Measuring secondary voltage (fwd)

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

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
>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