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Question about digital clamp ammeter reading technique/philosophy/principle
- Thread starter ArchieMedes
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gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
120208-1601 EST
ArchieMedes:
I have been gone for 4 days on an emergency service call on a machine that was close to shutting down an automotive plant. Long days and I am tired, but I will get back to you later on your question.
.
ArchieMedes:
I have been gone for 4 days on an emergency service call on a machine that was close to shutting down an automotive plant. Long days and I am tired, but I will get back to you later on your question.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
120208-2043 EST
T.M.:
2.85 is incorrect. Your other answers are correct. Rethink the analysis that produced the 2.85 . Something was lost.
.
T.M.:
2.85 is incorrect. Your other answers are correct. Rethink the analysis that produced the 2.85 . Something was lost.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
120208-2055 ESt
ArchieMedes:
An AC meter, such as a Simpson 260 or Fluke 27, is based upon full wave rectification of the AC input resulting in a DC output with some average DC value. For a sine wave signal with no DC component the average value is 0.636 times the peak voltage of the sine wave. If a half wave rectifier was used, then the average value would be 0.318 times the peak voltage. In essence this is the area under the curve. The easy way of getting a very accurate value for this constant is with calculus.
If you look at the definition of an RMS measurement from a mathematical point of view and then relate this to power in an electrical circuit you find that RMS measurements can be related to power in an AC or DC circuit like you would calculate power in a DC circuit.
Fundamentally an RMS measurement is the square of a variable, determination of its mean (average) value, then obtaining the square root of that average.
The average reading meter gets the average value of the absolute value (full wave rectified) of the signal multiplied by 1.11 and displays it.
The RMS reading meter multiplies the input signal by itself producing a non-negative result. This result is averaged, and the square root of that is obtained and displayed.
In one case you simply average the measurement. and in the other measurement the output is the square root of the average of the squared value of the signal.
When you look at different input signal waveforms and process them with the two different methods you will find some cases where the results are quite different. Yours being one illustration.
.
ArchieMedes:
With reference to the above highlighted text, how come the RMS reads higher than the average meter? What is the theory behind?
An AC meter, such as a Simpson 260 or Fluke 27, is based upon full wave rectification of the AC input resulting in a DC output with some average DC value. For a sine wave signal with no DC component the average value is 0.636 times the peak voltage of the sine wave. If a half wave rectifier was used, then the average value would be 0.318 times the peak voltage. In essence this is the area under the curve. The easy way of getting a very accurate value for this constant is with calculus.
If you look at the definition of an RMS measurement from a mathematical point of view and then relate this to power in an electrical circuit you find that RMS measurements can be related to power in an AC or DC circuit like you would calculate power in a DC circuit.
Fundamentally an RMS measurement is the square of a variable, determination of its mean (average) value, then obtaining the square root of that average.
The average reading meter gets the average value of the absolute value (full wave rectified) of the signal multiplied by 1.11 and displays it.
The RMS reading meter multiplies the input signal by itself producing a non-negative result. This result is averaged, and the square root of that is obtained and displayed.
In one case you simply average the measurement. and in the other measurement the output is the square root of the average of the squared value of the signal.
When you look at different input signal waveforms and process them with the two different methods you will find some cases where the results are quite different. Yours being one illustration.
.
T
T.M.Haja Sahib
Guest
Will you please highlight it?120208-2043 EST
T.M.:
2.85 is incorrect. Your other answers are correct. Rethink the analysis that produced the 2.85 . Something was lost.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
120209-0832 EST
T.M.:
When you remove the DC component from the original waveform by coupling with a capacitor, then the output waveform is a 9 V positive pulse for 10% of the time and a negative pulse of 1 V for 90% of the time. The squares of these are 81 for 10% and 1 for 90%, and the two areas to be added are 8.1 and 0.9 which equals 9.0 . The square root of 9 is 3. So the RMS value is 3 V.
.
T.M.:
When you remove the DC component from the original waveform by coupling with a capacitor, then the output waveform is a 9 V positive pulse for 10% of the time and a negative pulse of 1 V for 90% of the time. The squares of these are 81 for 10% and 1 for 90%, and the two areas to be added are 8.1 and 0.9 which equals 9.0 . The square root of 9 is 3. So the RMS value is 3 V.
.
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