 Today's session is 7-segment decoder. At the end of this session, student will be able to design a 7-segment decoder using 7447 IC. A 7-segment display is used to display decimal digit from 0 to 9 and hexadecimal numerals from 0 to F. A BCD to 7-segment decoder accepts a decimal digit in BCD and generate the corresponding 7-segment code. For example, in this slide we can see display of 7-segment decoder that is 757 it is displaying the number. As it is having 7-segment ABCDEFG and with the help of that format we can display any BCD number. For example, for displaying 0 only G will be off and for displaying 1 only B and C-segment will be on. Like this in 2 we are having AB, G, E and D. Now, while constructing this 7-segment display we need display circuit which will convert ABCD BCD number into the 7-segment. So here in 7447 IC we are having ABCD is BCD number where it is converted into the format of ABCDEFG that is 7-segment. That 7-segment will be provided to the display and that will display any BCD number. For this first step will be for truth table. In truth table we are having BCD number then output in the format of 7-segment and display. So in BCD number we are having ABCD and that number will be display over there. Now for 0 number all ABCD are 0 but in 7-segment only G is going to be off so ABCDEF all you can see here all are 1. For 1 as I said 1 we want only B and C should be 1 so B and C is 1 and remaining all segments are 0. So for 0 to 9 all segments are set at what number that should be on and what number that should be off. Accordingly we will go for designing its circuit. So this inputs will put in the first K-map. So for A we need that A-segment should be on for 0, 2, 3, 5, 7, 8 and 9. So accordingly we have put that 0th value then 2 value 3 value 5, 7, 8 and 9. So total 10 numbers are there and remaining 5 numbers we have put don't care. After simplification we are getting A-segment is having circuit ray A plus CD plus BD plus B-bar D-bar. So we have constructed this circuit for them. For B we are having simplified K-map is B-bar plus C-bar D-bar plus CD. So this is the circuit diagram. And as we have seen that C-segment in the truth table we are having all 1 only when 2 number is there that time it is 0. So accordingly at 2 place it is 0 and all other place that number segment is on. So after simplification only 3 input NAND gate is there and inputs are BCD. So for D, E and F again using that same criteria we have formed this K-map and the circuitry. This is for G. Now we will see one example how 7-segment decoder works and in that one segment how it works for all BCD numbers. So here ABCD that is the BCD number it will go through the 0 to 9. Then we will feel that K-map here we can see that at what position it should be on we have put 1 and other position we have put 0s. And wherever there is no 0 and 1 we are having don't care. So after simplification of this segment E we are having grouping format. So here there is no 8 group so we are doing the group of 4. So this is the group of 4 1 1 don't care and don't care. So this all a bar b bar a bar b a b and a b bar will be cancelled to each other. Only it will remain C and D bar so we are wrote here C and D bar. And remaining ones will cover into this corner group. So it will be giving us output is simplification output is b bar d bar. So for E we are having circuitry b bar d bar plus C d bar. So we are forming this circuitry for this segment E. So here b bar and d bar for this NAND gate we are having b bar through this inverter and d bar through the second inverter. And for C d bar we are having C directly from BCD input and D is coming with this inverter. Now when it will be on and when it will be off we will see one example. So when we are displaying number 2 that time E segment should be on. So BCD number 2 0 0 1 0. So we will see that how that will be input. So 0 is inverted it will be 1 then next 0 is inverted it will be 1. So b d bar is equal to 1 and C is 1 and D bar is 1. So output of this will be 1 and 1 so oring is 1. So automatically when 2 BCD input is there our output E will be on. In this second example when E segment should be off we will see that the BCD input is 0 0 1. So input 1 it is inverted so output is 0 here 0 is input from BCD to C so output is 0. So that both inputs 0 are going to be output 0 so E segment will be off. So have you seen any 7 segment application pause for 2 seconds and think about this. So we are having many examples of this 7 segment application is digital calculator, electronic meters, then digital clocks, odometer, clock radios and bank token displays. So these are references thank you.