Ic74154 decoder generate active low o/p i.e for example if input is binary 5 , 4 becomes low. I want a circuit which will give output when i/p is binary 5 then o/p is low for 0 to 4.It is possible by using very large no. of FET but it is very costly.
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\$\begingroup\$ Subtract 1 and OR the two together, unless the first output is all 0. \$\endgroup\$– Ignacio Vazquez-AbramsCommented Jan 3, 2015 at 12:20
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As you don't mention the values 6 and 7 (110, 111) you can simplify this problem to a simple AND gate (2^2 AND 2^0). Using a very inexpensive 4011 NAND gate gives the circuit below.
In view of Michael's comment (the OQ not being clear) the 16 outputs could be converted using three 7404 hex inverters.
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\$\begingroup\$ Similar comment here as on the other answer. I believe that the OP wants to be able to see the input vary even up to 6 or 7 and likely also include all the lower values and get the outputs to respond as a bar graph type display decoder would work. \$\endgroup\$ Commented Jan 3, 2015 at 16:26
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\$\begingroup\$ @MichaelKaras Thanks for that. I've added a solution that should cover that possibility. \$\endgroup\$ Commented Jan 3, 2015 at 19:38