Make an AND Gate with 7486IC: A Digital Logic Question Answered by Jason O

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The discussion revolves around creating an AND gate using a 7486 IC, which is a quad 2-input XOR gate. Participants explore various methods, including using diodes and resistors, to achieve the desired logic function, noting the complexity of the task. There is mention of the differences in output current capabilities of the 7486, which could affect circuit design. Suggestions include using voltage dividers and buffering techniques to manipulate the XOR outputs for an AND function. Ultimately, the consensus is that while it's theoretically possible, it may be more practical to use other components or methods for creating an AND gate.
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Hi Guys,

Quick question. If all I have is a 7486IC and I want to use it to make an AND gate, how woud I hook it up?

Thanks,
Jason O
 
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I'm not getting a quick answer on my Post-It. The 74xx86 is a quad 2-input XOR gate. I've been messing with combinations of two terms, including inversion by forcing one input of the XOR high, but not getting the AND truth table so far. Have you gotten close? Another approach might be to try manipulating the equation for the XOR to see if you can get it into an AND configuration with nested combinations of XOR. There's probably a cute solution, but I'm not getting it yet. I'll try again later if I have time.
 
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Build an AND gate out of XORs... It's actually quite hard. I was going for a latch approach but it's not obivous (to me) that this will work.

I think you'll probably make the drive there and back from Radio shack before anyone figures it out.

Is that really all you have?

Perhaps you have two diodes (two bjts wired as a diode are just as good) and a resistor, or maybe two NFETs and a resistor.
 
Hah, I think I got a way. I do need a resistor or two, in addition to a 74HC86. Or I can do it without the resistor(s), but I need to connect a couple of the HC86 outputs directly together, which is not normal design practice...
 
I notice that the OP said 7486.
There was a large difference between output source and sink currents in this family.
-800ua vs 16ma.
You could wire the outputs together and not burn one out.
Not good practice, but it would probably work.
 
Hi Everyone,

I was trying to help out a classmate who was taking Digital Logic design and for some reason, they had to make an AND gat using the 7483 (assuming I unerstand what they were telling me). I tried to find the solution using the boolean algebra and by sticking weird combinations of the gates together but nothing poped out at me either.

Thanks,
Jason O
 
Isn't a 74hc83 an adder? As opposed to the 74hc86, 4 xor gates.
 
Berkeman, what is your way? Were you using a voltage divider trick?

i.e. only the sum of the drive current from two gates would be enough to pump up the voltage on resistor across on input of the third gate so that it hits the vih_min spec? if one of the two gates is low it sinks current away from this resistor thus the input of the third gate is below vil_max. Hence, AND, i.e. both input gates must be driving high for the third XOR to pass the 1. Just wire the other input of the gates to ground so that they all act as buffers.

As for the sink and source differences. I think this could be mitgated by choosing output series resistors that are large enough to limit the drive current to, say 50%, of sink current.
 
Jdo300 said:
I was trying to help out a classmate who was taking Digital Logic design and for some reason, they had to make an AND gat using the 7483 (assuming I unerstand what they were telling me).
Well now, which is it young feller? (Quiz question -- which movie is that line from?). Making an AND function with an '83 adder is a whole lot easier than making it with an '86 quad XOR!

es1 said:
Berkeman, what is your way? Were you using a voltage divider trick?
Using the XORs you can tell if the two inputs are equal, but not if they are equal and both 1 (for the AND function). I was going to bring in two gates buffering one input, and wire AND those with the equal output, or something along those lines. :rolleyes:
 
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berkeman said:
Well now, which is it young feller? (Quiz question -- which movie is that line from?). Making an AND function with an '83 adder is a whole lot easier than making it with an '86 quad XOR!Using the XORs you can tell if the two inputs are equal, but not if they are equal and both 1 (for the AND function). I was going to bring in two gates buffering one input, and wire AND those with the equal output, or something along those lines. :rolleyes:

Hi Berkeman,

That was a typo, I did mean to say 7486. But after thinking through how complex it would be to do it that way, I'm beginning to think that my friend actually got the numbers mixed up.

- Jason O
 
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berkeman said:
I googled make AND function with XOR gate, and got a number of hits. Here's one of the hits, a thread where it looks like you do need to do something weird to get it done:

http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv004.cgi?read=7254

The suggestion was to just built a DTL 'and' gate and ignore the XORs. :)
http://en.wikipedia.org/wiki/Diode-transistor_logic

They are just using the XORs as buffers for isolation and to regenerate the level.

Which, admittedly, is not too different than my voltage divider suggestion. The difference is the DTL gate can generate the correct levels for CMOS while the resistor solution generates a swing that stays stuck in the transition region.
 
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