Measuring secondary voltage (fwd)

To: "'Tesla List'" <tesla-at-pupman-dot-com>
Subject: Measuring secondary voltage (fwd)
From: Tesla List <tesla-at-stic-dot-net>
Date: Thu, 22 Jan 1998 17:27:24 -0600
Approved: tesla-at-stic-dot-net

----------
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)


>From:  John H. Couture [SMTP:couturejh-at-worldnet.att-dot-net]
>Sent:  Wednesday, January 21, 1998 1:16 AM
>To:  Tesla List
>Subject:  Re: Measuring secondary voltage (fwd)
>
>At 01:19 AM 1/20/98 +0000, you wrote:
>>

>>
>><from John Couture's post of Mon, 19 Jan 1998 19:17:09 +0000>
>>
>>>It should be noted that the Tesla coil secondary voltage can only be
>>>approximated because connecting any metering instrumentation to the
>>>secondary circuit would decrease the voltage by an indeterminate amount.
>>
>>  If you really wanted to know the voltage and were willing to spend
>>  some time to find it, how about taking a series of measurements,
>>  using a voltage divider with a different impedance for each
>>  measurement?  By plotting measured voltage against voltage divider
>>  impedance magnitude and extrapolating to infinite impedance, it
>>  seems that a reasonably accurate value for the secondary voltage
>>  can be measured.
>>
>>  Jim Monte
>>
>--------------------------------------------------
>
>  Jim -
>
>  Wouldn't "extrapolating to infinite impedance" give you infinite voltage?
>
>  However, I believe your idea could be used by extrapolating to a very high
>impedance where the reduction in voltage could be negligible.
>
>  Maybe Robert Stephens would want to try this. It certainly could be used
>for small coils.
>
>  John Couture


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