Instrument, measurement, results

[SG-1]

Here are the results for the meters I tested using the square wave posted before.

The Ames Meter from Harbor Freight is a lot better meter that I expected when I bought it.

The final results for the CFL non-linear current tests are here to.

 

[gar]

180513-1520 EDT

SG-1:

Your results are interesting.

Your rectangular waveform swings between +5 V and -5 V and is +5 for 90% of the time.

The average value of this is 5*0.9 - 5*0.1 = 4.5 - 0.5 = 4.0 . On DC every one of the meters read 4 V as it should be.


The true RMS value of this waveform is 5 V as can be easily seen by inverting the negative pulse and the waveform is a continuous 5 V DC value.


Only the Fluke scope read the true RMS value correctly in AC mode as it also did in AC+DC mode.

Three others essentially read the RMS value of the AC component. Probably means there was an input coupling capacitor that removed the DC component.

The remaining two were grossly incorrect.


Peak to peak is only correct for the Fluke scope.


From these results it should be very clear to everyone that you must know how your instrument works and relates to what you want to measure.

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[SG-1]

In regards to the other meters that were grossly incorrect.
If you sum the AC & DC readings you get the correct RMS value.
The Fluke 289 read correctly in AC+DC mode.

Does your Fluke 27 AC & DC values sum to 5 ?

The biggest issue seems to be that out of the box you do not know if the AC setting will remove the DC component & it takes special equipment to find out, like a doorbell transformer & battery in series ?

If you know it does, then you can sum the AC & DC readings, but two measurements are always necessary.

How often would a residential electrician run into this ?

 

[gar]

180513-2356 EDT

SG-1:

For an arbitrary waveform with both AC and DC components you can not draw the general conclusion that the sum of the average DC value and the RMS value of the AC component equals the RMS value of the composite waveform.

Consider my square wave example, meaning 50% of the time one part of the square wave is more positive than the other 50%.

First, consider the + and - values to be 5 V. DC average is zero, and AC RMS is 5 V. So AC plus DC is 5 V RMS.

Second, add a +5 V bias. Now the waveform is +10 V 50% of the time and 0 for the other 50%. The average DC value is 5 V. The RMS value of the AC component is still 5 V, but the true RMS value of the composite is sq-root of ( (10^2)*0.5 ), or sq-root of 50 which equals 7.07 V. It is not 10 V.

A doorbell transformer and a battery are a good way to prove whether the AC mode includes a capacitor. Actually only the battery is needed.

Both my Fluke 27 and 87 have an input capacitor in AC mode.

In new times the ordinary electrician that does any troubleshooting probably needs a greater understanding of how their instruments, and circuit devices work.

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[SG-1]

For an arbitrary waveform may I draw the general conclusion that the sum of the RMS value of the AC waveform plus the RMS value of the DC waveform is the RMS value of the composite waveform ?

Don't miss the attached PDF.

It's interesting how good the (Harbor Freight) Ames meter is. It did not measure any frequency in AC for this waveform though.

I checked a Fluke 27II last night, it also has an input capacitor for AC mode.
This seems to be a standard on DMMs.

 

[gar]

180516-1015 EDT

SG-1:

For an arbitrary waveform may I draw the general conclusion that the sum of the RMS value of the AC waveform plus the RMS value of the DC waveform is the RMS value of the composite waveform ?

No.

But the power related to each separate RMS value can be added together, i.e. get total average power, and then take the sq-root to get an equivalent RMS value for the composite.

You see this, sq-root of the sum of the squares, done relative to harmonic content.

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