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Re: transformers
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!
Mad Coiler
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