I've tried to make a 24VAC triac switch driven by a triac optocoupler. It doesn't seem to be turning on fully. Direct connection of load to 24VAC measures .24amps but through the triac when on is only .13amps.
I've gone back to the drawing board using ST AN5114 but I'm having math problems figuring out R1 (and C1). Equations in AN5114 produce a negative resistance for R1.
Here's my circuit:
First. I calculated R2 according to V_line(peak)/I_surge. I'm using 24VAC so peak is 33.94V and the I_TSM from my optocoupler is 1.2V. R2 = 33.94/1.2 = 28.28. I've chosen a 33Ohm resistor for R2. No problem here, this seems very normal.
While AN5114 is based on a 220VAC circuit I didn't see a reason to change R_GK and left it at 390Ohms.
The triac's gate trigger current is I_GT=10mA and the triac gate voltage is V_GT=0.7V.
From Page 11, equation 2: I_RGK=V_GT/R_GK = 0.7/390 = 0.0018A.
From Page 11, equation 3: I = I_GT+I_RGK = 0.01+0.0018 = 0.0118A
And now the problematic part, calculating R1...
Page 11, equation 1: V_T = (R1+R2)*I+V_photo+V_GT
R1 = (V_T - V_photo-V_GT)/I - R2
= (1.3 - 1.7 - 0.7)/0.0118 - 33
= -126Ohms.
which makes no sense because it's negative.
If I use the exact numbers from the example in AN5114 it also doesn't make sense...
I_RGK = V_GT/R_GK = 1.3/390 = 0.0033
I = I_GT+I_RGK = .01+0.0033 = 0.0133
AN5114 doesn't reveal the value of V_T used. Straight from the T810T-8FP datasheet V_T (max) is 1.55V... so
R1 = (V_T - V_photo-V_GT)/I - R2
= (1.55 - 1.8 - 1.3)/0.0133 - 150
= -266 Ohms NOT the 51Ohms ST microelectronics comes up with.
To get 51 ohms V_T would need to be 5.77V which is far outside the V_T documented in the datasheet.
What the heck am I doing wrong? I'm totally wrong about what V_T is? What should the value of V_T be for a triac? How do you get V_T=5.77 for a T810T-8FP triac when it's datasheet says 1.55V. (if I knew that I could apply it to my triac).
Thank you.
