Timeline for Confusion regarding piece wise linear diode model
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 27, 2017 at 3:56 | comment | added | The Photon | You should get positive clockwise current, because the diode is in reverse breakdown. Current flows in to its cathode and out of its anode. This is clockwise according to your diagram. | |
| Oct 27, 2017 at 3:25 | comment | added | tapeside | But doesn't that then give me a positive clockwise current via KVL whereas I should be getting a negative current/anticlockwise current because the the diode is in reverse bias? | |
| Oct 27, 2017 at 3:00 | comment | added | The Photon | Yes, It should be +7.2 | |
| Oct 27, 2017 at 2:26 | comment | added | tapeside | So the Voltage should be saying +7.2V? | |
| Oct 27, 2017 at 1:36 | comment | added | The Photon | With your arrangement you should have gotten 1.1 A, not 0.11 A. | |
| Oct 27, 2017 at 1:35 | comment | added | The Photon | In your 2nd added question, you have the wrong polarity on the external voltage source. And \$(7.2 - 6)/12 = 0.1\$, giving the expected answer. | |
| Oct 27, 2017 at 0:51 | vote | accept | tapeside | ||
| Oct 27, 2017 at 0:51 | history | edited | tapeside | CC BY-SA 3.0 | added 255 characters in body |
| Oct 27, 2017 at 0:29 | vote | accept | tapeside | ||
| Oct 27, 2017 at 0:51 | |||||
| Oct 27, 2017 at 0:24 | history | edited | tapeside | CC BY-SA 3.0 | Edited to include more questions requiring clarification |
| Oct 26, 2017 at 22:13 | answer | added | The Photon | timeline score: 1 | |
| Oct 26, 2017 at 22:06 | history | asked | tapeside | CC BY-SA 3.0 |