# Re: transformers

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From: 	Peter Electric[SMTP:elekessy-at-macquarie.matra-dot-com.au]
Sent: 	Friday, August 29, 1997 3:03 AM
To: 	Tesla List
Subject: 	Re: transformers

Tesla List wrote:

> From:   Mad Coiler[SMTP:tesla_coiler-at-hotmail-dot-com]
> Sent:   Thursday, August 28, 1997 11:57 AM
> To:     tesla-at-pupman-dot-com
> Subject:        Re: transformers
>
> >From:  Edward V. Phillips[SMTP:ed-at-alumni.caltech.edu]
> >Sent:  Wednesday, August 27, 1997 4:16 PM
> >To:    tesla-at-pupman-dot-com
> >Subject:       Re: transformers
> >
> >"Don't hook together transformers that have different output
> voltages!
> >
> >"
> >       Why not?  Just look at the equivalent circuit of the
> >neon transformers connected in parallel.  Current will flow into
> >or out of the lower voltage transformer depending on the load
> >impedance.  For example, if the load is such that the voltage
> >equals that of the lower voltage unit, all of the current will
> >flow from the higher voltage one.  If the loaded voltage is
> >lower, both transformers will contribute.  Haven't thought through
> >the case of resonant capacitor charging, but bet there are no
> >real problems there which differ from paralleling transformers
> >of the same voltage.
> >Ed
> >
> >
> >
>
>   Well, I have never tried hooking different voltages in parallel
> becuase I have gone by the following:
>   If you hook 2 xformers together directly in parallel that have
> differing voltages then you have just made a complete circuit. It is a
>
> complete circuit because you have a complete, closed loop path between
>
> the two (or more) transformers, and you have an applied voltage wich
> is
> the difference between the two - I don't know what that was but I will
>
> use a 2000V difference for example (say a 7kV and 9kV neon). A 2000V
> difference will cause a current to flow (in the direction of the
> higher
> V xformer). Then the overall effect is the higher V xformer is
> applying
> 2000V across the other transformer. This would be OK maybee if you had
> a
> high resistance between them to drop the extra 2000V, but connecting
> them with wire you will have I=V/R were there is a (2000V difference /
> 2
> ohms???) were 2 ohms is arbitrarily the R of the wire. This will try
> to
> draw 1000 Amperes from the xformer. You must also always obey
> Kirchoff's
> Voltage Law in that the sum of the Vdrops and Vrises around a closed
> loop circuit must equal zero. Therefor following the path of current;
> +
> 9000V up through that neon, - 7000V down the other neon and -???V
> across
> the wire to get back to 0V. This again works out to be 2000V across
> your
> wires. Obviously something must give, and I would supect that would be
>
> something critical inside the transformer.
>
> Perhaps ASCII could help
>
>             ~1 ohm
>        +    1000V    -
>        ---~-~-~-~-~---
>        |             |
>        |             |
>     +  0             0  +
>        0             0
>  9000V 0             0 7000V
>        0             0
>     -  0             0  -
>        |             |
>      A |             |
>        ---~-~-~-~-~---
>        -    1000v    +
>            ~1 ohm
>
> I hope you can make sense of this and see that if you start at point A
>
> and go CCW to the right, add up all the volatges across the wires and
> other neon that it equals the 9000V neon.
>
> These laws are what they teach us in college and they say for example
> that KVL MUST ALWAYS be valid, it's impossible to defy KVL. If anyone
> has solid proof of being able to hook things like this up then I would
>
> be very interested in hearing them. The professors here have a
> tendancy
> of ignoring the exceptions in the rules because there are so many
> *below
> average* students that may get confused. I always enjoy using these
> exceptions and aditional stuff just to confuse the other students!
>
>
>

I think you forgot about the 30Ma current limiting! All that would
happen is the higher voltage transformer would current limit until the
voltage equaled the lower one. An inductive charging scenario would be
harder to calculate though.

Cheers,

Peter E.

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