- #1

DC12

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- 0

In a system that uses a triac to switch mains voltage, how can one calculate the resultant peak voltage, RMS voltage and power for a given phase (conduction) angle e.g. 50°?

Many thanks

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- Thread starter DC12
- Start date

- #1

DC12

- 4

- 0

In a system that uses a triac to switch mains voltage, how can one calculate the resultant peak voltage, RMS voltage and power for a given phase (conduction) angle e.g. 50°?

Many thanks

- #2

I_am_learning

- 686

- 16

RMS volatge = Sqrt of avg of square of instantaneous voltage. This you calculate by integration. And for finding the power, you need wave-form of current as well.

The you find power = Avg of ( V(instantaneous) * I(instantaneous) ) over a cycle.

In short, you need to do some work. :)

And, welcome of PF.

- #3

DC12

- 4

- 0

Drawing graphs is one way but I need a more technical solution. I need the formula!

Just a mathematical way to calculate the peak voltage based on a phase angle would be sufficient.

Here's hoping...

- #4

I_am_learning

- 686

- 16

Sorry, but we can't do your homework here.

- #5

DC12

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- #6

DragonPetter

- 831

- 1

I think he's saying just work out one or a few cases, and find the relationships. Then you can apply what you see to a general case. I would tell you, but I don't know an equation for this off the top of my head.

- #7

Jony130

- 627

- 138

I think that equation for RMS voltage across the load look like this:

[tex]V_{load} = Vpeak *\sqrt{\frac{2 \pi - 2 \varphi + sin2 \varphi}{4\pi}}[/tex]

Where [tex]\varphi - triac- angle- delay = triggering -angle[/tex] .

[tex]for-0\varphi - triac- is- full- on[/tex] .

[tex]And for- \frac{\pi}{2} V_{load} = Vpeak/2[/tex] .

Or use this graph

http://educypedia.karadimov.info/library/an1003.pdf

[tex]V_{load} = Vpeak *\sqrt{\frac{2 \pi - 2 \varphi + sin2 \varphi}{4\pi}}[/tex]

Where [tex]\varphi - triac- angle- delay = triggering -angle[/tex] .

[tex]for-0\varphi - triac- is- full- on[/tex] .

[tex]And for- \frac{\pi}{2} V_{load} = Vpeak/2[/tex] .

Or use this graph

http://educypedia.karadimov.info/library/an1003.pdf

Last edited:

- #8

Jony130

- 627

- 138

I thought that I shows some example.

For V = 230V we have 325V peak and conduction angle 50° so the triggering angle must be equal to 130°.

First we must convert 130° degrees to radians:

**r = 130 * ∏/180 = 130 * 0.0174 = 2.26 [rad] **

I use Wolframalpha

http://www.wolframalpha.com/input/?i=sqr(+(2pi+-+2*2.26+++sin(2*2.26))/(4pi)+)

but you could use Google too

**Vload = 325V * √ ( (2*pi - 2*r + sin(2r) ) / (4pi) ) = 325V * 0.249 = 80.9Vrms**

For V = 230V we have 325V peak and conduction angle 50° so the triggering angle must be equal to 130°.

First we must convert 130° degrees to radians:

I use Wolframalpha

http://www.wolframalpha.com/input/?i=sqr(+(2pi+-+2*2.26+++sin(2*2.26))/(4pi)+)

but you could use Google too

Code:

`sqr( (2pi - 2*2.26 + sin(2*2.26))/(4pi) )`

Last edited:

- #9

sophiecentaur

Science Advisor

Gold Member

- 27,725

- 6,295

The peak value is just the value of the sin(θ) at the turn on time.

This assumes a resistive load, of course.

- #10

DC12

- 4

- 0

Perfect. Thanks Jony130 and others!

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