Digital Led
Digital Led
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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!!
I sent you an email if you could please help I would be more than grateful.
you have to first reword it as 1s and 0s, or at least as a math expression.
start out with 4 switches, one each for a,b,c,d.
you need to define what "issue passes" means. I assume it means a majority, ie, 60 or more out of a total score of 100
how about
40a + 30b + 20c +10d
dividing by 10
4a + 3b + 2c + d
one way to do it is with a 10 input 1 bit adder.
(but see below for a better solution)
a switch on sets 4 inputs to a 1, otherwise they are 0
b switch on sets 3 inputs to a 1, otherwise they are 0
c switch on sets 2 inputs to a 1, otherwise they are 0
d switch on sets 1 inputs to a 1, otherwise they are 0
sum of the inputs will be a number between 0 and 10, on 4 lines.
now put those 4 lines into a digital comparator and compare it to 5. You will have a less than, equal to, or greater than outputs. Use the greater than to send to the LED.
for the 10 input 1 bit adder, you can use a ROM with 10 inputs and 4 outputs.
Another easier way is to look at the possible combinations, and feed that into a rom.
A...B...C...D...Value.. P/F
4....3...2....1...Value.. P/F (6 or higher is a pass)
0...0....0...0.......0.......F
0...0....0...1.......1.......F
0...0....1...0.......2.......F
0...0....1...1.......3.......F
0...1....0...0.......3.......F
0...1....0...1.......4.......F
0...1....1...0.......5.......F
0...1....1...1.......6.......F
1...0....0...0.......4.......F
1...0....0...1.......5.......F
1...0....1...0.......6.......F
1...0....1...1.......7.......P
1...1....0...0.......7.......P
1...1....0...1.......8.......P
1...1....1...0.......9.......P
1...1....1...1.....10.......P
So you have a 4 bit input and 1 bit output ROM, this can be done via diodes or jumpers.
truth table, from above table.
....A...B...C...D.....out
....0...0....0...0.......0
....0...0....0...1.......0
....0...0....1...0.......0
....0...0....1...1.......0
....0...1....0...0.......0
....0...1....0...1.......0
....0...1....1...0.......0
....0...1....1...1.......0
....1...0....0...0.......0
....1...0....0...1.......0
....1...0....1...0.......0
....1...0....1...1.......1
....1...1....0...0.......1
....1...1....0...1.......1
....1...1....1...0.......1
....1...1....1...1.......1
now all you need is some logic to synthesize this expression, or a ROM, or a bunch of diodes.
edit corrected for score must be 7 to pass.
You can use a LS138 to decode this.
put the A into the enable input, and the BCD signals to the 421 inputs respectively. Then the pass outputs are 3 thru 7 respectively. Run a diode to each of these 5 with the common tied to a LED and to a resistor to +5v.
.
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