1
\$\begingroup\$

I need to calculate the time a thyristor/triac needs to be on for a desired percentage of full power of each half cycle of the AC line input.

On for the full half cycle will give me 100%. On for 1/2 half the cycle will give me 50%. (sounds easy so far) But being on for 25% of the time does not give 25% of the power. And this is where I need someone with more brain cells than I still have. I have done some Google searching but have been unable to find the answer. Google doesn't understand what I really need.

Trying to think this through is stressing the remaining cells I have. And if I was confident I would get it right then it would be worth it. But I was never great at this kind of math even when I had cells to lose.

What I need is a an input variable, desired power percentage, and an output of the percentage of the half period of the sine wave.

\$\endgroup\$
  • \$\begingroup\$ Integrate for the area under a sine wave? \$\endgroup\$ – Tyler Apr 15 at 0:15
  • \$\begingroup\$ That does not help me. If I had the area then I would have already have the time. What I need is the time. I understand what you are saying. It is just math, but math that I am having difficulty with. I could do a trial and error method, calculate for X time and then see what the percentage is. Then build a table. But that is not a great solution. \$\endgroup\$ – Rudy Apr 15 at 0:19
0
\$\begingroup\$

If you really want power, which you probably do, you want v^2, since for most loads, power is usually proportional to v^2.

v(t) = (sin(t))^2

Integrating yields: t/2 - sin(2*t)/4

Evaluating from 0 to t is the same, since sin(0) = 0.

To get a percentage, you need to divide by the value at 100%.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ The numbers look good. But I am unable to work it backwards into a formula where I put in power and get out degrees. \$\endgroup\$ – Rudy Apr 15 at 1:26
  • \$\begingroup\$ Although I am trying. \$\endgroup\$ – Rudy Apr 15 at 1:32
  • \$\begingroup\$ Solving backwards is beyond my ability. If you can describe it as a purely math problem, someone in the math group might be able to help. Or, use a look-up table, or use Excel curve-fit functions to create an approximation (they can be very accurate). \$\endgroup\$ – Mattman944 Apr 15 at 1:49
  • \$\begingroup\$ Thank you for what you have done. If I can't get a formula then I will take what you have here and work it into a lookup table. I was never good at calculus. I passed but I really didn't get it. I will try and get something framed for the math group. \$\endgroup\$ – Rudy Apr 15 at 1:53
0
\$\begingroup\$

I'm going to approach this by taking the ratio of the average power, as a function of angle, to the average power for a whole half cycle. You can then equate this to the desired power percentage and solve for the angle.

Average power for a single half cycle is proportional to the following (the R and period will cancel out when we take the ratio)

\$ \int_{0}^{\pi}sin^2(wt)dwt = (wt/2 - \frac{sin(2wt)}{4})\Biggr|_{0}^{\pi} \$

\$ = \pi/2 \$

Therefore the ratio is

\$ \frac{2}{\pi}(wt/2 - \frac{sin(2wt)}{4})\Biggr|_{0}^{\theta} \$

\$ Power Ratio = \frac{\theta}\pi - \frac{sin(2\theta)}{2\pi} \$

This is a non-linear equation with no closed form solution (I believe). Similar to the previous answer you can plot the RHS for various values of the angle and see where it approaches your desired power ratio.

You can then find the required percentage of a half way by taking the angle, dividing it by pi, and multiplying by 100.

Edit:

Here's a plot of the power percentage against the angle

enter image description here

and here's a plot of the percentage of the period against the percentage of power required.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.