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4hv.org :: Forums :: Tesla Coils
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biggest rookie question in the league.

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LD LANGER
Mon Jun 06 2011, 09:10PM Print
LD LANGER Registered Member #3824 Joined: Sun Apr 10 2011, 08:29PM
Location: Edmonton, Alberta, Canada
Posts: 54
Hello, so I have been failing to grasp one concept that I assume to be very simple. In a DC charging coil, small components can be used in series to achieve a higher voltage rating. For example, I am using 25 x 1000V diodes for upwards of 25kV reverse voltage max. Here's what I don't get: How is the voltage shared between all of the components? I mean the drop across the diode should be negligible compared to the potential at that point, so how can each diode "share" the voltage drop? The same goes for the charging inductor (8 MOTs in series). Mainly, if for example the one end of the diode or inductor is at 15000V and the other end is at 14950V, then technically aren't all these devices seeing a very high voltage?

Thanks for anyone clearing this up.

Daniel
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Ash Small
Mon Jun 06 2011, 09:53PM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
The simple answer for the diodes is resistance. they all have 'some' resistance, which is approximately equal for all of them if they are 'identical'.

you can test this easily by connecting two identical resistors in series, doesn't matter what the values are, as long as the combined resistance is sufficient to limit the current to a suitable level, then apply a voltage accross them, and measure the voltage accross one, then the other. the voltage will be half the supply voltage. (I suggest using a small dry-cell battery).

The same applies if you connect two dry-cells in series, for example.
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LD LANGER
Mon Jun 06 2011, 10:21PM
LD LANGER Registered Member #3824 Joined: Sun Apr 10 2011, 08:29PM
Location: Edmonton, Alberta, Canada
Posts: 54
Ash Small wrote ...

The simple answer for the diodes is resistance. they all have 'some' resistance, which is approximately equal for all of them if they are 'identical'.

you can test this easily by connecting two identical resistors in series, doesn't matter what the values are, as long as the combined resistance is sufficient to limit the current to a suitable level, then apply a voltage accross them, and measure the voltage accross one, then the other. the voltage will be half the supply voltage. (I suggest using a small dry-cell battery).

The same applies if you connect two dry-cells in series, for example.


Right, although I don't think I'm at a misunderstanding of basic voltage division. However, by this logic, that would require that the supply voltage (potential at my tank capacitor) to be completely dropped across the diodes.

Obviously wrong scenario that is my understanding:

(~) --- -|>- --- -|>- --- -|>- --- -|>- --- -|>- --- -||- --- GND
15kV -- 15kV --- 12kV -- 9kV - - 6kV --- 3kV -- 0kV -- 0kV

Correct scenario which by my logic means the diodes are seeing very high voltage:


(~) --- -|>- --- -|>- --- -|>- --- -|>- --- -|>- --- -||- --- GND
15kV - ~15kV ~15kV ~15kV ~15kV ~14kV - ~14kV - 0kV


?

Daniel
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Nah
Mon Jun 06 2011, 11:53PM
Nah Registered Member #3567 Joined: Mon Jan 03 2011, 10:49PM
Location: USA, 1960s
Posts: 260
It isn't the diodes that are dropping the voltage, it is the load...

The diodes are dropping a constant .7 volts each, however, the load is "using up" the 15k

So, they are not really dropping 15kvolts, but rather letting it pass through, which is my understanding.

As Ash told you, as they all have (in theory) the same charateristics, they "see" the same voltage.

Think of it as it forming one big diode, don't see it as 25 diodes, but rather one big diode.
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Ash Small
Tue Jun 07 2011, 12:18AM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Think of it this way:


(~) --- -|>- --- -|>- --- -|>- --- -|>- --- -|>- --- -||- --- GND
15kV -- 3kV --- 3kV -- 3kV - - 3kV --- 3kV -- 0kV -- 0kV



BTW, I found this:

"Each diode should also be shunted with a 10 megohm resistor to keep the voltage distribution across them uniform"

This just ensures that the voltage accross each diode is the same, in case of differences between them.
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Dr. ISOTOP
Tue Jun 07 2011, 12:36AM
Dr. ISOTOP Registered Member #2919 Joined: Fri Jun 11 2010, 06:30PM
Location: Cambridge, MA
Posts: 652
When the diodes are reverse biased they have to collectively hold off 15KV, since their resistance is infinite.
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LD LANGER
Wed Jun 08 2011, 07:51PM
LD LANGER Registered Member #3824 Joined: Sun Apr 10 2011, 08:29PM
Location: Edmonton, Alberta, Canada
Posts: 54
Nah wrote ...

It isn't the diodes that are dropping the voltage, it is the load...

The diodes are dropping a constant .7 volts each, however, the load is "using up" the 15k

So, they are not really dropping 15kvolts, but rather letting it pass through, which is my understanding.

As Ash told you, as they all have (in theory) the same charateristics, they "see" the same voltage.

Think of it as it forming one big diode, don't see it as 25 diodes, but rather one big diode.

Yes, the diode forward voltage drop is 0.7V, but that isn't what i am misunderstanding. I guess I'm trying to figure out how stacking 25 in series I will get effectively a maximum rating of 25kV reverse voltage before breakdown occurs (each diode rated 1000V reverse voltage). As I see it, the first diode will "see" 15kV, and as it has no drop, its rating is surpassed by 15x.

Thanks
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Forty
Wed Jun 08 2011, 08:16PM
Forty Registered Member #3888 Joined: Sun May 15 2011, 09:50PM
Location: Erie, PA
Posts: 649
i think you're picturing the voltage the wrong way. it's a difference between two points, not so much a flowing thing. think of it this way, you put a weight onto a suspended bed of tissue paper and it rips through. then you put the weight onto 10 pieces of the same tissue paper and it doesn't rip through any of the sheets.
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doctor electrons
Wed Jun 08 2011, 10:13PM
doctor electrons Registered Member #2390 Joined: Sat Sept 26 2009, 02:04PM
Location: Milwaukee Wisconsin
Posts: 381
Yep! Voltage does not flow, current does. You can use the old beach ball analogy that tesla did.
Looking at the diode string as one big one is probably the easiest way to picture it. OR,
Think of the diodes as little check valves. If you had the same set up with plumbing, copper pipe
and check valves, the pressure could vary greatly and still yield the same results. You would have
the "backing" or "backup" from the other valves that were before the last one in the series. Hard
to explain i guess but i hope that helps a little bit wink
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Andyman
Wed Jun 08 2011, 11:44PM
Andyman Registered Member #1083 Joined: Mon Oct 29 2007, 06:16PM
Location: Upland, California
Posts: 256
Not sure if this is right, but lets see if i can give it a shot.
Since the diodes have some reverse standoff voltage, that means there's a resistance R to the reverse current flow. V=I*R but in our case, the standoff voltage V is not really dependent on amount of current, so we could pretend I=1 and say that V is roughly proportional to R.
Since resistors in series add up, the standoff voltage would add up in the same way. V1+V2+V3≈R1+R2+R3

Does that seem right? Or am i just pulling stuff outta thin air?

Or maybe like this: The voltage across the string is the sum of the voltages across each individual circuit element.
V_total = V1 + V2 + V3 + V4....
Now this would only be for the REVERSE voltage that's being blocked. The FORWARD voltage is passed thru like it's a solid wire..
When the current flows one way, the diode is a large resistor (with resistances added up in series), going the opposite way, the diode is a wire which doesn't care about voltage.
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