Timeline for Switching polarity of rectified AC power with a microcontroller
Current License: CC BY-SA 3.0
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| Apr 14, 2015 at 18:53 | comment | added | Adam Haun | The driver would be referenced to the negative end of the AC line through D3, but I supposed that's not the best ground. Depending on how much current the relay takes, you might be able to drive it from the cathode of D1/2. You'll probably need something bigger than a 1N4148, though. The FETs can then use the anode of D3/4 as a ground, just like the opto-coupler and driver. | |
| Apr 14, 2015 at 18:13 | comment | added | jon | Adam, thanks for your suggested circuit. I did actually consider a circuit like this. However, it seems that there may be a problem in the Vgs. Somehow the MOSFET needs to share a reference with the driver in order to drive the VGS, but I'm not sure how this can be accomplished. | |
| Apr 14, 2015 at 3:25 | history | answered | Adam Haun | CC BY-SA 3.0 |