then for the next 8 combinations, the W line is 1, which becomes 0 inside the 1st decoder(ACTIVE LOW) and the 1st decoder turns off, but it goes through a NOT gate and then ACTIVE LOW enable port of 2nd decoder, becomes a 1 and activates the next outputs 8 to 15. In this lecture, we are implementing 2:4 Decoder using verilog HDL.Channel Playlist (ALL). How? Because for the first 8 combinations, the W bit is 0, so it is a 1 for the first decoder, and enable line is on(ACTIVE LOW), but it goes through a NOT GATE and then to the ACTIVE LOW enable port of the second decoder, so it remains 0, so the second decoder doesn't activate. This lecture is part of Verilog Tutorial. For the values 0000 to 0111 ,the first decoder will turn on giving the decoded outputs 0 to 7, and for 1000 to 1111, the second decoder will turn on, giving decoded output 8 to 15. 2 to 4 decoder realization using NAND gates only. So, there are now 4 selection inputs i.e W,X,Y,Z. Write Verilog program for the following combinational design along with test bench to verify the design: a. 4.23 Draw the logic diagram of a 2-to-4-line decoder using (a) NOR gates only and. the three selection lines of each decoders are connected together as common line(X,Y,Z), the enable lines are ACTIVE LOW, they are also connected together with a common line W, but the second one having a NOT gate connected within. (b) List the truth table with 16 binary combinations of the four input. The two squares are two 3x8 decoders with enable lines.
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