Building an Inductive Load Simulator

You really do not give enough information. Putting an inductor in series with a lamp will result in reduced power to the lamp. In turn that means that the lamp will run at a lower temperature. In turn this means that the lamp resistance will be considerably lower than what it would be a design temperature.

The resistance of the lamp will be a nonlinear function of the inductance in series with the lamp. That makes calculation considerably more complicated than it would be for a resistor that is not temperature sensitive. If you cannot do it, do not expect it to be don for you for free.

Bill

I see it does get quite complex.

The inductance wouldn't have to be in series. I assumed we could get to where we need to be with minimal voltage drop across the inductor by over-sizing the wire used in the air-core inductor and that a paralleled inductor would be much larger.

I need to use a 500 watt Quartz lamp (because of the inrush characteristics) and must end up with a circuit power factor of .75 when connected to a 120V, 60Hz supply line.

I can't think of any other information to provide. What additional information would you like?

Steve

------------------------------- If you have a series inductor such as to get the desired power factor, it will definitely impinge unfavorably on the lamp voltage even if the wire used has negligable resistance. Oversizing the wire won't help with the voltage drop. Assuming the lamp resistance was constant then the voltage across the lamp for a 0.75 pf would be 90V- however as pointed out by Salmon Egg, the lamp resistance would not be constant-so lower lamp power, cooler filament, lower resistance and lower power factor, etc -you could end up with a very dim lamp.

You are concerned with the inrush characteristics but a series inductor will change these as well. A parallel inductor would be a better choice. In either case, an air coil inductance will be quite large.

There is also the question of why you want a pf of 0.75. If you are trying to simulate an inductive load then why use a lamp?

If it is absolutely necessary to have a real power of 500 Watts at a pf of

0.75 then use a resistor which doesn't have the temperature variation of a lamp and consider that you want 90 V across this with a series inductor and a 120V supply. This is 5.6A so the resistor is 16.2 ohms so the inductor must be 14.3 ohms or 0.038H at 60 Hz. A bit large for an air core coil. If you use a lamp rather than a relatively constant resistor- then you have a bit of a messy problem. --

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer Steve

I am conducting an in-house test of a switching device prior to submittal to a testing lab...trying to simulate the labs tests.

The paramaters of the test require the load be characteristic of Tungsten, achieving steady state current within 4.1mS with an inrush of at least 10x steady state current. Resistors could be used, but if they are, we would still need to simulate the inrush by shunting part of the resistance upon switch closure. I would rather just stick with Tungsten lamps. The test parameters require 500W lamps be used.

It seems we need to calculate a parallel inductor. That's what I need help with.

Let me suggest that you go electronic. Use a triac with phase control to limit inrush current. As your lamp heats up and its resistance increases, you use 100% conduction angle. You could also use a relay to short out the triac. That way you get unity power factor except at the start. It is likely to be cheaper because an inductor of the size you need is going to be expensive and much larger.

Bill

The worst case is no inductor at all. It shouldn't be too hard to reliably turn on a 500 watt lamp. Am I missing somthing?

Thanks SE,

You're on the right track. You gave me an idea. I'll report back.

Your stated specs, 500 watts resistive load, 0.75 power factor and

120 volt 60 Hz supply, are enough to caculate a parallel inductor. There are several steps and I'm not sure which one you're missing, so here's the whole process:

1. Calculate reactive load: Draw a right triangle. The horizontal leg is the

500 watt resistive load (real power), the vertical leg is the inductive load (reactive power) to be determined and the hypotenuse is the apparent power. Power factor, 0.75 in this case, equals real power divided by apparent power. Solve for apparent power then reactive power.

1. From reactive power, solve for impedance at 120 volts.
2. From impedance, solve for inductance at 60 Hz. I get a number under 100mH.
3. Work out the inductor construction from the formulas. Verify with a meter and protect with a fast-acting fuse. Any wire large enough for the current will have minimal DC resistance and quickly turn into a demonstration of a short circuit or a fusible link.

Good luck, Mike

------ If you want the load test to be characteristic of tungsten then model the v-i -time characteristic of a tungsten lamp. This will depend heavily on the thermal balance involved and conventional R-L or R-L -C circuit models won't work. Forget the inductance because the inrush characteristic is dependent on the thermal behaviour of the lamp- not upon some unknown inductance You have, initially, a cold resistance which is cold. and a resulting high inrush current (sure some filament inductance is present but this is typically a pretty minor effect. You are trying to model a thermal effect by a model which doesn't apply (and a steady state "power factor" model is even less applicable. I suggest that you look carefully at what is actually occuring and try to model that. Conventional (linear) circuit theory doesn't apply. Ohm's Law is up shit creek without a paddle.

Or...

Consider an inductor in parallel with the lamp. An inductive load of 440 VAR in parallel with 500 watts should give a total apparent power of about

667 VA and a power factor of 0.75 (if I got my sums right). That way the lamp voltage stays pretty much constant (assuming negligible line losses)

For 120V 60Hz, I think that would be about 0.087 Henry

daestrom

That's the way I calculated it too, but I got more than 100 mH. With an inductive load of about 440 VAR, you need an impedance of about 32.7 ohms. At 60 Hz, that's more than 100 mH.

daestrom

-------- I am a bit confused- in an earlier message you got 87mh and now indicate over 100mh. The 87mh (0.087h) figure was correct.

I do agree that the parallel construction is the way to go but I have a problem in that if the objective is to simulate a tungsten lamp-why use an inductor-just go with a tungsten lamp? A linear resistor/inductor combination, series or parallel will not give the desired inrush characteristic or the tungsten lamp steady state behaviour.

Would steve-gatz clarify the specific parameters to be met? Inrush as for a tungsten lamp- OK. 0.75 pf (steady state-meaningless otherwise)-why? Is the lamp 0.75 pf- if so- why? It appears that the intent is to use an inductor (series or parallel-doesn't really matter) and resistor (even switched) to simulate the actual characteristic- it won't.

If you want a simulation using linear circuit elements, then use a parallel capacitive circuit where you can jig resistance in series with the capacitor to get a high inrush current with about the right time constant and jig the total for the right pf except capacitive This too has problems as the duty on the switch is different than it would be for an inductive load.

His primal instinct to actually use a tungsten lamp is right on target. What do they do in the lab tests that he is trying to simulate and is the simulation an actual test or a computer model test?

No, I'm afraid I'm the one that was confused. I read 100 mH, but somehow was thinking 100 uH (microHenries). Guess too much 'holiday cheer'. Sorry 'bout that.

Yes, that is a question. I was focusing on wanting 500 watts with a pf of

0.75 and was just thinking of a way to 'simulate' that load. Although it's hardly a 'simulation' since what I proposed was to connect the exact load the OP spec'd.

daestrom