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# Why use RMS

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RMS produces the same power dissipation as it's DC equivalent. 6vrms produces the same power dissipation as 6vdc. The mathematical representation is as follows:

square the equation, integrate, then take the square root. There must be zero offset to get the result Vrms = .707Vp.

It is the job of the microprocessor to calculate RMS. Each value is squared in the microprocessor. The RMS is found by integrating and then taking the square root.

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Kevin,
Are you talking about a True-RMS meter that uses a special chip to do the mathematical function, or a simple meter that average or peak rectifies the signal, and assumes that it is a pure sine wave during its adjustment calibration?

The simple meter does not remove the DC, its input capacitor blocks any input DC. Its recifier converts the AC to its equivalent DC for it to measure.

The True-RMS mathematical chip does not change the wave by converting it into a square wave. It mathematically multiplies the number by itself.
This mathematical function does not occur first, but last. First, the average is mathematically square-rooted, then integrated, then squared.

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I am certain that the mathematical order of operation is to square the function, integrate, then take the square root. If you have the output from the A/D, you can determine peak to peak, average, and RMS.

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Kevin,
Thank-you for correcting me. I am sorry that I got the sequence wrong.
A good description of RMS, simple meters and true-RMS meters is here:
http://www.ee.unb.ca/tervo/ee2791/vrms.htm

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It is my opinion that we should find a way to measure waveforms without relying on an A/D process. You have to remember that even an unknown waveform can be electronically deciphered using a circuit. For instance, a full wave rectified sine wave can be filtered to give you a ripple which can be used to determine the average. How could one determine the average using the ripple?

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Kevin,
If the filter uses equal charging and discharging resistances then its output is the average.
Ripple can be reduced by using a slower charge and discharge of the filter capacitor.

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It does not give you the average. It gives you the peak of the voltage with ripple, which can be used for determining the average. The ripple contains the information about what happened to the filtered part. For instance, a reduced ripple would indicate a greater average.

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Kevin,
No, a circuit that equally charges and discharges a capacitor cannot end up with the peak voltage, but will end up with the average.
As an example, take a perfectly symmetrical square wave that goes from 0V to 5V. Feed it through a 10K resistor to a DC-coupled scope with a high input resistance and it will show the square wave from 0V to 5V. Now put a big capacitor across the scope's input to form a filter. The scope will show a very small ripple waveform that is at exactly 2.5VDC, the average voltage of that square wave. Increase the value of the capacitor and the ripple will disappear, and the scope will still show the 2.5V average voltage.
The ripple is the remainder of the signal that has had most or all of it filtered away.
Maybe you are thinking of a power supply filter capacitor and its ripple. The very low resistance transformer winding and low resistance rectifier quickly charge the capacitor to a voltage that is near the peak voltage of the transformer's sine wave. The load discharges the capacitor slowly until the next peak voltage quickly charges it again. On a scope the ripple will show a sawtooth shape of the fast rising charging and the slow ramp-down of the discharge. If you increase the load then the ripple will also increase because the capacitor is discharged faster.

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The capacitor filters the wave, thus changing it. The peak of the input allows for DC charging of the capacitor, then input voltage drops and you are left with the DC charge that will not drop because of the rate of change. What I had in mind is a full wave rectified sine wave whose average is less than what is produced after filtering.

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