# What are the possible states that these diodes could be in?

This is the problem circuit:

Im confused on how to treat this circuit the problem asked to graphically plot the voltage V_s = f(V_e), but i stumbled on the case where both the diodes are conducting -assuming they are ideal diodes- V_s = E and V_s = -E because they are parallel, but this is absurd right?

• If V_s = E the diode is going to start to conduct, it's current, and that through R/2, is still zero. Therefore Ve(t) = 2 V_s = 2 E. The same goes for the opposite polarity. Jun 4 at 12:03
• I understand that the voltage of R/2 is non-zero because there is a current going thru it because the left diode is conducting, but i dont see why Ve(t) = 2V_s? @HarryH Jun 4 at 12:09
• Oh wait are you saying that V_s(t) = E = Ve(t)/2? due to them being parallel branches? Jun 4 at 12:10
• responding to your last comment, no i get it when both the diodes arent conducting we study the voltage divider because its the only active circuit that is also parallel to V_s Jun 4 at 12:26
• Vs(t) = -E = Ve(t)/2 is where the right diode starts conducting. If Ve(t) goes further negative (it's the independent variable here) Vs(t) will stay at -E. Jun 4 at 12:29

this is absurd right?

Yes, it's absurd.

There is no situation when both diodes (assumed ideal) conduct at the same time: -

• $$\D_{LEFT}\$$ conducts when $$\V_e\$$ is positive and more than twice $$\E\$$ in magnitude
• $$\D_{RIGHT}\$$ conducts when $$\V_e\$$ is negative and more than twice $$\E\$$ in magnitude
• how did you know the condition when each diode is conducting in relation to V_e and E? Jun 4 at 10:34
• I looked at the circuit and considered it @HellBoy Jun 4 at 10:36
• I did too but unfortunately i didnt get any idea, so what i did is consider the 4 cases first case which was absurd i had V_s = E- V_d and for the right diode, V_s = -E + V_d, which triggered the absurdity, second case D_left is conducting so V_s = E and V_e = ? idk how to use the basic laws to derive it because im confused by the parallel resistance which reside to the left to that diode. Jun 4 at 10:39
• @HellBoy use Thevenin's theorem to convert Ve and the two resistors to a single voltage source (Ve/2) and one series resistor. That gets rid of the parallel resistor and un-confuses you. Jun 4 at 10:45
• I've rolled-back your question to its original form @HellBoy Jun 4 at 12:17