Dual Inverter
Dual Inverter
How would I go about creating a digital circuit that turns on a led if the issue passes and turns off if doesn?
The vote has to be weighted if someone votes yes they vote all of their shares and vice versa. Mr. A has 40 shares, Mr. B. has 30 shares, Mr. C has 20 shares and Mr. D has 10 shares. How would I go about creating a circuit on a protoboard that will do what is specified. Please help. I'm stuck!!
And by the way the equipment i have to work with is on a protoboard using NAND gates, NOR gates, AND gates, NOT gates OR gates, inverters, BCD-to-7 segment decoder/driver Dual D flip-flop XOR gate, Expandable 3 to 8 decoder 8x1 multiplexer, and an arithmetic logic unit. Thanks again.
Since you need a proportion, let's divide by 10, so A=4, B=3, C=2, D=1. Now, let yes=1 and no=0. You also need switches to input the votes.
There are lots of ways to do this problem. Here are my fav's:
Solution 1:
First part of problem is to add the votes together. So, how can you add them together? Easiest to use the ALU, put B into one input port (0011), and A, C, and D on the other input port (0100 + 0010 + 0001 = 0ACD). You can also make adder from gates, won't need many for such a simple adder.
Second part is to compare to minimum passing. Is 50% passing or not? Just use the gates to compare the ALU or adder output to (value) 5 or 6 to determine if passed. For 60%, 6,7,8,9,10 are passing: (4's bit AND 2's bit) OR 8's bit, so that. For 50%, (4's AND 2's bits) OR (4's AND 1's bits) OR 8's bit.
In either case, drive LED with the final gate output or a buffer or inverter with adequate current capacity. This is probably the cheapest way to do it.
Solution 2:
Assume 60% means to pass. From basic arithmetic, combinations that pass are:
light = A(B+C) + BCD
Implement this in gates.
Sol'n 3:
Use 2 decoders, put ABCD into them, OR the outputs that are true for passing vote together.
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