Secondary voltage electronic halogen transformer

Thread Starter

YokoTsuno

Joined Jan 1, 2013
43
I am trying to derive a scaling factor for an analog voltmeter for the purpose of measuring the secondary voltage of an electronic halogen transformer.

http://www.ledbenchmark.com/faq/Transformers-Output-and-Compatibility.html

The output voltage of these things is a "high frequency" square wave (30-100kHz) with a 100 Hz (line frequency) envelope (see link). Digital multimeters often have problems measuring this unless you opt for high bandwidth, True-RMS models with a considerable price tag, so why not try an analog meter.

The RMS specs of the DUT (Osram HTM70) is 11.5V. Tried 3 different types including 2 wideband audio VOMs one with a BW of 1MHz and all 3 show a 10V reading. Analog meters measure the average value and apply a scaling factor of Pi/2sqrt(2) to adjust the reading to the RMS value of a sine wave. Since the average value of the above waveform is the same, at least to my understanding, I expected the same reading. This is obviously not the case.

Where's the flaw in my logic?
 
The "average" of a periodic waveform that spends an equal area above and below the time axis is ZERO.

RMS AC is the the same as an equivalent amount of DC that dissipates the same power in a fixed resistance.

Ave = (1/t2-t1) ∫ f(t) dt; See: http://archives.math.utk.edu/visual.calculus/5/average.1/

RMS = Root Mean Square or https://en.wikipedia.org/wiki/Root_mean_square

The fudge factor means that you know the shape. The fudge factor works for sine wave inputs. Usually,, there is another creiteria used and that's the crest factor. Note here https://en.wikipedia.org/wiki/Crest_factor there is a reference to the applicability to multimeters. The frequency response of the meter also plays a part.

I made a quick box to do UV intensity measurements of a UV source that basically used a neon light transformer. The waveform was pretty weird and lop-sided. I then took the output of an appropriately filtered reference photodiode and did a current to voltage conversion and applied that to an RMS reading meter. We actually had a real UV meter, so we calibrated our quick and dirty one to that.

Intensity will be proportional to power. With power measurements, there are losses. Some system may look at P = A * V^2 from P=V^2/R

Some TRMS meters allow measuring the AC portion and the sum of the DC+AC portions. It might be better to call it the steady state voltage and the time varying voltage.

So, either have real equations for your voltage or numerically apply the formulas to the values of v(t). You can easily numerically integrate. https://en.wikipedia.org/wiki/Numerical_integration
 

Thread Starter

YokoTsuno

Joined Jan 1, 2013
43
Thanks for the effort you put in. Unfortunately it doesn't shed any new light on the problem.

The "average" of a periodic waveform that spends an equal area above and below the time axis is ZERO.
Obvious, but that is not how an analog meter works. The needle position is proportional to the average value of the rectified ac voltage. Since most people are only interested in the RMS value, a fixed scaling is applied.
RMS AC is the the same as an equivalent amount of DC that dissipates the same power in a fixed resistance.
This is clear to me.
So, either have real equations for your voltage or numerically apply the formulas to the values of v(t). You can easily numerically integrate.
These tools only confirm my understanding. You can fill out the following equation: (2/3.14*sin(100*x) + 2/(3*3.14)*sin(300*x)+ 2/(5*3.14)*sin(500*x)+2/(7*3.14)*sin(700*x))*2*sin(x). Boundary [0,314] in this tool for instance:

http://www.zweigmedia.com/RealWorld/integral/integral.html

The equation is the mathematical representation of the voltage coming out of the transformer (square wave with an envelope of a sine). The result produced by the tool deviates 1-2% from the average value of a sine wave (2/Π*Vamp), this can be explained by the finite rise time of the square wave(finite number of harmonic terms). Even for a real transformer the rise time is finite (plus the fact that there's small gap in the waveform, internal oscillator needs minimum voltage to start), that's why the amplitude of the secondary voltage is 19V (I verified this with an oscilloscope and not 16,2V as one would expect.
 
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