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I have an inquiry about the circuit below: When a high signal is continuously given to the transistor, it enables the optotriac with zero crossing. The optotriac enables the power triac, and the AC signal goes to the rectifier, and gets rectified. What is the DC rectified voltage?

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    \$\begingroup\$ This sounds like a homework problem. Even if it's not, do show what you've tried so far or your question won't likely get answered. \$\endgroup\$ – Hearth Aug 1 '18 at 15:09
  • \$\begingroup\$ Hi. It's not a homework question. It's from a project that I'm working by myself about a DC motor controller. I think that the DC voltage should about 170VDC, since I'm not doing phase control, right? \$\endgroup\$ – Blue_Electronx Aug 1 '18 at 15:11
  • \$\begingroup\$ Show the math you've done so far; where are you getting stuck? \$\endgroup\$ – Hearth Aug 1 '18 at 15:12
  • \$\begingroup\$ I just thought that the rectified voltage would be the peak one, I mean, 120x1.41 = 169.2VDC since I'm not triggering the triac at a specific time. I think I'm confused about the concepts of phase control and zero crossing control. \$\endgroup\$ – Blue_Electronx Aug 1 '18 at 15:28
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If there is no load on the output, the voltage will approach the peak line voltage of \$\sqrt{2}V_{RMS}\$, minus maybe a bit from the triac drop and two diode drops. If there is a load, the voltage will droop between peaks as the capacitor discharges and you will get a lower average voltage.


You have a zero-voltage switching opto, which "should" prevent the triac from turning on mid-cycle, however there may be odd conditions under which that won't be true. Such a turn-on could damage the triac, capacitor or diodes.

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