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[SOLVED] Replacing 5 to 32 line decoder with four 3 to 8 line decoder???

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moonnightingale

moonnightingale

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I am atatching the Pic.
Kindly tell me when we are having 5 to 32 decoder, we are having all 32 outputs as unique outputs
but when we are using 3 to 8 decoders, the outputs are not unique. from figure u can see that D0 and D8 will be same
similarly D15 and D23 will be same. Each output is repeated four times.
So how can we use it practically.
 

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Each 3 to 8 decoder outputs are not unique, however the E input to the 3 to 8 decoder means that only one decoder is enabled (active) at any one time. There is no case when more than one 3 to 8 decoder is active. Thus signals A4 and A5 determine which 3 to 8 decoder is active correctly decoding the 32 lines.
 

Yes exactly that i know that output of 2 to 4 decoder acts as enable.
I just want to ask that in practical world can we replace 5 to 32 decoder with 3 to 8 as u also mentioned, its outputs are not unique.
So practically i think it has no use for implementation.
 

Your input is 5 bits from 00000 to 11111 which is equal to decimal value 0-31 (32 different values), so when you have a 5 bit line decoder the result will be 32 different output lines.

In your schematic consider it like a number from 0 to 7 (A0,A1,A2) and 2 more bits (A3,A4) as carry

when A4=0 and A3=0 output is (0*8) + (A2,A1,A0)
when A4=0 and A3=1 output is (1*8) + (A2,A1,A0)
when A4=1 and A3=0 output is (2*8) + (A2,A1,A0)
when A4=1 and A3=1 output is (3*8) + (A2,A1,A0)

Alex
 
The one is as practical in the real world as the other. All we are talking about is signal flow. Whether we are talking 5-to-32 or 3-to-8, each path IS UNIQUE. Try simulating the circuits: **broken link removed**.
 

alexan_e said:
Your input is 5 bits from 00000 to 11111 which is equal to decimal value 0-31 (32 different values), so when you have a 5 bit line decoder the result will be 32 different output lines.

In your schematic consider it like a number from 0 to 7 (A0,A1,A2) and 2 more bits (A3,A4) as carry

when A4=0 and A3=0 output is (0*8) + (A2,A1,A0)
when A4=0 and A3=1 output is (1*8) + (A2,A1,A0)
when A4=1 and A3=0 output is (2*8) + (A2,A1,A0)
when A4=1 and A3=1 output is (3*8) + (A2,A1,A0)

Alex
Click to expand...

Thanks alexan
ur explanation is really very good andnowi have understood
 

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