Where is mean voltage used? - Page 1
Forum section: electronics / general

author: mustafayilmaz
author_rank: Contributor
timestamp: November 09, 2023, 11:01:24 am
content: I know what RMS voltage is. RMS sine voltage does the same job as dc voltage.In full wave rectifier circuits there is something called average dc voltage. Its formula is 0.6366*Vpeak.I don't understand what this means. Where is this formula used?I have a guess; What would sine voltage be like if it were a straight line?Why do we need average voltage when there is rms voltage?Does average voltage represent pure dc voltage? (0V ripple)

author: ebastler
author_rank: Super Contributor
timestamp: November 09, 2023, 01:04:02 pm
content: The average voltage is what you canmeasurerelatively easily, by sending the signal through an integrator or low-pass filter.

author: retiredfeline
author_rank: Frequent Contributor
timestamp: November 09, 2023, 01:10:20 pm
content: Mean voltage is when you don't want kind voltage.

author: ebastler
author_rank: Super Contributor
timestamp: November 09, 2023, 02:31:13 pm
content: Quote from: retiredfeline on November 09, 2023, 01:10:20 pmYeah, anything above 100V or so can get pretty mean.

author: TimFox
author_rank: Super Contributor
timestamp: November 09, 2023, 02:33:40 pm
content: Mean or average voltage: some voltmeters, especially older analog units, measure themean absolute value(average of the full-wave rectified signal), but the display is calibrated for RMS voltage.Just as the mean is 0.6366 xVpeak, the RMS is 0.707 xVpeak. Therefore, the scale factor between them isVRMS= 1.11 xVavg.-hp- was careful to label the meter "average responding" for such circuits, even though the calibration was RMS.Note that these values apply to a sine-wave input. For DC or a square wave,Vpeak=Vavg=VRMS.The diagram posted above is an ideal rectifier, where the op amps eliminate the voltage drop across the diodes to produce the waveforms shown; the filter extracts the mean voltage of the FW rectified waveform.Another place where mean voltage appears: in an old-fashioned "choke-input" filter after a rectifier, the rectified waveform goes through a large inductor to a capacitor.Neglecting losses in the inductor, for a sufficiently large current, the output voltage is the mean voltage (as discussed above) with a ripple voltage.For small currents, the inductance is insufficient and the output voltage approaches the peak voltage (as in a capacitor-only filter).

author: Zero999
author_rank: Super Contributor
timestamp: November 09, 2023, 09:04:13 pm
content: The mean voltage is just the DC component, which should be zero for the mains. If it's non-zero, then you have a problem.

author: TimFox
author_rank: Super Contributor
timestamp: November 09, 2023, 09:05:39 pm
content: The OP is referring to full-wave rectifiers. They put out a waveform that is positive-definite, with large DC level.

author: Zero999
author_rank: Super Contributor
timestamp: November 09, 2023, 09:33:49 pm
content: A multimeter set to DC will measure the average voltage.

author: coppercone2
author_rank: Super Contributor
timestamp: November 09, 2023, 11:05:54 pm
content: the point of average voltage is to determine a DC offset or a average DC offset . at least on a scope. to see for instance what bias level your coupling capacitor is at.

author: Nominal Animal
author_rank: Super Contributor
timestamp: November 10, 2023, 03:21:06 am
content: I'll offer a mathematical approach, just in case anyone finds it useful. I don't generallylikemath that much, but in this case, I do find it intuitive, explaining well the difference between the two.Quote from: mustafayilmaz on November 09, 2023, 11:01:24 amIt comes from the fact that if you take a sine wave, flip the negative half-waves to the positive side, the average voltage you get over each full wave or cycle is$$\frac{1}{2 \pi} \int_0^{2 \pi} \left\lvert V_{pk} \sin t \right\rvert dt = V_{pk} ~ \frac{4}{2 \pi} \approx 0.63661977 ~ V_{pk}$$The integral calculates the area under the curve, and if we divide that area by the width (\$2 \pi\$, one full cycle), we get the average "height", the average DC voltage. Thus, it indeed is the average value of a rectified sine wave.Quote from: mustafayilmaz on November 09, 2023, 11:01:24 amIf you recall Ohm's law, (instantaneous) power is \$P = V I = I^2 R = \frac{V^2}{R}\$ (for a purely resistive circuit). If you are more interested in the work (or power) you can get out of some sinusoidal voltage, also assuming that the current will be in the same phase as the voltage (so purely resistive load), then you want to look at the average of the squared voltage instead. Of course, units of "voltage squared" don't mean much, so you do a square root afterwards. This is exactly what root-mean-square is! Such an average for a sinusoidal voltage is$$\sqrt{ \frac{1}{2 \pi} \int_0^{2 \pi} \left\lvert V_{pk} \sin t \right\rvert^2 ~ d t } = \sqrt{\frac{1}{2 \pi} V_{pk}^2 \pi} = \sqrt{\frac{1}{2}} ~ V_{pk} \approx 0.70710678 ~ V_{pk}$$Thus, average voltage tells you what the corresponding constant DC voltage would be.RMS voltage tells you what DC voltage would yield the same amount of power, assuming the alternating current is always in the same phase as voltage (which is a sensible assumption, since we're comparing to DC voltage and current).The same logic expands toallwaveforms, it's not limited to just sinusoidal ones; only the numerical scale factor changes, depending on the shape of the function. Essentially, for any time-dependent voltage signal \$V(t)\$ we have$$V_\text{AVG} = \frac{1}{T} \int_0^T \left\lvert V(t) \right\rvert ~ d t$$and$$V_\text{RMS} = \sqrt{\frac{1}{T} \int_0^T V(t)^2 ~ d t}$$A good meter that can tell the average does so by repeated (random) sampling, because$$V_\text{AVG} \approx \frac{1}{N} \sum_{N} V(t)$$(Even though the rectified sinusoidal varies between zero and only slightly above \$V_\text{AVG}\$, it stays longer at the higher values than in the lower values, giving the expected average.)Approximating integrals with sums, to any desired accuracy or resolution, is a common mathematical technique. There are rules to when and how it can be done, but when done so, it is absolutely valid, and does not incur any kind of systematic error. Here, the main thing is that the samples are not taken in phase with the signal, so either at much higher frequency than the signal itself, or randomly.For RMS, there are two approaches. One is mathematical, using repeated (random) sampling, just doing a bit more calculation (squaring and square root)$$V_\text{RMS} = \sqrt{\frac{1}{N} \sum_{N} V(t)^2}$$and the other is by measuring the actual average power, by using a (low resistance) resistor that gets heated by the voltage, with the temperature difference carefully measured.Thus, I'd say your intuition or guesses were on the right track. Average voltage is used when you are concerned about the voltage itself, and not power. RMS voltage is used when you are putting the voltage to work, as it represents the DC voltage that would do the same amount of work (i.e., yield the same power on a purely resistive load; power as in instantaneous watts, or during some interval as watt-seconds or Joules: 1 W s = 1 J.)

author: EPAIII
author_rank: Super Contributor
timestamp: November 10, 2023, 04:55:17 am
content: Mean and average are pretty much the same thing. If you take the arithmetic average of any pure, sinusoidal waveform, it will be zero because there will be a negative instantaneous value for every positive instantaneous value and vice-versa. That is exactly the same thing for mean and average because those two words mean the same thing.Now, if you rectify that AC waveform and all the values are positive, the situation changes. The mean or average value is not half way between zero and the instantaneous, maximum, peak value. If you perform the integration over the positive going, half sine wave, you will get an average or mean value that is 0.637 times the peak value. That is in the nature of the shape of the sine wave.From a practical point of view, the mean or average value of a rectified, AC waveform is the value that a mechanical meter (moving coil mechanism) will respond to. This factor is built into almost every analog VOM (Volt Ohm Meter). Of course, there is an almost constant Voltage loss in a bridge rectifier which effects every AC Voltage scale on those meters. The fact that it is a constant amount of lost Voltage means that it has a greater effect on a low Voltage, AC scale than on a higher Voltage one. This is why most VOMs will have separate scales for 0 to 10 VAC and 0 to 10 VDC while higher AC and DC Voltages will be read on the same scales.As others have said, RMS values are often considered more useful because they represent the amount of energy present or work that can be done. And it is easier to build RMS scales into digital devices than analog ones. True RMS scales are present only on the most expensive analog meters due to the cost of the circuitry needed. But in digital meters the true RMS value can be calculated with math from the individual, instantaneous samples. The common and less expensive analog meters were simply given an additional correction factor to translate their average readings into the equivalent RMS values when the waveform is a pure sine wave. That, of course meant that they were inaccurate when other waveforms were being read.So the answer to the OP's question of where is mean Voltage used, it is primarily used in analog VOMs. And it is used there by necessity because getting a true RMS value with analog circuitry would be prohibitively expensive in most of those meters.Quote from: Zero999 on November 09, 2023, 09:04:13 pm

author: EPAIII
author_rank: Super Contributor
timestamp: November 10, 2023, 05:03:34 am
content: Not necessarily true. If a scope's gain is set to display a rectified sine wave full scale (zero at the very bottom and the peaks at the very top) and then the DC offset is set so that the average or mean value is at the center line (normally used for zero), then the negative peaks will be driven below the bottom of the screen and not be seen and the positive peaks will be below the top.In order to display any AC signal, rectified or not, from peak to peak and using the full vertical space of the screen, you need to set the DC offset to the value that is half way between the positive and negative peaks.Quote from: coppercone2 on November 09, 2023, 11:05:54 pm

author: EPAIII
author_rank: Super Contributor
timestamp: November 10, 2023, 05:19:17 am
content: I also find the remarks about a low pass filter somewhat concerning. There are many types of low pass filters so I don't want to get into any arguments about specific types. But I don't think that simply saying "after a low pass filter" is always true.Most power supplies have a (bridge) rectifier followed by some filter capacitors and, since the output is assumed to be primarily DC, those capacitors must be called a low pass filter. But many power supply designs, dare I say most, do not operate at the average or mean Voltage nor at the RMS Voltage. They usually are designed to operate at higher Voltage levels by having capacitors that will maintain a higher percentage of the peak Voltage for the discharge interval between the half sine waves. In fact, in theory those filter capacitors would remain at the peak value if the current being drawn were equal to zero: each peak would charge them to that level and with no load current there would be no draining current to discharge them.So, while some low pass filters may indeed only pass a DC level equal to the average or mean value, not all low pass filters operate in that manner. And I will leave it to others to name and give examples of low pass filters that do, as a design rule, output only the average/mean Voltage level.

author: ejeffrey
author_rank: Super Contributor
timestamp: November 10, 2023, 05:49:14 am
content: Quote from: EPAIII on November 10, 2023, 05:19:17 amIt's not really about the filter (capacitors and possibly inductors and resistors) but the source (diode bridge) which has a nonlinear impedance. If you look at a buck converter, if you are in continuous condition mode (either the low or high switch always conducting) the filter output is just the mean voltage. But when it goes into discontinuous mode and the inductor current drops to zero the low switch turns off and is high impedance then you get the same behavior as a filtered bridge rectifier where it's no longer an average.

author: SiliconWizard
author_rank: Super Contributor
timestamp: November 10, 2023, 06:46:10 am
content: Just a note to add to what's been already said: in practical use, one must define over which period of time we consider the "mean". If you use a low-pass filter, for instance (assuming that it's the proper approach in your given context), it gives you some kind of moving average (over a period proportional to its time constant). So, short point being, when we talk about mean (or average) voltage (or current), we very often talk about a moving average, in practice.One way to get a 'true' average from a precise start point to a precise end point would be to use an integrator, reset it at the start point, let it run until the end point, stop it. And divide the accumulated value by the duration. Making sure the integrator is sized properly so as not to saturate till the end point.But in practice we usually "settle" for a moving average. (Or a moving RMS if RMS is what we're after.)