how did you calculate the 50 mA current ?Surge Current Rating means the maximum current allowed briefly, under certain specified conditions, e.g. one full cycle at 60Hz.
BTW, R3 needs to be increased to avoid frying the opto-LED. 400 Ohms wold allow ~50mA to flow!
Voltage across R3 = 24 - 1.2 (bridge rectifier drop) - 1.5 (opto-diode Vf) - 2 (green LED) = 19.3V. 19.3V/400 Ohms = 0.048A = ~50mA.how did you calculate the 50 mA current ?
As the datasheet says, "Peak Non-Repetitive Surge Current (One Full Cycle Sine Wave, 60 Hz,TC = 25°C) 250A".what do you mean surge current is the maximum current allowed ?
It's not so much the input voltage as the current through the opto-diode that matters and must be kept below the rated maximum (which I haven't checked). LEDs need to be current-driven, not voltage-driven. Often only a few mA are needed for an opto. What the datasheet is telling you is that at the rated current the diode forward voltage will be no more than 1.4V (I just guessed 1.5V in post #28).what is the maximum input voltage on the optoisolator ? should it be the forward voltage that is 1.4v in datasheet?
I very much doubt it. If we assume, for argument's sake, the triac (whatever type) drops 1V when 'on', then at 20A it will have to dissipate 20W. It will get very hot!! Use a heatsink.If i want to use triac without heat sink at 20A should i increase maximum rating current to 40 A? will that be ok.?
Re read the thread from the beginning. You will quickly find nothing he has said or asked for adds up.I thought you said your load would be 50 Ohms? That would draw < 5A. Where did you get the 20A figure?
Re read posts 2 - 32. All of this has been well explained plus alternate functional concepts have been give as well.i don't want big size PCB , so i will prefer the Mosfet . what do you think about the first circuit in post 1?will it work fine?
recommend Mosfet please that don't need heatsink.I thought you said your load would be 50 Ohms? That would draw < 5A. Where did you get the 20A figure?
If it really will be 20A then perhaps a FET as you originally proposed would be better than a triac from a power dissipation viewpoint. Perhaps something like this? But an SSR would likely be a better solution.
Quite right. My bad. So a LARGE heatsink would be needed. Good luck finding a 500V FET with a lower Rds(on).i think that you mean the dissipated power in 20A will = 20x20x0.078=31.2 watt