ECN Forum Forums General Topics of the Trade General Discussion Area "power miser" lightbulb power reducers

Assuming that these things are diodes: If all light bulbs and diodes conduct on only the positive side of the voltage waveforms, then the neutral I think will get hammered, having to supply all that single direction current, for all 3 phases. Normally, if bulb on phase A is at the positive peak of the cycle, bulbs B and C will be drawing negative current of half the peak current, per bulb. Adding up to be equal but opposite of bulb A. The diodes will block that, and the neutral has to pick up the slack. Now, if you have hundreds of bulbs in the building, and the diodes are randomly pointed (so some bulbs conduct on the negative side, and the others on the positive side, then the neutral will see less burden. So tell the installers to not be consistent when they install the diodes. Tell them to alternate the diode direction from bulb to the next bulb.

No current flow at all in the negative 1/2 wave. No current on the grounded conductor still 120 degrees off the next phase. Current flow will still be the same magnitude in the positive 1/2 wave. No hammer or overload anywhere. In fact only 1/2 the possible energy and still in balance if all phases have equal load.
Sorry I don't see a problem at all except for the brown light output.

Huh?

The neutral doesn't enter the equation at all. It's the hot that supplies the power. What doesn't enter the bulb can't possible enter the neutral. It's a non-issue.

wa2ise is right.

Balancing the neutral depends on the current waveforms, positive or negative, adding up to zero through the whole cycle. With pure resistive loads, that's exactly what happens. In that case the worst situation is all the loads on one phase. The neutral will have no more current than the hot phase involved. When these loads are evenly distributed over three phases, the neutral ends up with zero current (and the same happens when each phase has the same power factor on lagging loads).

When the loads do not present a pure sine wave current, then you begin to see problems. This is harmonics. And half-wave diodes make a LOT of harmonics.

If each phase has lots of half-wave load, all in the same direction, then there is very little current overlap time, and the neutral ends up conducting at separate times for all three phases, in the peak 33% of each phase's cycle (and maybe more, depending on the diode characteristics). The current in phase A (at the times its diodes are conducting) cannot return through phase B or phase C when the diodes in those phases are not conducting, and so the neutral gets the current just as if phases B and C had zero loads (which is the case when their diodes are in blocking state).

If the diodes are installed half positive and half negative within each phase, then you may have some approximation of a full waveform, depending on when the diodes actually conduct. They won't conduct all the way to zero crossing, so there will still be some amount of harmonics and added load on the neutral.

Other hazards exist with half-wave loads, too.

On single phase, if all the diodes on pole A are positive and all the diodes on pole B are negative, the neutral will see the sum of current for both poles. If someone decided that because the diodes reduce the overall load, that they can now double the loads, then the neutral could be running as high as 200% of rating.

On any system, when the diodes do not have an equal amount of current in the positive and negative directions, you then have a DC component of current. That doesn't play well with transformers, especially when you have a lot of it.

I have not analyzed what this would do to current transformers, either in power meters or in ground fault detection devices.

Thanks pdh. Anyway, I wondered if I got it wrong or not, so I ran a computer simulation to see what would happen. See the result over in the photo gallery forum. Yes, the neutral gets a lot of DC current.

Just in case people didn't notice, that waveform is upside down. The zero line is the 2nd one, just under the top line. So there's a lot of DC current in this, around 0.95 amps, plus the added 0.1 amps triplen current, mostly 3x, but some higher harmonics like 9x are also present. That DC+triplen current would be approximately equivalent to 1.44 amps of AC. Given that 100 watts through 144 ohms is 0.833333 amps, we're seeing about 72% overcurrent on the neutral compared to plain 100 watt resistive load without a diode on just one phase alone ... even before we consider the saturation effects the DC will have on the supplying transformer.

These things might be OK when the amount of load they are part of is a very small percentage of the transformer and wiring ratings. But on a grand scale, it can be very unsafe. Just use a smaller wattage bulb or switch to CFL (or both).

I'm wondering if any cheap CFLs have DC and/or harmonic currents.

Putting aside the issue of the actual current flow in the neutral wire, if the current is 1.44 amps with the 'Economiser' fitted, in comparison to 0.833 amps without it, does that mean it actually costs more to run and yet gives out less light? This is very confusing, because I know if I were foolish enough to fit an 'Economiser' unit to an Edison Screw type bulb here, on an EU 230v 50hz single phase supply, the effect described would not occur.

I have a lamp holder with a CT on it I was playing with for my energy monitor experiment I will see if I can see any current anomalies with a CFL in it.

If you chase the various patents from the Power Mizer you eventually get to this.
Scroll past the ads.

http://www.freepatentsonline.com/3823339.html

Alan ... in wa2ise's simulation, once you account for the fact that you have 3 lights operating, and figure it in turns of real power used, a diode will in fact economize, even in a three phase system. You can only get so much power to be dissipated by each load, and for resistive loads, the diode cuts that in half. Actually, it's probably just a bit more than half because the diode won't start conducting immediately when the zero crossover transitions to forward voltage (resulting in less overlap and more current diverted to the neutral), and because of the lower operating temperature of the filaments resulting in more current.

The problem is that because of the timings of the intermittent (half cycle) current, much or most of the return current goes by way of the same shared common neutral conductor. If the diodes were to conduct only during the center 66% of the half cycle (33% of the whole cycle), then at no time could any line-to-line current flow happen, and all of the current will be forced through the neutral. Regular dimmers can cause this problem, too. Because regular dimmers do (there may be a few cheap ones that don't) conduct in both directions, the dimmer would have to be adjusted so it is conducting only 33% of the time for this effect to be at its worst.

So if you have a multi-wire branch circuit on a common neutral, protected at 20 amps (North American standards), with 15 amps of load (for example, 30 60-watt incandescent bulbs) on each phase, you would expect essentially zero current on the neutral (ignoring slight bulb wattage differences due to manufacturing variations). Now put these diode type power reducers in place on each bulb, all in the same polarity. Ignoring the effects of the reduced filament temperature (which would lower resistance and raise the current, slightly), you now have 7.5 amps of load on each phase. During the highest part of the conducting cycle of each phase, neither of the other phases are conducting. This might be as little as 33% of the conducting time, but it is during the current peaks, so it will be most of the total averaged current. This could be an average of say, 6 amps per phase. This 6 amps conducts through the respective phase and the neutral. And this happens in sequence for all three phases, giving a net average on the neutral of 18 amps (at 180 Hz in North America).

That's at least within the designed in tolerance-error forgiving deratings that NEC requires. But, that narrows the forgivable tolerance errors in other aspects (for example, the 60-watt bulbs might actually be 67 watts, due to variations in manufacturing that batch they all came from).

But someone might be thinking "I now only have 7.5 amps of load on each phase of this MWBC ... so I can double up and join another branch in with it and still be well within ratings". So just double all the figures I gave above and that someone would end up with 36 amps on the neutral which is likely to be AWG #12 because it is a 20 amp circuit. Don't forget I-squared-R means you are dissipating 4+ times as much power in that neutral as it should be in a normal worst case (all loads within rating on one phase). Each phase conductor will still be around 15 to 16 amps, which should never trip the breaker. So the 36 amp neutral current will continue at that level until something bad happens.

Three phase power can easily be a problem with non-linear harmonic loads like this. But the diodes operating in half-wave mode (conducting in one direction only) can also pose a similar risk to two pole single phase circuits with a shared neutral. For that to happen, (most of) the diodes would have to be polarized one way on phase A and the other way on phase B. So with 30 60-watt bulbs on each pole of a two-wire shared neutral circuit, these diodes would cause all the current on the neutral (because there's zero conduction overlap time between poles at 180 degrees). Ignoring the slight rise in current due to the lower filament temperatures, that's the same as the worse case (all loads on one pole) and not yet a hazard. But the apparent reduced load on each pole could lead people to double up the loads, resulting in double the current (30 amps) on the neutral. It would be almost as bad as the three phase scenario.

The single phase scenario would have very little DC current on the neutral. if both poles are equivalent windings on the same transformer core, then the effect should still be an alternating magnetic field. But if the 120/240 system is derived from two separate 120 volt transformers in series, each transformer will be dealing with DC current. They won't be happy. Fortunately in North America, a setup like that is unheard of because 120/240 transformers are "everywhere", so the primary hazard is someone doubling up the neutral.

The three phase scenario will have lots of DC current. Making sure that half the diodes are one way, and half the other way, on every phase, should at least eliminate the direct current issue. I'd just avoid these things on a big scale. They could be OK for a few lights. But a smaller wattage bulb would still do better (more efficient in terms of usable light per watt because they would be running at design temperature). If you have a large scale lighting system and need to save power, you should already be doing something other than incandescent.

See NEC 210.4(A)FPN and 310.10 FPN(2) and of course 310.15(B)(4)(c).

You can run the calculations yourself for European standards on a 400/230 volt three phase system and a rare 460/230 volt 2-pole single phase system. A single pole single phase system won't have the same kinds of problem. It can still pose problems to the source transformer, especially if it is three phase and customers on other phases are doing the same thing, in significant amount.