Running 115v device on 230v

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

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

Peter,

I don't think so!

Two 115v 1.5KW heaters wired in parallel and run off 230 each have

If I substitute SERIES for parallel in the above paragraph I agree.

AWEM

I think you will find that the DEFINITION of power is the product of Volts and Amps

With a diode in series, your AVO is measuring DC current (mean) An AVO by virtue of its mode of operation (moving coil ), on its AC ranges is also measuring mean current. It just happens to apply scaling factor so that for a sine wave its indication correlates with the RMS value. It is not an RMS measuring instrument.

NO NO NO, I'm sorry but its RMS value will be 230/SQRT(2). The mean Voltage will be 230*SQRT(2) /PI

They may very well have done but it wouldn't reach the same temperature on the different supplies

Did you measure your experimental resistor temperatures ?

Ian

OK, I can get it to Peter on boxing day if Charles is coming your way, or I can weight it and post it

pinging Charles....

Dave

Yes, of course. Sorry.

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

If you dig back through "Radio Constructor" issues and the "In Your Workshop" articles, there is one where 'Smithy' teaches his 'lad' (Dick I think IIRC) to do the self same thing as a reduced power device while the iron is not in use.

A small micro-switch on the iron rest switched the diode in and out of circuit.

It would have been in the early 1960's.

That's showing my age :-))

Peter

-- Peter A Forbes Prepair Ltd, Luton, UK snipped-for-privacy@easynet.co.uk

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

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

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

AWEM

think IIRC) to

Blimey, that makes you an old codger then too

AWEM

Not quite confusing mean and RMS, just forgetting about the impact of (V**2)/R! A good lesson in engaging brain before committing fingers to keyboard. Given my profession I really shouldn't have made that mistake!

Dave

HM Govt just started paying my winter heating allowance!

Peter

-- Peter & Rita Forbes Email: snipped-for-privacy@easynet.co.uk Web:

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.

1. 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

regards

Ian

I remember the feature in the mag, not that particular one though. I've got a few years to go before I get my heating allowance, though!

Cheers Tim

Dutton Dry-Dock Traditional & Modern canal craft repairs Vintage diesel engine service

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

Just a bit of guesswork to add to the enthusiastic confusion surrounding this discussion.

All the posts so far have assumed that the 1.5 KW rating of the resistor is fixed value. There is no description of the resistor but it sounds very much like a standard water heater resistor - a nichrome element embedded in powdered magnesium oxide surrounded by a copper tube.

IF this is the case the 1.5KW rating is with the resistor in a water cooled environment. Dunked in oil the appropriate power rating may be very different.

Damage to the resistor is unlikely because this construction can quite happily withstand dull red heat. However the lower specific heat and conductivity of oil may lead to local overheating.

I've no figures to quantify this. It it might be worth a small experiment with a variac and the heater an open bath of oil to see what happens. Damage to the resistor is almost impossible - the power limit is likely to be local boiling of the oil.

Jim

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

Been a bit busy (always lots of work in the run up to Christmas) so I missed a chunk of this but if I'm still needed to collect from Peter in Luton and drop to Andrew in Bromley that's fine with me.

Keep me in the loop by email.

Regards

Charles