Question

Results Circuit (a) Circuit (b) Circuit (c) Circuit (d) B F Table 10-2: Truth Table for Circuit in Figure 10-2 2. D Symbol 4 Table 10-3: Results of the 7-segment display using circuit shown in Figure 10-3

2. The circuits in Figure 10- 2 (a) and (d) constitute a half-adder. Explain why this is so.

3. Logic outputs are said to be “asserted” high or low depending upon which level (high or low) is “active” (that is, produces the desired effect. Look at your truth table for the 7447 (Table 10- 3). Are its outputs asserted high or asserted low? What about the inputs to the 7447?

4. How do you explain the strange symbols that are displayed on the 7 segment display when the numbers 1010, 1011, 1100, 1101, 1110, and 1111 are applied to the input terminals DCBA?

5. Which is the most significant bit (A, B, C, or D) of the inputs to the 7447 and which is the least significant bit? Could that be reversed? If not explain why not. If so, explain how

Answer 1

Answers:

2. The truth table of a Half-adder is

A B    Sum    Carry_out

0 0 0    0

0 1 1    0

1 0 1    0

1 1 0    1

You would notice that Carry_out is nothing but ANDing of A and B. Therefore the circuit in 10.2b) computes the carry_cout

You would also notice that Sum is XORing of A and B. This is done by circuit of 10.2d).

So circuits of 10.2b and d together constitute a half-adder.

Answer 3:

As per the truth table 10.3 the outputs of 7447 are asserted high. That is when a segment of the 7-segment display needs to glow, the corresponding output from 7447 goes high. The inputs are also asserted high (BCD 0 is input as 0000, BCD 1 is input as 0001 and so on).

Having said this, if you check out the truth-table of 7447 in its datasheet (http://html.alldatasheet.com/html-pdf/80216/NSC/7447/187/4/7447.html) you will find that outputs are asserted low. So I think the truth-table in 10.3 is actually a complement of actual 7447 truth table.

Answer 4:

These six symbols correspond to inputs 1010, 1011, 1100, 1101, 1110 and 1111. In hex these are A, B, C , D, E and F respectively. These letters cannot be adequately displayed using just the seven-segments.

Answer 5:

D is the most significant input bit, A is the least-significant input bit. Yes, these can be reversed by inverting DCBA to /D /C /B /A before inputting them to 7447. So you will need to use four additional NOT gates.

Hope this helps.

PS: There seems to be another mistake in truth-table in 10.3. The output corresponding to DCBA = 1000 should be all high. This table shows e output as 0, which is incorrect.