Diode to half power

[MrAl]

Hi Pauldreed,

Ok that's fine, but i'd still like to know where you are measuring the power, how you are measuring it, and with what instrument you are using and i asked if the power meter was the one outside on the wall that the power company provides as that one is pretty accurate i think.

You may be right that the power company wont care about a 500 watt half wave load if you happen to be the only one in that power block that is using a half wave load. I cant answer for them though so your best bet is to use a triac or lamp dimmer or else try the half wave load and see if you get any complaints in the future
It is also possible that there are regulations in your area that do not permit half wave loads of more than some wattage value.

 

[pauldreed]

Hi Pauldreed,

Ok that's fine, but i'd still like to know where you are measuring the power, how you are measuring it, and with what instrument you are using and i asked if the power meter was the one outside on the wall that the power company provides as that one is pretty accurate i think.

You may be right that the power company wont care about a 500 watt half wave load if you happen to be the only one in that power block that is using a half wave load. I cant answer for them though so your best bet is to use a triac or lamp dimmer or else try the half wave load and see if you get any complaints in the future
It is also possible that there are regulations in your area that do not permit half wave loads of more than some wattage value.

It's a Landis + Gyr 5235a meter as fitted by the power company outside on the wall.
1 LED blink = 1Wh.
But doesn't a triac create more distortion/harmonics than a diode as it chops the 360 deg sine twice instead of once and whilst at peak power instead of zero crossing?

I still don't understand where the harmonics originate with a resistive load?

 

[MrAl]

Hi again pauldreed,

Ok that partly explains what happened with the power measurement. The power in the load has to be measured at the load, not at the diode+load because that way we actually measure the power of the diode plus the power of the load, which is not really what we were after. Since it's only a difference of maybe 5 watts or less and you arent worried about it, lets concentrate on the other aspects. Originally i was just pointing out that the load changes with temperature, but since it really isnt that bad with your load we dont really have to worry about it here.

I didnt do a harmonic analysis myself (yet) but the harmonics of a half wave load come from cutting the bottom half off. In particular though, there will be a DC offset which is not present in a full wave sine. This means there is a net DC current in the line all the way back to the transformer. This can cause excessive buzzing and other effects like waveform distortion, but yeah it does depend on how big the transformer is and how over rated they made it.

The triac creates higher order harmonics by turning the load on fast at somewhat high level voltages. As i said i didnt do the analysis so i cant comment yet on what to expect. If you do a Fourier analysis on the half wave and on a triac type wave you'll see that they both contain harmonics, but the half wave has a DC component where the triac DC component is much much smaller almost negligible.

BTW you are right that a resistive load does not change the harmonics, but a non sinusoidal waveshape (including a half wave) does contain harmonics already with any load.

 

[4pyros]

WOW this has been oging ON and ON.......................

I was told in high school that using a diode in series with a soldering iron would reduce the power to 1/4. I have not been able to find any proof of this theory. Have I been living a lie all these years?

OP; I did read in some places on the web that some people have expereanced some hum/distortion in some equpment sometimes using a diode. So it can happen. If you are not having any problems than good.

This hole thing is not commonplace and may give you troubles down the road.

Maybe you would be better off with a full wave rectifier.

Andy

 

[pauldreed]

WOW this has been oging ON and ON.......................

I was told in high school that using a diode in series with a soldering iron would reduce the power to 1/4. I have not been able to find any proof of this theory. Have I been living a lie all these years?

OP; I did read in some places on the web that some people have expereanced some hum/distortion in some equpment sometimes using a diode. So it can happen. If you are not having any problems than good.

This hole thing is not commonplace and may give you troubles down the road.

Maybe you would be better off with a full wave rectifier.

Andy

A full wave rectifier would not give me a 500W load!

 

[4pyros]

A full wave rectifier would not give me a 500W load!

Why not its still 1/2 the peak to peak voltage of the full signwave?

 

[BrownOut]

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.

I get 1/√2 of the voltage of the full sine wave. Also 1/2Vpeak.

 

[Reloadron]

This weekend if time allows I'll try a few test.

Something interesting with regard to the elements I mentioned. The resistances I measured on both 500 watt and 1 KW elements was exactly what they work out to be (give or take a few ohms) for 240 volts at room temperature. These elements under load typically run in the 700 degree F. range. I would have expected the room temperature resistance to be much higher than the 115 and 57 ohms I measured respectively.

Ron

 

[crutschow]

I get 1/√2 of the voltage of the full sine wave. Also 1/2Vpeak.

Of course. A little typo on my part.

 

[crutschow]

Why not its still 1/2 the peak to peak voltage of the full signwave?

But now there are twice as many positive (or negative) cycles. You don't change the power by simply inverting half of the waveform. The load doesn't care whether one cycle consists of one positive peak and one negative peak, or two positive (or negative) peaks. They all will generate the same power in the resistive load.

 

[Diver300]

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.

Just going back to this, because there was doubt that the figures would be exactly half.

If you are counting 1 Wh pulses for 1/10th of an hour, and multiplying by 10, how did you get to 545?

Anyhow, you counted 109 pulses, it seems without the diode, and 54 or 55 pulses with the diode. I don't see that as particularly remarkable. It only means that whatever errors would cause the power ratio to vary from 50% are smaller than the resolution of this measurement, which should have been 10 W, if there area whole numbers of pulses.

 

[MrAl]

This weekend if time allows I'll try a few test.

Something interesting with regard to the elements I mentioned. The resistances I measured on both 500 watt and 1 KW elements was exactly what they work out to be (give or take a few ohms) for 240 volts at room temperature. These elements under load typically run in the 700 degree F. range. I would have expected the room temperature resistance to be much higher than the 115 and 57 ohms I measured respectively.

Ron

Hi there Ron,

Are you saying that you measured the resistance of two heaters at room temperature (say 20 deg C) and at hot temp (700 deg F) and you found the resistance to obey Ohm's Law exactly (ie have the same resistance at both temperatures)? And what do you mean by "what they work out to be"?

I measured my soldering iron a while back and it definitely changes resistance as it heats up. Yes, this isnt a heater though Actually, i dont know of any metal that doesnt change resistance with temperature.

 

[The Electrician]

This weekend if time allows I'll try a few test.

Something interesting with regard to the elements I mentioned. The resistances I measured on both 500 watt and 1 KW elements was exactly what they work out to be (give or take a few ohms) for 240 volts at room temperature. These elements under load typically run in the 700 degree F. range. I would have expected the room temperature resistance to be much higher than the 115 and 57 ohms I measured respectively.

Ron

Wouldn't the fact that the resistances at room temperature are what the power ratings would have them be, suggest that the resistance at operating temperature is about the same? That is, that the temperature coefficient of resistance is very low?

 

[MrAl]

Hi Electrician,

Hey you're up early today too

The way NiChrome wire designs go is to figure out the change in resistance with temperature and that gives the true operating power. In other words, with a change of some percentage with some known excitation voltage, what is the final value of the resistance so we know what the power is during normal operation. The room temperature operation isnt that important except maybe as a starting point.
Im starting to wonder what kind of material is used in these heaters being talked about here. Maybe most people arent worried about 2 percent changes, and maybe the designs use lower final temperatures.

For a temperature change from 315 deg C to 20 deg C we would see a 3.3 percent change in type A and a 5.2 percent change in type C. A 1000 watt heater made with type C would actually use 1052 watts at room temperature, until it heated up at which time it would use 1000 watts. Rough calculation for this heater would be 512 watts at half 'power'.

 

[Reloadron]

Wouldn't the fact that the resistances at room temperature are what the power ratings would have them be, suggest that the resistance at operating temperature is about the same? That is, that the temperature coefficient of resistance is very low?

Yeah, which I found a bit peculiar. One of those things that I just never gave much thought to. I assume they are nichrome wire elements and as I said, these typically run about 700 degrees F. I guess I just wanted to believe there would be a larger delta in resistance.

Ron

 

[MrAl]

Hi Ron,

I did some calculations for type C NiChrome in the post just before yours. Maybe you can compare.

 

[The Electrician]

I was told in high school that using a diode in series with a soldering iron would reduce the power to 1/4. I have not been able to find any proof of this theory. Have I been living a lie all these years?

You've been living a lie.

Here is experimental evidence; whether you consider it proof is up to you.

I used a Weston wattmeter to measure the power consumed by an electric toaster. The toaster is essentially just a nichrome wire connected to a line cord.

The first image shows the wattmeter scale with the toaster connected and no diode in series. The reading must be multiplied by 2 because there are 2 current ranges on this wattmeter--5 amps and 10 amps. I was using the 10 amp range. You can see that the reading is 437 which multiplied by 2 is 874 watts. You can also see below the scale that the wattmeter can read both AC and DC. You can also see the Fluke clamp-on ammeter in the image.

The second image shows the reading with a diode in series. I used one diode of a bridge which can be seen in the image. The reading is 216 which when multiplied by 2 is 432 watts. Notice that this is a little less then 1/2 the value with no diode. The diode drop reduces the power a little.

Hair dryers use this diode-in-series technique to reduce the blow dry power when you select a power level less than full power. The dryer is drawing several amps of DC current from the outlet when on medium power. This current is not enough to hurt the pole pig; it's just not enough to push the core into saturation very far.

 

[4pyros]

You've been living a lie.

Here is experimental evidence; whether you consider it proof is up to you.

I used a Weston wattmeter to measure the power consumed by an electric toaster. The toaster is essentially just a nichrome wire connected to a line cord.

The first image shows the wattmeter scale with the toaster connected and no diode in series. The reading must be multiplied by 2 because there are 2 current ranges on this wattmeter--5 amps and 10 amps. I was using the 10 amp range. You can see that the reading is 437 which multiplied by 2 is 874 watts. You can also see below the scale that the wattmeter can read both AC and DC. You can also see the Fluke clamp-on ammeter in the image.

The second image shows the reading with a diode in series. I used one diode of a bridge which can be seen in the image. The reading is 216 which when multiplied by 2 is 432 watts. Notice that this is a little less then 1/2 the value with no diode. The diode drop reduces the power a little.

Hair dryers use this diode-in-series technique to reduce the blow dry power when you select a power level less than full power. The dryer is drawing several amps of DC current from the outlet when on medium power. This current is not enough to hurt the pole pig; it's just not enough to push the core into saturation very far.

Hay thanks for taking the time to do that experiment and for destroying my high school 1/4 watt theory.

While you have the stuff out, Would you mind trying it again with full wave rectification and see what you read?

Andy

 

[Reloadron]

Thanks for taking the time. Truly a classic watt meter. Back in the day I loved those units. Weston made some really great meters like the one shown. Unfortunately I saw many go to a dumpster before I could dumpster dive and rescue them. Thanks again for taking the time and the images. I was going to do it this weekend, now I can do real important stuff like getting the beer cleaned out in my fridge.

Thanks also MrAl for the nichrome data. I just had it in my head the change would be so much greater.

Ron

 

[The Electrician]

Hay thanks for taking the time to do that experiment and for destroying my high school 1/4 watt theory.

While you have the stuff out, Would you mind trying it again with full wave rectification and see what you read?

Andy

OK. Don't read too much into this value. The first images were taken close together in time; this one was taken about 3 hours later. The line voltage may have changed (probably decreased) because it's later in the afternoon. This is the same toaster load with a full wave rectified grid voltage applied.