digital-logic
Right now, I've got:
(x1 and y1)
(x2 and y2)
(x3 and y3)
(x4 and y4)
So the outputs from these gates will give me 1 when the bits from both numbers are the same. The issue here is that I will get positive results when they are the same and when each bit is different from each other bit (because 0,0 will still give a 1 in an and gate).
I'm probably missing something very obvious here.
Thanks
For the logic that you descibed you need a 32 input OR gate followed by an inverter. You do not want the XOR that you show in your diagram.
If you cannot get an OR gate with 32 inputs then work with some smaller sizes. For example, if all you can get are 4-input OR gates then it will take 8 of these gates to accept the 32 total inputs. Next use a second bank of two more of the 4-input gates to pickup the 8 outputs of the first stage. Finally use a 3rd stage of one final gate to accept the two outputs of the 2nd stage. The two remaining inputs of that 3rd stage gate can be tied to GND. The output of that 3rd stage gate connects to your inverter to produce the final output.
The concept looks sound.
The main thing I see is that your 5.6K resistor is way too high to guarantee a low level for a 'LS04. The 100 ohm resistors you used elsewhere are okay (but they'll get warm if the switch is held).
Suggest you use ~150 ohms for the pull down resistors and maybe 4.7K for the LED resistors.
It would be better to use a spare inverter for each of the LEDs (so the output is low for LED 'on') and connect the LED + resistor between Vcc and the output. That's because you're loading the outputs with the LED and TTL can't source current that well. Then you can use 470 ohms or even less for the LEDs and they'll be much brighter.
Best Answer
4 x Xor gate... and 4 input nor gate
Xor each pair of inputs then use a 4 input nor gate...
Xor gate gives 0 when both inputs are the same.
a 4 input nor gate gives a logic 1 if all inputs are 0, and 0 if any input is 1.
or
Not((X1 XOR Y1) OR (X2 XOR Y2) OR (X3 XOR Y3) OR (X4 XOR Y4))