# Divide power using triac ; formula ; calculation

Discussion in 'Embedded Systems and Microcontrollers' started by Vindhyachal Takniki, Dec 15, 2014.

1. ### Vindhyachal Takniki Thread Starter Member

Nov 3, 2014
357
6
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 .

File size:
30 KB
Views:
45
2. ### JWHassler Member

Sep 25, 2013
204
34
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

3. ### JWHassler Member

Sep 25, 2013
204
34
..... 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

4. ### JWHassler Member

Sep 25, 2013
204
34
... and then the analytic solution pops right out! (I think)
power_fraction = ( arccos(sqrt(cycle_fraction)) ^ 2)/π

Apr 24, 2011
7,436
1,626
6. ### Vindhyachal Takniki Thread Starter Member

Nov 3, 2014
357
6
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??

File size:
37.5 KB
Views:
26
7. ### JWHassler Member

Sep 25, 2013
204
34
@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