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I'm looking to calibrate an energy meter.

As a part of that process, I need to cheaply find a device that can create a power-factor of 0.5, in order to calculate the required phase correction.

The best, or worst, device that I can put my hands on has a power factor of 0.78.

Everything else appears to be much better than that and, not being an electrical engineer, I'd prefer it if there was a device that wouldn't be too involved electronically.

Can anyone suggest, perhaps an off-the-shelf device, that would perform this badly?

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    \$\begingroup\$ Tou forgot to specify how much current tou need. Hit the edit link ... \$\endgroup\$ – Transistor Apr 9 at 12:48
  • \$\begingroup\$ Just make one; teh calculations aren't that dificult. \$\endgroup\$ – user_1818839 Apr 9 at 12:54
  • \$\begingroup\$ Assuming you are on the inductive side of unity power factor, add an inductor in parallel to lower the PF? \$\endgroup\$ – winny Apr 9 at 12:54
  • \$\begingroup\$ Have you tried an electric motor (without load)? \$\endgroup\$ – Hot Licks Apr 9 at 21:07
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An unloaded power transformer is a pure inductance. As you increase the load, the power factor improves.

So, take a Variac, plug an incandescent light into it, and you should be able to dial up any power factor you like.

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    \$\begingroup\$ Yes, it is inductance, but it is a large inductance -- so the load current (although with a poor power factor) will be small. \$\endgroup\$ – jp314 Apr 10 at 6:06
  • \$\begingroup\$ Thanks very much for your comment. We have a variac, so I will give that a go. I will accept this comment once I've found out that it works =). \$\endgroup\$ – Alex Apr 12 at 9:00
  • \$\begingroup\$ @jp314: Well, I'm surprised. I finally got a chance to measure the magnetizing current on my larger variac, and it's only about 16 mA, which implies that the inductance is on the order of 20 H! I really expected the current to be about an order of magnitude larger. \$\endgroup\$ – Dave Tweed Apr 17 at 23:34
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The cheaper LED lamps - the ones that use a capacitive dropper rather than a switch-mode power supply - can have a very poor power factor. The power factor could be as low as 0.2. Buy a bunch of them, and wire them in parallel to get a significant current draw.

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You can buy inductors with a wide range of current and inductance ratings. An unloaded induction motor has a low power factor. If you use a single-phase motor, you can probably disconnect the capacitor if it has one. Unloaded, it should start with a little manual twist of the shaft. You may have difficulty with the distorted magnetizing current waveform if you use a transformer. That could also me an issue with a motor. Using something with a higher voltage rating than your measurement voltage would probably help.

You may want to consider how the meter performs with a distorted waveform. An energy meter should indicate real power and/or energy without regard to power factor or waveform. If the meter provides a power factor reading, it should probably indicate total power factor combining the effects of distortion and displacement.

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Poor power factor in practice is usually inductance-based. On the other hand, inductors for testing purposes look rather bulky and expensive.

If you are OK with capacitive load, the capacitors for asynchronous motors come in various capacitances and are rated above mains voltage. They are also rather cheap.

You can combine them with incandescent lamps, space heaters or other known near-unity power factor loads and get anything from 0 to 1.

You can even borrow them from a local A/C parts supplier and return them once you are done with the measurements.

p.s. be sure to check the actual capacitances. They vary a lot.

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"Off the shelf" requests for specific devices make for shopping questions, which are forbidden, but it's ok to ask what the name of such a device is. In this case, "Power Factor Load Bank", "Resistive/Reactive Load Bank" or separate resistive and reactive banks will work for you. They look expensive.

A large inductor has a terrible power factor. A resistor bank in series with an inductor has an arbitrary power factor. You can buy or build an inductor or build your own heater to use as a resistor bank of arbitrary resistance and power. A fan for the heater can either be separately powered or contribute to the reactive load.

If you have an arbitrary power level, you can calculate an inductance/resistance combo with ample current rating, reasonable size and correct power factor, you can parallelize it to achieve the desired power level. An inductor is just loops of wire on a core of desired permeability, and a heater is just wire (in your case with a small temperature coefficient looped back and forth to achieve desired form factor, so both can be made at home, especially if you have a suitable place to go junk diving for transformer cores in particular, either an ample size one to get you started or a number of smaller matching ones like doorbell transformers. Already wound transformers wouldn't be good inductive loads for you as their inductance is so high as to limit no load current and losses to a small % of full load value, although that small load has low power factor (0.2 was the number I found for a transformer with 3A no load but not from a datasheet), you would need huge transformers to get significant currents. Putting the right size series resistive load on the primary and parallel resistive load on the secondary would allow you to either use maximum power at an arbitrary power factor, or produce a lower power at same or lower power factor. The power factor of the load is transferred to the primary, so there's no reason you can't incorporate a transformer, but there's a limit to how much you can load it and have it still contribute to lowering power factor, even with reactive loads, as the loss increases due to reactive currents are true power.

Calculating the resistor sizes for the input and output heater banks could be difficult though if you wanted to build the heaters in one shot without multiple taps. The necessary transformers may still be huge. On the other hand, a single wind your own inductor on a doorbell transformer with appropriate series heater would be much easier to calculate and give you an idea of how large a modular design would have to be for a desired power level. You would want to aim for 10% more inductance than you think you need, and use the largest size of wire (or parallel set of wires) you can wind the appropriate number of turns with to maximise the available power factor ratio of the inductor you create. Again, a big project to calculate, but easier and you can produce experimental estimates at a low cost.

If you prefer, you can make tapped inductors by winding single or inductors with layers stacked before adding a new turn to the core. Use large wire on a rod, toroid, or regular core. Cement the windings in place (I can't remember the product for this but you can look it up) and on one side of the winding, file a flat face of exposed conductor along the path of the tap arm.

Either inductors or heaters, if designed to exceed the target value, can be adjusted down by removing turns or length.

One last thing, be very conscious that your power company may be quite displeased with you connecting large low power factor loads to their grid, and they may reflect that in their billing. For a large building to blow power factor, the cost can be in tens or hundreds of thousands of dollars for a single event.

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  • \$\begingroup\$ Indeed, I put in a purchase request for a power factor load bank or grid simulator, and unfortunately it was too expensive for the company to purchase. Hence how I ended up asking the question on here. \$\endgroup\$ – Alex Apr 12 at 8:59
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I think you need a fix load to calibrate. If capacitive load is suitable, a capacitor and a resistor is the more easy way I could say. But some calculations must be done to select components.

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One of those off the shelf plug-in power-bill savers - they are capacitors, they have a near zero power factor.

a transformer wilth no load (basically an inductor)

A small switched mod power supply with no load or a light load (they have a very high crest factor) it must be small enough to not have power factor correction.

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That's easy. LED Christmas lights.

Those are comprised of series strings of ~30 LED emitters, plus some simple thing like a resistor and a plain diode to stop reverse current. The string is only lit half the time; when the AC power is one polarity. As such, it has a power factor slightly under 0.50. If the string has more than 30-ish lights, they will have another ~30 LED string facing the opposite direction, to even out the power factor, giving PF around 1.0.

Note that if it has an odd number of series strings, its PF will be less than 1.00.

So it's a simple matter of getting *any LED Christmas lights, really, and either disabling or reversing certain series strings to make the power factor as bad as possible (0.50).

For a larger load, easy -- just make one of these things, a junction box on the end of an extension cord, except stick a 20A/240V diode between the hot wire and the receptacle. This will deny half the AC cycle to the loads plugged in. This will make motor loads go insane. However resistor/heating loads such as a toaster or "radiator style" heater will work fine, and give 50% PF.

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