бұрын

Why you said 2.5 volt Won't there be some voltage drop in resistor itself. Your kvl will fail if this happens

@ytaccount9420 Жыл бұрын

Sir please post more interview questions!!!

@analogeasy-peasy7559 Жыл бұрын

We are a bit late, but finally, there is one new video

@hemantj675 Жыл бұрын

Isn't it a second order circuit?

@amitbar7266 Жыл бұрын

I think the discharge and charging current are different so in negative half cycle the Vx will be less than 0.5 volts Please correct me if I am wrong !!

@hakuna3132 Жыл бұрын

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

@analogeasy-peasy7559 Жыл бұрын

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

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