Dataflow Modeling | #12 | Verilog in Hindi | VLSI Point

[@sagarvishwakarma5727]

In whole KZbin I watched this types of video in vlsi and I am thankful to mam which provides in Hindi languages.

[@harshdeshwal18]

By your channel I am learning verilog easily.Please don't stop uploading videos and upload frequently... thank you very much..🤗☺️😊

[@vlsipoint]

Thank you so much Harsh, videos will be coming back to back. Stay connected, keep supporting and share with your friends as well✌✌

[@Coursekro]

You have given very good information, we appreciate your talent. Thank you very much. ❤️❤️❤️❤️❤️❤️❤️

[@vlsipoint]

Thank you so much! Stay connected ✌

[@Yashwant_Chavan]

Very thank you mam❤

[@nikitagalar5638]

Thank you ❤️

[@pratapbasia4975]

Mam test bench ke liye video banao

[@vlsipoint]

For testbench, system verilog videos will be uploaded soon.....

[@yogeshrathour6705]

Mam 2 to 4 decoder coding nhi samajh aya

[@aryasingh1861]

Hello Ma'am After solving for example number 2 (on paper and according to your logic) I am getting D3 as "0" and D2,D1,D0 = 1 if I gave AB= 00 for active low Enable(that you have considered) Again for AB= 01, I am getting D2 as "0" and others as 1 and similarly for D1 and D0 also. But according to my knowledge - if AB = 00, D0 - high AB = 01, D1 - high AB = 10, D2 - high AB = 11, D3 - high . but here I am getting them as low i.e. "0" (according to the assign statements given in the video). Do let me know where I am doing wrong.

[@moulikchouhan7958]

you are right but in the given "example 2" you can observe that it is for active low signals. As we know active high decoder will generate minterms and this given example can be used to generate maxterm. Hope so you got it. You have any doubt please feel free to ask.

[@vlsipoint]

Yes Arya, moulik is correct, check it once. You can ask these small doubts in our telegram group, there you'll get quick reply.

[@chandrikachowdary6155]

Mam I didn't understand decoder code

[@Yashwant_Chavan]

it just like simple 2:4 decoder but here mam usese for active low, means you just consider like this without not operator - output [0:3] D; assign D[0]=(~A & ~B & E), D[1]=(~A & B & E), D[2]=(A & ~B & E), D[3]=(A & B & E); endmodule My way is Right mam?

[@ayushbhargava4998]

Mam why u used negation sign in example 2 we can write the code without it also?? Also you have assigned output [3,:0] D...why u used comma its [3:0] i guess

[@SandeepGupta-re2cn]

It's a long time to upload the next video ma'am....please can upload a video too early they might be help me.

[@vlsipoint]

Sure Sandeep, videos will be uploaded frequently.

[@deepanshiroy3726]

Just reading the ppt ... sorry to say

[@stemaryan9470]

Hello sis please ak ak video dark web ke uper lado 🙏🏼 next video I riquest you ❤️👍🙏🏼

[@Rohitsahu0718]

2-4 decoder explain again

[@greeneyesboy7534]

App toh reading karthe ho yr reading hum bhi kar sakty Hy .