Raindrops creating electricity?

"Commander Kinsey" snipped-for-privacy@nospam.com wrote in news:op.17ehayehmvhs6z@ryzen:

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low and brief enough not to melt it?

violently explode when it lost a cell and became a 10V battery, sucking a huge amount of current from the rest. I wasn't at home at the time, but came back to a very strong smell in the driveway and thought there was a dead animal somewhere. I later found a battery missing from the shelf in the garage, and found pieces scattered everywhere.

As my kids would say, "LOL".

If you are standing there when the lightning strikes, you will be instantly dead, so the amount of power involved is infinitesimal.

Trust John Larkin to miss the point

Surely it only gets used when you need need hundred of kilovolts?

An electric-wind driven windmill could convert some of the power to mechanical movement and rotate a magnet which could produce harvestable power, but the efficiency would be dire.

Two series resistors in the right ratio ? :)

Or four hundred.

It makes sense for an amp, but not for raw power. For example, there's 240V at that socket there, with virtually no resistance. If I connect a resistive heater to it to it to draw power, I get all the power as usable heat pretty much, even though the resistance of the heating element is vastly more than the supply. What if I connected a f****ng big element to it, such that it's resistance was equal to the supply line? Half the power would be dissipated in the supply line. I'd drop from 99.9% to 50% efficiency.

Just take a normal one but moreso. Do you really think it's much harder to convert 50V to 5V than 10V to 5V?

Look at going the other way. A photomultiplier power supply uses a inverter to step up 5V to 1000V.

You need a specially wound transformer with a high turns ratio, and the stray capacitance across the secondary makes it hard to run it at a high enough frequency to let you avoid saturating the core. Boost that by another factor of 400 and you are in an entirely different ball game.

Insulating the secondary winding well enough so that it won't break down at 400kV would be an interesting problem on it own.

If you wanted to make the point that you are a pig-ignorant wanker you've succeeded in spades.

Yes, that's one implication of the theorem. You could really conjugate match your AC line for a cycle or two and get a lot of power before the breaker trips.

People don't impedance match audio amps either. "Damping factor" is one of those quaint terms that audiodudes use to express the mismatch. A good power amp has milliohms of output impedance.

A "600 ohm" audio output is usually much less.

RF types use 'S11' and 'S22' to express matching. A horrible mismatch sounds better expressed in dB.

My NMR gradient amps were very good current sources, which is why they worked better than hacked audio amps.

Yes, fewer switcher chips are available at 50 volts. I'm doing a bunch of designs now with +48 in and chips are relatively rare and sometimes weird.

But I specified 400 KV. Give it a shot.

What is a breaker? Is that a modern version of a fuse? :-)

I'm sure I was told you do to get the most power out. I guess that's only true for rubbish amps with a high resistance, where connected low ohm speakers to it gets less power out by dropping the voltage. IF you impedance matched a milliohm amp, you'd get more current than it was capable of producing and break it.

A little coil?

That's too much like real sound. What am I hearing in that clip?

Just string a modest long-wire antenna and run it through a hi-z input audio amp and listen to the world.

One round-the-world sound is

BEEEOOOooooo that lasts several seconds. And lots of chirps and pops.

Yes, that's right.

If you impedance match any amp the efficiency will be below 50%.

Give it a shot.

Don't. The one thing we can be confident of is that a little coil won't stand off 400kV.

A 15mm gap in 5 atmospheres of SF6 will only stand off about 120kV. For 400kV you'd need 50 mm. That's not a small coil.

Transformer oil does better - you only need 10mm to just stand off 400kV.

Of course you'd need a lot of inductance to keep the current from saturating the core and you can't use a multilayer winding to get lots of turns.

So you need a big core and have to bank wind it, as well as sinking it in transformer oil.

The windings will have a lot of capacitance so you wont be able to switch them fast , which means that you will need even more inductance to stop the core saturatiing

Mindless experimentation isn't going to get you anywhere.

It the impedances are the same, half the power is in the load, this is the best case scenario, and why it says on the back of the amp "use 8 ohm speakers". If the speakers are a lower impedance, most of the power is heat in the amp. If the speakers are higher impedance, you're dropping the current.

Where exactly in the circuit do you perceive the problem?

Your confidence is misplaced.

Most audio amplifier are designed not to break. Typically they are designed to source enough current to drive an 8 ohm speaker. If you use one to drive a 4 ohm speaker, you will likely get only half the voltage swing. The crudest protection would blow a fuse if you try to get more voltage swing because that would make it try to deliver more current than the output devices were capable of supplying. More subtle protection tends to sound horrible when the amplifier goes into current limit.

Wrong.

It isn't.

<snip>

Sketch up a 400 KV power supply, post it, and we can discuss.

The Scottish wanker is evading the problem.

And so is John Larkin.

The most practical solution I could come up it to suspend magnet on a needle bearing in a sealed (and largely evacuated) chamber and use the 400kV source to create an electric wind to rotate the magnet. Put figure of eight coil outside the chamber under the magnet, and rectify the AC current induced in the coil.

It would be a pathetic power source, but it is practicable.

So you don't know then, I thought it was an empty point.

Yes, we know. Thank you for repeating his message, and telling him that his post has arrived correctly.

Please just ignore these in future. Instead of following up TWICE...

Andy

The traditional method aggregates the raindrops in a dam and drives a turbine. Commonly called hydroelectric generation.