0 Replies Latest reply on Sep 16, 2013 8:40 AM by krapahon

    Need help understanding Optocoupler - Triac schematic

    krapahon

      Hello everyone,

       

      I'd like to use an optocoupler-triac to drive a 220VAC load (an ac/dc switching power supply, to power up additional gear).

       

      I read alot of documentation and datasheets about optotriacs and decided to go with Fairchilds recommended design for a MOC3063:

      moc3063.png

       

      My electronics skills being a little bit rusty (especially with AC), i'm having a hard time determining the power consumption of the 2x 360ohms resistors.

       

      This is what i came up so far:

      - When the optocoupler is off, no significant current flows into the 360ohm resistors (let's forget about leakage current in the Triac an optotriac). The RC snubber however, will leak more current, if we use a very low impedance load, the current will be less than 1mA through the 39ohm and 0.01uF cap and they will dissipate less than 1W of power. So using 1W components for the RC snubber is perfect.

       

      - When the optocoupler is on and we reach permanent / stable AC. The Triac acts as a short-circuit, so no current will flow inside the RC snubber, so far so good. And i think this is where i'm starting to get things confused: If the Triac act as a short this would also mean that no current is flowing between pin 4 and 6 of the optocoupler, since the open Triac is the path of least resistance ? For this to work, it would mean that on every zero cross over of the AC supply, (when the current changes direction) the Triac locks and a tiny amount of current is diverted to the Triac gate until it reaches the gate threshold (Igth) and the Triac is now open for another half cycle, all the current now flows through the Triac until the current reverses (or drops below the hold current value ?)

      This would mean that on every half cycle, we have Igth flowing through a single 360ohm resistor. If we take a Triac with a gate threshold of 30mA. This would mean that the resistor dissipates P = R*I^2 = 0.3W ? Now since the current doesn't flow through the resistor continuously but only for a fraction of the time, using a 0.25W component should be enough ?

       

      My first question is, is my analysis correct ? Am I missing something ? Is it correct that the Triac needs to be switched on every half cycle using the gate current ?

       

      Second question is about making the circuit a bit more fail safe. The power supply I intend to drive consumes max 2A @ 220VAC, so i'll be adding a quickblow 2A fuse on the hot wire. But if for some bizarre reason the Triac fries off in "always open" mode, so no matter the gate current, it will always act as an open switch. This means that if the optocoupler is triggered a significant amount of current may pass between pin 6 and 4 as well as through the two 360ohm resistors. The worst case scenario would be a low impedance load, meaning we get most of the 220VAC voltage across the resistors in series, and this will dissipate P = U² / R = 67W ! At best some smoke, at worst a fire !? And the 2A fuse wouldn't help at all.

       

      Is it possible to add a 50mA quickblow fuse  between gate 6 and the 360ohm resistor to prevent this catastrophic failure ? The datasheet of the MOC3063 says that it can support an inrush current of 1.2A (this is how the 360ohm were sized) would this inrush current be sufficient to blow up fuse ? Should I be looking at a resettable fuse/thermistor instead to protect this line ?

       

      And finally, the power supply i intend on driving can deliver a 75A inrush current when turned on (http://www.meanwell.com/search/hlg-240h/default.htm). How will this impact my schematic is there anything i need to look out for when choosing other components ? Can i drop the snubber completly for a switching power supply if i use a transistor advertised as snubberless ?

       

       

      This was alot for a first post. I hope you won't find my questions too dumb and that a kind soul will help me see the light in this silicon mess that i got my self into.

       

      Thank you for reading and taking the time to help!

       

      Best regards,

      Kra