reduce 125VAC to 120VAC, small form factor, and clean 60Hz s

[P E Schoen]

"John Fields" wrote in message
news:thdtp8dvog23s26tiej8aav5tc4kt9kbsu@4ax.com...

On Mon, 13 May 2013 16:28:09 -0400, "P E Schoen" <paul@peschoen.com
wrote:

I don't think a variac is like two magnetically coupled inductors -
except, maybe, for the case where the winding is extended in order to
get 140V out 120V in - otherwise it's more like an inductive voltage
divider.

It's an inductive voltage divider in both cases, which are identical except
that the tap is soldered while the slider is connected by means of pressure
through a carbon brush.

So I did a transient analysis with the input at 240 VAC (which required
346.6V rather than 339 as expected for sqrt(2) ).

I don't understand...

If you did a transient analysis with a stiff 240VRMS source, why would
you lower its peak to 346.6V?

It must have been some anomaly with LTSpice. I ran a simulation with just an
AC source of 339.4V peak and the RMS value was 234.75. Perhaps I chose an
interval which was not an integer multiple of half the period.

That's the nice thing about simulation; you get to change what's
keeping your predictions from being accurate, regardless of whether
those changes are real in the real world.

I adjusted them because I did not know their real world values, but I know
something about the characteristics of variacs from personal experience.

So, what is the inductance of a variac coil?

Mine, a Thordarson Meissner VAR104, exhibits a measured 311mH with the
output switch in the 120V position and 227mH with the output switch in
the 140V position

I changed the load resistance to 7 meg to find out the magnetizing
current,
and it is about 100 mA at 2 kHz. This corresponds to 3.5 amps at 60 Hz,
which is pretty obviously too high. I would expect more like 500 mA. So I
change the inductances to 500 mH, which gives 480 mA at 60 Hz.

Again, change the model to suit your expectations, regardless of the
constraints dictated by the real world.

But I changed the model to coincide with what I know about variacs, and my
WAG of 500 mH is reasonably close to your measured 311 mH.

OK, so I'll try K=0.999 for the coupling factor. I get 119V at 60 Hz,
118.6V
at 400 Hz, 108.6V at 2 kHz. This "seems" about right.

Well, perhaps because that's what you want to see, but what is it in
the real world?

I relied on the manufacturer's specification to come up with a reasonable
value.

For my purposes, however, I plan to use a fixed V/Hz to get higher power
from the coil. It seems to reason that if I apply ten times the voltage
at
ten times the frequency, I should get ten times the power into ten times
the
resistance. Sure enough, with a 70 ohm load at 600 Hz and an input of
2400
V, I get 1198V output and 20.5 kW.

Obviously it will now require some actual measurements on a toroid variac
core to determine the magnetizing current and leakage inductance, and
also
the core losses as a function of frequency.

So do the work and post what you find in the real world, instead of
just posting conjecture.

I have done something quite similar, as posted previously, but I would like
to repeat those tests more rigorously with a larger core to see if they also
apply.

------------------------------------------------------------------
Using this setup: Vin

/
.. +--[AMP]--+ +-----+--------+ Vout
.. | | | | | /
.. | S||P | [VARIAC]<--+------+
.. [GEN] E||R [DVM1] | | |
.. | C||I | | [LOAD] [DVM2]
.. | | | | | | |
.. +---------+ +-----+--------+-------+------+

Where the transformer primary is rated at 120VAC 60Hz and the
secondary is rated at 12V 35A, is used backwards, and:

GEN Goldstar FG8002
AMP Denon POA800
DVM1 Fluke 8060A
VARIAC Thordarson Meissner VAR 104
LOAD 100 ohms 300 watts
DVM2 Wavetek 27XT

The Variac's spec's:

120V 60Hz in, 0 to 120V or 0 to 140V out at 12 amperes.

Empirical data taken with the output voltage switch in the position
shown and the variac's output unloaded:

OUT Rs L I(mag)
VOLTS ohms H A

120 1.59 0.311 0.0325

140 1.48 0.227 0.0561

With the variac's output switch in the 120V position and the output
cranked to 50% we have:

freq Vin Vout Vout/Vin
Hz volts volts
-----+--------+-------+----------
60 120 60 0.5
100 121 60.3 0.498
1000 120.15 58.8 0.489
2000 119.37 57.7 0.483
4000 120.11 58.4 0.486
6000 121.49 60.5 0.498
10k 119.65 65.5 0.547

Vin changed as the frequency was varied, so it was leveled by
changing the output voltage of the generator to get Vin close to 120V.

So, the output voltage variation with a 0.6A (36W) load stayed within 2%
from 60 Hz to 6000 Hz, and only showed significant variation at 10 kHz. I am
surprised that the ratio increased rather than decreased. Is your 100 ohm
load non-inductive? I would expect a reduction of voltage at higher
frequencies, especially at higher load currents, due to the core losses and
series inductance of the upper half of the variac winding.

But I think you have proven that a variac will operate quite acceptably far
above line frequency and well into the audio range, which IIRC you disputed,
and resulted in this "challenge". Thanks for doing the real world research.
I think many of us may learn from your results.

Paul

 

[John Larkin]

On Sun, 26 May 2013 17:06:04 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Mon, 13 May 2013 16:28:09 -0400, "P E Schoen" <paul@peschoen.com
wrote:

"John Fields" wrote in message
news:aj41p8dd4b7b0snn97d5shofintb055jvc@4ax.com...

On Sun, 12 May 2013 21:16:19 -0400, "P E Schoen" <paul@peschoen.com
wrote:

What do you expect to find with your setup? 190 watts is barely enough
for
even the smallest variacs which are like 0-140V at 2A.

Whether, with a fixed input level and a fixed load, the output level
will change as frequency is varied.

I ran a simulation. The results depend on the inductance of the transformer
and the coupling factor K. I used 500 mH for the two coupled inductors,

---
I don't think a variac is like two magnetically coupled inductors -
except, maybe, for the case where the winding is extended in order to
get 140V out 120V in - otherwise it's more like an inductive voltage
divider.
---

and
K of 0.995, which I think may be reasonable for toroids. The result is a
flat Vout/Vin up to 2.5 kHz at which point the output drops by 0.5 dB. This
is small signal analysis, however:

http://enginuitysystems.com/pix/Variac.png

So I did a transient analysis with the input at 240 VAC (which required
346.6V rather than 339 as expected for sqrt(2) ).

---
I don't understand...

If you did a transient analysis with a stiff 240VRMS source, why would
you lower its peak to 346.6V?
---

With a total of 0.2 ohms
ESR at 60 Hz I got 120 VRMS into a 70 ohm load (200W), and 118.9V into 7
ohms (2.02 kW).

At 2 kHz I got 48.6 VRMS. That is surprisingly low. Setting the coupling
factor K=1 gives 119V. Hard to say what it really is. K=0.998 gives 87.2V.
If I change the inductance of the two halves to 50 mH, I get 118.6V.

---
That's the nice thing about simulation; you get to change what's
keeping your predictions from being accurate, regardless of whether
those changes are real in the real world.
---

So, what is the inductance of a variac coil?

---
Mine, a Thordarson Meissner VAR104, exhibits a measured 311mH with the
output switch in the 120V position and 227mH with the output switch in
the 140V position
---

I changed the load resistance to 7 meg to find out the magnetizing current,
and it is about 100 mA at 2 kHz. This corresponds to 3.5 amps at 60 Hz,
which is pretty obviously too high. I would expect more like 500 mA. So I
change the inductances to 500 mH, which gives 480 mA at 60 Hz.

---
Again, change the model to suit your expectations, regardless of the
constraints dictated by the real world.
---

OK, so I'll try K=0.999 for the coupling factor. I get 119V at 60 Hz, 118.6V
at 400 Hz, 108.6V at 2 kHz. This "seems" about right.

---
Well, perhaps because that's what you want to see, but what is it in
the real world?
---

For my purposes, however, I plan to use a fixed V/Hz to get higher power
from the coil. It seems to reason that if I apply ten times the voltage at
ten times the frequency, I should get ten times the power into ten times the
resistance. Sure enough, with a 70 ohm load at 600 Hz and an input of 2400
V, I get 1198V output and 20.5 kW.

Obviously it will now require some actual measurements on a toroid variac
core to determine the magnetizing current and leakage inductance, and also
the core losses as a function of frequency.

---
So do the work and post what you find in the real world, instead of
just posting conjecture.
---

Let's see what you got!

Paul

---
Using this setup: Vin
/
. +--[AMP]--+ +-----+--------+ Vout
. | | | | | /
. | S||P | [VARIAC]<--+------+
. [GEN] E||R [DVM1] | | |
. | C||I | | [LOAD] [DVM2]
. | | | | | | |
. +---------+ +-----+--------+-------+------+

Where the transformer primary is rated at 120VAC 60Hz and the
secondary is rated at 12V 35A, is used backwards, and:

GEN Goldstar FG8002
AMP Denon POA800
DVM1 Fluke 8060A
VARIAC Thordarson Meissner VAR 104
LOAD 100 ohms 300 watts
DVM2 Wavetek 27XT

The Variac's spec's:

120V 60Hz in, 0 to 120V or 0 to 140V out at 12 amperes.

Empirical data taken with the output voltage switch in the position
shown and the variac's output unloaded:

OUT Rs L I(mag)
VOLTS ohms H A

120 1.59 0.311 0.0325

140 1.48 0.227 0.0561

How did you measure L? Inductance meters are notoriously
unpredictable.

With 120 volts AC at 60 Hz applied, and a magnetizing current of 0.032
amps, L figures out to be around 10 henries. Some of Imag is actually
brush loss, from the partly shorted turns, so L may actually be even
higher. You could lift the brush and separate that out.

With the variac's output switch in the 120V position and the output
cranked to 50% we have:

freq Vin Vout Vout/Vin
Hz volts volts
-----+--------+-------+----------
60 120 60 0.5
100 121 60.3 0.498
1000 120.15 58.8 0.489
2000 119.37 57.7 0.483
4000 120.11 58.4 0.486
6000 121.49 60.5 0.498
10k 119.65 65.5 0.547

Vin changed as the frequency was varied, so it was leveled by changing
the output voltage of the generator to get Vin close to 120V.

Cool; a Variac that works to 10KHz.

Did Imag change much with frequency?

 

[P E Schoen]

"John Fields" wrote in message
news:r325q85gmd66dtcucr7qco0m0ocsonlh82@4ax.com...

Oops...

Why would you _raise_ its peak to 346.6V?

I found that, for a sine wave AC source with even 100 uOhms of internal
resistance, into a 100 ohm resistive load and 10 nH of inductance, the
measured voltage is 234.95 VRMS with 339.4 volts peak.

With 346.6V, I get 239.9 VRMS. I don't know why. Even 10 pH with 1 uOhm ESR
gives the same result. Try it. ASC file follows.

Paul

---------------------------------------------------------------

Version 4
SHEET 1 880 680
WIRE 368 128 96 128
WIRE 464 128 448 128
WIRE 464 208 96 208
WIRE 96 240 96 208
FLAG 96 240 0
SYMBOL voltage 96 112 R0
WINDOW 123 0 0 Left 2
WINDOW 39 24 124 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 339.4 60 0 0 0 100)
SYMATTR SpiceLine Rser=100u
SYMBOL res 448 112 R0
SYMATTR InstName R1
SYMATTR Value 100
SYMBOL ind 352 144 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 5 56 VBottom 2
SYMATTR InstName L1
SYMATTR Value 10n
TEXT 62 232 Left 2 !.tran 1

 

[John Fields]

On Sun, 26 May 2013 22:18:28 -0400, "P E Schoen" <paul@peschoen.com>
wrote:

"John Fields" wrote in message
news:r325q85gmd66dtcucr7qco0m0ocsonlh82@4ax.com...

Oops...

Why would you _raise_ its peak to 346.6V?

I found that, for a sine wave AC source with even 100 uOhms of internal
resistance, into a 100 ohm resistive load and 10 nH of inductance, the
measured voltage is 234.95 VRMS with 339.4 volts peak.

With 346.6V, I get 239.9 VRMS. I don't know why. Even 10 pH with 1 uOhm ESR
gives the same result. Try it. ASC file follows.

Paul

---------------------------------------------------------------

Version 4
SHEET 1 880 680
WIRE 368 128 96 128
WIRE 464 128 448 128
WIRE 464 208 96 208
WIRE 96 240 96 208
FLAG 96 240 0
SYMBOL voltage 96 112 R0
WINDOW 123 0 0 Left 2
WINDOW 39 24 124 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 339.4 60 0 0 0 100)
SYMATTR SpiceLine Rser=100u
SYMBOL res 448 112 R0
SYMATTR InstName R1
SYMATTR Value 100
SYMBOL ind 352 144 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 5 56 VBottom 2
SYMATTR InstName L1
SYMATTR Value 10n
TEXT 62 232 Left 2 !.tran 1

---
That's cheating!

If you have a stiff 240V source, then when you look at the voltage
across the load it'll be whatever it is when the losses between the
source and the load are taken into consideration.

That is to say, in the real world you can't just arbitrarily change
the mains voltage in order to get 240V across your toaster.

--
JF

 

[Fred Abse]

On Sun, 26 May 2013 19:04:22 -0400, P E Schoen wrote:

er.

It's an inductive voltage divider in both cases, which are identical except
that the tap is soldered while the slider is connected by means of pressure
through a carbon brush.

It isn't an inductive divider. It's a *transformer*. The voltage ratio is
determined by the turns ratio. The inductance is determined by the
*square* of the number of turns.

Modeling an autotransformer isn't as simple as two coupled inductors in
the ratio; voltage^2.


Modeling a 99H inductor, unity coupled to a 1H inductor won't give you
10:1 voltage stepdown at the tap. Try it. It'll give you 1/(sqrt(99)+1) =
0.0913, or 10.949:1.


Assuming unity coupling, reasonable with a toroid, the "primary"
inductance will be L1+L2+2*sqrt(L1*L2) - Mutual inductance has to be
considered.

Your example circuit isn't a 50:50 divider. With your choice of K=0.995,
it's 0.577:1.


--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[Fred Abse]

On Sun, 26 May 2013 22:18:28 -0400, P E Schoen wrote:

I found that, for a sine wave AC source with even 100 uOhms of internal
resistance, into a 100 ohm resistive load and 10 nH of inductance, the
measured voltage is 234.95 VRMS with 339.4 volts peak.

With 346.6V, I get 239.9 VRMS. I don't know why. Even 10 pH with 1 uOhm ESR
gives the same result. Try it. ASC file follows.

Reducing min timestep to 1us gets it pretty close to 240. 100ns even
better, but slow.

Numerical integration is only as good as the time between samples.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[Fred Abse]

On Sun, 26 May 2013 16:30:52 -0700, John Larkin wrote:

How did you measure L? Inductance meters are notoriously
unpredictable.

With 120 volts AC at 60 Hz applied, and a magnetizing current of 0.032
amps, L figures out to be around 10 henries. Some of Imag is actually
brush loss, from the partly shorted turns, so L may actually be even
higher. You could lift the brush and separate that out.

I did something similar with a Staco small variac (actually a Tek 576
curve tracer spare part).

With the wiper as near central as possible, the end-to end inductance
should be four times the end-to-wiper inductance. Nowhere near, apparently
indicating lots of leakage inductance, which is plain silly.

Adjusting the exciting current (HP4274A LCR meter, at 100Hz, which is as
low as it goes) for the same amp turns in both cases, I get the expected
4:1 result. The core is highly nonlinear.

If/when I get time, I must take some B-H curves.

Lifting the wiper made negligible difference to either the inductance, or
tan delta, at least not with the quick and dirty connection setup I was
using.

It may be significant that most Staco variacs are characterized up to
2kHz. Except these very small ones.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[Fred Abse]

On Mon, 27 May 2013 09:32:29 -0700, Fred Abse wrote:

I did something similar with a Staco small variac (actually a Tek 576
curve tracer spare part).

Forgot to mention; inductance of whole winding 10.3 henries, measured at
100Hz.


--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[P E Schoen]

"Fred Abse" wrote in message
newsan.2013.05.27.16.09.49.949860@invalid.invalid...

On Sun, 26 May 2013 19:04:22 -0400, P E Schoen wrote:

It's an inductive voltage divider in both cases, which are identical
except
that the tap is soldered while the slider is connected by means of
pressure
through a carbon brush.

It isn't an inductive divider. It's a *transformer*. The voltage ratio is
determined by the turns ratio. The inductance is determined by the
*square* of the number of turns.

Modeling an autotransformer isn't as simple as two coupled inductors in
the ratio; voltage^2.

An autotransformer is a special case, but the center tap is still equivalent
to two inductors of equal value connected in series. For two 1 H inductors
in series, K=0.995, and a 100 ohm load, the voltage ratio is 2:1. The
current in the top inductor is 642 mARMS, and the current in the bottom
inductor L2 is 640 mARMS but opposite phase. The current in the 100 ohm
resistor is 1.175 ARMS, or the sum of the two currents. Thus, each inductor
sees half the output current, so the voltages are equal.

Modeling a 99H inductor, unity coupled to a 1H inductor won't give you
10:1 voltage stepdown at the tap. Try it. It'll give you 1/(sqrt(99)+1) > 0.0913, or 10.949:1.

Yes, I agree

Assuming unity coupling, reasonable with a toroid, the "primary"
inductance will be L1+L2+2*sqrt(L1*L2) - Mutual inductance has to be
considered.

Your example circuit isn't a 50:50 divider. With your choice of K=0.995,
it's 0.577:1.

Perhaps my term of "inductive divider" is a misnomer. The inductance
measured from the slider to GND will vary according to the square root of
the turns, but the voltage is based on turns ratio.

Paul

===================================Version 4
SHEET 1 880 680
WIRE 464 128 96 128
WIRE 464 208 96 208
WIRE 96 240 96 208
FLAG 96 240 0
SYMBOL voltage 96 112 R0
WINDOW 123 0 0 Left 2
WINDOW 39 24 124 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 339.4 60 0 0 0 100)
SYMBOL res 448 112 R0
SYMATTR InstName R1
SYMATTR Value 100
TEXT 64 232 Left 2 !.tran 1

 

[John Fields]

On Mon, 27 May 2013 09:32:27 -0700, Fred Abse
<excretatauris@invalid.invalid> wrote:

On Sun, 26 May 2013 19:04:22 -0400, P E Schoen wrote:

er.

It's an inductive voltage divider in both cases, which are identical except
that the tap is soldered while the slider is connected by means of pressure
through a carbon brush.

It isn't an inductive divider. It's a *transformer*. The voltage ratio is
determined by the turns ratio. The inductance is determined by the
*square* of the number of turns.

Modeling an autotransformer isn't as simple as two coupled inductors in
the ratio; voltage^2.


Modeling a 99H inductor, unity coupled to a 1H inductor won't give you
10:1 voltage stepdown at the tap. Try it. It'll give you 1/(sqrt(99)+1) =
0.0913, or 10.949:1.

---
Actually, it seems like it'll give you 100:1, as one would expect.

Version 4
SHEET 1 880 680
WIRE 176 48 32 48
WIRE 176 80 176 48
WIRE 32 144 32 48
WIRE 176 208 176 160
WIRE 32 320 32 224
WIRE 176 320 176 288
WIRE 176 320 32 320
WIRE 32 368 32 320
FLAG 32 368 0
SYMBOL ind 160 64 R0
SYMATTR InstName L1
SYMATTR Value 99
SYMBOL ind 160 192 R0
SYMATTR InstName L2
SYMATTR Value 1
SYMBOL voltage 32 128 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 100 60)
TEXT 200 184 Left 2 ;K L1 L2 1
TEXT 40 344 Left 2 !.tran .1
---

Assuming unity coupling, reasonable with a toroid, the "primary"
inductance will be L1+L2+2*sqrt(L1*L2) - Mutual inductance has to be
considered.

Your example circuit isn't a 50:50 divider. With your choice of K=0.995,
it's 0.577:1.

---
Unless I made a misteak, I defaulted to unity coupling, which makes
the voltage at the junction of the two equal-valued inductances half
of the excitation voltage.

--
JF

 

[Jamie]

P E Schoen wrote:

"John Fields" wrote in message
news:r325q85gmd66dtcucr7qco0m0ocsonlh82@4ax.com...

Oops...


Why would you _raise_ its peak to 346.6V?


I found that, for a sine wave AC source with even 100 uOhms of internal
resistance, into a 100 ohm resistive load and 10 nH of inductance, the
measured voltage is 234.95 VRMS with 339.4 volts peak.

With 346.6V, I get 239.9 VRMS. I don't know why. Even 10 pH with 1 uOhm
ESR gives the same result. Try it. ASC file follows.
Distorted wave.

Jamie

 

[John Larkin]

On Mon, 27 May 2013 12:52:29 -0700, Fred Abse <excretatauris@invalid.invalid>
wrote:

On Mon, 27 May 2013 09:32:29 -0700, Fred Abse wrote:

I did something similar with a Staco small variac (actually a Tek 576
curve tracer spare part).

Forgot to mention; inductance of whole winding 10.3 henries, measured at
100Hz.

10H! Just about what I estimated from JF's magnetizing current.

Just to add some complexity, many DVMs aren't true RMS, and many have low AC
bandwidths, and the magnetizing current is probably not sinusoidal.


--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators

 

[Michael A. Terrell]

Fred Abse wrote:

On Mon, 27 May 2013 09:32:29 -0700, Fred Abse wrote:

I did something similar with a Staco small variac (actually a Tek 576
curve tracer spare part).

Forgot to mention; inductance of whole winding 10.3 henries, measured at
100Hz.

There's no need to Hertz yourself. ;-)

 

[Fred Abse]

On Mon, 27 May 2013 18:34:01 -0500, John Fields wrote:

Unless I made a misteak,

Yes, you did:

TEXT 200 184 Left 2 ;K L1 L2 1

Coupling is effectively commented out.

I guess you put it in as "text", instead of "Spice Directive"

Effectively, there's no coupling.

I have things set to put directives and comments in in separate colors,
precisely to spot such errors.

If you put the coupling in properly, (You may need to add some series
resistance,to avoid a singular matrix), you'll get what I said you should.

I defaulted to unity coupling, which makes
the voltage at the junction of the two equal-valued inductances half
of the excitation voltage.

You actually defaulted to no coupling

As Paul pointed out, the only ratio where an autotransformer behaves as an
inductive divider is 50%.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[Fred Abse]

On Mon, 27 May 2013 17:26:28 -0400, P E Schoen wrote:

An autotransformer is a special case, but the center tap is still equivalent
to two inductors of equal value connected in series.

You spotted my howler.

50:50 is the only ratio where the pure inductive divider model is true.

Extra marks if you can tell me where my math went wrong

Hint: factor of 2.

I did write it at 3AM. I guess I better go to bed earler ;-)

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[John Fields]

On Tue, 28 May 2013 01:30:01 -0700, Fred Abse
<excretatauris@invalid.invalid> wrote:

On Mon, 27 May 2013 18:34:01 -0500, John Fields wrote:

Unless I made a misteak,

Yes, you did:

TEXT 200 184 Left 2 ;K L1 L2 1

Coupling is effectively commented out.

I guess you put it in as "text", instead of "Spice Directive"

---
Nope; I entered it as a spice directive again, just now, ran it, and
the results are the same: 100Vin, 1V out at the junction.

I even tried it with a dot in front of the "K", and LTspice barfed.

It seems the semicolon in their circuit list doesn't mean the same
thing that it does on the schematic.
---

Effectively, there's no coupling.

I have things set to put directives and comments in in separate colors,
precisely to spot such errors.

---
I don't need to do that.
---

If you put the coupling in properly, (You may need to add some series
resistance,to avoid a singular matrix), you'll get what I said you should.

---
Since I didn't get a singular matrix error notification, I assume that
I put the coupling in properly.

Can you post an LTspice example of what you mean?

Please?
---

I defaulted to unity coupling, which makes
the voltage at the junction of the two equal-valued inductances half
of the excitation voltage.

You actually defaulted to no coupling

---
Nope; see above.
---

As Paul pointed out, the only ratio where an autotransformer behaves as an
inductive divider is 50%.

---
And, yet, the dial is linear.

--
JF

 

[Fred Abse]

On Tue, 28 May 2013 13:35:27 -0500, John Fields wrote:

Can you post an LTspice example of what you mean?

Please?

Try this: Note the syntax of the last line. The
exclamation point shows it to be a directive.
This example gives 9.118 volts peak at the tap.
Take the 1 milliohm rser out of the voltage source,
the matrix goes singular.

Version 4
SHEET 1 880 680
WIRE 176 48 32 48
WIRE 176 80 176 48
WIRE 32 144 32 48
WIRE 176 208 176 160
WIRE 32 320 32 224
WIRE 176 320 176 288
WIRE 176 320 32 320
WIRE 32 368 32 320
FLAG 32 368 0
SYMBOL ind2 160 64 R0
SYMATTR InstName L1
SYMATTR Value 99
SYMATTR Type ind
SYMBOL ind2 160 192 R0
SYMATTR InstName L2
SYMATTR Value 1
SYMATTR Type ind
SYMBOL voltage 32 128 R0
WINDOW 3 -141 53 Left 2
WINDOW 123 0 0 Left 2
WINDOW 39 -137 77 Left 2
SYMATTR Value SINE(0 100 60)
SYMATTR SpiceLine Rser=1m
SYMATTR InstName V1
TEXT 40 344 Left 2 !.tran 0 1 0 1u
TEXT 208 200 Left 2 !K L1 L2 1


--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[P E Schoen]

"Fred Abse" wrote in message
newsan.2013.05.28.19.58.35.835800@invalid.invalid...

Try this: Note the syntax of the last line. The
exclamation point shows it to be a directive.
This example gives 9.118 volts peak at the tap.
Take the 1 milliohm rser out of the voltage source,
the matrix goes singular.

I used the RMS values and I get a ratio of 0.0914, rather than 0.1 as might
be expected. 100 H should be 10 times the number of turns for 1 H, but
because the inductors are coupled it does not quite work that way. 99 H
corresponds to 9.95 turns while 1 H is 1 turn. Thus the turns ratio is
1/10.95 = 0.0913.

Changing the K factor to a comment shows a ratio of 0.01, proving that
non-coupled inductors function as pure reactance and their inductances may
be used for the voltage ratio, just like resistors.

A coupling constant of 0.09 yields a ratio of 0.084.

0.5 shows a ratio of 0.054

0.1 => 0.0196
What sort of transformer might have such a low coupling factor?
Could two isolated inductors be arranged mechanically to have an adjustable
coupling factor, and thus function as an isolated variac?

0 => (what do you think?)

Paul

 

[Fred Abse]

On Tue, 28 May 2013 13:35:27 -0500, John Fields wrote:

And, yet, the dial is linear.

See curve posted to a.b.s.e,

Approximately linear in the middle,

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[Fred Abse]

On Tue, 28 May 2013 17:25:36 -0400, P E Schoen wrote:

I used the RMS values and I get a ratio of 0.0914, rather than 0.1 as might
be expected. 100 H should be 10 times the number of turns for 1 H, but
because the inductors are coupled it does not quite work that way. 99 H
corresponds to 9.95 turns while 1 H is 1 turn. Thus the turns ratio is
1/10.95 = 0.0913.

More succinctly, the total inductance of 1H, tightly coupled to 99H is:

L1+L2+2*sqrt(L1*L2) - mutual inductance must be considered.
=99+1+2*sqrt(99*1)
=99+1+2*9.9499 = 100+19.8998 = 119.8988

Hence the inductance ratio is 1:119.8988
And the turns ratio is sqrt(1/119.8988) = 0.0913

Could two isolated inductors be arranged mechanically to have an adjustable
coupling factor, and thus function as an isolated variac?

0 => (what do you think?)

It *is* possible to make a double-wound variable transformer ("isolated
variac"). Grundig, In Germany used to make exactly that. Wound on a toroid
about 9" dia, giving a 5 amp rating at 0-270V, from a 220V supply.
Originally sold under the Hartmann & Braun trademark, and sold as a
workshop unit, cased, with current and voltage meters, they are still
highly sought after. Lighter and less bulky than the usual 1:1
double-wound-plus-conventional-variac that most makers offer.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)