Design Article

IMG1

Power Trip: Dealing with kVA issues, power factor, and smaller boost vs. buck regulators

Dave Ritter and Tamara Schmitz, Intersil Corp.

2/15/2010 6:00 AM EST

(Editor's note: this is part of an on-going series of "dialogues" between the authors; there are links to the previous installments at the end, immediately above the "About the Authors" section. Also below are also two "Desert Island Design" articles, by one of the authors.)

[Clouds are parting after recent afternoon thunderstorms. Dr. Tamara Schmitz is out walking her dog, passing a small substation at the end of the block, where Dave Ritter seems to be busy at something near a company service truck]

Dr. T (Tamara Schmitz): Hey Dave, I seem to bump into you everywhere!

Dave (Dave Ritter): Oh hi, Dr. T–out for a walk after the storm?

Dr. T: The lights flickered a few times during the storm, and then I saw the trucks down the street, so I thought I'd nose around to see what was happening. What are you doing here?

Dave: I worked for these guys (gestures to the truck and the guys operating some equipment out of the back of it) as an intern a long time ago, and they still give me a call if they're in the neighborhood. It's fun, that power troubleshooting. That guy over there is Ralph. He's running a TDR, trying to find a short in a buried high-voltage line.

Dr. T: A Time Domain Reflectometer, pretty sophisticated for a power problem. It provides a profile of the impedance of a transmission line versus length. How does that help?

Dave: In this case, most of the cable looks like a giant coax cable, except when a line breaks. The fault causes a short and the impedance drops at that point. If Ralph does his calculations correctly, they'll dig a hole and find the bad cable. If not, they just end up with a hole.

Dr. T: Very interesting. What's that gizmo you're running?

Dave: This is an infrared scanner. It's a TV camera that sees heat. I scan the equipment looking for trouble spots. [Dave aims camera at a triplet of overhead lines.] See those wires? They're very-high-voltage lines feeding that transformer over there.

Dr. T: They almost appear to glow in the infrared. That must be due to the heat from resistive loss in the cable. They all seem to be about the same, so I guess the load is evenly distributed.

Dave: Right. It's important to notice if one goes dark, it indicates something called 'single phasing' which is a big problem in power systems. But I digress, look at this shot.

Dr. T: Wow, that looks like it's almost on fire, almost like we have three huge knife switches right over our head. In your IR camera, the middle one glows like a light bulb, but–by the eye–it looks just fine.





Figure 1: High-voltage substation switch–
(top image) View in Normal Light;
(bottom image) View in Infrared (note top-of-middle contact on the left).

(Click on image to enlarge)

Dave: Here, try looking through the binoculars.

Dr. T: They still look about the same wait a minute the center one, right around the contacts, looks discolored. It's a bit darker than the other two. I'll bet it's oxidized and that increases the resistance, heating it up.

Dave: Very good, professor! I didn't know anybody paid attention to power in the academic world anymore. It's actually a run-away effect: the heating causes oxidation, which increases resistance, which in turn causes more heating and more oxidation and more heating. It's best to catch it early before it fails completely.

Dr. T: How much power are we talking about here?

Dave: I think those switches are rated at a few hundred amps. The lines are something like 30 kV (Editor's note: the power industry uses both "k" and "K" here; we'll use the "k".)

Dr. T: So we've got mega watts streaming right over our heads.

Dave: Yes, we sure do. But the switches only see the current, since they are in series with the load.

Dr. T: Makes sense. But what about that monstrous transformer over there: the one that the switches feed?


Figure 2: 5000 kVA substation transformer
(Click on image to enlarge)

Dave: To be honest, I'm not sure, but the smaller versions are rated in kVA (Kilovolt Amperes), which is something like power. I'll bet that big box is rated at several thousand kVA.

Dr. T: KVA implies volts times amps, which is power. Why not just say watts?

Dave: In a word or two, Power Factor.

Dr. T: That means phase matters also, right?

Dave: Right. A lot of the load ends up being inductive: other transformers, lighting ballasts, electronic power supplies, and motors to name a few. Of course, part of the load is always resistive. That's the part that does the work. But there is also a large part of the current that just swishes back and forth between source and load at 60 Hz. That's the inductive part, and it requires more capability to handle it. So equipment must be sized for the total load, not just the power. The industry has settled on the simple kVA rating as a reasonable way to accomplish that.

Dr. T: But if we just put the right capacitor at the right spot we could cancel the inductive part, couldn't we?

Dave: Right, and it's done all the time. But capacitors are expensive, so we try to trade-off correcting it and just handling it.

Dr. T: All of this power talk reminds me of a question from a customer. He uses our switching regulators to power his FPGAs. The FPGAs are rated in power, but our regulators are rated in current. Kind of like the difference between the transformer and switch ratings in this station. Why can't we rate them both the same way?

Dave: Good question. In normal operation, the overhead switch only sees the current flowing through it, so it is rated in amps. A buck regulator is similar. It has switches that are always in series with the load. To prevent overheating during overloads, the switches are current-limited, making the whole regulator limited in output current.

Dr. T: Just to give an example, we have one buck regulator, the ISL8011, that can be adjusted to output anything from 0.8 V to the input voltage level–anywhere from 2.7 V to 5.5 V. Does the maximum output current remains the same?

Dave: Yes. If a buck regulator is rated at 1 A, it is guaranteed to supply that current regardless of output voltage. At 0.8 V it is only supplying 800 mW, but at 3.3 V it is supplying 3.3 W.

Dr. T: Makes sense. You then design your system for maximum power conditions.

Dave: And optimize as best you can for any other typical operating conditions.

Dr. T: That makes power calculations easy for each supply. The FPGA may have many power supplies serving the inner core logic, the I/O bank, the auxiliary logic and more. I guess I should remember superposition and add them up.

Dave: Now you are getting it. . . .

Dr. T: Does that apply to all of our switching regulators?

Dave: Unfortunately, no. A buck regulator has switches in series with the load current. A boost regulator has switches in series with the input current. So the current limit on a boost regulator is at the input side, and limits the total power available at a given input voltage. The net effect is a limit on the output power. So boost regulators are output-power limited, while buck regulators are output-current limited.

Dr. T: What about buck-boost? Does it just switch between output-power limited and output-current limited, or are their other constraints?

Dave: Ah, the famous buck-boost. I worked on one a few years ago. We needed to supply a portable circuit with a steady 3.0 VDC from a battery that varied from 3.2 to 2.5 V. The interesting thing is that we are actually using the internal switches to channel around the inductor current. So in all cases we are limiting the inductor current, which is the same as the switch current. How that appears to the user depends on the operating mode. So, the answer is, yes, the buck-boost switches between the two, but it does it naturally by limiting inductor current. When boosting, the inductor current is the same as the input current. When bucking, it is the same as the output current.

Dr. T: And since the output voltage is constant, we'd have to keep track of the input voltage to see what power and current can be supplied to the load. It's a little more complicated, but I think I can handle it.

Dave: I am sure you could. . . .Watch out for that puddle! [Loud splash as Dr. T's foot disappears into a recent, rain-filled hole.] Gee, the IR camera says your foot is really cold now.

Dr. T: You needed an IR camera to tell you that?!

Previous "dialogues" in this series:

Other related articles by Dave Ritter:

About the authors
Dave Ritter grew up outside of Philadelphia in a house that was constantly being embellished with various antennas and random wiring. By the age of 12, his parents refused to enter the basement anymore, for fear of lethal electric shock. He attended Drexel University back when programming required intimate knowledge of keypunch machines. His checkered career wandered through NASA where he developed video-effects machines and real-time disk drives. Finally seeing the light, he entered the semiconductor industry in the early 90's. Dave has about 20 patents, some of which are actually useful. He has found a home at Intersil Corporation as a principal applications engineer. Eternally youthful and bright of spirit, Dave feels privileged to commit his ideas to paper for the entertainment and education of his soon to be massive readership.

Tamara Schmitz grew up in the Midwest, finding her way west with an acceptance letter to Stanford University. After collecting three EE degrees (BS, MS, and PhD), she taught analog circuits and test-development engineering as an assistant professor at San Jose State University. With 8 years of part-time experience in applications engineering, she joined industry full-time at Intersil Corporation as a principal applications engineer. In twenty years, she hopes to be as eternally youthful as Dave. .


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