Diode to half power

[pauldreed]

I am using a diode to half the power used by a 1kW AC immersion heater, do I need some sort of snubber circuit to suppress possible transients, or does the zero crossing nature of a diode make this unnecessary?

 

[Diver300]

There shouldn't be any transients.

The peak power won't be changed, and the rms current will only be reduced to 70.7% of what it was.

 

[carbonzit]

What's zero crossing got to do with it? The heater is a non-inductive source, so you don't have any inductive "kick-back" to deal with, so no need for a snubber.

I think you're thinking of triacs regarding zero crossing. A diode simply blocks half the AC cycle.

 

[pauldreed]

OK, I thought that it wouldn't be necessary as the diode runs lukewarm and no other issues have become apparent, but wanted to check to avoid future problems.

 

[Sceadwian]

If it's run from 60hz AC you'll never have a problem with it conducting in the wrong direction, that's only an issue at high frequencies. Keep in mind some type of filter on the input would be recommended if not required, simply cutting half the AC cycle on a 1000watt heater will introduce massive noise on the power line.

 

[pauldreed]

If it's run from 60hz AC you'll never have a problem with it conducting in the wrong direction, that's only an issue at high frequencies. Keep in mind some type of filter on the input would be recommended if not required, simply cutting half the AC cycle on a 1000watt heater will introduce massive noise on the power line.

It's run at 50 hz AC, but don't follow the rest of your post. Are you suggesting that some sort of filter will be required?
If so, what do you recommend?

 

[Sceadwian]

Paul, normally the heater coil is fully resistive throughout the entire AC cycle so it loads each side of the AC cycle equally, the diode causes the circuit to suddenly become VERY non linear, current will only flow during one half of the AC cycle, and this will create noise on the power line, lots of it. A large inductor, used as a choke and capacitor would help smooth out the sudden missing half cycle and reduce the noise created. A good capacitor to use might be one from a microwave oven, the inductor (choke) could be perhaps the primary winding of a microwave oven transformer (with the secondary not connected to anything) Be warry of MOTS (microwave oven transformers) they're incredibly useful devices but the secondary puts out several thousand volts and more than enough current to kill nearly instantly.

 

[4pyros]

I am using a diode to half the power used by a 1kW AC immersion heater, do I need some sort of snubber circuit to suppress possible transients, or does the zero crossing nature of a diode make this unnecessary?

You will only get 1/4 power it will not half the power

 

[crutschow]

You will only get 1/4 power it will not half the power

How do you figure?

If you remove half of each cycle you will get half the power. A full cycle gives full power. A half cycle gives half power.

But the power company won't be happy with a 500W half-cycle load since DC current tends to saturate the power transformer and it generates harmonic losses in the power line, as well a large amounts of noise due to the harmonics.

Edit: You would be better off using a lamp dimmer to reduce the power.

 

[4pyros]

How do you figure?

Do the math for AC power

 

[carbonzit]

You don't need to do the math; you only need to look at the waveform.

There are some places--not that many, but some--where inspection is all that is necessary to come to the right answer. This is one of them.

Take an AC waveform, any symmetrical waveform, and rectify it (ideally). You automatically have half the power. No math required (although math will prove this).

 

[awright]

4pyros, I think you are confusing cutting the voltage in half with cutting the duty cycle of the waveform in half.

If you cut the voltage applied to a resistive load in half, for example, by means of a transformer, the current is also cut in half (by Ohm's law) with the result that the power (the product of voltage and current) is cut to 1/4. However, when you cut the duty cycle in half (in this case, by eliminating half of the half-sine waves) without changing the amplitude of the remaining half-sine waves, you are simply eliminating half the power applied to the load, similar to turning the power switch off for 50% the time 60 times per second.

I fully agree with the posts above that criticize the use of a simple diode to reduce power to a heavy load. The simple diode is quite acceptable for reducing the power applied to a small lamp because the unbalanced current will be small and of little consequence. However, a 1 KW nominal load driven by a half-wave rectified waveform can cause problems with a transformer that is not specifically designed to handle the DC component of the half-wave rectified current. And, of course, the power company may not be happy with you.

The best solution is an autotransformer or a 2:1 transformer of 250W power rating used in a line bucking configuration. Second best, but least expensive, would be a simple home-brew triac circuit set up for the fixed power output you want or a higher than normal rated lamp dimmer. A common lamp dimmer is typically rated for a 600 watt load (at full power) and your 1KW load may be excessive for it. However, you can probably find a dimmer rated for 1KW or more at a premium price at an electrical supply house.

The transformer will generate negligible noise on the line. The triac circuit or dimmer will produce noise on the line but at least it is balanced and should not cause problems unless you have sensitive electronics on the same circuit or nearby.

awright

 

[pauldreed]

I did consider a Thyristor circuit to reduce the power, but was advised that they also generate a lot of noise/harmonics.

 

[canadaelk]

1/2 the voltage = 1/4 the power. That pesky inverse-square law again! E

 

[Diver300]

1/2 the voltage = 1/4 the power. That pesky inverse-square law again! E

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

The inverse square law is about electromagnetic radiation. Power in a resistor is proportional to voltage squared, there is no inverse anywher.

Anyhow, the voltge is not halved, depending on how you average it, but there are several ways of calculating the average voltage.

You could simply measure the average voltage with a DC meter. It would show zero with a full AC waveform, and quite a lot with a half wave rectified waveform, so that would be increased.

If you measure AC volts r.m.s. (route mean square), which is how mains AC voltages are normally measured, the voltage will be reduced to 70.7% of what it had been. RMS voltage and current are often used because it allows simple calculation of power.

The positive peak voltage would be unchanged when the diode is added.

The negative peak voltage would be the same magnitude as the positive peak without the diode but would be zero with the diode.

The peak to peak voltage would be halved when thd diode is added.

Also, how voltmeters, especially cheap ones, measure non-sinusoidal voltages is anybody's guess. A true RMS voltmeter would show a half wave rectified voltage as being 70.7% of the votlage.

Anyhow, the average power is halved, because the same power is only consumed but for only half the time. It really is as simple as that. There is no square law for running stuff for a fraction of the time. If you use an device for half of the time, it uses half of the energy. I would guess that you have noticed that if you drive half as far, your car uses about half the amount of petrol, not a quarter....

If you had a square wave, on for half of the time, and off for half of the time, it would show half the voltage on a DC voltmeter. However, if that square wave were loaded with a resistor, the power would not be the same as it would if the voltage were steady and half of the peak.

Say 10 V, 100 Ω

Full power = 1 W

square wave, 50% duty cycle. Average voltage is 5 V
When on, 1 W. When off 0 W, so the average is 0.5 W

Half voltage, 5 V,
power is 0.25 W.

 

[MrAl]

Hi,

I have to second Carl's idea of using a triac or higher powered lamp dimmer rather than a single diode for the same reasons as others mentioned here about the line power problems that come up. Of course the added advantage is that it can be adjusted to whatever power level is needed too that way.

It's easy to prove that a half wave sine produces one half the power in a resistive load. That's for a perfect resistor though too (which the heater is not). The calculation for power involves an integral, but after that integral is taken we have to divide by the period. Since a period of 0.5 will yield full power (for a sin squared wave) and for half wave we would use a period of 1 (twice that of 0.5) and we are simply dividing the same number in half at that point, we would end up with one half as much power as with a full wave source. Note we use a period of 1 because we integrate over half the period normally which is really one period of sin squared.
In other words, we integrate over one half period for the full wave and use a total period of 1/2, and integrate over one half period for the half wave but use a total period of 1 which leads to one half as much as using a total period of 1/2.

There is a slight catch here however. That is, the resistance of the heater element is going to change with temperature. This means that as the unit is set at lower and lower power levels the resistance goes down somewhat. This means that for a half wave source we would get slightly more heating than with a perfect resistor that doesnt change resistance with temperature. The amount of heating depends of course on the characteristics of the heater element and these can vary quite a bit between different types of heaters.
This makes Carl's idea of using the triac even better, because we would be able to adjust for a perfect 500 watts if that's what we really needed, whereas with the single diode we might get 600 watts instead.

 

[alec_t]

But the power company won't be happy with a 500W half-cycle load since DC current tends to saturate the power transformer and it generates harmonic losses in the power line, as well a large amounts of noise due to the harmonics.
Edit: You would be better off using a lamp dimmer to reduce the power.

I've just been playing with LTSpice. Using a diode to halve the power apparently results in ~ 43 % harmonic distortion, 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.
Of course, you could always persuade a neighbour to use the diode method on their immersion heater; with the diode reverse-connected, of course, to balance things out as far as the power company is concerned .

 

[MrAl]

Hi alec,

Oh ok nice idea. Maybe you can find a heater element model so we can see how much difference the half wave power becomes with a real heater as i mentioned in my post just before yours, #16.

 

[alec_t]

Maybe you can find a heater element model

I hadn't thought to search for one on the net, but I doubt we'd be lucky. I like the theory, but I think it's rather academic as the resistance difference between the two methods is likely to be very small, given the thermal inertia of the heater and the thermal-insulating effect of the mineral electrical-insulator material within it.

 

[crutschow]

.....Of course, you could always persuade a neighbour to use the diode method on their immersion heater; with the diode reverse-connected, of course, to balance things out as far as the power company is concerned .

That would only work well if both heaters were on at the same time and only if you were sharing the same phase on the same transformer.