Re: Capacitor bank pretty well complete
- From: "Tim Williams" <tmoranwms@xxxxxxxxxxx>
- Date: Thu, 15 Dec 2005 14:42:15 -0600
"Ignoramus8558" <ignoramus8558@xxxxxxxxxxxxxxxxxxx> wrote in message
news:MEiof.43628$rg3.35215@xxxxxxxxxxxxxxxxxxxxxxxx
> It looks very nice, but was it really only a few hours?
Yeah, probably five, maybe more, but the kind of hours where you spend one
or two each day and work goes real slowly, you just take your time.
> So, forgive me for an ignorant question, but what are you planning to
> do with your caps? Charge them with line voltage and discharge into
> your heating coil? Or what? I do not have a mental picture of what you
> will do, and am very ignorant about electronics.
Nah, nothing so crude. ;) Although...
The capacitors are rated for 400V/us peak, so I can safely, repeatedly pull
8kA peak out of the bank.
There's probably 0.1uH or more there in the wiring, which constitutes a
resonant frequency of 110kHz. For a lossless connection and 630VDC initial,
that would constitute 445Vrms across the inductance for a current of 6300A,
within ratings.
Or better yet, short the terminals through a coil. 5kA through a few turns
will make some generous magnetic field, maybe I can shoot an aluminum plate
into the air a few feet (not much since it's only four joules stored).
But anyway, enough of the fantasy, I'll tell you how it works.
Know anything at all about radio? Tuned circuits and whatnot?
(You should anyway- when you're switching 200A on and off, a lot of pieces
of wire suddenly become inductors. Bypass capacitors can ring with these
bits and potentially make things worse!)
The basic idea is, to heat metal with an AC magnetic field, you need a
pretty strong field. A good way to generate this is an iron-cored
transformer, but that needs big scales to work on, so ain't gonna cut it
here. So we use a coreless inductor.
When you run amps through a coil, it makes a magnetic field, which induces
voltage in the work, which has resistance, P = V^2/R and you get heating.
In the coil, it takes a lot of amps to build that field, though. If you
figure 1" insulation around a 4" dia slug of iron, that's a 6" coil. The
cross-sectional area is 16 to 36, or 44% of the primary is coupled to the
secondary (coupling factor 0.44, roughly speaking). It'll be a bit lower in
practice due to stray field and whatnot. So if we run 100kVA (such as, say,
316V * 316A) through the primary, 44kVA is coupled to the work, which
reflects back some of this as back EMF (silver, copper, gold and aluminum
being the worst load metals for this reason), and dissipates perhaps 20kW of
that. Overall, you have a 5:1 ratio between volt-amps in the primary and
actual power consumed.
If you run transistors (or SCRs, or tubes) into that primary alone, you have
to handle all 100kVA as if it were real power. If your inverter itself is
90% efficient, you'll be burning (100k / 0.9) - 100k = 11kW in the process!
That's 66% overall (31kW input, 20kW output, 11kW dissipated) for what
should be a 90% efficient system. There's a better way.
If you cancel the inductive component of the primary, you are left with real
current that does real work. A capacitor is the opposite of an inductor.
You can connect them in parallel or series to accomplish this. In series,
currents must be equal; in parallel, voltages must be equal. In either
case, the reactive components cancel at resonance (you may remember an
inductor's reactance rises as frequency goes up, while a capacitor's falls;
when equal, they resonate).
That leaves the problem of driving it. With 20uF anchoring the tank
circuit, you aren't going to make the voltage change much faster than the
sinewave on it. It's perfectly okay to drive with a sine wave, but it needs
either a much faster class D amplifier, or an inefficient linear amplifier.
Class D amps just aren't cheap or easy to make in the 200+kHz, 10kW range,
and aren't very efficient either. (Class A through C means linear
amplification with a varying amount of conduction time; class D is with the
transistors always either ON (low voltage, high current) or OFF (low
current, high voltage). Since voltage or current is low at any given time,
power loss is low, except during switching.)
Inductors are class D's friend, so what I did is connect an inductor
("Lmatch") in series with the tank. The inductor provides "springiness" so
I can swing 300V in a microsecond without shattering my transistors,
allowing me to get from one side to the other with a minimum of power loss.
The only problem is it causes a series-resonant peak, where current can rise
to a hundred amperes or more (and it's worse when the coil is lightly
loaded: Q multiplication effect).
So in summary to your question, it cancels the reactance of the work coil,
allowing me to drive it with 10-20kVA instead of a full 100-200kVA.
Tim
--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
.
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