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@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@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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