Timeline for How should I analyze this zero crossing circuit?
Current License: CC BY-SA 4.0
9 events
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| Dec 5, 2022 at 15:49 | comment | added | WalterPH | Thank you very much for taking the time to explain these things to me, now I understand this circuit better. | |
| Dec 5, 2022 at 15:47 | vote | accept | WalterPH | ||
| Dec 3, 2022 at 20:32 | comment | added | jonk | @WalterPH For design purposes, you could just assume the differential behavior (in other words, that it somehow 'magically' works at the zero-cross for reasons you are not prepared for), and then just work on a design. But before a design procedure could be developed for you, you'd need to specify what criteria are important. For example, is specifying the width of the opto current pulse important? Increasing the resistor values reduces the width. Is specifying the peak of the opto current pulse important? Increasing the resistor values reduces the peak current. So there's a conflict. | |
| Dec 3, 2022 at 20:23 | comment | added | jonk | @WalterPH You said (as Bruce points out) that you don't want to involve differential equations. Yet it actually is the fact that a capacitor's current peaks when the slope of the voltage changes across it are at a maximum. And this peak occurs once every half-cycle of the AC input. The fact that the two diodes (one actual and one from the BJT) "clamp" one end of the capacitor while the other end is exposed to the rectified AC is what yields this peak right at the zero-crossing. It is exactly this differential understanding which explains why it works right at the AC zero-crossing. | |
| Dec 2, 2022 at 23:20 | comment | added | Bruce Abbott | Which reminds me that the value of R3 should be chosen to provide a suitable amount of Base current. Yet another variable! | |
| Dec 2, 2022 at 23:17 | comment | added | Bruce Abbott | Remember too that while this current flows the voltage at D5 cathode is negative relative to 'ground', turning Q1 off. Only when V1 falls below the voltage on C1 does the current reverse, causing positive voltage at D5 cathode. Q1 is turned on from C1 via current going through R3 (V1 being effectively disconnected at this point because the blocking diode is reverse biased). | |
| Dec 2, 2022 at 23:16 | comment | added | Bruce Abbott | It is. Follow the current path from V1+ through R1+R2, C1, D5 and back to V1-. Don't be fooled by the 'ground' - it's just an arbitrary reference point chosen to make the circuit look more conventional. | |
| Dec 2, 2022 at 22:48 | comment | added | WalterPH | Your answer really helped me understand a bit more, actually I would think diode D5 would be in series with capacitor C1 but I'm not really sure, could you tell me why diode D5 is not in series with capacitor C1 in your simplified model ? If you can't, I'm still very grateful for taking the time to help me with your answer. | |
| Dec 2, 2022 at 5:13 | history | answered | Bruce Abbott | CC BY-SA 4.0 |