Linearize Phase Control SCR from Angle to RMS

Archie

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Join Date
May 2002
Location
Orangeburg, SC
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This is a bit of an academic exercise and I wanted to check if anyone has already developed a formula before I start diving into some integral calculus.

I am experimenting with precision heat control using a phase control SSR that uses a 4-20mA signal. A phase control relay controls AC voltage output by switching on at a certain point in cycle then back off at zero crossing. At 50% it will switch on at the peak of each sine wave cycle, therefore giving an RMS output of 50% of the source (e.g 60Vrms from a 120Vrms source).

When the signal goes on either side of the 50% mark, it is not linear. For example at 75% (16mA control) you may expect to get 90Vrms, but you will get higher, let's say roughly 110Vrms.

What I want to do is to pass my control signal through a linearizing function so if I call for 75%, it will switch on at the correct angle in the sine wave to give 90Vrms. This would be formula to calculate the area under sine wave based on the angle.

I have attached an image I got from Wikipedia to give a visual of a phase control relay output.

PhaseAngleControl.gif
 
I think this is right.

The integral of the sin(a) is -cos(a)+C. The definite integral of the sine function is the difference between the values of the integral at the two points of interest:

(-cos(b) - -cos(a) +C, where "a" is the basis angle and "b" is the end angle. In our case each of the control periods is its own unit and we know the power starts at zero for the control period. This will simplify the definite integral as we know C is zero and the basis angle will be zero. So the total power as a value between zero and 1 in one half of the sine wave is:

Power = (-cos(b) + 1)/2.

The "+1" part is just "- -cos(0)" from the basis angle portion above. The "/2" is to normalize the result to 1 since the integral of 1/2 a unit sine wave is 2.

At this point, solving for "b" will give you the angle required for a desired relative power level:

b = arccos(1 - (2*Power))

Remember, the way this is written, "Power" is a relative value between 0 and 1.

Keith
 
Vrms=Vpeak*Sqrt((2pi - 2alpha + sin(2alpha)/4pi)


Where alpha=0 for full conduction and alpha=pi for no conduction.


I would either generate a look-up table of conduction angles against rms and use that, or solve the equation numerically by starting alpha at zero and incrementing it until the result is close to your desired rms.
 
I started with kamenges formula:

b = arccos(1 - (2*Power))

and turned it into a function that gives me a raw analog output 0-32767 from the percentage of target Vrms (0-1):

LinearRMS:=REAL_TO_INT(ACOS(1-2*Voltage)/3.141592654*32767)

Looking at the output on an oscilloscope and roughly counting the area under the curve, this appears to be right on at the different percentages.

Brilliant math! Thank you!

This is a part of a project to develop tight temperature control for a press. Another part of this is to write the software that will learn the heat propagation rate through the die, heat losses in different ambient temperatures, energy required to bring material introduced into the mold up to the target temperature, etc. Using this data a predictive algorithm will then control the heater to minimize temperature fluctuations.
 
I guess it's the average versus RMS difference. I'll do some test with both formulas and see if I can accurately measure power output. It's the actually wattage into a resistive load that I am truly interested in linearizing the control.
 
I'm fairly certain L D[AR2,P#0.0]'s formula is the correct one. I solved what was asked without really thinking about what it meant. My bad.

Keith
 

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