At 06:15 PM 12/26/2003 +0000, you wrote: >On Fri, 26 Dec 2003, Spehro Pefhany wrote: > > At 11:42 AM 12/26/2003 +0000, you wrote: > > >TR2's data sheet says it needs a max 50mA of gate current to turn on, > > > > Unless you are operating in the pesky quadrant IV (in which case,= 100mA), > > which you are not. > >I believe the moc3041 has a built-in zero-crossing detector, so I guess >that means it can only turn on at the beginning of the 1st quadrant and >the end of the 2nd? There are (not surprisingly) 4 quadrants. Quadrant IV is negative voltage on MT2 and positive gate current. As the current comes from the power line and not some external supply, it can only possibly be in Quadrant I or III. (+Ve/+Ve or -Ve/-Ve) > Peak voltage is 1.41 * 240 =3D 340V, so current is 61mA at the peak > > of the AC line before the load turns on. > >Errmm, what's the 1.41 all about then? Peak voltage of a sine wave of 240 VRMS voltage is sqrt(2) times the RMS voltage. The waveform goes from -340V to +340V. v(t) =3D 340 * sin(2 * pi * 50* t) (assuming 50Hz) > > Note: You really should be using more like 100R than 5K6. You want it to > > turn on near the beginning of the AC cycle, not some poorly determined > > time around 1/4-1/2 way through (getting earlier as the thyristor heats > > up), right? Note that the 100R will BURN UP if the triac does not turn > > on so you need to use a flameproof type for safety. This will also > > take out the MOC, most likely. > >The max forward current of the output photo-triac is rated at 1A, so >surely if I use a 100R resistor in there it'll get 3.4A (using your peak >voltage above)? And if the gate current is spec'd as a max of 50mA at the >triac, wouldn't it break too? No, because the triac will turn on at 50mA and there won't be any voltage left... > > >Any ideas of what I'm doing wrong? > > > > I suspect you have MT1 and MT2 swapped. > >I was under the impression it didn't matter which way round they were? I >didn't see it mentioned in the data sheet? Ah. It very much matters. The center lead is MT2. > > > Also, how did you calculate the 200R? You need to guarantee 15mA (@25=B0= C, > > MORE at lower temperatures) through the LED for this part to be= guaranteed > > to work. > >It was an incorrect, hasty decision :) ... I did 5v/200R (what I could >find in my junk box at the time) and got the 25mA, but now I realise that >the double-diode drop would make it more like 2/200, so there's only 10mA >flowing. But there's still 240vAC at the triac's gate, and I didn't see >any mention of coupling-transfer ratio's like in normal opto-isolators, so >I assumed that if I can see voltage there, I'm getting all the current >too? The trigger current is rated at 15mA, so you need to guarantee that for it to work at 25=B0C. If it might be colder, you need more current. >I'm starting to think I'll stick with relays instead. You know where you >are with a switch. There are plenty of gotchas with relays too. >What sort of application would an optically driven triac be better than a >relay in? I can think of dimmer circuits and motor-speed controllers, but >that's all. A typical relay is good for 100,000 operations at full current. At one per 10 seconds, it could wear out in a week and a half. That might not be acceptable. ;-) (derating can extend the life, but not always by as much as you'd guess). So, if there are a lot of operations (eg. a temperature control where the process time constant is small), the triac is better. The acoustic and electrical noise from a relay can be bothersome. OTOH, relays are rugged electrically, seldom fail on, and generally are quite efficient. Neither is perfect. >Thanks for the input, Spehro. Sure. Hope it helps. Best regards, Spehro Pefhany --"it's the network..." "The Journey is the= reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body