1. I am using a pure resistive load & have to divide output power by using triac. Ckt has 220V ac

2. I have a zero crossing detector output signal at every zero so I get two signal in once complete cycle.

3. User can select 0.0 – 1.0 value & according to that I have to output the power .
e.g if user have selected 0.5 then I have to output half power during 0-180 phase of ac signal.

4. I have some calculation on when to turn on triac which i have attached.

5. Plz check if I doing it right.

6. If I am doing it right, currently I am calculating angle by hit & trial. But I have to do it in MCU, I don't know how to calculate it otherwise because formula contain arithmetic & trigonometry function as in attached .

 

The output power will be proportional to the integral of the voltage, so the shape of the output is cos(t).
If the triac is triggered at t=0, the power will be maximum. If triggered at just short of ∏ radians, the power will be zero
I think a lookup table would do fine for what you show, a resolution 10% of full-power

 

..... except that power is proportional voltage squared: d'oh.
An analytic solution is now somewhat harder, yes sir, but a lookup table is still doable.
Graph the function (cos(α)/2 +.5)^2 in the region [π:0] to get power-fraction vs cycle-fraction

 

... and then the analytic solution pops right out! (I think)
power_fraction = ( arccos(sqrt(cycle_fraction)) ^ 2)/π

 

Littelfuse has several great ap notes on the subject.

 

1. I have attached a file, in which it is explained how to do phase control.
2. First one is equal phase division. Which I have done correctly.
3. Second is equal power division. this is what I am trying to achieve, its calculation I had described in post#1.
I want to know if I am doing it correctly.


@JWHassler: How do you have calculated the formula? Is my method wrong?

@ErnieM: Figure AN1003.4 in app note have this method I guess. Do my calculations are ok??

 

@JWHassler: How do you have calculated the formula? Is my method wrong?
I'm not sure mine is right.
Here is a graph of your results, though:


Looks quite like the curve in the Littelfuse app-note cited above (except reversed, showing start-time instead of conduction-angle.)
Excel is helpful in these things