Decoder Part 2 | GATE 2023

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GATE CSE LECTURES BY AMIT KHURANA

GATE CSE LECTURES BY AMIT KHURANA

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🏷️ TAGS ►decoder in digital electronics in hindi,decoder in hindi,decoder circuit,decoder by study extent,decoder truth table,2 to 4 decoder truth table,decoder in digital electronics,digital electronics lectures,digital electronics,डिकोडर,What is Decoder,decoder logic circuit,what is decoder in hindi,binary decoder in hindi,decoder,what is decoder explain in hindi,encoder and decoder in hindi,decoder and encoder in hindi,encoder and decoder problem in hindi

Пікірлер: 8
@adityaSingh-wk9jk
@adityaSingh-wk9jk 5 ай бұрын
Loved it sir ♥️♥️
@pawanchoure1289
@pawanchoure1289 5 ай бұрын
great😊👏👍
@shaikhoney4974
@shaikhoney4974 Ай бұрын
@huzaifam7060
@huzaifam7060 4 ай бұрын
VIDEO NOTES 1) We can make a big decoder using multiple small decoders. To make a 3x8 using 2x4, we need 2 2x4's. Now suppose the three input lines are ABC, we will add the Most Significant Bit ( A in this case) as enable pin to both 2x4's. One will have A value and the other will have A' value. The other inputs ( B & C in this case) will be inputs to both decoders simultaneously. Now finally use this data to make Block diagram first, then truth table, minimization of equations, then circuit diagram. 2) There is also another way to make a 3x8 using multiple 2x4's. The only difference here from the previous that will use one 2x1 decoder for A instead of making it input to one decoder and inverted to other decoder. This way A will work the same like earlier. 3) Now to make a 4x16 using 2x4's. We will be needing 4 2x4's. CD will be input to each decoder simultaneously. AB will pass to one additional 2x4 decoder before and it's different output lines will be enable pins for the front 2x4 decoders. Therefore, will be needing 4+1=5 2x4's to implement a 4x16. 4) Now use the same logic to know 2x4's needed for 6x64. Final 2x4 decoders will be 16. Before it there will be 4 2x4. And before it 1 decoder. Total 2x4's needed for 6x64=1+4+16= 21 5) To make 5x32 using 2x4, 8 final decoders, 2 before it, 1 even before it( make one input as 0 fixed here cuz there will some unused outputs). Total=1+2+8=11 2x4 decoders.
@guy.with.moustache
@guy.with.moustache Ай бұрын
Thanks buddy
@Thakur_Abhays
@Thakur_Abhays Жыл бұрын
28:04 i think it should be 2^m/4^k someone verify it...
@sujoykrhaldar
@sujoykrhaldar Жыл бұрын
Hi Abhay, it will be m/4^k in terms of log -> cause we have m no of outputs from ( m * 2^m ) or in terms of log ( log2 m * m). So in step -1 -> (from extream right as we scan from right to left ) - m/4 (let it be X) step -2 -> X/4 or (m/4)/4 = m/4^2 step -3 -> m/4^3 ...... so on to step k -> m/4^k and it should be =1 or
@mrinaltantubai7341
@mrinaltantubai7341 2 жыл бұрын
Sir pls 1 November se aap DSA resume karna...pls sir
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