Reactive Loads

or "How my 480V network scared my wallet"


Our family's water well -- which was never very good, and has been downright pitiful the last few years -- died, and I had to drill a new one. (My 3rd, actually, but that's another story.) The new site was at the other end of the property, nearly 1,000 ft away, and wire (esp. the size needed to prevent an insane voltage drop) is expensive. I could've ordered separate PUD service for it, but that's also expensive, and I wanted the well tied into my existing power backup system (eventually to be full-time solar-powered with my other outbuildings). I wanted flexibility.

The answer: transformers!

Remember those Ohm's Law & power charts from the Technician exam? By doubling the voltage, you get double the power with the same current (and thus avoid exceeding the safe current capacity of a wire). Better yet, doubling the voltage also cuts the voltage drop % in half, and that's the main concern over a long run. (12AWG copper's good for 20A, right? Only if your gear doesn't mind a 30% voltage drop at full load, and you want to use 7A to gently warm the ground!)

I found a great deal on 2AWG aluminum wire (more work than copper to do right, but when you can save $1k , it's worth it, right?) and started looking at transformers, capacities & efficiencies from idle to full load, and doing some back-of-the-envelope cost comparisons. I determined that my proposed system was roughly similar in cost (10 years to catch up) to either using bigger wires or getting secondary service from PUD. The flexibility is outstanding, however; I can size my transformers appropriately for my loads, but if I want to increase available power, I only need to upgrade the transformers; with 600V (the max the wire is rated for) 75kVA transformers, I could run my entire 300A PUD service from one end of the property to the other with 5% drop -- no need to dig up trenches to upgrade wire.

(I also ran 8 pairs of CAT6 and 3 pairs of 50nm fiber. Everything's in conduit, in spite of the fact that it more than doubled the price. I want this to survive, and I don't want to have to trench again -- the past 3 months have been brutal!)

Besides, running your own 480V distribution network is geeky cool for a rural ham homeowner. You can't deny it, and I'm not ashamed to admit that both the challenge and the nerdiness factored just a little into the direction I took this project....

So, I did the deed...

Power on

When it was finally time to turn the network on, I flipped the breaker, the step-up transformer hummed to life, I tested everything with and without loads on it, was gratified to see that turning on a power tool at the bottom only dropped the outlet voltage from 125.5 to 125.1, and was pleased with myself -- until I clamped an ammeter on the 240V feeder. Then I was scared.

I was shocked to find that my little starter network of 3 transformers (15kVA + 2x 5kVA) was pulling over 8A without any load (except a couple of nightlights) on it. Doing the math on that 2kW load (My transformer didn't specify core losses or idle current draw, but this was much more than the typical industry numbers I had found in my research!), I saw that I could be looking at $100 a month without even running anything!

Except that it wasn't really 2kW; it was 2kVA...  What's that, you say? Isn't VA (Volt-Ampere) just a fancy name for a Watt? Look at the chart above-- Volts * Amps = Watts!
That is true -- for DC, or purely resistive AC loads.  Ohm's law applies, and the current is always directly proportional to the voltage, i.e. if you graphed voltage & current together on an a dual-channel oscilloscope, one waveform would line up perfectly with the other on the X (time) axis, and by adjusting the Y scale, you could get them to line up perfectly; multiply the RMS values (RMS? Hmmm...) and you get power. A reactive (inductive or capacitive) load, however, messes with it, and the current ends up leading (if capacitive) or lagging behind (inductive) the voltage.

The reason is somewhat intuitive. Picture a capacitor placed across a battery or other DC power source. When power is first applied, the capacitor is empty, and it takes all the current it can get, presenting an effective short (and thus keeping the voltage near zero). As it charges, its resistance increases, and as the current flow slows the voltage can finally climb. Once charged, no current flows, and full open-circuit voltage is present.  An inductor, on the other hand, resists changes in current flow, so when you first connect it, the sudden rush of current through the relatively short length of wire quickly hits a brick wall (presenting an effectively open circuit) and it takes a while before it slowly (slowly being a fraction of a second) starts letting more and more current flow. Eventually, full DC current flows, overloading the supply and dropping the voltage to near zero.

With alternating current, before the inductor's magnetic field has a chance to settle down, the voltage has changed again, so it never gets a chance to get to that point of being a short circuit. The higher the frequency or bigger the inductor, the more this is so. Good thing, too, an ideal transformer primary would draw no power if nothing was attached to the secondary side, but the primary winding of my step-up transformer has a DC resistance of less than 0.1 Ohms -- a major short!

Side trip: In-rush current

There's still a major inrush of current for several 60 Hz cycles  when power is first applied to the transformer, until the magnetic field stabilizes. This is the worst if power is applied right at the zero crossing; if the switch is turned on right at a peak, the "shock" of the instantaneous voltage rise will in effect trigger the inductance and stop the inrush quicker.  This inrush is why building codes allow circuit breakers feeding transformers to be oversized (esp. if there's appropriate overload protection on the secondary side or in the transformer itself), and why sometimes you need a special inrush-limiting circuit (for smaller loads a thermistor may be sufficient, perhaps with a timed relay to bypass it after 2 seconds) to protect everything. If you "right-size" a breaker, you'll be playing dice, with the breaker tripping as soon as you turn it on most of the time, and it taking several attempts to get it to "stick".  (I confess to personal practical experience.) This is not good for any of the equipment -- the breaker, the transformer, the loads -- so should be avoided.

Power Factor

A perfectly inductive load (a theoretical construct, but an unloaded transformer is hopefully pretty close!) will cause the current to lag by a full 90 degrees, so that when voltage is max, current is zero, and visa-versa. Anything times zero is zero, right? Starting to get the idea? In either of those conditions, there's virtually no power. Thus, a highly reactive circuit draws very little power (and also does very little work, including generating very little heat -- this is why using capacitors / inductors instead of resistors to limit A/C power to something can be a great idea in some cases.)

Thus, the real power (expressed in Watts) drawn by any reactive load is always less than its apparent power (expressed in VA).  The ratio of the two (real / apparent) is called the power factor, and is often written as a percentage.  The question I faced was: What was my network's power factor? It was obviously > 0 because (a.) this is real life and (b.) the transformers are all warm. How to answer this question?

Practical method #1: Watch your power bill.  Two weeks after turning on the transformer network (and a few days after activating the new well) I used the PUD's online tool to generate a graph of my daily usage for the last 2 months. I marked the two days that should see a power increase, and carefully examined the graph without finding any visible sign (except my 2 marks) that anything increased.

Practical method #2: Put a real power meter on it.  (Think Kill-a-Watt for 240V.) I don't have one, and don't want to buy one at the moment; I have to pay the well guy.

Nerdy method: Use a dual-trace oscilloscope to display EMF & current simultaneously! It may not provide an easy number, but measuring the phase shift should allow you to calculate the power factor -- it's just the cosine of the angular difference!

What the 'scope shows

First off, I apologize for the poor pictures. No screenshots from a fancy digital storage oscilloscope (DSO) -- which might also do the integration of the waveforms through software and calculate the real power for you -- these are photos of a classic analog scope. I tossed a camera to KK7FNI and was busy teaching the kids how to use a scope, and didn't even think to address picture quality -- or even to turn up the intensity of the CRT to make it easier to see.

#1: A 120 VAC RMS 60 Hz reference voltage (each major vertical line is 100V, so yes, about the expected 170V peak)

For convenience, I measured at a 120V outlet, instead of across the same 240V circuit I measured current on.

#2: This is what the current waveform looks like with the well pumping water. In spite of the flat tops & spikes, the trapezoidal waveform isn't too far from a sine. Each major vertical line is 5A.

#3: No-load current. Very distorted! Note that much of the time (the shallow cups) is spent right around the X axis, indicating near-zero current flow.

I say "no load", but the secondary has the two step-down transformers -- also inductive -- drawing power. It would be interesting to see the waveform with nothing attached to the secondary side of the transformer being measured.

#4: Ah, now we have them together. One cycle (plus a little extra) is shown. You can see that the current spike occurs 90 degrees after the voltage peak, right as the voltage is crossing the zero line. Most of the time, the current is is under 2A.

Unfortunately, our new well quickly emptied its underground reserve and is now delivering about 1/4 gpm. As a result, the pump only runs for about 3 minutes, then the pump races as it sucks air and has less load, then the protection device shuts it down for 30min and it tries again.

The next 10 pictures, taken from 30 FPS video, show what it looked like after I shut the breaker off for a while, then turned it back on. On power-up, you have the step-up transformer, the two step-down transformers, and misc. loads including the control electronics (and very quickly, the pump) drawing power, so again, the reactance is complex.

#5a: Frame 1 - the first current spike seen

#5b: Frame 3 - current spike is falling

#5c: Frame 5 - Interesting that, based on the trace still visible from the last scan, it looks like the positive side got the big spike, and the negative side got a more constant current (which looks like it'll be reduced this frame compared to the last cycle). It may well be an indicator of which half of the phase it was in when the switch connected.

#5d: Frame 7 - continuing to settle

#5e: Frame 10 - This was an interesting development (with the in-phase current rising above the trailing spike), esp. in light of the next picture.

#5f: Frame 13 - returning back to "normal". Still a lot of in-phase current.

#5g: Frame 15 - motor starting up. (The motor's start capacitor may explain much of what we're seeing in the frames after #7.)

#6: The pump is running under load. Still inductive, but much less as actual work is being done.

#7: Water ran out; pump is running dry for a second. See the real power (current under the voltage hump) starting to drop. 

#8: Pump is off again, w/ current low when voltage is high.


Given the complexity of the waveform (I was naively expecting a sinusoid), I didn't make ant attempt to measure real power. Clearly, though, it's more reasonable -- both idling and under load -- than what a glance at my digital ammeter first suggested.

Probe bias

Something I didn't consider until after writing the article: The current probe itself is an inductive clamp pickup w/ a circuit that converts amperage to voltage (1A --> 1 mV). From China, it's designed to be plugged into a digital VOM and doesn't advertise a use case involving a scope. The inductive pickup itself undoubtedly introduces some distortion (duh, inductive?) though my uninformed opinion is that it's likely much less than the much larger equipment being measured in this case. I would do well to test it sometime on a mostly-resistive load.