AND-OR Gate
#1
Posted 19 June 2012 - 02:01 PM
#2
Posted 19 June 2012 - 02:08 PM
There is no image attached (?)See they attached image.
#3
Posted 19 June 2012 - 02:25 PM
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#4
Posted 19 June 2012 - 02:39 PM
I think I got the idea, although the circuit will not work as shown (there is nothing connected to the gate inputs which are wired together). You'd probably want to select some of the numerous 74xx series logic ICs, there are AND+OR and NAND+NOR gates with different number of inputs, which can be combined together, add inverter if needed (for example, NAND or NOR gate with all inputs wired together is inverter too).Sorry, here you go. Not sure if I drew it right, but hopefully you get the idea.
To control the pumps or valves, you'd probably need some kind of power switch, if you need to control them directly - you could use a relay, solid state relay, power MOSFETs etc. It depends on the device characteristics - voltage, current, etc.
#5
Posted 19 June 2012 - 03:05 PM
Apply De Morgan's theorem and you can create a logically equivalent circuit using just one sort of gate.
Nak.
#6
Posted 19 June 2012 - 05:59 PM
#7
Posted 19 June 2012 - 06:07 PM
#8
Posted 19 June 2012 - 06:15 PM
#9
Posted 19 June 2012 - 06:59 PM
So this is all good stuff but not being a hardware guy, I am not sure how how I am supposed to get "( A and B ) or C" out of this. Can I link to NAND gates together?
Yes you would effectivly link the output of a nand to the input of another nand until you built the boolean expression you wanted. Or you could use the device suggested below, or just order a and gate and an or gate, you typically get 4 devices per chip so would have no problems creating your circuit.
Nak.
#10
Posted 19 June 2012 - 07:38 PM
You can make anything from NAND gates, and as Nak said, the nice thing is they come in fours.
I am attaching a badly drawn diagram of connecting up an AND - OR circuit using NAND gates. It needs three out of the four that come in a 74xxx00.
The logic is as follows:
Normal AND-OR D = A and B E = C (E is just in the table to make H a bit clearer) H = D or E A B C D E H 0 0 0 0 0 0 0 0 1 0 1 1 0 1 0 0 0 0 0 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 1 1 1 1 0 1 0 1 1 1 1 1 1 1
NAND version F = A nand B G = 1 nand C H = F nand G A B C 1 F G H 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 0 1 1 0 0 1 1 1 0 1 0 1 1 1 0 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1
In the diagram, I have shown the two logic diagrams, how the insides of the 74xxx00 are laid out, and how to connect it up.
Note that the inputs to the unused gate (pins 1 and 2) and tied high, this stops the gate floating.
Hope this makes sense - Paul
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#11
Posted 01 July 2012 - 02:39 PM
#12
Posted 01 July 2012 - 02:41 PM
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#13
Posted 01 July 2012 - 04:04 PM
#14
Posted 01 July 2012 - 07:08 PM
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#15
Posted 01 July 2012 - 07:28 PM
#16
Posted 01 July 2012 - 08:44 PM
#17
Posted 03 July 2012 - 12:01 PM
There is definitely something wrong.
You said before it would not turn off "how ever my Led seems to always stay on", did you change anything so that it goes off now?
What is the part number of the chip you have, I can only read the start CD74...
Is there a notch at the top left of the picture? (Notch is at the Pin 1 end of the chip)
Might be worth driving the LED from pin 6 instead of pin 8 so you can test the AB inputs are being NAND'd correctly. (Column F in the table a posted a while back.)
I am using chip CD74AC00E.
If I drive the LED from pin 6, it is off when I have both pins positive.
#18
Posted 03 July 2012 - 04:45 PM
Maybe there is a problem with the last NAND gate's input from pin 6.
Instead of taking the output from pin 6 to pin 9, can you try using a wire on pin 9 and taking that high and low to see what happens when C is also low. If it does not behave, the gate at pins 8,9,10 maybe damaged, try using the gate at pins 1,2,3 instead.
This is the datasheet, I can't see anything obviously wrong. The gates are all laid out as in the HC(T) part.
Don't give up, and remember its supposed to be fun!
Paul
#19
Posted 03 July 2012 - 04:57 PM
#20
Posted 03 July 2012 - 05:35 PM
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