# AND OR logic gates

1. Oct 7, 2008

2. Oct 7, 2008

### nottheone

My electronics are pretty rusty so I could be wrong but,

For the AND assume that the inputs must be either high or low (not really true). Assume 6V is high. The truth table for an AND gate is the output will only be high if both inputs are high. If either or both inputs are low (ground) it will forward bias their respective diodes and effectively put a low resistance in parallel with the 4.7k pull-up resistor and ground the output (low). Putting a high on an input in this circuit will do nothing because the diode will be reverse biased. In this circuit no inputs will also leave the output high. The main point is that a low on either input will ground the output.

The OR is a little different because there is no connection to power. The truth table for the OR is that the output will only be high if either or both inputs are high. Because of the direction of the diodes the output will go high even if one input is low. With one input low and one input high the only current path is through the resistor because the diode on the low will be reverse biased and block current. You can think of a reverse biased diode as an extremely large resistor which will drop all the voltage across it. If both inputs are high the current path will still be through the resistor. In this circuit there will be no output if the inputs are not connected to high or low.

3. Oct 8, 2008

### h0dgey84bc

So let me think about this. We know for an AND gate that if both inputs are 0 then out put, if one input is 1 and other 0 (or vice versa) the output is 0, and if both inputs are 1, the output is finally 1.

So in the pic in the link I posted, if say A is 1 (6V) and B is grounded at 0. Then current is going to flow from V to B, but not V to A (since both are at a high). But the fact current is free to flow from V to B means that the out point and B must be at the same potential, i.e. zero? (and similary if it was A high and B low).

But if both A and B are high, then no current will flow from V to A or B, since theyre high potentials, and thus out must be high too.

4. Oct 8, 2008

### nottheone

Either that AND gate is different than when I looked at it last night or I must have been seeing things because it isn't the way I remember it. My explanation of the AND is wrong regardless. I will rephrase my answer.

For the AND assume that the inputs must be either high or low (not really true). The truth table for an AND gate is the output will only be high if both inputs are high. If either or both inputs are low (ground) it will forward bias their respective diode(s) and effectively ground the output (low). Putting a high on both inputs will hold the output high because the diodes will be reverse biased. In this circuit no inputs will also leave the output high. The main point is that a low on either input will ground the output (low).

5. Oct 8, 2008

### chroot

Staff Emeritus
You've pretty much got it, h0dgey84bc.

In the AND gate, if either or both of the diodes conducts, the output will see a voltage that's one diode-drop above ground, which is considered logic 0. (A typical forward-biased diode drops about 0.7 V.) If neither diode is conducting, the output is nearly V (but somewhat less if you load it and pull significant current through R).

In the OR gate, if either diode conducts, the output will see a voltage that's one diode-drop below A or B, which is considered logic 1. If both diodes are turned off, the output sees a voltage equal to ground (or a little more if you pull significant current through R).

- Warren