decoderlogic-gates
I have an idea of how to get started, but not sure where to go from there.
Since I am using two 3-8 decoders to develop a 4-to-16 decoder, I want to use 4 inputs out of the two 3-8 decoders. So I'll use all three of the first and the first of the second, and connect the last two inputs to ground, since they won't be used.
Any pointers on where to go from here are appreciated.
Your approach is correct.
This is probably an error, or oversight, in the textbook: the authors forgot to allow for selecting the final decoder, which is what your 9th decoder does.
There's not a standard formula. First, you have to stack up enough decoders to give you the number of output pins you need. Then, you have to "tree up" enough additional decoders to select the output-layer decoders.
That's exactly the way it's done, as long as you don't want to turn off all the LEDs at any time. If so
simulate this circuit – Schematic created using CircuitLab
will allow a master inhibit to override the select line.
Best Answer
you have to design a 4x16 decoder using two 3x8 decoders. here is the schematic that may help you.
simulate this circuit – Schematic created using CircuitLab
the two squares are two 3x8 decoders with enable lines. 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. So, there are now 4 selection inputs i.e W,X,Y,Z. 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. 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. 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. so you get a total 16 outputs (0-15)
If you use two 3x8 decoders you need not use 16 AND gates, and If you want to use AND gates only, then think about the internal circuitry of a decoder, think of this diagram , and you can easily construct using AND gates.