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Quad 63 (and later) Delay Line Inductors
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Old 7th June 2021, 01:37 PM   #391
jan.didden is offline jan.didden  Europe
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But you more than 3mA to drive the panels, and a resistor would mess up the freq response with the capacitive load part.

Jan
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Old 8th June 2021, 09:56 AM   #392
Hans Polak is offline Hans Polak  Netherlands
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3mA at 2500V means 7.5Watt.
Probably enough to be lethal, especially with Janís superconductive shoulders.

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Old 8th June 2021, 10:07 AM   #393
Lampie519 is offline Lampie519  Netherlands
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Old 8th June 2021, 02:33 PM   #394
mattstat is online now mattstat  United States
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Quote:
Originally Posted by Hans Polak View Post
3mA at 2500V means 7.5Watt.
Probably enough to be lethal
Watts aren't the typical unit of concern when it comes to electrocution. Current is where the action is. Impedance of the body and voltage, of course, influence the amount of current flowing through you, but a focus on simple watts isn't one that's going to provide the best prediction of lethality.

This link gives a basic rundown of the typical hazard ranges
Electric Shock Hazards

Last edited by mattstat; 8th June 2021 at 02:40 PM.
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Old 8th June 2021, 03:09 PM   #395
esl 63 is offline esl 63  Sweden
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If you mean 2500V on ONE plate the power output is 4 times that = 30 watt. So 3mA in one plate is in the ballpark. It is also considered "safe".

Quote:
Originally Posted by Hans Polak View Post
3mA at 2500V means 7.5Watt.
Probably enough to be lethal, especially with Janís superconductive shoulders.

Hans
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Old 8th June 2021, 04:14 PM   #396
jan.didden is offline jan.didden  Europe
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Two times. Double the voltage at the same current is 2 x power. But it is separate anyway, you either touch one panel or the other, both at the same time is physically very hard, so its 2500V @ 3mA in your suggestion.

But anyway, as I explained above, 3mA limit is not feasible.

Jan

Last edited by jan.didden; 8th June 2021 at 04:19 PM.
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Old 8th June 2021, 04:16 PM   #397
esl 63 is offline esl 63  Sweden
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P=U^2/R
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Old 8th June 2021, 04:28 PM   #398
MarcelvdG is offline MarcelvdG  Netherlands
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2500 V peak per stator is 5 kV peak differential

The impedance plots earlier in this thread show magnitudes of the order of 400 kohm

5 kV/400 kohm = 12.5 mA, so if there were a current limiter in the amplifier, it would have to be set to at least 12.5 mA

If you want to use a series resistor as current limiter, you would presumably want its impedance to be much smaller than that of the loudspeaker, so most of the 2.5 kV peak ends up in the loudspeaker and not across the current limiting resistors.

Accepting 10 % voltage loss, you can allow resistors of 22 kohm per stator, resulting in a perfectly lethal Norton current of just over 110 mA.

Where on Earth does that 3 mA come from?
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Old 8th June 2021, 04:29 PM   #399
jan.didden is offline jan.didden  Europe
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P=U^2/R is not applicable here; I is constant, V varies.
P = I x V; 2 x V same I is 2 x power. Do we have to spend time on that here??

Jan

Last edited by jan.didden; 8th June 2021 at 04:33 PM.
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Old 8th June 2021, 04:29 PM   #400
jan.didden is offline jan.didden  Europe
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Quote:
Originally Posted by MarcelvdG View Post
Where on Earth does that 3 mA come from?
Just a random number I guess.
My design is for 12mA peak current. Should be just sufficient.

Jan

Last edited by jan.didden; 8th June 2021 at 04:52 PM.
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