Timeline for What are the possible states that these diodes could be in?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 4, 2023 at 12:19 | comment | added | Andy aka | Two resistors of equal value form a 2:1 potential divider @HellBoy | |
| Jun 4, 2023 at 12:17 | comment | added | Andy aka | I've rolled-back your question to its original form @HellBoy | |
| Jun 4, 2023 at 12:05 | comment | added | HellBoy | Ah thought i could do so, sorry my bad i didnt know that | |
| Jun 4, 2023 at 12:04 | comment | added | HarryH | en.wikipedia.org/wiki/Voltage_divider | |
| Jun 4, 2023 at 12:04 | comment | added | Andy aka | Read my full comment. | |
| Jun 4, 2023 at 12:03 | comment | added | HellBoy | i dont understand what you mean by 2:1? | |
| Jun 4, 2023 at 12:02 | comment | added | Andy aka | Because of the 2:1 potential divider formed by the two resistors @HellBoy also, why are you editing your question? This is a Q and A site and you don't go making changes to your question based upon what is provided in answers. | |
| Jun 4, 2023 at 11:09 | comment | added | HellBoy | Also why specifically twice more than the magnitude of E from which equation did you derive that from? | |
| Jun 4, 2023 at 10:53 | comment | added | HellBoy | Sorry i quite stumble on this confusion on each problem , matter fact i did read about this in several books but i still cant understand it | |
| Jun 4, 2023 at 10:52 | vote | accept | HellBoy | ||
| Jun 4, 2023 at 10:50 | comment | added | Andy aka | @HellBoy this is a Q and A site and not a forum and I'm not going to be dragged into a forum-like conversation. If you don't know how to convert a voltage source and two resistors to a new voltage source and series resistor, you should either ask a new question or return to your text books or google it. | |
| Jun 4, 2023 at 10:48 | comment | added | HellBoy | I swear my only problem is this, we start by two parallel resistances which two different current traverse them, somehow we make an equivalent resistance in which a unique current traverse it and we do our analysis based on this new resistance and somehow the results are valid in both cases? and for the thevenin voltage how would i extract it do i just disconnect the two diodes and anything that hangs on their linear circuit and just leave the two resistances in the original circuit? | |
| Jun 4, 2023 at 10:45 | comment | added | Andy aka | @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, 2023 at 10:39 | comment | added | HellBoy | 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, 2023 at 10:36 | comment | added | Andy aka | I looked at the circuit and considered it @HellBoy | |
| Jun 4, 2023 at 10:34 | comment | added | HellBoy | how did you know the condition when each diode is conducting in relation to V_e and E? | |
| Jun 4, 2023 at 10:33 | history | answered | Andy aka | CC BY-SA 4.0 |