Calculating true RMS current

A rough guild
convert
RMS to Peak is 1.414
peak to RMS is .707

This is where meters that are not true RMS go wrong: they essentially measure the peak voltage and display "RMS" based off a similar calculation. That's only accurate when measuring a pure sine wave. With dirty power or a dimmer at anything other than full (without any voltage regulation), the number it gives you is meaningless and not directly related to the actual RMS value.
 
This is where meters that are not true RMS go wrong: they essentially measure the peak voltage and display "RMS" based off a similar calculation. That's only accurate when measuring a pure sine wave. With dirty power or a dimmer at anything other than full (without any voltage regulation), the number it gives you is meaningless and not directly related to the actual RMS value.
Most non-true-RMS meters measure the average voltage (rather than the peak voltage), and scale that to give the RMS equivalent assuming a sine wave. The general concept is the same, in so far as the displayed value is only correct for a very few select waveforms, and incorrect to various degrees for any other ones. (Of course, for a given specific waveform, the relationship between displayed values and the actual RMS value is consistent, but to use that fact you have to know in advance what the waveform is and what conversion factor it requires.)
 
I think most (decent) switch mode power supplies these days have some form of power factor correction built-in, as I believe is required by many certifications/codes. For ones that do, the current waveforms for the supply will be approximately sinusoidal, and the neutral currents well-behaved.

The Source 4 Series 2 Lustr, for one example I chose more or less random, has a claimed power consumption of 167W, and a current consumption of 1.44A at 120V, giving an apparent power of 172.8 VA, and a power factor of 96%. That's really close to being a purely resistive load so far as the AC supply is concerned.
 
Power factor is not the issue, the issue is that switch mode supplies tend to draw more current at the peak of the waveform and this will lead to higher neutral currents, so has anyone with a large LED rig actually measured it.
 
Power factor is not the issue, the issue is that switch mode supplies tend to draw more current at the peak of the waveform and this will lead to higher neutral currents, so has anyone with a large LED rig actually measured it.
Consuming excess current at the peak of the waveform will necessarily lead to a poorer power factor. Switch mode power supplies without power factor correction have pretty poor power factors due to the peaky nature of their current consumption, and power factor correction works by smoothing out the current peaks--either through the use of passive components such as a filter inductor, or (far more commonly these days) by having an active device modulate the incoming line current throughout the AC cycle, forcing it to be a nice sine wave that is in phase with the line voltage.
 
Consuming excess current at the peak of the waveform will necessarily lead to a poorer power factor. Switch mode power supplies without power factor correction have pretty poor power factors due to the peaky nature of their current consumption, and power factor correction works by smoothing out the current peaks--either through the use of passive components such as a filter inductor, or (far more commonly these days) by having an active device modulate the incoming line current throughout the AC cycle, forcing it to be a nice sine wave that is in phase with the line voltage.
OK could you link to this "active device' for my information? I totally fail to see a mechanism for an inductor to fix this problem but I would like to be enlightened, I recently had a case of an harmonic filter, a very clever device which completely stuffed up the dimmers it was supplying. Why does consuming excess current at the peak of the waveform necessarily lead to a poorer power factor, which relates to apparent power vs real power ? I'm always keen to learn something new.
 
Here's one (comparatively) readable application note on power factor correction for switching power supplies, courtesy of Toshiba. It briefly mentions the use of passive PFC circuits (basically sticking an inductor in the front end of the power supply), and goes into a reasonable amount of detail on various active PFC schemes. The active devices used are typically MOSFETs of some description, though there's no inherent reason other more specialized switching transistors (perhaps IGBTs?) could be used in odd situations.

As to why peaky current causes a poor power factor, that's a bit tricky to explain without delving into more calculus than I care to tackle without either earning a salary or college credit for the work. However, I think I might be able to show the basic concept in a simplified form without (much) explicit calculus, and only a moderate amount of hand waving.

A few reminders and definitions, first. Ohm's Law: V=IR, voltage = current times resistance; and the fact that power (in watts) is the product of the current and the voltage. Power, in terms of physics, is the rate of energy consumption or transfer; so for a given rate of work (such as light production), the power going into the fixture, averaged over time, must be some constant value.

RMS values of measurements (voltages, currents, etc.) are defined as being the square root (the "R") of the mean or average (the "M") of the square (the "S") of the instantaneous measurement. For a DC signal, this works out to be the measurement itself, ignoring signs--its absolute value, if you like. For other waveforms that vary periodically, it's a bit more complicated to work out. Power factor is defined as the ratio of actual power (well, the average actual power, if it varies cyclically for a periodic AC wave) and apparent power; and apparent power is the product of the RMS value of the applied voltage and the RMS value of the current.

(As to why RMS values are used and defined as they are, that's because they have the really handy property that, for purely resistive loads, you can apply Ohm's Law and the "normal" definition of power being volts times amps, etc. to the RMS values and obtain useful averaged results, results that directly correspond to those of DC circuits, without having to analyze everything at every point in the voltage and current waveforms. It reduces the analysis along the waveform as a whole to an analysis of a single value.)


DC Circuit (power factor = 1.0)

1635306902731.png


For the DC case, everything is very straightforward. The voltage is constant, and the current is constant, with their ratio being determined by the DC resistance of the widget, and the power is constant, with the average power always being equal to the (unchanging) instantaneous power. For this example, it is 1 volt, 1 amp, a 1 ohm load (per Ohm's law), and hence 1 watt of power dissipated.


Pseudo-AC Square Wave Circuit (power factor=1.0)

1635306957426.png


Now, instead, let's consider a square wave, sort-of-AC situation. The voltage switches between +1V and -1V periodically. The RMS value of the voltage, the square root (the "R") of the mean or average (the "M") of the square (the "S") of the signal's instantaneous value, is pretty obviously 1V: the voltage is always either +1V or -1V, and either +1 or -1 squared is 1. The mean of 1, no varying, is likewise 1, and the square root of that is 1 V rms. Similarly, the current in our 1 ohm load also varies identically between +1A and -1A, and we see 1 A rms. The actual power consumed at any given point in time is exactly 1W, and the apparent power--the product of the RMS voltage and the RMS current--is also 1W, so the power factor is 1.0.


The Interesting Case (power factor < 1.0)

1635307002709.png


Finally, for the example that we're really discussing. Let's change the load so it's not a purely resistive load, but instead something that consumes current for only part of our cycle. I'm arbitrarily saying the first half of the square wave, but that detail doesn't really matter. The average (actual) power should be the same as before, so we need to use twice the power--two watts--for the half of the time the current flows, to make up for the zero watts it consumes the other half of the time. In order for that to work, the current during the "on" part of the cycle must be 2A rather than 1A. Computing the RMS current, we can see that half the time the square of the current measurement will be 4 and the other half of the time it will be 0, so the mean of the square of the measurement is 2, and the current is 1.41 A rms (i.e. the square root of 2). The apparent power here is 1.41W and the actual power only 1W, giving a power factor of 1/1.41 or about 71%.

For a realistic scenario of a non-PFC switching power supply, where the voltage is a (nominal) sine wave and the current a sort of-but-not-really square wave sort of thing, it requires performing integration to compute the RMS values explicitly. Because of that, I'll just wave my hands a bit and say the same basic concept holds. Intuitively (if this stuff can be said to be intuitive) the increase from the squaring part of the RMS computation will outweigh the corresponding decrease in the mean part of the RMS computation.

Power factor can be thought of as a sort of metric for how much a load deviates from being purely resistive--whether that's because it is (partly) reactive or capacitive, or has other non-linear characteristics. If a widget has a high power factor, it behaves like a good resistor and therefore the current and voltage waveforms are in phase and have similar waveforms, and for that reason will not lead to unexpectedly high neutral line currents.
 
Firstly I love calculus, second I get all the data on P.F., however my reservations concern the distortion of the waveform leading to non cancellation in the neutral. However this week I will set up an experiment with 3 LEDs on a 3 phase system and see what actually occurs, if it's interesting I will post a video.
 
Here's one (comparatively) readable application note on power factor correction for switching power supplies, courtesy of Toshiba. It briefly mentions the use of passive PFC circuits (basically sticking an inductor in the front end of the power supply), and goes into a reasonable amount of detail on various active PFC schemes. The active devices used are typically MOSFETs of some description, though there's no inherent reason other more specialized switching transistors (perhaps IGBTs?) could be used in odd situations.

As to why peaky current causes a poor power factor, that's a bit tricky to explain without delving into more calculus than I care to tackle without either earning a salary or college credit for the work. However, I think I might be able to show the basic concept in a simplified form without (much) explicit calculus, and only a moderate amount of hand waving.

A few reminders and definitions, first. Ohm's Law: V=IR, voltage = current times resistance; and the fact that power (in watts) is the product of the current and the voltage. Power, in terms of physics, is the rate of energy consumption or transfer; so for a given rate of work (such as light production), the power going into the fixture, averaged over time, must be some constant value.

RMS values of measurements (voltages, currents, etc.) are defined as being the square root (the "R") of the mean or average (the "M") of the square (the "S") of the instantaneous measurement. For a DC signal, this works out to be the measurement itself, ignoring signs--its absolute value, if you like. For other waveforms that vary periodically, it's a bit more complicated to work out. Power factor is defined as the ratio of actual power (well, the average actual power, if it varies cyclically for a periodic AC wave) and apparent power; and apparent power is the product of the RMS value of the applied voltage and the RMS value of the current.

(As to why RMS values are used and defined as they are, that's because they have the really handy property that, for purely resistive loads, you can apply Ohm's Law and the "normal" definition of power being volts times amps, etc. to the RMS values and obtain useful averaged results, results that directly correspond to those of DC circuits, without having to analyze everything at every point in the voltage and current waveforms. It reduces the analysis along the waveform as a whole to an analysis of a single value.)


DC Circuit (power factor = 1.0)

View attachment 22344

For the DC case, everything is very straightforward. The voltage is constant, and the current is constant, with their ratio being determined by the DC resistance of the widget, and the power is constant, with the average power always being equal to the (unchanging) instantaneous power. For this example, it is 1 volt, 1 amp, a 1 ohm load (per Ohm's law), and hence 1 watt of power dissipated.


Pseudo-AC Square Wave Circuit (power factor=1.0)

View attachment 22345

Now, instead, let's consider a square wave, sort-of-AC situation. The voltage switches between +1V and -1V periodically. The RMS value of the voltage, the square root (the "R") of the mean or average (the "M") of the square (the "S") of the signal's instantaneous value, is pretty obviously 1V: the voltage is always either +1V or -1V, and either +1 or -1 squared is 1. The mean of 1, no varying, is likewise 1, and the square root of that is 1 V rms. Similarly, the current in our 1 ohm load also varies identically between +1A and -1A, and we see 1 A rms. The actual power consumed at any given point in time is exactly 1W, and the apparent power--the product of the RMS voltage and the RMS current--is also 1W, so the power factor is 1.0.


The Interesting Case (power factor < 1.0)

View attachment 22346

Finally, for the example that we're really discussing. Let's change the load so it's not a purely resistive load, but instead something that consumes current for only part of our cycle. I'm arbitrarily saying the first half of the square wave, but that detail doesn't really matter. The average (actual) power should be the same as before, so we need to use twice the power--two watts--for the half of the time the current flows, to make up for the zero watts it consumes the other half of the time. In order for that to work, the current during the "on" part of the cycle must be 2A rather than 1A. Computing the RMS current, we can see that half the time the square of the current measurement will be 4 and the other half of the time it will be 0, so the mean of the square of the measurement is 2, and the current is 1.41 A rms (i.e. the square root of 2). The apparent power here is 1.41W and the actual power only 1W, giving a power factor of 1/1.41 or about 71%.

For a realistic scenario of a non-PFC switching power supply, where the voltage is a (nominal) sine wave and the current a sort of-but-not-really square wave sort of thing, it requires performing integration to compute the RMS values explicitly. Because of that, I'll just wave my hands a bit and say the same basic concept holds. Intuitively (if this stuff can be said to be intuitive) the increase from the squaring part of the RMS computation will outweigh the corresponding decrease in the mean part of the RMS computation.

Power factor can be thought of as a sort of metric for how much a load deviates from being purely resistive--whether that's because it is (partly) reactive or capacitive, or has other non-linear characteristics. If a widget has a high power factor, it behaves like a good resistor and therefore the current and voltage waveforms are in phase and have similar waveforms, and for that reason will not lead to unexpectedly high neutral line currents.
Great stuff, Drew! Thank you for taking the time to write, edit and illustrate this. Sticky-dom for this one?

Guy Holt from ScreenLight & Grip in the Boston area has a great thread over at ProSoundWeb on power factor, harmonics, voltage drop, and power quality. He's also an IATSE member in the studio mechanics craft specialty.

 
Sorry people but a true RMS meter will NOT give you an accurate reading of dimmer load if there are many dimmers at different levels, it is just too complex a waveform, if you don't believe me call the experts at Fluke, a cheap moving iron meter is actually more accurate, you also have the complication of DC content to stuff things up.
Sorry, David--I don't agree.

Modern True-RMS digital meters offer a wide variety of "crest factor" specifications. That is the ability of the meter to maintain accuracy in the presence of a severely non-sinusoidal waveform. Bottom line: for measurements on phase-control dimmer systems, we don't need to be sourcing hard-to-find and obsolete iron-vane meters (although I have a favorite Weston on my bench, mostly as decoration!). High quality true-RMS meters from Fluke and the like provide more than enough high-crest-factor accuracy, no matter how many dimmers and how many phase-angles are being simultaneously measured.

Some more summary data on accuracy here:


ST
 
Some years ago I asked an expert at Fluke about this question and he told me that with all the factors in multi dimming situations and D.C. content then it was not possible to be super accurate. Either he was wrong or things have changed.
Now a moving iron meter is a true RMS meter AC/DC and costs around $30 which is a little different to a Fluke.
 
Last edited:
It seems to me that the goal here is to make measurements good enough to know that circuit breakers and wires are not overloaded. Wire with a rating of 200 amps isn't going to burst into flame if the load is actually 209.6 amps, so absolute accuracy isn't required. A true RMS measurement gives us a good value of how a particular load is heating the equipment, which is what we need to know.
 
It seems to me that the goal here is to make measurements good enough to know that circuit breakers and wires are not overloaded. Wire with a rating of 200 amps isn't going to burst into flame if the load is actually 209.6 amps, so absolute accuracy isn't required. A true RMS measurement gives us a good value of how a particular load is heating the equipment, which is what we need to know.
Just what he said.

ST
 

Users who are viewing this thread

Back