Timeline for How is the extinction angle for a half wave rectifier with RL load calculated?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2018 at 16:30 | comment | added | Dirceu Rodrigues Jr | You can use that I provided or build it yourself, for example through a Matlab script (fzero). Some books of Power Electronics included it, too - when presenting single phase uncontroled rectifiers (RL load). | |
| Aug 31, 2018 at 16:29 | vote | accept | riccs_0x | ||
| Aug 31, 2018 at 16:10 | comment | added | riccs_0x | @DirceuRodriguesJr And where can be finded the prebuilt plots? | |
| Aug 30, 2018 at 20:20 | comment | added | Dirceu Rodrigues Jr | In other words: The OP approximation could be something valid for a small time constant (L/R). | |
| Aug 30, 2018 at 18:57 | comment | added | D.A.S. | I got that part but missed the fact that extinction goes to 0 when R goes to 0 meaning 360 deg or Continuous conduction and phi angle of 75 deg becomes 180+90 deg for Beta, so my simulation is correct too. | |
| Aug 30, 2018 at 18:26 | comment | added | Dirceu Rodrigues Jr | The botton left point on curve refers to "pure resistance" and upper right point refers to "pure inductance". | |
| Aug 30, 2018 at 18:11 | comment | added | D.A.S. | For R=1, I get 5.08ms extinction or 180+109 deg. due to group delay? when phase angle approaches 90 deg. extinction exceeds 90deg. | |
| Aug 30, 2018 at 17:32 | history | answered | Dirceu Rodrigues Jr | CC BY-SA 4.0 |