Timeline for Class D inductor current calculation
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2015 at 18:35 | comment | added | Dwayne Reid | Note that if driving the load from a H-bridge, splitting the inductor into two pieces (one per leg of the load) allows for EMI reduction. | |
| Sep 1, 2014 at 15:09 | comment | added | hjf | Oh, yes! Of course, the voltage across the inductor is never more than Vsupply. I should have realized that. I think I mixed up H-bridge and BTL. Here's an example of the several output configurations this amplifier offers: i.imgur.com/NXm5JaE.png Later I mixed up the figure of 40Vrms with 40Vpp, you're right, I need Vpp/2=113/2=56.2V (ideal, of course). | |
| Sep 1, 2014 at 14:56 | history | answered | Dave Tweed | CC BY-SA 3.0 |
