# AND gate using diodes

Discussion in 'Homework Help' started by ihaveaquestion, Mar 24, 2010.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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Hi,

I'm lost when analyzing the following circuit to find V and I... the answer is under f)

http://img151.imageshack.us/img151/1910/95958391.jpg

I'm pretty sure you have to make an assumption as to which diode is forward and reverse biased... but I don't know where to start...

thanks

2. ### Wendy Moderator

Mar 24, 2008
20,766
2,536
No assumption needed, just calculate the diode drop for each diode, the lowest voltage on the output wins (it is an AND gate after all). I routinely use AND and OR gates in my designs, they work very well.

3. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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So I think what you're saying is just take the 1V source (since it's the lowest of the three), calculate the current, I = (5 - 1)/ 1k = 4 mA, and somehow that makes V = 1V as well... I'm not sure why. Could you explain a little more please? And also why this would be an AND gate?

4. ### jlcstrat Active Member

Jun 19, 2009
58
3
I'm still not sure about the 1 volts, but in order to get 5 volts out each diode needs to not be forward biased. You need 5 volts on each input to open each diode. Hence the AND.

May 1, 2009
314
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Anyone else?

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Because the book amuses ideal diode Vf=0V.
So last diode gives 1V on Vout.
Diode in the middle will see 1V-2V=-1V so it is reverse biased.
The diode on top will see 1V-3V=-2V so it also reverse biased.

You need 5V at each input to get 5V on the output.

7. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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I still don't understand why only the 1V is forward biased....

Also, if we're going to be more specific, all 3 inputs need 5 volts or GREATER to have them all forward biased, correct? I'd really like to understand this problem inside and out

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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If all inputs were at >=+5V then all diodes would be reversed biased.

9. ### Wendy Moderator

Mar 24, 2008
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Figure the voltages out, write them down. Don't forget the .6V diode drop. If a diode has less that the needed .6V drop it is effectively an open switch.

10. ### Wendy Moderator

Mar 24, 2008
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It occurs to me you are missing a fundimental. A diode can not drop more than the 0.6V when it is forward biased. This is what determines which diode is conducting at any one time.

11. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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How do I find the voltages out if I can't determine the current? To determine the current I need to know which diodes are open or closed which is my goal in the first place... I think I need some step by step help... thanks

12. ### Wendy Moderator

Mar 24, 2008
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With diodes current isn't the issue, they drop 0.6 V when forward biased. Go with that assumption and all else follows, including current. Out of the 3 diodes only one can be forward biased, and will drop 0.6V, the others will be back biased by definition.

The forward dropping voltage is the key.

13. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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OK I will do the analysis and show my work, but before I do, are you saying that without even doing any kind of analysis/assumptions, I should already KNOW that only one of the three diodes will be forward biased?

14. ### Wendy Moderator

Mar 24, 2008
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If plus is on one side and the cathode is more negative, and the difference is more than 0.6V, yes. This is fundimental to how diodes work.

The bottom diode will have 1.6V on the anode. This means the other two diodes are back biased, and are out of the circuit.

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15. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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I'm sorry but I still don't understand... could you elaborate or maybe draw some figures in the attached picture? Very confused

16. ### Wendy Moderator

Mar 24, 2008
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Do this in steps, assume the top diode is conducting. The output would be 3.6V. This means the other two diode are dropping a lot of voltage. This is an impossible condition (remember, I keep dwelling on the 0.6V forward voltage drop). This eliminates the top diode.

Assume the second diode is conducting. The output would be 2.6V. Again, the bottom diode would be dropping 1.6V, an impossible condition. So this scenario is out.

Assume the bottom diode is conducting. The output would be 1.6V. The other two diodes are back biased. This is a self consistent senario that follows the rules. Since we've eliminated the other two diodes it is a winner.

Once you have the voltage out, all else follows, you can calculate currents for example.

The thing to remember about a diode AND gate, the output is always set by the lowest voltage. This is what makes it a logical AND gate.

17. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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That helped some, Bill...

Are the following two statements correct?

1-If there's a voltage drop across the diode from right to left, then there's a voltage rights from left to right... this is why the output would be 3.6V with only diode 1 on...

2-If the first diode is on (3V input) and the others off, this could not be the case because as you mentioned, there would have to be a voltage drop across the bottom diode example of 2.6V to make 1V on the left side... but since the diode drop is only 0.6V, 3.6 - 0.6 = 3V which is not 1V...

18. ### Wendy Moderator

Mar 24, 2008
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Instead of left to right, think anode to cathode. If the anode is positive to the cathode it is forward biased, and 0.6V is its number. Using designations like left to right are really confusing.

There can only be one voltage on the output, this is pretty obvious. So what voltage doesn't violate how diodes work (the 0.6V drop)?

The only way statement 2 is true is if the other two diodes are off, which would mean the input voltage is extremely positive on both those diodes.

Again, it is the polarities that are important. If 0V is shown then that is a voltage, if the input is left open then the diode is left out of circuit.

Reread the end of statement number two, you are trying to think of this in other terms than diode polarity, and it gets convoluted and unecessarily complex from there. If a diode is forward biased the max it can have is 0.6V. Use this as a guide.

I have used the forward dropping voltage as part of a voltage regulator. It is not absolutely stable, up the current and the numbers creep up, but in the small milliamps range 0.6V is a good number to use.

When ever I draw a schematic I add designators. This is because unless you can define exactly what parts you are talking about it can be extremely confusing.

I'll be back with an illustration with part numbers.

So, if the output is 1.6V, you have 2V on the cathode of CR2, and 1.6V on the anode. The cathode is more positive than the anode, so the diode is off. CR3 is even more extreme in the voltages, so it is also off. Off in this context means nonconducting.

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19. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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Thanks, Bill... I think I have a better understanding of it all... but based off the book's answer (Vout=1V), this means that they are using an ideal diode model for this? i.e. no voltage drop across diodes? Because if they were then the output voltage would indeed be 1V + Voltage rise going from cathode to anode over bottom diode.

If that's the case, it's a little weird to solve a problem using an assumption but then you're final answer doesn't use that same assumption, no?

20. ### Wendy Moderator

Mar 24, 2008
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Yes, they are using an ideal diode, so 1V would be the answer. Polarity is still the key. Learn to think in those terms and you can't go wrong.

Schottky diodes drop considerably less, 0.2 to 0.3 volts. They use a slightly different mechanism. In the real world there are no ideal diodes, though there are some op amp circuits that compensate nicely for a diodes deficiencies.