# Diode bias assumption problem

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

May 1, 2009
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2. ### jlcstrat Active Member

Jun 19, 2009
58
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I can't quite figure it out, but you're applying node voltage to basically a series circuit which has no "nodes." With the one diode open you have 20 volts in series with two resistors.

3. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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'but you're applying node voltage to basically a series circuit which has no "nodes." '

That's an incorrect statement... there are nodes.

4. ### jlcstrat Active Member

Jun 19, 2009
58
3
I never got this deep into diode models, so it must just be beyond my grasp. I still just see it as a simple series. Hopefully someone can straighten it out.

5. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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I appreciate your input, but the definition of a node has nothing to do with diodes... a node is a node.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The resulting values arising from your assumptions (re: the diode states) made in each case aren't at all surprising. The two cases are distinctly different, so one would expect a different result. The solutions are both "right" for the given diode states (on or off) you have imposed. The assumed model used in the circuit solution has no inherent means of predicting the true outcome in the real circuit with diodes, other than highlighting the potential inconsistency of any assumptions you initially made.

Last edited: Mar 25, 2010
7. ### jlcstrat Active Member

Jun 19, 2009
58
3
I realize that, but I thought a node had to be an intersection of more than 2 elements.

8. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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thanks, tnk.... are you saying there isn't one correct answer then?

also jlcstrat, yes, that's the correct definition of a node... so if you look at my circuit there's 3 nodes.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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In the example 3.2 and in figure 3.6b you got voltage divider.
10+--->5K--->10K-->-10V
So voltage on 10K resistor is equal:
20V / (5/10 +1)=13.33V so V=3.33V

10+--->10K--->5K-->-10V
So voltage on 5K is equal
20V/ (10/5 +1) =6.66V so V=-3.33V

10. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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I don't understand how you go from 13.33V to 3.3V and 6.66V to 3.3V

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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In either case those voltages are the potential drop across the resistor in question. However the lower end of the resistor in each case is sitting at a potential of -10V, which then gives rise to the final result being the potential drop across the resistor offset by -10V.

So

13.33-10=+3.33V

or

6.66-10=-3.33V

12. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Were you saying earlier there is no 1 right answer?

Mar 6, 2009
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14. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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So I suppose my next question, then, would be:

Was the assumption I made in the bottom case shown in my work in the 3rd link correct or incorrect?

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Here is the problem with the assumption.

Were D1 assumed to be off (as you did) then the voltage at node B would be -3.33V (as you found). But D1 anode is connected to ground (0V) and D1 cathode would be at -3.33V. So we would have the paradox that D1 is forward biased but not conducting. The assumption that D1 is off with D2 on is therefore incorrect.

Assuming that D2 is off and D1 is on also produces a paradox.

16. ### Wendy Moderator

Mar 24, 2008
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In many ways this is similar to the diode AND problem.

Remove diode D1 from the circuit, and you will have:

ID2 = (20V - .6) / 15KΩ =1.293ma

Point B would be -3.53V

Diode D1 (when put back into the circuit) would be forward biased, and Point B would be -0.6V instead. The 5KΩ resistor would now have 1.88ma flowing through it.

I = (-10V - -0.6V) / 5KΩ = -1.88ma

D2 and the 10KΩ resistor would have 10.6V across them, so the current would be 1ma.

ID2 = (10.6V - 0.6V) / 10KΩ

The constant voltage drop of the diode sets up all other conditions, as with the AND gate. You have to go through the logic steps to calculate it out. If D1 had be less than 0.6V, or reverse biased, it is not in the circuit. The fact it was forward biased made all the difference.

17. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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tnk, that does make some sense...

Bill, I think I'm a little too novice which is why I often get confused by your responses... if you could be a little more elementary with me.

Is there any strategy to make good predictions on which are on/off in the first place when first looking at the circuit?

18. ### Wendy Moderator

Mar 24, 2008
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Well, removing the parts and seeing what the voltages would be is one strategy. Another is go through the steps, make assumptions and see if they are consistent. If a voltage exceeds the forward dropping level of a diode and the diode is on then that is a red flag.

There hits a level where it can be hard to explain things simply because it is "obvious". I had another member complaining because I intuitively picked the capacitor value he needed. I couldn't explain how I did it, other than 30+ years of experience.

Reading will compensate nicely for experience. I did that too when I was new.

19. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Bill,

Coming back here after looking at the AND gate thread, which I understand better now, I can understand the calculations you did a few posts above... but based on all those calculations, I fail to see at what point specifically you were able to identify fallacies to determine the bias of the diodes.

Thanks

20. ### Wendy Moderator

Mar 24, 2008
20,764
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How about the one where the diode is dropping 3.3V and is forward biased? This is the key point.

I took the diode out of the circuit, showing a 3.3V. With the other diode back in you have the condition I just mentioned. At this point the other circuit is irrelevant, a forward biased diode might as well be a straight piece of wire, and we are dealing with a power supply scenario, in that the voltage is fixed and unchanging.