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Re: Re:High speed Tesla spark photographs - polarity indicator
Original poster: Vardan <vardan01@xxxxxxxxxxxxxxxxxxxxxxx>
Hi Peter,
I made a different version.
http://drsstc.com/~streakcam/instruments/Indicator-00.gif
http://drsstc.com/~streakcam/instruments/Indicator-01.JPG
http://drsstc.com/~streakcam/instruments/Indicator-02.JPG
http://drsstc.com/~streakcam/instruments/Indicator-03.JPG
Mine works off secondary base current. Sort of 90 degrees out of
phase, but easy if you know what's going on ;-)) It seems to give
very sharp points of light to the streak cam. Maybe good or bad
depending on how one looks at it. Positive is closest to the spark
per your standard ;-)
Mine looks like this to cam V-2.0.
http://drsstc.com/~streakcam/instruments/StreakCamera2-03.jpg
One will notice the "1/4 wave" leader there ;-)))
StreakCam V-2.0 is "here":
http://drsstc.com/~streakcam/instruments/StreakCamera2-01.JPG
http://drsstc.com/~streakcam/instruments/StreakCamera2-02.JPG
The mirror was spinning at 7500 RPM in those photos but "cheap cam"
made it look "still"... I have been pushing higher RPMs now to
separate the lead up leaders... If anything gives way, I get to go
buy a "new" camera ;-))
I think the polarity indicator needs to be moved to the top terminal
point. I think we both can do that...
Cheers,
Terry
At 07:04 PM 9/21/2006, you wrote:
The circuit diagram mystery of the polarity detectng LED's is very
simple. Just two 10 ohm 1 watt resistors and the two red LED's.
One of the resistors is connected to the ground electrode on one end
and the spark hits the other end of the resistor. So if 2000 A
passes then a 10 ohm non inductive resistor should develop 20,000 V
across it, which is enough to spark right across it on the
outside. I have seen this happen with higher value resistors.
The other 10 ohm resistor is in series with the two LED's which are
in parallel but of opposite polarity. This then goes across the
first 10 ohm resistor. So why should this work when theoretically
the LED's should be receiving half of the 2000A. ie 1000A in each
LED in each cycle. We know that we can push current limits in an
IGBT to perhaps 100 times the continuous current rating, but this is
50,000 times overload (1000A vs 20mA).
So what is happenning? I suspect that the actual current is a lot
lower. Added to that is the higher inductance of the loop with the
LED and secondary 10 ohm resistor compared with the main 10 ohm on alone.
I knew I wouldn't be able to work out real performance with out some
trial and error so that's how I came up with this. I did try the
resistors separately to see if they would flash over due to
inductance issues but they were OK. The resistor values are bizarre
though. For example you cant run the LED's off DC without putting
about 4A through the 10 ohm resistor (=40W) and burning it up the
resistor up instantly.
It really is bright considering the fact that it is only pulsing in
several microsecond pulses for brief periods. I think something is
different about the red LED's and not just the lower forward voltage
compared to the blue/green LED's.
What is also interesting is to use spark gap arrestors which also
show the relly fast stuff without burning out. Interestingly,
focally they are brighter than the arc itself and a bit more of a
point source. In the end I used the LED's but certainly worth
thnking of the arrestors for > 1MHz fast stuff.
Peter
Original poster: Vardan <vardan01@xxxxxxxxxxxxxxxxxxxxxxx>
Hi Peter,
Cool idea!!
I will "guess" two LEDs a resistor and a MOV. The strike current
might be 2000A for a few nS which will kill most TVSs and such. It
really helps to keep the device voltage low so the instant power is
lower too. A capacitor might also work if one could guess the right value.
Cheers,
Terry
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