Diode to half power

[pauldreed]

To settle the discussion, I have tried measuring the power, with and without the diode, by counting the LED flashes on the calibrated (electric company) electric meter over 6 minute periods, and multiplied the results by 10 (1 flash = 1Wh). Without the diode I got 1090 Watts, and over the same period with the diode I get 545 Watts, which surprisingly is exactly half.

 

[crutschow]

To settle the discussion, I have tried measuring the power, with and without the diode, by counting the LED flashes on the calibrated (electric company) electric meter over 6 minute periods, and multiplied the results by 10 (1 flash = 1Wh). Without the diode I got 1090 Watts, and over the same period with the diode I get 545 Watts, which surprisingly is exactly half.

No surprise. Love it when theory matches practice.

 

[MrAl]

To settle the discussion, I have tried measuring the power, with and without the diode, by counting the LED flashes on the calibrated (electric company) electric meter over 6 minute periods, and multiplied the results by 10 (1 flash = 1Wh). Without the diode I got 1090 Watts, and over the same period with the diode I get 545 Watts, which surprisingly is exactly half.

Hi,

That's not only interesting that is also too hard to believe. You're saying that the voltage drop of the diode exactly compensates for the slightly lower resistance of the heater when operated near 500 watts. What a coincidence
I suppose it's possible but that's too amazing.

No surprise. Love it when theory matches practice.

Yes, but only when it really does (see above)

 

[Sceadwian]

Sorry, but I have to agree with MrAl, look at the diode drop out voltages. It will never be 50% power perfectly. It will however NEVER be 1/4 power except under some obscure harmonic conditions not defined in the previous posts.
Half cycles used for AC current limiting in a resistive element can practically be said to be half power.

 

[crutschow]

Since the heater is immersed in water it likely doesn't change temperature much from dissipating 500W to dissipating 1000W, thus its resistance wouldn't change much either between the two power levels. And the 0.7V drop of a diode is quite small when compared to the mains voltage (<1%). So to me it's not that surprising that the power with the diode is 1/2 the power without. (Although it is probably lucky that it happened to measure exactly 1/2.)

 

[4pyros]

It's not the inverse square law, and it's not half the voltage.

Why would it not be half the voltage if rectified?

 

[Reloadron]

OK, looking for takers. I'll send anyone in the Continental US their own, yes their very own 500 watt, 1000 watt or even 1500 watt 240 volt heater strip to experiment with. It will be yours to keep and postage is on me. Just run the experiment and post the results.

You will need obviously a big diode, a 240 volt supply and the ability to measure with some degree of accuracy 240 VAC and the current.

I have been following this thread and I agree the power will be about half. I say about because heater elements and diodes are not quite an exacting science.

Today at work I managed a few min and measured the resistance of a 500 watt and 1,000 watt 240 volt strip heater element. Interestingly Ohms Law held true. The 500 watt elements were about 115.2 Ohms and the 1 KW units were about 57.6 Ohms. I say about because they sure as hell were not exact, a variation of +/- 5 ohms should be expected. Does anyone think for a moment that an immersible hot water heater 1,500 watt element will be exact? Will the water really know if it is delivering 1475 watts or 1525 watts of heating? I doubt it as the water is not that bright.

Based on the above I can see where the OP got the numbers he did. I mean how much difference would the forward voltage drop of a diode matter? Considering the actual quite flexible resistance of the load?

I do have a 240 volt 20 amp bigger than life variac and the means to measure voltage and current. My only source of 240 volts is the electric dryer outlet and I am not about to head to the basement to seek 240 volts. Using the drier outlet means getting another plug but even worse explaining to a patient and loving wife why I am again expanding my work/hobby into yet another room. She calls it encroachment?

Takers?

Ron

 

[carbonzit]

Why would it not be half the voltage if rectified?

It's half the peak-to-peak voltage, but that doesn't result in 1/4 the power.

This is one of those cases where you can really just look at the waveform and know the answer, in this case that the power will be cut in half.

Draw, or visualize, a sine wave. Let's assume it's the power wave, not a voltage or current wave.

The total power is the area under the curve, right? This being a symmetrical sinusoid, the total power is the sum of the areas between both halves of the wave (positive and negative) and the X-axis, right?

Now erase the bottom half of the wave. Bingo--half the power (and exactly half) is gone.

Q.E.D.

 

[awright]

Interesting how a simple question gives rise to a protracted discussion in these forums. No criticism - I think we are all enjoying the discussion.

I was careless in mentioning use of a 2:1 step-down transformer in bucking configuration to cut power in half. I should have referred to a 10:3 step-down transformer in line bucking configuration for 50% power reduction. Note that the bucking transformer need only have a power rating of about 210 watts (.7 of the full current * 0.3 of the full voltage).

Reloadron said, "Will the water really know if it is delivering 1475 watts or 1525 watts of heating? I doubt it as the water is not that bright." Actually, water is quite brilliant in this specific area of expertise. Accurate calorimetry would reveal exactly the power dissipated in the water. It's not the water that's not that bright, its our inaccurate or careless measurement techniques that are not that bright.

awright

 

[pauldreed]

Hi,

That's not only interesting that is also too hard to believe. You're saying that the voltage drop of the diode exactly compensates for the slightly lower resistance of the heater when operated near 500 watts. What a coincidence
I suppose it's possible but that's too amazing.

Yes, but only when it really does (see above)

Mr Al, are you looking for a conspiracy theory?
Those were the results that I got over 2 six minute trials and yes it was a surprise that it was exactly half, but I posted as I found it, I have no call to massage the figures....

 

[alec_t]

yes it was a surprise that it was exactly half

Not so surprising IMHO. The effects of small differences in heater resistance and diode drop are probably swamped by inaccuracies and limited resolution of the power company's meter which was used to get the result. What accuracy is claimed for consumer power-use meters?

 

[pauldreed]

Not so surprising IMHO. The effects of small differences in heater resistance and diode drop are probably swamped by inaccuracies and limited resolution of the power company's meter which was used to get the result. What accuracy is claimed for consumer power-use meters?

Can't tell you in % terms. but the meter spec sheet shows accuracy to IEC 62053-21 class 1 or 2.
There may have been some drift if I had timed it over 2 x 1 hour's, instead of 6 mins each. Who knows, but not that important IMO.

 

[MrAl]

To settle the discussion, I have tried measuring the power, with and without the diode, by counting the LED flashes on the calibrated (electric company) electric meter over 6 minute periods, and multiplied the results by 10 (1 flash = 1Wh). Without the diode I got 1090 Watts, and over the same period with the diode I get 545 Watts, which surprisingly is exactly half.

Hi again,

Your conspiracy: he he.

We would need to look into this a little more...
Where is this 'calibrated electric company electric meter' located? Is that the good (accurate) one on the outside wall of the building or something?

You have to realize that even a 1 percent change in resistance would result in a difference of 5 watts. A diode drop of 0.5v would result in an approximate difference of 2 watts. NiChrome wire (like what heaters are usually made out of) changes resistance with temperature quite a bit.

We all agree however that it wont be 1/4 power that it will be "close" to 1/2 power.

 

[crutschow]

Why would it not be half the voltage if rectified?

The average voltage may be 1/2 but if you measure the voltage with a true RMS meter or calculate the RMS voltage you will find that a half-wave rectified sine wave has √2 of the voltage of the full sine wave.

 

[ronsimpson]

If I remember from 30 posts ago.......You have a diode and a heating element.
Some (many) watt meters can not measure DC power on the power line.

If the current is measured via a resistor then the meter is OK.
If the meter used a current transformer to measure power it will not read good when the +current does not equel the -current. (diode)

 

[The Electrician]

I've just been playing with LTSpice. Using a diode to halve the power apparently results in ~ 43 % harmonic distortion,

I get 43.5% for this case using a purely mathematical method.

whereas using a lamp-dimmer set for half power (i.e. chopping off the leading half of each half-cycle) results in 95 % harmonic distortion !! You pays yer money, yer takes yer choice.

I get 65.05% for this case. Can you check your result?

 

[The Electrician]

NiChrome wire (like what heaters are usually made out of) changes resistance with temperature quite a bit.

We may also disagree about what the phrase "quite a bit" means. Why not give an actual number for the change so we can decide for ourselves whether the change is "quite a bit"?

Since this is an immersion heater, the temperature of the nichrome heater won't be much higher than the surrounding water which won't be above 100 degrees C.

According to the temperature coefficient of nichrome varies from .00013 to .00017. Let's take it to be .00015.

If the room temp is, say, 20 degrees, then the temperature rise of the heater will be 80 degrees if the water is boiling. Then the resistance of the nichrome in the heater will increase by a factor of (1+.00015*80) = .012 = 1.2%. I don't consider that to be "quite a bit" of resistance change.

 

[pauldreed]

I get 43.5% for this case using a purely mathematical method.

I get 65.05% for this case. Can you check your result?

I can understand that using a diode would create an imbalance in the supply line (as it is only drawing current for 180 deg of the cycle) but is the supply company even going to notice this? after all - it's 1kW, when they supply megaWatts?

What makes the harmonic distortion anyway? I assumed that with a resistive load, the diode would use the zero crossing of the waveform to shut off the supply, with no kickback from the load??

I have run the 'diode' circuit for about 2 hours today, I have tried the TV, radio, VOIP phones etc, and have noticed no detrimental effects whatsoever.
After 2 hours, the diode was just warm to the touch.

 

[MrAl]

Hi again,

pauldreed:
You neglected to read post #33 and tell us HOW and WHERE you are measuring the power and what exactly you are using to measure it. I asked, is it the power companies accurate meter outside the house on the wall or something, or something else?

Electrician:
Yes there are too many variables, but what i was saying basically was that it would be quite a coincidence if the power was EXACTLY 1/2 the full power. Even with a 1 percent change in resistance we might see a 5 watt difference.
Immersion heaters are not all created equal either, and many of them have the coil inside where there still is a large thermal gradient between coil and outside case with some insulation between. It's hard to say exactly what this gradient is without measurements.

 

[pauldreed]

MrAl, you neglected to read post #32 - 'it's not that important!!' I don't really care if it's 540 Watts or 545 Watts, it's roughly half a kiloWatt...

The aim of this thread was to explore whether it was viable to use a diode in this manner and explore potential problems such as transients.