# Running 115v device on 230v

wrote:> . . . . . Avo's do not show true RMS on the AC

How about ..... for a full wave rectified sine wave
PEAK = 1.4 * RMS ( SQRT(2) )
MEAN = 0.9 * RMS ( 2 *SQRT(2)/PI )
The avo will have this latter factor built into its scaling, Suppose we have, a full wave rectifier feeding a resistive load. If the rectifier output is 10 amps.(mean) DC
An AVO in series with the output should read 10 amps on the DC scale. An AVO measuring the AC input should read 11.1 amps rms I believe within measurement accuracy this would be the case.
Now feed the same resistor via half wave rectification. Both meter readings will halve. (the meter is an averaging one and is seeing the same current for half the duty cycle)
But what are the new Actual currents?. Mean DC current is actually halved, so the DC reading is correct However the true RMS current has reduced by SQRT(2) , only down to 7.9 amps rms. So the good old well trusted AVO is under-reading.

could be ILLUMINATING
. . . . . Briefly.
Ian
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wrote:

You are right, I was converting RMS to peak and not RMS to mean. (form factor being 1.11 for a sine wave)
My bad, I said the right thing wrong and confused the maths even more!
Mark Rand RTFM
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wrote:

That was why I suggested it as an experiment to prove the issue :-)
'Course, you've got to allow for the fact that 12V car bulbs are actually rated at 14V, since that's what 12V cars run at...
Mark Rand RTFM
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try
actually
Mark,
The subjective bulb test has just been carried out ! a/ Take three 50v 60watt light bulbs (just happened to have a box!) b/ Wire two bulbs in series across 100v AC from Variac c/ Wire one bulb in series with a power diode across the same 100v at the same time d/ Compare brightness of the three bulbs
Result: all three are different brightnesses !!!!!!! In the series chain of two bulbs the upper one appears dimmest, it's lower brother brighter and the one in series with the diode slightly brighter still, but the variation is not very much.
Trying to take the subjective element out of it I whipped out my infra-red thermometer. Dimmest bulb is 195ish middle one 200 ish and the brightest one with the diode 205ish all in degs C of course.
OK the two in series obviously have slight manufacturing differences as will the third but just possibly there may be something in this rms/average lark - just possibly <G>
AWEM
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I have to seriously admire you Andrew for putting theories to the test, and have to smile that you happen to have so much "kit" just hanging around :)
It's certainly the best way to learn.
I must say the differences between temperatures are less than I would have guessed.
There are , as you say manufacturing differences between the bulbs. (You could try each bulb in turn in the diode leg ?) But I wonder if the temperature coefficient of resistance is reducing the power dissipation of the bulb fed via the diode? (this effect would give a lower current in the diode fed leg than in the 2 series bulbs fed directly)
It's getting late now but maybe some time you might try the following?
Voltmeter across output of variac
Use the same bulb for each of the following tests.
1. No diode in circuit, take the voltage up to the rating of the bulb and measure the temperature.
2. Diode in circuit, take the voltage up to a level that gives the same bulb brightness ( as confirmed by reaching the same temperature ) and measure the variac output voltage.
(And hope that the thermometer reading is not affected too much by the 50Hz/100Hz temperature ripple in the respective tests)
regards
Ian
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Problem with this is that it isn't the same mA !
The AVO8 doesn't really measure RMS current. The Fluke might, it depends on the model. This is because RMS requires that you integrate the area under the curve of the waveform. Traditional RMS meters heated a thermistor bead to measure the heating effect of the waveform. Modern meters digitise the waveform and make an approximation. Their ability to get a good approximation depends on the shape of the curve and the rate of oversampling.
The AVO is even cruder - it will rectify the AC signal, take the mean value of the result, and apply a fiddle factor such that a sine input will give the right number. For non-sinusoidal waveforms this will give a predictable result but it won't be RMS.
As some others have said, to do the test properly you need to measure the temperature rise of the heater element.
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On Thu, 21 Dec 2006 17:31:14 -0000, "Andrew Mawson"

Correct experiment but wrong conclusions.
Your experiment confirms that a series diode halves the AVERAGE voltage and AVERAGE current reaching the load.
But power dissipation is proportional to RMS values NOT average values.
The simplest way of looking at this is to imagine 230v applied to your 115V 1.5Kw load resistor. Both the RMS voltage and the RMS current are each twice the rated value so the resistor dissipates 6KW - 4 times its rated power.
If you now add your series diode, the load still dissipates 6KW during one half cycle and nothing in the next half cycle. The average power disipation is now reduced to 3KW - still 100% overload.
Jim
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Andrew Mawson wrote:

A ggod experiment but with a lack of understanding of measuring instruments. AC volt and amp meters come in two flavours, the true RMS types that you need for this experiment and average reading types such as the AVO. Average reading instruments will give the same result as true RMS reading instruments ONLY when the waveform is purely sinusoidal, with other waveforms such as we have here they will give large errors, as you have found.
Is your Fluke true RMS?, if so use that to measure voltage across the load and you will see it isn't half, use it to measure current and that won't be half either. Bear in mind that some meters are limited in resolution on the current range so 1mA may be too little for them to correctly show you that it's actually 1.4mA. Greg
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Cliff Ray wrote:

I'm sorry Cliff but you're talking utter rubbish, as are those talking about nonesense like thermall mass, comparing thermal mass with period etc etc.
The whole point of RMS is that it's a measure or ANY repetitive waveform, not just sinewaves, so applies equally to rectified sinewaves. By definition (in this context) the RMS value is the equivalent constant DC value that will deliver the same power to a resistive load as the waveform being measured.
Effectively a heater is a resistive load, although there will be a tiny inaccuracy due to a typical heater having a small inductance. So 230V RMS 50Hz sine wave will deliver the same power as 230V DC, or any one of an infinite number of waveforms with the same RMS value.
Half wave recification reduces the RMS value of a sinewave by root 2, thus 162.6V RMS will be delivered to this 115V heater which will attempt to deliver twice the power it was intended to. I say attempt because it will of course overheat and so it's resistance will increase, but in reality the element will almost certainly melt!.
A simple way to understand why rectifying 230V produces 230x root 2 is to realise that you are removing alternate half cycles of the 230V waveform, therefore you will deliver half the heat to a load. Now power is proportional to voltage squared, so if the RMS voltage had halved the power would have quartered, but as it's only halved the voltage must have reduced by root 2. Greg
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wrote:

Mr Diplomacy strikes again I see ...<G>
Regards, Tony
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Tony Jeffree wrote:

I thought I was being quite restrained really 8-), lets face it there are some here who think they can invent their own physics and argue themselves into being right!. Asking questions and trying to understand the answers is healthy and to be encouraged, but yer canna rewrite the laws o physics 8-)
Greg
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I thought I'd let this run a while before I put my two-pennyworth in. Gents; we must remember that RMS in the conventional sense is measuring a sinusoidal waveform. A half-wave rectified sinusoid is anything but sinusoidal. Its full of harmonics which all contribute. Nobody who understands what's really going on would suggest the RMS of a square-wave is root 2 x peak.
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Utter rubbish!, RMS is all about measuring ANY repetitive waveform.

And as such RMS is an ideal way to measure it.

I wonder where you think square waves come into this... Greg
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Ah, the killfile is a wonderful invention... :-))
Peter -- Peter A Forbes Prepair Ltd, Luton, UK snipped-for-privacy@easynet.co.uk http://www.prepair.co.uk
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Andrew Mawson wrote:

NO - the current is the same, but the voltage is doubled, and as power equals voltage times current, the power is also doubled.
Note, a diode halves the power to an ac resisitive load - but running at twice the voltage _quadruples_ the power, so overall doubling the voltage and using a diode doubles the power.
Two 115V 1.5kW heaters wired in parallel and run from 230V would each have 115 V across them, and would run at 1.5kW each, so overall you would have 3kW total - you could then half the power by adding a diode, giving 1.5 kW total.
--
Peter Fairbrother

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Peter Fairbrother wrote:

~~~~~~~~ Series?
Nick
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each have

have
1.5 kW

Peter,
I don't think so!
Two 115v 1.5KW heaters wired in parallel and run off 230 each have 230v across them !!!!!!!!!
If I substitute SERIES for parallel in the above paragraph I agree.
AWEM
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Andrew Mawson wrote:

Yes, of course. Sorry.
--
Peter Fairbrother

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Dave, Charles, Peter, and all others who contributed many thanks.
Charles has dropped off the transformer and it looks splendid
Regards,
AWEM
news:...

AC
running
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