[Amps] HV transformer and Variac

[Larry Benko]

Gerald,

I just saw your post and Ron's comments are in line with what I believe 
happens.  We know for a fact that a turn or 2 are connected together by 
the wiper.  If that was not the case there would be dead spots as you 
rotated the variac voltage control.  If we assume the variac has 120VAC 
across about 200 turns there are 0.6V per turn and if you short out 2 
turns that is only 1.2V.  If the carbon brushes represent .3 ohm the 
"shorted turn(s)" current is 4A or a power of about 4.8W.  This current 
is localized and the heat will spread out due to the copper wire and the 
coil.  I just took a 10A open frame variac and powered it up, set the 
tap to about 50%, and put NO load on the output.  In about 10 minutes 
the variac was powered off and the tap point area was slightly warm 
while far away from the tap point the turns were cooler.  It appears 
reasonable that the tap is a "partially shorted turn(s)" with a tradeoff 
between variac output impedance and temperature rise.

Larry, W0QE

On 7/24/2015 10:33 AM, TexasRF at aol.com wrote:
> Hi Larry, thanks for adding that analysis.
> Your approach clears up something that has bothered me about Variacs 
> for a long time: When you examine the tap/brush part of the device it 
> is obvious the more then one turn of the autotransformer is being 
> connected to the sliding tap at the same time. That means that one or 
> more turns are shorted.
> A shorted turn in a conventional transformer will result in a large 
> current flow, copious amounts of smoke and likely some sparks as well.
> Using the inductive analysis, a shorted turn simply reduces the total 
> inductance and I assume only a small amount of IR loss in the shorted 
> turn(s). At r.f. a shorted turn coupled to an inductor acts to reduce 
> the total inductance. There have even been some published designs 
> using a tuning device made of a rotated shorted turn to resonate the 
> plate circuit in a 6 meter amplifier.
> Perhaps your Spice program can analyze the loss caused by a shorted 
> turn in an autotransformer?
> 73,
> Gerald K5GW
> In a message dated 7/23/2015 4:43:18 P.M. Central Daylight Time, 
> xxw0qe at comcast.net writes:
>
>     Gerald and others,
>
>     To add to the auto-transformer (variac) discussion I did a quick
>     simulation in LTSpice to make sure my memory was correct before
>     making a
>     fool out of myself. :-)
>
>     There are 2 currents in the variac: the magnetizing current and the
>     current due to the load.
>
>     1.) The magnetizing current is generally set to be a couple of
>     percent
>     of the maximum load current.  If we view the variac as 2 coupled
>     inductors in series the magnetizing current is in phase in the 2
>     inductors and the junction of the 2 inductors is the voltage out
>     point.
>
>     2.) The current due to the load  does not add in one of the
>     windings as
>     people often believe.  The current due to the load is out of phase in
>     the the 2 series coupled inductors.
>
>     To make this clearer here are a couple of examples assuming an AC
>     input
>     voltage of 120VACrms and a 10 ohm load.  The magnetizing current is
>     150mArms (~2H total variac inductance) and the windings have .001
>     ohm of
>     resistance.
>
>     1.) Assume a tap point of 90%, the output is 120*.9 = 108Vrms and the
>     load current is 10.8Arms.  The inductor that connects to the AC input
>     has a current of 9.73Arms and the current in the inductor that
>     connects
>     to AC common is 1.13Arms.  The inductor currents are OUT OF PHASE
>     due to
>     the transformer action and they sum to 10.8A.
>
>     2.) Assume a tap point of 50%, the output is 120*.5 = 60Vrms and the
>     load current is 6.0Arms.  The inductor that connects to the AC
>     input has
>     a current of 3.02Arms and the current in the inductor that
>     connects to
>     AC common is 3.01Arms.  The inductor currents are OUT OF PHASE due to
>     the transformer action.
>
>     Notice how the currents in the winding of the transformer are less
>     than
>     the load current which might seem impossible.  If anyone wants the
>     simple LTSpice circuit email me.
>
>     Larry
>
>
>     On 7/23/2015 8:14 AM, Gerald Williamson via Amps wrote:
>     > Hi Ros, to split hairs here, one would say the autotransformer
>     would loose
>     > very little efficiency. As you know, when there is copper wire
>     and there is
>     >   current, there will be a loss of current squared times resistance.
>     >
>     > The majority of the current would be in the turns of the
>     autotransformer
>     > between the input and output terminals so the loss would be
>     related to the
>     > input  to output voltage ratio.
>     >
>     > That is my personal understanding and subject to correction.
>     >
>     > 73,
>     > Gerald K5GW
>     >
>     >
>
>     _______________________________________________
>     Amps mailing list
>     Amps at contesting.com
>     http://lists.contesting.com/mailman/listinfo/amps
>