Full wave rectified RMS voltage

[mpegjohn]

I have been trying to get my head around this little problem for some time, and decided I need help from those that know.

i have a simple full wave centre tapped diode rectifier circuit.
The transformer secondaries are spec'd at 12V RMS each winding. Measured at 13.87V RMS.

I have a 1K load resistor, no smoothing cap.

If I use my scope to look at the output across the load, it looks like a full wave rectified waveform, as you would expect. It has a peak voltage of 19.4V. The DC voltage, according to the scope, and my Fluke is 12.43v. Close to 0.63 X 19.4 =12.2V
However the RMS voltage readings I get are a bit strange.
According to me the RMS voltage should be 70% Vpk = 13.7V.
But on my fluke I get 6.49V AC RMS, and this agrees with my copes calculations.

What am I doing wrong with regards to my RMS values?

John.

[Radio Wrangler]

Great question!

The 'square' bit of RMS means that the sign of the input is always going to be made positive in the squaring process, so the RMS value of an unsmoothed full wave rectified UNSMOOTHED sine will be the same as the unrectified sine.... 0.707 of the peak voltage.

The difference is that the rectified signal has this energy capability in the form of DC and AC components.

Many multimeters on AC use an AC coupling capacitor to remove any DC component, true RMS ones try to read the RMS of the AC ripple, neglecting the DC component.

David

[GrimJosef]

Is the 6.49V AC measured with the meter on an AC range ? If it is then the DC component of the voltage waveform may be being stripped out and all you're measuring is the alternating component which will be assumed to be going 'equally' positive and negative (I've put equally in inverted commas because to be frank I'm not sure how that's calculated for a +/- asymmetric waveform).

EDIT: Beaten to it by David .

Cheers,

GJ

[ionburn]

I am not sure I understand you here, but two things I will mention are:

1: RMS voltages with meters are generally only valid for sine waves so a full wave rectified waveform will be invalid / inaccurate.

2: Perhaps far less relevant here, but there will be a rectifier voltage drop.

[Refugee]

It very much depends on the design of the RMS converter in the meter you are using.
The only reliable readings with a randomly selected meter are the un-rectified AC and the smoothed DC.

[terrykc]

I'm not sure that I agree with your calculations here. You say that each half of the winding produces a measured 13.87V RMS

This will be equivalent to a peak voltage of 13.87 x 1.414 = 19.61V

However, assuming you are using silicon diodes, they will introduce a drop of ~700mV, so you should see a peak output of 18.91V, so your figure is ~500mV high.

The RMS value is 18.91 x 0.707 (where does your figure of 0.63 come from?) which is 13.37V

You now start using 70% (OK, only 1% out) to calculate an RMS figure of 13.7V, which is 330mV high.

But you started off with 19.4V peak and 19.4 x 0.7 = 13.58 not the 13.7V which you claim! However, using the correct figure of 0.707, 19.4 x 0.707 = 13.71 which is as near as dammit to the answer you got using the figure of 0.7!

Now we get to the RMS measurement part. Assuming it was AC, the peak to peak voltage in your case would be 2 x 19.4 = 38.8V which a true RMS reading meter would 'see' as 13.71V RMS but your meter isn't seeing a 38.8V pk-pk voltage it is only seeing a 19.4V pk-pk input, which it should calculate as 6.86V RMS.

I say 'should' because it isn't seeing the symmetrical sine wave it expects but an extremely asymmetrical one, so exactly how it would handle it, I don't know.

I think you are making two basic mistakes here.

The first is that you are taking voltage readings to the nearest 10mV as 'gospel' without making any allowance for the accuracy tolerance of your instruments.

The second is that you have overlooked the fact that a True RMS meter doies precisely that - it calculates the RMS power of the applied voltage but that will only agree with the theoretical 0.707 of peak voltage if the voltage is a pure sine wave and the chances of getting a pure RNS signal from a 13A socket is extremely unlikely!

If you 'scope the mains, the distortion is often sufficiently bad to be instantly visible to the naked eye - what does your's look like?

If your meter is not a true RMS one, it will simply rectify the incoming voltage and apply a correction factor to get an accurate RMS level from it on the assumption that the input really is a sine wave.

Take this as an extreme case: CATV PSUs use a saturated core technique to produce an output which closely resembles a square wave. A True RMS meter will measure the output as 60V - an 'average reading' type of meter will measure it as 66V!

[mpegjohn]

Thank you I think things are a bit clearer for me now.
So if I understand correctly my True RMS meter is only measuring the AC part of the waveform.

terrykc, the 0.63 value is the average, DC, value of the rectified output.
It is true that I haven't taken into consideration the accuracy of my instruments, and measurements to 10mV accuracy is indeed a bit daft, I admit.

So how would you measure the RMS voltage of the output that is un-smoothed?

[terrykc]

I got distracted whilst composing that post so missed the three previous posts which had basically covered the same ground!

[mpegjohn]

Thank you for your answer though Terry.

[terrykc]

Originally Posted by mpegjohn

So how would you measure the RMS voltage of the output that is un-smoothed?

Why would any one want to? Why would someone generate an unsmoothed supply? The only applications I can think of off the top of my head would be for driving DC motors - but it's a very, very long time since I played with model trains!

If I really wanted to be that accurate I can measure the peak-to-peak input and subtract 700mV (or 1.4 volts if it is a bridge rectifier) and calculate it from that.

Modern silicon rectifiers do't go high resistance and deteriorate like older selenium types - if they did the would get very hot very quickly and self destruct! A valve rectifier would be best checked by substitution.

If I really wanted to know the answer to the question you've posed, I thing simply doubling the reading taken with a true RMS meter would be sufficiently accurate but this begs another question - why do YOU want to measure the RMS output of an unsmoothed DC supply?

[GrimJosef]

Originally Posted by mpegjohn

... So how would you measure the RMS voltage of the output that is un-smoothed?

If you've got a scope trace from which you can take measurements then you can work out pretty much any number you might want from that.

Cheers,

GJ

[mpegjohn]

I don't NEED to do it, I am just interested how one would go about actually measuring the RMS value.

Yep, I could calculate it from a scope trace.

I could measure the temperature of the load resistor, I imagine, and compare with known currents/voltages.

John

[cmjones01]

Originally Posted by mpegjohn

I don't NEED to do it, I am just interested how one would go about actually measuring the RMS value.

Yep, I could calculate it from a scope trace.

I have two True RMS voltmeters in the lab: an HP 34401A and a Datron 1061A. When switched to their AC ranges but connected to a DC source, they measure zero. Their inputs are AC coupled so they'll only measure the AC component of the input. I think this is how yours is behaving, too.

With your full-wave-rectified-unsmoothed waveform, this will lead to a slightly odd result, because the AC coupling will shift the waveform down to its mean point, and the voltmeter's RMS converter will then full-wave rectify that, leading to a very odd-looking waveform. See attached sketch.

It is this odd-looking waveform that the meter will then calculate the RMS value of. It's fairly spiky so different 'true RMS' meters may manage this with varying degrees of accuracy. However, the sketch shows that what starts as a waveform that's recognisably half of a 12V sine wave, finishes up as something about half as tall once it's been AC coupled and full-wave rectified. Hence the meter's reading of about 6V seems about right. I don't have an unsmoothed DC supply here to test my meters on, otherwise I'd try and reproduce the result.

It would be possible to do the same measurement from first principles by coupling the unsmoothed DC via a (large) capacitor into a load resistor, then using the resistor to heat an object of known heat capacity and measuring the temperature rise.

Chris

[mpegjohn]

Thank you Chris and everyone for your answers.
I can see now what the True RMS meter will "see", and it ain't pretty.

I might have a go at the resistor heating up though, might be fun to do.

John.

Why would any one want to?

I have an ultrasonic cleaner that uses a rectified and not smoothed supply.

[Herald1360]

If you run an incandescent bulb or want to know the dissipation in a resistive load fed with full wave rectified unsmoothed dc, knowing its rms value could be handy.

[stuarth]

You could be running LED lights from the transformer.

Stuart

[stuarth]

Seems to me that a “true RMS” meter which ignores the DC component isn’t measuring true RMS!

Stuart

[terrykc]

Originally Posted by stuarth

Seems to me that a “true RMS” meter which ignores the DC component isn’t measuring true RMS!

As RMS stands for the Root Square Mean of a sine wave, what has any DC offset to the sine wave got to do with it?

If you had a 2V Pk-Pk hum ripple superimposed on a 200V HT rail, what would you expect your true RMS meter to measure?

Hint: 'true RMS' only applies to the AC ranges of your meter.

[ms660]

Worth a watch, excuse the odd language/misinterpretation:

https://www.youtube.com/watch?v=LZEie3nAo1E

Lawrence.