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@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
>
>
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