Triac - Opto drive question - Page 1
Forum section: electronics / projects

author: Jester
author_rank: Frequent Contributor
timestamp: February 05, 2017, 08:46:01 pm
content: I have little to no experience with Triacs.Please see circuit below......I simply want to switch thehotleg of a 240Vac (incandescent light) circuit with a maximum load of 10A, (could be as low as 1A.)No dimming required.It's not obvious how I calculate R1. I have also seen similar circuits with R2, however I think R2 is only needed with zero cross optos (but I'm not sure?)How do I calculate R1?If there are obvious improvements to be made, I'm all ears. Links to data sheets are belowhttp://www.mouser.com/ds/2/389/bta16-954517.pdfhttp://www.mouser.com/ds/2/239/S_110...IES-337181.pdf

author: Benta
author_rank: Super Contributor
timestamp: February 05, 2017, 09:36:38 pm
content: It's quite simple.You calculate it from the allowed peak current, either through the MOC3052 TRIAC, or the the maximum gate current of your BTA16.The MOC3052 allows 1 A peak, the BTA16 4 A peak.So 1 A peak it is.Take thepeakAC voltage and divide by 1 A.For 230 V systems: 330 ohms.For 120 V systems: 180 ohms.Than multiply by 2 or a bit more (this is my own rule of thumb) to have some margin230 V: 680 ohms120 V: 390 ohmsDone.

author: james_s
author_rank: Super Contributor
timestamp: February 05, 2017, 11:13:52 pm
content: Or just look at the datasheet for the opto-coupler and use the values it suggests. What you're trying to do is a very common use of triacs and the datasheets will often have application notes that show suggested circuits.

author: Benta
author_rank: Super Contributor
timestamp: February 05, 2017, 11:29:15 pm
content: QuoteExcellent idea!Unfortunately, the datasheet doesnotsuggest any values, but only the calculation I did above.Checking facts before hitting the keyboard does avoid thread clutter...

author: Jester
author_rank: Frequent Contributor
timestamp: February 06, 2017, 12:01:16 am
content: Benta,Thank youFollow up question....Assuming 230V and 680 ohms, the peak current is 478mA. Sopeakpower rating of R1 needs to be > 110W. If I understand, the continuous power rating of R1 would be depended on the load because a lower power load would require a longer time to reach the latching current and therefore the gate current would need to be applied for longer. That being said I'm guessing that any resistor that can meet the instantaneous peak power will likely be larger than needed for the continuous power rating. Does that make sense?It seems that through hole carbon resistors are often used for R1. I was hoping to use a SMT resistor, perhaps I can find a reasonably priced SMT resistor with high peak power rating

author: BrianHG
author_rank: Super Contributor
timestamp: February 06, 2017, 02:24:47 am
content: These datasheets have values for their example schematics:https://www.fairchildsemi.com/datasheets/FO/FOD410.pdfhttps://www.fairchildsemi.com/datasheets/MO/MOC3061M.pdf

author: james_s
author_rank: Super Contributor
timestamp: February 06, 2017, 02:48:49 am
content: Quote from: Benta on February 05, 2017, 11:29:15 pmI only suggested that after looking at a project I did some time back and noting that I used exactly the same values you calculated and I got mine directly from the datasheet. Obviously knowing how to calculate it is a useful skill, but so is knowing what to look for in the datasheets. I was simply trying to provide an alternate approach, not saying yours was wrong.

author: bktemp
author_rank: Super Contributor
timestamp: February 06, 2017, 08:09:41 am
content: If you do not need dimming, use an optotriac with zero crossing detection: Because it turns on only near zero crossing, the maximum voltage seen by the resistor is much lower (10-30V), therefore the resistor value can be lower, reducing the peak current and the peak power dissipation.

author: Jester
author_rank: Frequent Contributor
timestamp: February 06, 2017, 01:47:20 pm
content: Quote from: bktemp on February 06, 2017, 08:09:41 amUsing a zero-cross opto makes sense for my application (no dimming and incandescent load).In the Fairchild notes for the FODM3083 it states"If a three quadrant "snubberless" TRIAC is driven by the opto-TRIAC, the calculations are different. When the main power TRIAC is driving a resistive load, it shuts off during the fourth quadrant. If sufficient holding current is still flowing through the opto-TRIAC, the opto-Triac will turn on and attempt to carry the power TRIACS load. ..... For this reason, using an opto-TRIAC to drive a three quadrant "snubberless" power TRIAC is not recommended."I want to use a D2PAK device, going back to the ST data sheet for the BTA16, T1610 (see attached). It appears that both of the D2PAK devices are 3 quadrant, so this looks to be a poor choice. When I do a Digi-Key search (>=600V, >=15A, standard Triac in D2PAK package only two parts remain one of them being the ST T1610. I'm getting the feeling I'm missing something here?http://www.mouser.com/ds/2/389/bta16-954517.pdf

author: bktemp
author_rank: Super Contributor
timestamp: February 06, 2017, 02:18:15 pm
content: That note sounds odd to me:Fourth quadrant is negative voltage on A2 but positive voltage/current on gate. How can there be a positive current flowing into the gate when there is a negative voltage on A2 with the optotriac connected between A2 and gate? If the load is inductive it seems plausible, but not for a resistive load.When using optotriacs the TRIAC is driven only using quadrants 1 and 3. That should be fine even for 3 quadrant TRIACs.Littlefuse has many 4 quadrant SMD TRIACs:http://www.littelfuse.com/products/switching-thyristors/triac.aspx