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Measuring secondary voltage (fwd)
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From: John H. Couture [SMTP:couturejh-at-worldnet.att-dot-net]
Sent: Saturday, January 24, 1998 2:14 AM
To: Tesla List
Subject: Re: Measuring secondary voltage (fwd)
Jim -
You will have to use more resistance for your dividers. A few years ago I
tested some voltage dividers to determine the secondary voltage of a small
TC (450 watts). I found that when the divider caused about 10 ua of load the
voltage drop was around 10 KV. This was with a 1000 meg divider.
Vs = R x I = 10^9 x 10^-5 = 10 KV
It is obvious that voltage dividers for Tesla coils are a big problem. As
Malcolm Watts said you need large coils. But even with large coils if the
loading is 50 ua a divider of 1000 megs gives
Vs = R x I = 10^9 x 5 x 10^-5 = 50 KV
Greater loading only makes things worst. The Tesla coil puts out a high
sec Inst. voltage but not much RMS current.
However, I did not pursue this very far. I will leave it up to other
coilers to do more research to find the secondary voltage.
Maybe developing a curve and extrapolating will help.
John Couture
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At 11:27 PM 1/22/98 +0000, you wrote:
---------------------------------- Big snip
>----------
>From: Jim Monte [SMTP:JDM95003-at-UCONNVM.UCONN.EDU]
>Sent: Thursday, January 22, 1998 1:11 PM
>To: tesla-at-pupman-dot-com
>Subject: Re: Measuring secondary voltage (fwd)
>
>Hmmm, it looks like I didn't explain what I meant well enough ...
>Here's a a more concrete example that should clarify things, I hope.
>
>Suppose you have the following set of resistive voltage dividers
>divider 1: R1= 1, R2= 999
>divider 2: R1= 10, R2= 9990
>divider 3: R1= 100, R2= 99900
>divider 4: R1= 1000, R2= 999000
>divider 5: R1=10000, R2= 9990000, with the unit of resistance being
>whatever is appropriate for the problem.
>
>The voltage across R1 will be 1/(1+999) = 1/1000 of the total voltage
>across the divider. By placing each divider across some unknown
>and measuring the voltage across R1, we would get 5 voltage measurements.
>Unless the unknown is an ideal voltage source, the measurements would
>differ depending on how much the voltage divider loaded down the
>unknown.
>
>Next plot measured voltage vs total voltage divider resistance. For the
>wide range of resistances above, a semi-log plot would probably be best.
>Then, based on the shape of the plot, estimate what the voltage would
>be as the voltage divider's effects became insignificant.
>
>measured voltage
> A
> | _____estimated unloaded
> | X X X X X voltage
> | X X
> | X X
> | X
> | X
> | X
> |
> | X
> |
> |X
> +----------------------------------------------------------->
> total resistance
>
>
>This approach should be scalable to any size coil. I don't know how
>many points would be required to get accurate results, and as I said
>originally, this will not be quick, but I think it would work.
>
>Jim Monte
>
>
>