Is there such thing as "DC true RMS" ?

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If you're driving a light bulb with a battery, you have a steady state, DC, so in this case there's no reason to discuss "true RMS", but is there such thing as "RMS DC voltage/current" with a varying unidirectional current/voltage?

I read that "true RMS" refers to the area under curve in relation to DC voltage within the same limits of integration.

Let's say you have 10% duty cycle 300V pulse from t=0 to 10. If you integrate the area under curve, it is the same as steady state 30V from t=0 to 10, so can you call that particular signal a "300pk, 30v rms" ?

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Yes. RMS is applicable. It is simply the root { average of the square of the voltage or current}. You have calculated the average or pure DC component of the voltage. For pure DC the rms value =average =peak but that is not true for your duty cycle

Initially the use of rms was defined in conjunction with power and the magnitude of the rms current for any waveform is the same as the magnitude of pure, steady DC for the same power. That is for 120Vrms at 60Hz, the average power dissipated in a given resistor will be the same as would be dissipated at 120VDC. This "equivalent heating" concept is more convenient than use of average or peak voltage. For your 10% duty cycle example, the rms value is given by: root[{(300*300*0.1) +0*0*0.9)}/1] = 94.9V rms

You have calculated the average DC component. Your 10% duty cycle will consist of a DC component which has an average of

30V and a superimposed AC signal which will be rich in harmonics. The added AC signal will be +270V for 10% of the cycle and -30V for the rest of the cycle. Note that the power dissipated into a 100 ohm resistor during a duty cycle will average (300^2)*0.1/100 =90 watts of which 9 watts will be due to the DC component and the other 81 due to the AC components .
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Don Kelly

So, 4v 0-pk 50% (calibration output on a tek scope) duty cycle square wave measures as 2.0v DC and 2.0V AC on a meter, but using the "DC + AC true RMS" function, it is reported as "2.82v". Does this mean that its "equvalient heating" is comparable to 2.82v steady DC?

Also, how exactly is it "AC contents" when there is no alternation (changing polarity) ?

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4V for 1/2 cycle, 0 for the other half. rms value 2.82V

Or consider 2VDC average +2Vpeak square wave rms value of DC =average value =2V rms value of square wave =2V total rms value =square of sum of squares of rms components =2.82V

Now consider heating: Assume a 1 ohm load with 4v for half cycle, and 0 for the second half cycle the ave power in the load is (16*0.5 +0)=8 watts at 2.82V rms the average power =(2.82^2)/1 =8 watts -the same

at 2VDC the average power is 4 watts for a 2V peak (AC) square wave the average power is 4 watts sum of the latter two =8 watts

As for having an alternating component, add, at each point in time, a DC voltage of 2 V and a symmetrical alternating square wave of 4V peak to peak (i.e. 2V max and -2V minimum. Draw it. All that you have is the AC Square wave shifted upward by the 2V DC offset. Note that in electronic amplifiers, all that you have is a varying DC in the transistor but by blocking the DC component with a capacitor, you can obtain the AC signal that you want (you don't want to put DC into your speaker- I had it happen- magic smoke escaped so the speaker no longer worked -the clue is when the noise stops and the smoke appears).

Reply to
Don Kelly

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