How to calculate resistors for triac diac array?

[diegoterc3]

Hello. I have some trouble understanding and calculating resistor and capacitor values for a little reflow oven proyect I am trying to make.

The thing is this. I have a 700W electric oven and I am trying to control the power that goes through them with a triac diac array (the same that they use for light dimmers). I have attached a picture for the circuit.

Well I have the usual data: 2 heating elements that use 700W at 200VAC - 60Hz would use Irms=700/220=3.1818A. The resistive heating elements would have R=V*V/P=48400/700=69.14. Lets say 70Ohm.

Then I have chosen an optoisolator TLP160G and a BTA25-600C TRIAC. The values that I have chosen are acording to this app notes.


But I would like to know the WATT values for my resistors since my goal is to use everthing SMD but the triac. I need some info of how to calculate the resistance/voltage across some elements of the circuit. Any ideas?

Thank you in advance.

[Paul Price]

There are no diacs involved in the circuit, but you are using an optically coupled triac to trigger the power triac controlling the heating element.

There is only one resistor in the HV side of the trigger circuit. When the circuit is not triggered, the current trough the resistor is ~0.0mA  No power is being used.

When the triac is being triggered, the instantaneous current is (VAC/R) and certainly less than an amp, but since the triac trigger triggers in a few microseconds after the trigger isolator LED gets the trigger signal, very little average power is  dissipated.

Once the main triac is triggered, the voltage across this resistor is less than the ON voltage of the main triac (<4V ).

So, the answer is the resistor could be quite small in wattage, .5W would be quite conservative, but large enough in size to handle the HV applied across it.

The power dissipated by any resistor with DC or RMS VAC applied across it is:             W=(V*V)/R

So, I will leave it to you to calculate the wattage of the resistor feeding the optotriac LED.

[diegoterc3]

Hello Paul, according to this app note:

https://www.fairchildsemi.com/application-notes/AN/AN-3003.pdf

The minimum resistance for an optotriac is R=Vpk/Max surge current(1.2A for TLP160G)=230*sqrt(2)/1.2=271. That's why I chose the 300 ohm resistor.

Now, if I understand you correctly, for a brief instance between the led trigger signal and the optotriac trigger signal this 300 ohm resistor has to withstand the  Instantaneous current VAC/R=220/300=0.73A. After this, it only disipates less than the Peak On-State Voltage which is 1.55V for BTA25-600C.

In the worst case W=V*I=220*0.73= 160W?
I understand that I can change the resistor to 100k or something like that, but will this affect the activation for the HV Triac?

Please correct me if I am wrong. Thanks.

[Zero999]

The resistor only dissipates 160W for a few microseconds. The instantaneous power may be quite high but the over all power dissipation is low. Suppose it triggered within 5x10^-6s and the repetition rate is 120Hz: a 0.5W resistor is more than enough, work it out.

[diegoterc3]

Ok. I did a little research and found out that the average power for resistors is:

http://www.ee.ic.ac.uk/hp/staff/dmb/courses/ccts1/01400_AcPower.pdf  on page 5

Then used this code in matlab to find power dissipated in a 100us:

t=0:0.0001:0.25;
R=300;
f=60;
T=1/f;
w=2*pi*f;
Vrms=220;
Vpk=Vrms*sqrt(2);
fun=@(t)(((Vpk*sin(w*t)).^2)/R);
Teval=100*0.000001;
Pavr=(integral(fun,0,Teval))/Teval

The answer is around 150mW of power. If this is right, instead of 100us I would need to calculate it for the time that the TRIAC needs to turn ON. I can´t find this information on the datasheet. Is there any good approximation?

Thank you.

[Paul Price]

The time required for the TRIAC to be triggered depends on the LED current, LED pulse width and the polarity and amplitude of voltage across the TRIAC, but a .5W resistor would in any case be a best choice because it must handle a peak HV (>340V with 230VAC circuits) across it prior to triggering.

With just a few volts across the TRIAC it will  trigger in a few microseconds, certainly <100uS.

[diegoterc3]

Ok I have a new question:
The average power in a resistor is Pavr=(1/R*T)*integral(0,T,Vpk*sin(2*pi*f*t)). That's what most textbooks say.
 Since the time it takes the TRIAC to activate is <100us. This would mean that the power dissipated in the resistor is less than the time of half cicle of the sinusoidal wave (See picture).

If I wanted to evaluate the average power dissipated in this period of time the formula would be (MATLAB):

To=0.003;
Tact=0.000001;
Teval=To+Tact;
t=To:0.0000001:Teval;
R=300;
f=60;
T=1/f;
w=2*pi*f;
Vrms=220;
Vpk=Vrms*sqrt(2);
x=Vpk*sin(w*t);

fun=@(t)((Vpk*sin(w*t)).^2);
Pavr=(integral(fun,To,Teval))/(Tact*R)


Where
To is the initial time.
Tact is the TRIAC activation time.
Teval is the evaluated time or final time.
R is the resistor value.
f is grid frequency,
and the rest is self explanatory.

The answer to this is 264.2179W which is too much. For this I am assuming that:

Pavr=(1/R*(Tact-To))*integral(To,Teval,Vpk*sin(2*pi*f*t)), since I am only solving for a short segment of the sine wave.

Is this correct? Thank you in advance.

[oldway]

SMD resistor can't work with peak voltages higher than 100V.
I have had such problem with SMD resistors in snubbers.
You have to use at least 3 SMD resistors 1206 of 100R in serie.

Since the time it takes the TRIAC to activate is <100us.

Wrong, 100µs seems to be Toff and not Ton.
Ton is very short, 1 or 2µs.

[diegoterc3]

Yes. In the code I evaluated for 1us and I will work with a 1/2W resistor. Anyway this half watt is nothing compared to 264. My question is wheter the watt calculation is correct or not.

[oldway]

Yes. In the code I evaluated for 1us and I will work with a 1/2W resistor. Anyway this half watt is nothing compared to 264. My question is wheter the watt calculation is correct or not.

You don't know what is the right waveform of voltage across the resistors....How would you calculate the power than?
You try to guess, not to calculate... 
First measure the waveform with an oscilloscope and than you can calculate the power.
Don't worry, with 3 x SMD 1206 resistors (3 x 0.5W = 1.5W) in serie, that's OK.

[diegoterc3]

I know the wave across the resistor is the sine wave that comes from the electric grid. I showed the circuit diagram above.

[oldway]

That is not right: this sine wave will be apply to the resistor only a very short time when TLP160G is triggered.
Max. dissipation in resistor will occur if triac is triggered at max. peak voltage of sine wave.
But you don't know how long this sine wave is applied to the resistor (probably more or less 2µs), nor what are the rise and fall times of this waveform.

[diegoterc3]

Yes that's what I am assumming. As you say I can not now that rise and fall times of the wave form, also we agree that the max dissipation in the resistor will occur if triac is triggered at max peak of the sine wave.

So keeping this in mind. A segment of the sine wave will pass through the resistor for the TLP160G trigger time. In the code I am calculating the average power for this segment at the worst case scenario. Since the freq is 60Hz I know that the peak is on [(1/60)/4] seconds and I am assuming that the trigger time is 1us. So having this of the average power formula gives me 264W.

What I don´t know is if the T of the basic formula becomes (Tf-To) for a segment that doesn´t start at zero for 1us.

[diegoterc3]

Ok. I hope this image is helpfull. The formula above is the formula we all know from textbooks.

The second image at the bottom demonstrates the period of time that I am trying to evaluate. I am not sure if 1/(Tf-To) is correct.

Thank you.

[oldway]

If you accept to simplify the calculation and you don't worry about rise and fall times, you can also consider that during a short time (2µs), voltage is constant and equal to the peak value if the sine wave.
Than you have to calculate the average and rms voltages applied to the resistor.
It is a rectangular pulse wave of peak voltage = 220V x square root of 2 = 311V and of 2µs every 1/120 = 8.33ms (T).
Duty cycle (D) is : 2µs/8.33ms = 240 E-6
Than, average voltage is 311 x 240 E-6 = 74.6 mV
rms voltage = 311 x square root of 240 E-6 = 4.82V
With resistor = 300R,
Power = V²/R
Average power: 18.6 E-6 or 18.6µW
rms power: 77.4 E-3 or 77.4mW
The rms value is the true power dissipated in the resistor.
This power is very low because de duty cycle is also very low.

NB: you should measure the true duration of the waveform with an oscilloscope...We estimated this duration at 2µs, but it can be greater.

[diegoterc3]

Ok that just gives me a better idea.
I won´t be turning the TRIAC on and off periodically. I will just turn it on once and after some sensing off. But it won´t be synchronized with the grid. I think is a good idea to assume that it is a squared wave form but in that case the answer would be 220sqrt(2)^2/300=322W using the average power formula. Is this correct? I am just evaluating for 1 instance of activation.

[oldway]

But it won´t be synchronized with the grid. I think is a good idea to assume that it is a squared wave form but in that case the answer would be 220sqrt(2)^2/300=322W using the average power formula. Is this correct? I am just evaluating for 1 instance of activation.

It is not correct.

If you feed the opto diode of the optoisolator TLP160G with a dc current or a long and not sincronised pulse, you still have a very low dissipation in the gate resistor.

At every half cycle, as soon as voltage between MT1 and MT2 is high enough (probably 3 or 4V), triac of the optoisolator will conduct and the main triac BTA25-600C will trigger as soon as Igt is reached (voltage of more or less 15V on 300R resistor).

Voltage between MT1 and MT2 drop then to more or less 2V and there is no more any dissipation in the gate resistor.

Again, power dissipation in 300R resistor stay very low, nothing to do with 322w ...