HI BOB, I am trying to build a test circuit that will produce a pulse current from a capacitor. My target is around 200 A at 100 ms. Is this possible? We have an instrument called a PVI that does the same only at lower current and at shorter duration, but I don’t know how it is being controlled. I hope you can give me advice or a basic control circuit that I can start working on.
–ROMMEL C. VILLON
HELLO, MR. VILLON, You have not told me what the load voltage or resistance is, but I made a first guess of a couple volts. I started to try a capacitor of 20,000 µF at 20 V. I almost computed that the cap voltage would drop 1 V. Wrong. Per I = C dv/dt, it would drop 1000 V. If 200,000 µF, it would drop 100 V.
So, you might need 2,000,000 µF. (Farad supercapacitors would not put out that current.) You then would need 100 20,000-µF caps in parallel, charged up near 20 V. This is not quite the same as a car battery, but it deserves the same kind of respect. Don’t drop any wrenches near the high-current terminals.
You’d need more than a dozen transistors with emitter ballast to get the currents to share. Put them in a Trarlington. Get a fast op amp to turn them on crisply—and off. I think you ought to start at 20 A and scale up by making the transistors 10 or 20 wide. This is a lot of charge! Be careful. Things may blow up. Get your head and body 20 feet away behind a tempered glass wall when you trigger it.
Maybe 12-V batteries will work best, with two or three in parallel. I mean, every time you start your car, you have transients like that, but into a 9-V load (the starter). Make sure you have fuses that are appropriate, because if the transistors fail, you will have a vicious current flow through the melted transistors. And, use a remote-control disconnect switch. Get the idea? Have fun! (You might set up a big relay to automatically disconnect it all after 0.15 s just in case there is a failure.)
You did see my column (June 12, p. 60) on current sources, right? But the art of extending a 2-A current to 20 and then 200 A is not simple. It is not just electronics. It is the art of management of high currents.
–RAP
DEAR RAP, Remember that current-monitor circuit you drew up in your error-budget article (June 8, 2006, p. 18)? First, if I invert the pnp, it still seems to work. Can you explain? Second, suppose I would like a negative voltage current monitor. How would you change the circuit for that?
–PETER BERG
HI, PETER: First, some pnps have some reverse alpha, but not as good as 99%. Maybe 40% or 70%. So, the gain might not be right. The gain might be lousy. Also, the VBE breakdown is certain to be much smaller, so it would not work at 10 or 20 or 30 V. It might seem to work at 4 or 5 V. Think about it.
Second, if you wanted to bring up a signal from nearly –15 V, representing an I × R drop near –14.95 V, that is easy. You would need an npn transistor, and you would need an op amp whose common-mode (CM) range goes near the negative rail. This would bring the signal to a volt or so below the ground bus.
What if you wanted to bring this signal near a volt above the ground rail? If you want that, let me know. What do you have for a positive rail? What do you want to do with the signal? Feed it to an analog-to-digital converter? This would become an analog problem, and you should start writing down what you want. Needing and wishing are two different animals. Defining what you need is a good idea.
–RAP
BOB, I saw your note about the LM4562 (May 22, p. 80), and I read the datasheet. I’m curious about the application circuit from page 25 of the datasheet for a sinewave oscillator. Is that an incandescent lamp used as a circuit element? (Exactly. /rap) Can you tell me why this would be used?
–LLOYD SLONIM
DEAR LLOYD, A lamp’s filament has a big positive temperature coefficient. This oscillator will start oscillating and the amplitude will grow—until the lamp gets warm and then stabilizes at a certain amplitude. Hewlett & Packard started out their whole business with their HP200 oscillator, which used this principle, designed in 1939. Ancient history. Not very modern.
–RAP
HI BOB, Solving an electronic circuit design problem while keeping the component count as small as possible is similar to solving a puzzle (May 8, p. 64). I like it! Good work! –”HAMURO”
DEAR HAMURO, Thanks!
–RAP
Comments invited! rap@galaxy.nsc.com —or: Mail Stop D2597A, National Semiconductor P.O. Box 58090, Santa Clara, CA 95052-8090
The high current pulse circuit requirements sounds a lot like a discharge spot welder. Some Google searching will probably lead to how they do it.
Also depending on application think about using a big hunk of iron and copper (transformer) to allow reduced current (at higher voltage) to be controlled. This could reduce the capacitance bank size since internal resistance would not be as significant.
Anonymous -July 29, 2008
A good (i.e., easy) way to get a high power pulse is to use a sheet of lead insulated with a thin layer of teflon. Use a brass or copper rod and pound on the end with a hammer to punch through the teflon. OK, it's not high PRF and the jitter is terrible, but you can dump a lot of energy (and voltage) through a low impedance with this. Just insulate it well enough so it doesn't dump through you.
Anonymous -July 29, 2008
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