# RMS voltage calculation in phase firing angle

1. ## RMS voltage calculation in phase firing angle

I have this formula to calculate the load RMS voltage: I'm working on 220 VAC 50 Hz.

I try delay = 0 degree ==> Vo = 220 V this is right, the RMS

delay = 90 degree ==> Vo = 155.5635 why ?? It should be 110 V right ? 2. ## Re: RMS voltage calculation in phase firing angle

No. 90° is half the power or 71% of the RMS voltage. 3. ## Re: RMS voltage calculation in phase firing angle

No. 90° is half the power or 71% of the RMS voltage.
Ok, can you give me more explanation please so I can understand clearly the theory here. 4. ## Re: RMS voltage calculation in phase firing angle

It's according to the definition of RMS value: average the square of input value over interval of interest, take the square root. Try with a cut sine wave and you'll get the result. http://en.wikipedia.org/wiki/Root_mean_square

1 members found this post helpful. •

5. ## Re: RMS voltage calculation in phase firing angle

How do you find 71% ?

So you mean delay 90 will give half power (Watt) not half of the RMS or peak voltage ?

Actually I want to make a comparison between calculated RMS value with measured RMS value of the load, but by using the formula in post #1 give me too big difference. E.g. delay 90, measured RMS: 92 V while calculated RMS: 155.56 V. 6. ## Re: RMS voltage calculation in phase firing angle

There is a nice graph and some explanations in this thread. 7. ## Re: RMS voltage calculation in phase firing angle

The 92V value suggests that you are not measuring true RMS. You get 110V averaged rectified value, a cheap non-RMS meter can be expected to display it. 8. ## Re: RMS voltage calculation in phase firing angle

Well, If I want to make a comparison between the calculated load voltage with the measured load voltage, how I should make it?

Should I keep comparing the RMS voltage and find the difference ? or
Should I compare the average voltage ?

Or maybe I should compare the power ?

I'm getting more confuse now :( 9. ## Re: RMS voltage calculation in phase firing angle

You can compare whatever quantity you want, but should take care the calculated and measured value are comparable. E.g. if you calculated RMS quantity, you should measure RMS. If your instrument can't, you have a problem. 10. ## Re: RMS voltage calculation in phase firing angle

OK.

Seem like it is exactly what I need. Take a look at the output voltage calculation at the bottom. It says Vo=Vs.D where D is the total conducting time divided by period. My measurement result is close to that.

1. Now what voltage is it? Is it RMS ?
2. Did you know how the writer can derive that formula (Vo = Vs.D) ? •

11. ## Re: RMS voltage calculation in phase firing angle Originally Posted by ArdyNT Take a look at the output voltage calculation at the bottom. It says Vo=Vs.D where D is the total conducting time divided by period.
Sorry, that's wrong. Originally Posted by ArdyNT Did you know how the writer can derive that formula (Vo = Vs.D) ? 12. ## Re: RMS voltage calculation in phase firing angle

Sorry, that's wrong.
Wow, really? why ?
Can you suggest me another formula to measure the load voltage ? 13. ## Re: RMS voltage calculation in phase firing angle

Did you go look at the link godfreyl recommended?

Here's another chart: The red curve is the RMS value of the voltage versus conduction angle. That is the voltage that a true RMS meter will read (assuming the 220 VAC grid waveform is undistorted). The blue curve is the actual value of the average of the absolute value (full wave rectified) of the waveform, but this is not what a "non-RMS" meter will read. The black curve is what an average responding (non-RMS) meter will read; it's greater than the blue curve value by a factor of 1.1107.

So, your meter should read according to the red or black curve, depending on whether it's a "True RMS" meter or not.

- - - Updated - - -

Here's a derivation of the formulas for the RMS and average value of the triac generated waveform with 220 VAC RMS as the input grid voltage. The standard definition for RMS is used, and the average calculation is standard also. These formulas were derived using Mathematica and the formulas are not in the form a human would leave them, but they are correct and give the proper result.

Don't forget that you must multiply the average result by 1.1107 to get what an average responding meter would actually read. 1 members found this post helpful. 14. ## Re: RMS voltage calculation in phase firing angle

Ahaa, thank you very much.

I still have questions:

1. Why do you just integrate a half cycle (up to 180) ? Why don't integrate both positive and negative half cycle and then add them together in the end?
2. Could you please tell me how do you get 1.1107 ? What is the reason ? (My meter gives me the average voltage x 1.1107 and the derived formula you show is helped me a lot but I just want to know how to find 1.1107) 15. ## Re: RMS voltage calculation in phase firing angle

Why do you just integrate a half cycle
Because the 2nd half cycle has the same area.
how do you get 1.1107
It's pi/(2*√2) 16. ## Re: RMS voltage calculation in phase firing angle

Just in case you want more details of how the 1.1107 number is derived, here you go.

If one has a meter that is a so-called "average responding meter", but you want to determine the RMS value of a waveform, it will be necessary to multiply the reading by a factor which is the ratio of the RMS value to the Average (absolute) value of the waveform. This factor will be different for different waveforms, but since the usual waveforms measured are sine waveforms, the desired ratio is the RMS to average for a sine wave.

Here is how it's calculated: the numerator is the RMS of a sinewave, and the denominator is the average absolute (full wave rectified) value of a sine wave.  17. ## Re: RMS voltage calculation in phase firing angle

Ok I got it. I did search also and found that it is called as form factor.

Sorry if I am asking too much, from your explanation I can say that: Average voltage= (RMS Volt/1.1107) , am I right ?

Then I just try to calculate the RMS Volt (i.e. conduction angle 90) Vrms = 155.5 V.
Next, Average Voltage=155.5/1.1107 = 140 V

From the curve in post #13, RMS voltage of 90 degree conduction angle seems right around 155.5 V, but why I got the average voltage is 140 V and did not match with the average voltage curve. Average should be around 110 V right ? 18. ## Re: RMS voltage calculation in phase firing angle

The factor 1.107 is valid only for (full wave) sine voltages. Strictly speaking the factor will be observed between RMS measurement and average rectified value. The simple average of a sine wave is zero.

A second comment reagrding factor 1.107. A multimeter that measures average rectified value instead of RMS already uses it as an internal correction factor. So the reading is equal to a true RMS meter for sine voltages, but different for other waveforms, e.g. a phase angle controlled voltage. 19. ## Re: RMS voltage calculation in phase firing angle

Is is only true that "Average voltage= (RMS Volt/1.1107) , am I right ?" for a sine wave. When you have triac control of a load, the waveform is only a pure sine wave at a conduction angle of 180°. You'll notice that the red curve and the black curve coincide at 180°.

At any other conduction angle, the output waveform is most definitely NOT a sine wave. You understand that, right?

The actual average response of a non-RMS meter is multiplied by 1.1107 under the assumption that you are measuring a sine wave, and want the RMS value. If you're not measuring a sine wave (which is what is going on at any conduction angle other than 180°), then the multiplication by 1.1107 is not appropriate. That's why the black curve does not coincide with the red curve except at 180°.

One could use the formulas for RMS and average of the triac output waveform and calculate the ratio of RMS to average versus conduction angle (and also divide by 1.1107 to get a curve of the factor which relates the ACTUAL reading of a non-RMS meter to the true RMS voltage). Remember that the READING of a non-RMS meter is not the true average value of a waveform; it is 1.1107 times the average value, so that for a sine wave the actual reading of the non-RMS meter is the RMS value. This graph shows the result: Notice that the ratio is ONLY equal to 1.000 for a conduction angle of 180°. For other conduction angles, it is different. In particular, at a conduction angle of 90° the calculated ratio of RMS to the ACTUAL reading of an average responding meter is 1.41421 (which I recognize is Sqrt(2)).

1 members found this post helpful. 20. ## Re: RMS voltage calculation in phase firing angle

Ok, I see. Your explanations are very clear One last thing:

In here : The integration is only up to 180 and the omega is only pi/180 while here: it is up to 2pi.

What I want to clarify, is that ok to integrate only a half cycle of full wave like this: in order to find the average voltage ? •

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