William Beaty wrote:
On Sat, 20 Oct 2007, Stephen A. Lawrence wrote:
William Beaty wrote:
I totally missed any announcement that self-acting or "closed-loop"
operation was achieved.
WHOA slow down, that's not what was said.
That's exactly what was said. Or at least strongly implied... and then
if we made the wrong conclusions, he didn't correct us.
I've lost my sound again on this system
Then you can't hear what was said!
Yes, you're right; I apologize. I should have said "That's not what was
_shown_".
Never occurred to me it was a "ground" wire being connected.
None the less a self-powered circuit which isn't drawing _anything_ from
the environment shouldn't need to be grounded either. The behavior of
this circuit makes it obvious that it is interacting strongly with the
environment through that wire.
It's also sitting in a metal pie pan, which may be significant; no
mention is made of what, if anything, the pie pan is tied to, but
certainly its presence could have a large impact on the capacitive
coupling between the circuit and the rest of the universe.
******************************
ISSUE for Bill:
Tesla-coil type circuits tend to have high impedance, "looking back"
into the circuit, or so I have been led to believe. In contrast, LEDs
have (nearly) ZERO impedance when forward biased (they have a fixed
A Tesla coil with floating leads has a fairly huge current in the center
of the coil. It's an LC resonator, after all, so the AC pulses of many
kilovolts are associated with high current pulses inside the floating
resonator. If you insert an LED in series with the center of the coil, it
might burn out! If you insert it closer to one end of the coil or the
other, it will see proportionally lower current. But if you hook an LED
directly across the coil terminals, it will ruin the resonance effect
(lower the Q) and make the output drop to a very low value. Tesla coils
require very high impedance loads in parallel with the output... but they
need low impedance loads if the load is inserted in series in the center
of the (floating) coil.
I can do a quick test.
I have a tiny fluorescent inverter; a high-freq supply of 5,000V at 50KHz,
with 5VDC supply required. It's basically the same as a CW Tesla coil.
I solder a white LED to the HV terminal, put a 1N4148 diode across it
backwards so it can still light up even with a series capacitor under AC
drive. I apply 5VDC power supply. And the LED lights brightly! Yay.
But I recognize this as a familiar "antenna effect" where the LED's leads
function as an antenna with a current in the floating conductor. The LED
is in series with this tiny antenna and it 'intercepts' this current. So
if I snip the LED's floating 1" lead off, will it go dark? I do it.
Nope, but now the LED glow is barely visible now, whereas before it was
intense. Now if I should bring my fingertip near the very short floating
LED terminal, I'll restore some "antenna" effect via capacitive coupling.
Will the LED get brighter? Yep, several times brighter. If I grab the
entire assembly and touch the floating LED terminal to the 1" piece of
wire sitting on the insulating table top, will the LED get bright again?
Yep, it does. It's amazing how short an "antenna" is needed to get the
LED to glow brightly. I would have predicted that a several-inch wire
would be needed, but only about 1/2" is needed.
There's one obvious way to make this snipped-leads LED very much brighter,
even with the leads trimmed back to about 0.2 inches: increase the
operating frequency. That will make all the coupling capacitances appear
larger. Suppose I could increase the frequency by 100 times? Run it at
5MHz as Ron did, rather than 50KHz? In theor the current in the floating
LED should increase by a factor of a hundred, and so should the LED
brightness. The present dim glow should get intense. Hooking several
LED/diode pairs in series should add far more "antenna," and make all the
series LEDs glow very brightly, with the ones nearer the floating tip
being dimmer.
Also... this tiny 'inverter' puts out 5KV, while Ron's ferrite-rod
transformer might be far less (or it might not.)
At lower voltages the single-wire LED-driving effects are much smaller.
But at higher frequency the effects are much larger. If I could increase
my frequency by 100x from 50K to 5M, but lower the volts by 10x from 5K to
500, wouldn't the LED still glow 10x brighter than I'm seeing? Yet it's
already very bright with only a 1" floating lead to act as a capacitive
"antenna." With the lead clipped off, and with 5MHz at 500V, I'd expect
it to glow very brightly.
I hope Ron figures out how to measure the output voltage across his two
floating coil terminals. Ground one, then approach the other with a scope
probe. Then crudely calibrate this setup by using a known voltage and
freq on a similar floating wire, then approaching the scope probe by the
same distance. It might allow measurments within 5% of the genuine
value.
voltage drop, but no impedance on top of that). This is why they must
normally be driven by a current source, not a voltage source.
To answer a question put elsewhere, by someone (Jones, ?maybe?): No,
you can't light an LED with "voltage only" (high voltage and extremely
low current) -- you need an mA or two to light it, at somewhere between
1 and 3 volts depending on the particular LED. 3,000 volts from a high
impedance coil which can't source more than, say, 0.05 mA won't do the
job; the impedance mismatch is too large.
That 0.05mA is for a parallel connection, which throws 3KV across the LED.
By instead placing a load in series with the center of the floating coil,
the load intercepts the very large resonance current, and only grabs a
tiny percentage of the full output voltage. Now that you mention it...
when placed in series, the LED should see a constant-current supply (so
lots and lots of LEDs could be placed in series with the floating coil
before they'd start to pull down the output.)
Thanks much for this. The amount I had not understood about Tesla coils
was far larger than I realized.
