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Can somebody please explain to me how you exactly determine whether a diode conducts (shorted) or is cutoff (opened)? I found a couple of posts on this but I wasn't able to justify the answers myself, probably something I'm doing wrong. It was said that the anode should be greater than the cathode for the diode to conduct but it's not making any sense to me in these examples.

The answers are: for (a), D1 is cutoff and D2 is conducting. V = 2V, I = 3.5mA (b) D1 is conducting, D2 is cutoff. V = 1V, I = 2mA

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  • \$\begingroup\$ Have you done all the diode math yet? \$\endgroup\$ – Ignacio Vazquez-Abrams Mar 10 '15 at 1:27
  • \$\begingroup\$ Do you have trouble verifying the answers or getting to the answers? (or both?) \$\endgroup\$ – Spehro Pefhany Mar 10 '15 at 1:30
  • \$\begingroup\$ Pretty much both. I think I need a run through of it lol \$\endgroup\$ – Martyax Mar 10 '15 at 4:51
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The model you are using is:

  1. If \$V_{anode} < V_{cathode}\$, \$I_D = 0\$ (cutoff)
  2. If \$I_D > 0\$, \$V_{anode} = V_{cathode}\$ (conducting)

The general procedure for solving systems with diodes is:

  1. Make an assumption about the state of every diode (either cutoff or conducting).
  2. Try to solve the circuit with those assumptions.
  3. If you have a solution, check to make sure that solution is consistent, i.e. all of your original assumptions were correct. If there is no valid solution, your original assumptions were probably wrong. Pick a new set of assumptions and repeat.

For example, take circuit a. Let's assume D1 and D2 are both conducting. By definition then, the node labeled \$V\$ must be at 1V and at 2V at the same time, which is nonsense. So we know that both diodes can't be conducting at the same time. Now pick new assumptions and try again (left as an exercise for the reader).

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  • \$\begingroup\$ So then when D1 is cutoff and D2 is conducting, I understand there's no current going through the D1's cutoff. And in D2, there is a current since we're assuming it's conducting. So now you have to check to make sure Vanode is equal to Vcathode? Vanode I'm pretty sure would be 2V, but then how do you calculate Vcathode to also equal 2V? Edit: Also, for D1 since we're assuming it to be a cutoff, Vanode is 1V, but what would Vcathode be? I think this is where I'm having the real issue. \$\endgroup\$ – Martyax Mar 10 '15 at 4:53
  • \$\begingroup\$ I wrote the conditions specifically: if <blah>, then <something>. So for conducting, you assume <something> is true (Vanode=Vcathode). After solving, you check <blah> (Id>0, or current flows from anode to cathode). You don't assume anything about current when solving. Likewise, do something similar for cutoff: assume Id=0, and solve as normal. Then check that Vanode < Vcathode at the end. \$\endgroup\$ – helloworld922 Mar 10 '15 at 5:11
  • \$\begingroup\$ Ahh I think I got it. I tried both possibilities with both circuits and was able to confirm the right answers. Thank you! \$\endgroup\$ – Martyax Mar 10 '15 at 5:51
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From the answers, it appears the question is using ideal diodes, which have no forward voltage drop. (Real silicon diodes have about a 0.7 volt forward voltage drop, which will change the output voltage, but not affect which diode is conducting.)

In (a), if D1 is conducting, it will pull the output up to 1 volt. However, the anode of D2 is at 2 volts, so D2 will pull the output up to about 2 volt - that will make D1's cathode more positive than its anode, so D1 will not conduct.

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If the p terminal of the diode is towards the p terminal of the battery then the diode is ON. Similarly, if the n terminal of the diode is towards the n terminal of the battery then also the diode is ON. But if battery and diode is facing opposite sides of each other then the diode is OFF. I hope u got my point. :)

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    \$\begingroup\$ Welcome to EE.SE. (1) How does your answer help establish whether D1 is conducting or not? (2) Did you read the accepted answer from two years ago? I hope u got my point. :) \$\endgroup\$ – Transistor Nov 11 '17 at 17:21

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