# TRIAC motor control. Why is my resistor burning?

I have designed a digital motor controller using the following circuit. One of my resistors keeps overheating. There is a zero-cross detector elsewhere in the circuit used to time the firing of the TRIAC, and the circuit does control the motor, but resistor R36 keeps burning up after being on for 2-3 minutes.

I took this circuit from the MOC3022 datasheet here: https://www.onsemi.com/pdf/datasheet/moc3023m-d.pdf

I am using is this TRIAC.

This is the motor I am controlling.

I question the following:

1. Perhaps the 470 ohm resistor value is correct but I didn't give it the right power rating and that is why it keeps burning.
2. Perhaps the resistor should be a higher value.
3. I used the "typical application circuit" from the datasheet for the MOC3022. Perhaps I should have used the schematic examples for inductive loads which place the load above the TRIAC not below like here. Also for the inductive loads, R36 is a higher value. I ignored them because they say for sensitive gate TRIACs which I am not using, but now I am concerned that was a mistake.

Any input you can give on where I messed up and how to approach the proper design of this circuit would be much appreciated.

• What is the power rating of R36?
– vir
Nov 23, 2021 at 21:44
• It's pretty strange, since the resistor is unloaded if the triac is on all the time. You should tell us the power rating, capacitor type and PCB layout as well. Nov 23, 2021 at 21:55
• The only explanation I can see is if you have a wiring mistake. R36 and C6 are in series across the 115V (from motor rating data sheet) presumably 60Hz supply. There combined impedance ¦Z¦ is about 56,440 ohms giving a current of some 2mA and a power dissipation in R36 of 2mW. So it should be safe by a 2 orders of magnitude. Nov 24, 2021 at 0:11
• Are you sure that C6 is actually 0.047 uF? If it were, by mistake, say 1uF then the dissipation in R36 would be about 860mW and cause a 1/4 W resistor to burn out as you describe. Nov 24, 2021 at 0:30
• @vir power rating is 1/8W Nov 25, 2021 at 4:45

You can find a more detailed explanations and calculations on this site: Also you can see the respective power ratings if you compare the resistor sizes from a ready made SSR: The resistor in question is the bigest one in Melf package its length is close to Triac width (6.6mm), probably it's 0207 size (5.8mm) @ 1W. Note how smaller the R1 and RGK are.

EDIT:

$$R_2>\dfrac{V_{line_{peak}}}{I_{MOC_{surge}}}>\dfrac{120V\cdot \sqrt{2}}{1A}>170 \Omega$$

$$V_{T_{ripple}}=(R_1+R_2)\cdot I_{GT} + (1+\dfrac{R_1+R_2} {R_{GK}})\cdot V_{GT}$$

no RGK:

$$V_{T_{ripple}}=(R_1+R_2)\cdot I_{GT} + V_{GT}$$

$$R_1+R_2 = \dfrac{V_{T_{ripple}}-V_{GT}}{I_{GT}}$$

$$\dfrac{dV_{C_1}}{dt}(max)=\dfrac{V_{line_{peak}}}{R_1\cdot C_1}$$

dV/dt(max) for LiteON MOCxx is 1000V/us - link

It comes out that optima values of elements are R2=180..270 Ohm, R1=47..68 Ohm, C1=22..47 nF

• Thanks Marko, I saved the doc. for a future reference. It is interesting to see how we approach to a solution, application side vs. development side. I recalled one of my friends (retired) at work. While I was tapping my calculator, he brings "how about this?".
– jay
Nov 24, 2021 at 14:06
• Yes thank you Marko for this very useful document. I am going over it. I think I should redesign my circuit to have the load above the TRIAC. The MOC3022 datasheet also recommended it for inductive loads. Nov 25, 2021 at 4:56

Throwing a guess. Only condition to see what you describe, assuming everything else ok, is: Q3 turn on lags.

Q3 turn on lags, most likely from large current (heavy load). Motor inrush current is one that contributes to. Zero crossing may not help as hope, since the inrush condition lasts over many cycles. You can see it on oscilloscope.

Sensitive TRIAC will help.

Edited.

I just went through the datasheet. And, my first guess likely was incorrect, I guess again. However, the sensitive TRIAC theory will still work.

Here is what I suggest: Try lower resistor value. And, the design parameters and procedures:

Design parameters example, just for the trigger circuitry:

1. Gate trigger current, Igt = 70mA (max of ++ & -- quadrant).
2. On-state voltage Vt = 1.7V
3. Gate trigger voltage, Vgt - 1V
4. Latching current, IL = 30mA (max of ++ & -- quadrant)
5. Holding current, IH = 15mA
6. Gate trigger resistor wattage Pr = 0.25W
7. Line voltage Vac = 120V

Procedure:

1. Decide the peak wattage limit on the resistor: Prp <= Pr, Prp = 0.25W
2. Peak voltage across the resistor: Vr = Prp / Ig = 0.25W / 70mA = 3.57V
3. Resistance Rg = V/I = 51 ohms
4. Calculate actual average wattage on the resistor:
W = {[asin(3.75V/120V * sqr(2)) / asin(1)] / [1/(60Hz * 4)]} * 0.5 * 0.25W =
...., where..
-. asin() part is to calculate the percentage time of the resistor takes current.
-. 0.5 is to approximate the sine curve to ramp.
I am not doing the work to prove my calculation is incorrect. The point is the actual average wattage on the drive resistor is significant smaller than the peak wattage and the resistor rating. I forgot how the peak wattage of the resistors are spec-ed, likely from mechanical aspects.

There are other design parameters you may want to verify, if you are interested.

The calculation says, you can use "R35 + R36 = 51ohm" resistor safely, for 1/4W resistor. If the resistance gets higher, the trigger voltage gets higher, the wattage gets larger, the load gets sharp current increase at every cycle of the line.. etc.

Try something much lower value than what you have now. I would start from a single resistor, about 82ohm 1W. I did not calculate what the resulting Vt & Ir will be.

Meantime, we can calculate how the resistors are burning, if you can offer an enough motivation.

Wait a moment.. there is a definition for Tgt, gate-controlled turn-ontime. But, nothing for rated impulse peak gate current. Wonder if it matters.