# How many steps in a variac

Discussion in 'Electronic Design' started by Gene, Feb 10, 2009.

1. ### GeneGuest

For a lab bench variac that has a dial connected to a brush that
slides across the winding, does anyone know how many winding/taps
there typically are? Or another way of putting it, what is the typical
voltage step between windings? Say there are 100 windings, meaning
about 1.2 V per winding (ignoring for the moment that it might have
taps for output voltages > input voltage). Would it really go down to
1.2 volts (very first tap/winding)? I'm interested in voltage control
at low voltages.

Gene

2. ### Jon SlaughterGuest

Thats simply not true(for the most part). It will take ignore any windings
after the tap. You are essentially shorting out the windings with the brush
so how could any of them after the first matter? (they do matter in that
there is non-zero resistance but 99% of the current will go through the
first winding.

It imagine if you think about it for a second you'll realize your error.

|
|
-----
(think of it as a variable resistor if you like)

The bush contacts several windings. Any windings after the closes one
contacted will be ignored(again, for the most part). Simply the past of
least resistance in action. (the windings are not in parallel.

similarly you have something like

|
|
+----+----+
| | |

The current will ignore R3,R4, and R5 assumine the wire thickness is much
lower than the resistor/winding resistance.

i.e., it is identical to

|
|
+----+
|

I imagine it might be possible for the windings to share the current do some
degree because of the contact resistance(specially if lots of current is
going through it causing it to heat up locally). Obviously the contact(in
my diagram) forms a voltage divider at the points.

If I took lessons from the assholes on here I'd bitch you out for being
wrong... luckily I'm not that much of an asshole and my life doesn't totally
suck. After all, I could be wrong(something people like Phil never think
of... even if they did it they would assume they are correct).

In any case I think if the OP wants to get better resolution he'll need to
use some other method such as another transformer or add some load.

3. ### Phil AllisonGuest

"Gene" <>

** Connect an AC volt meter to the output and find out for yourself -

BTW

Only an utter moron use a variac to derive a low voltage AC supply.

Cost it is stupidly dangerous.

...... Phil

5. ### BenjGuest

I sure as hell hope YOU aren't giving anyone any lessons! First off,
you might take a close enough look at a variac to notice that it's a
TRANSFORMER and not a bunch of resistors. A shorted turn on a
transformer has a very different effect from shorting out a couple of
resistors on a voltage divider. The only reason you don't get smoke,
heat and melted parts because of the bridging brush is because of the
resistance of the carbon and the quite low voltage generated by any
given single turn. The BRUSH is what forms the "voltage divider"
because it is the resistive material. And the device give you
resolution rather than steps because of the GEOMETRY of the brush to
copper wires. Think about it asshole.
A low voltage transformer IS the way to get low voltage output from a
variac. Not only are you wasting a lot of resolution by only running
the variac on low settings, but also they are [usually] wound with the
same gauge wire over the length of the turns. Thus if you are trying
to get a ratio between the line voltage and the [low voltage] output
you will be using the same gauge wire for both windings. This make the
wire gauge for the low voltage section too light and output current is
thus limited by that deficiency compared to your usual two winding low
voltage transformer where the low voltage secondary is wound with very
heavy wire.

Plus, since a Variac is an autotransformer, there is no isolation from
the power line which can be a safety factor in some situations. A
secondary low voltage transformer solves all these problems as others
have suggested.

6. ### GreegorGuest

As a sideline issue, the comment about the wiper arm
shorting one winding intrigued me.

That one winding would produce low voltage and
high current, does that have a history of causing failure?

If you've got a variac that outputs the range
0-line voltage and you output to the primary
on a 12v transformer you get very fine
control of the range 0-12 v ac.

With a 6v transformer instead you get very
fine control for the range 0-6 v ac.

How is this so dangerous, aside from cost?

7. ### Phil AllisonGuest

"Greegor"

As a sideline issue, the comment about the wiper arm
shorting one winding intrigued me.

** There is no hard short.

See the other replies.

That one winding would produce low voltage and
high current,

** There is no hard short.

See the other replies.

does that have a history of causing failure?

** There is no hard short.

See the other replies.

** Good safety practice is to *never connect* a load to a variac that CANNOT
handle the FULL output voltage - connecting a load that can only handle a
tiny fraction of that voltage ( like 1%) is plain NUTS !!!

PLUS a variac provides NO isolation from the AC supply - so a missing or
reverse connected neutral conductor means that both connections to the
variac are at a dangerous ( ie full line ) voltage.

If you've got a variac that outputs the range
0-line voltage and you output to the primary
on a 12v transformer.....

** That was never the OP's Q nor relevant to my comment.

Please learn to read before posting here again and wasting people's time.

....... Phil

8. ### Jon SlaughterGuest

Asshole uh? Why can't I say people are wrong? Why can everyone else? Why is
it ok for Phil, Nobody, Charles, and others like yourself able to reply to a
post without reading it and say it's wrong and bitch and call people
assholes(which you seem to be good at) without reading the post correctly?
Why is there a double standard that allows morons to get away with such
things?

Just like what you have done!! I did not say anything incorrect in my post
yet you try to blast it out of the sky. Also, in your own "blasting" you
make numerous logical mistakes. One is that you assume you can't replace
resistance by impedence! This is basic electronics but your allowed to get
it wrong?

Cannot a variac be modeled by

Contact
+------+--....--+-------+
| | | |
| | | |
R1 R2 ... Rn ... R(n+1)
| | | |
V-->---Zi---+--Z1--+--... --+--Zn---+---Zf

?? Surely if you thought about it more you would see it is correct and that
what I have said in my original post is completely correct. (I never said

Note also that I put this little caveat:

"The current will ignore R3,R4, and R5 assumine the wire thickness is much
lower than the resistor/winding resistance." (the wire thickness is the
contact resistance!!! But I guess all those brain cells you have couldn't
figure that out)

Which you completely ignore(as expected).

You people are awefully arrogant. It's quite entertaining. I hope I didn't
get your panties in a bunch?

Trust me... there's plenty more where this came from! I'll keep it up as
long as you people keep up your shit. As long as you guys can read posts
half-ass and respond by calling people assholes and make worse logical
mistakes than you assume the other person made I'll call you out on it. (I
was hoping phil would be the first to respond but I guess he's still trying
type up his own paper)

Test: FIND ONE logical mistake I made in the post... not spelling or grammer
or typo's. If you can do that then I'll submit that you are correct. I've
electronics.

Now, as usual we are both right(and that almost always is the case when
someone is calling someone wrong). Your fallacy is you simply fail to read
the entire post and when you misunderstand something you have to show your
ego. (again, find anything I said as being wrong)

Note I say

"I imagine it might be possible for the windings to share the current do
some
degree because of the contact resistance"

Which in all your glory you have completely missed and it says exactly what
you have said. So you have said I am wrong yet I said what you have just

The only thing I'm wrong about is expecting more from a bunch of second
class humans.

Of course I'll expect to be hearing from you soon. Completely ignore
everything I have said and call me wrong and an asshole some more. I guess
that is what makes your pathetic life interesting?

9. ### Jon SlaughterGuest

Why do people make aggressive, authoritative, and totally idiotic
Now that's something I would really like to know!!! I would give my let nut

10. ### Phil AllisonGuest

"Palinurus"

** Is that set the dial to a "fraction of a volt" OR set the AC voltage
output within such a fraction ??

** So the latter ??

Pointless - as the AC supply does not remain so steady for more than a few
seconds.

...... Phil

11. ### Jon SlaughterGuest

huh? A turn generates power and dissipates power? As far as I know a winding
doesn't generate any power.

The electric field is converted into a magnetic field which is then
converted back into an electric field at a different intensity.

http://en.wikipedia.org/wiki/Autotransformer

So you disagree with wiki? Maybe you should rewrite the page?

Look at the first figure... do you see that big squiggly line? Do you know
that is called an inductor? Do you see the taps?

Do you realize you can draw it as a series of inductors where connected by
wires andt he taps taken at the wires?

i.e.

----/\/\/\---tap1---/\/\/\/\---tap2---\/\/\/\/\---

??? Surely you do get that much?

Do you realiez that one can conceptualize it as resistors? Maybe this is the
leap of faith you have? Luckily good basic run of the mill impedence comes
to the rescue:

http://en.wikipedia.org/wiki/Electrical_impedance

OMG!!!!

"just as impedance extends Ohm's law to cover AC circuits, other results
from DC circuit analysis such as voltage division, current division,
Thevenin's theorem, and Norton's theorem, can also be extended to AC
circuits by replacing resistance with impedance."

Do you understand that? The autotranformer is just one big voltage divider.
abstractly at least... it does this a bit different than a resistive network
but it works out to be the same. After all, we were talking about the
current instead of the voltage reduction. For resistors the voltage is
reduced by the drop in potential across a resistive element. For the
inductors the voltage is converted into a magnetic field but then converted
back into an electric field. Different principles at work but same abstract
analysis. (This is because the current through the windings is exactly
that... regardless of what the inductor is doing to limit the current.

In fact since it seems to escape you guys:

V---L1--A--L2---gnd

What is the voltage at point A? Is it L1*L2/(L1+L2)?? Assuming the inductors
have the same construction so there inductance per turn is the same then it
works!!! Again, wiki proves this!

A = V*L1*L2/(L1 + L2)

If L1 = L2 then we have

A = V/2*L!!!!

Which is exactly which one expects!

What is that? Voltage divider law? Same principle holds for resistors too,
and capacitors? right? Get it now? (Note L really is X_L = jwL but in this
case the magnitudes are identical except for the factor of w!!!!!! Or just
think of the symbol L as a shorthand for X_L.. nifty huh?)

I know it's hard but think!!! Or you can just jump on the band wagon... I
hear Phil "FUCKWIT" Allison is taking applications. (You just have to go
around calling people Fuckwits and saying they are wrong and prove it by
changing the hypothesis. Benji is doing a good job of that and I think
larkin is taking notes.)

Waiting on Benji to reply and call me an asshole again. This time he
probably won't even read a tenth of the post.

12. ### J.A. LegrisGuest

Slaughter murders yet another analysis!

Here's a puzzle for you:

Apply your inductive voltage-divider "theory" to an autotransformer
set anywhere else except at the halfway point - say, at the 1/3 point.
Then the inductance of the short side is 1/4 of the long side (recall
that L is proportional to the number of turns SQUARED). Therefore the
output voltage should be 1/5 not 1/3. Where did you go wrong?

13. ### GreegorGuest

On Feb 10, 12:37 am, John Larkin wrote
Thanks.

14. ### Rich GriseGuest

When was the last time you set a variac on your bench and actually
MEASURED the output?

Thanks,
Rich

15. ### Tim WilliamsGuest

So anyway...

The number of steps is either the number of turns, N or double, 2N
(give or take the width of the brush and how the ends are handled).
Which one depends on the width of the brush; if it spans an integral
number of turns, it will move off one turn just as it moves onto the
next, so that the number of contacts is constant and you get N steps.
If it spans a half-integral number of turns, however, it will move off
one, then move onto the next, and so on, so that the number of
contacts alternates between, say, 2 and 3, then you get 2N, double the
resolution.

Width of the brush is a continuous variable, so it's worth observing
what happens at non-half-integral widths: in this case, the phase
between moving-onto-a-wire and moving-off-a-wire changes as something
like (Width MOD Spacing) / Spacing.

Tim

17. ### Rich GriseGuest

Maybe he means a do-over, like a let serve. ;-)

Cheers!
Rich

18. ### Jon SlaughterGuest

Your right... lets figure it out then, shall we(as if your going to
following along)?

Since the point was to find a law relating the voltages we have

V1 = L1*I1' + M*I2'
V2 = L2*I2' + M*I1'

Now I1 = I2 + I3 where I3 is the current into the tap: Z2 = 2nd inductor
impedence, Z3 = load impedence. The current divder law gives

I2 = Z3/(Z2 + Z3)*I1 = Z*I1 (Z is unitless)
I2' = Z*I1' + Z'*I1

or

V1 = (L1 + M*Z)*I1' + M*Z'*I1
V2 = (L2*Z + M)*I1' + L2*Z'*I1

Canceling the I1's gives

L2*V1 - M*V2 = [L2*(L1 + M*Z) - M*(L2 + M)]*I1'

= [L1*L2 + M*Z*L2 - M*L2 - M^2]*I1'

With me so far?

We know that M = k*sqrt(L1*L2)

so

L2*V1 - M*V2 = [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'

Now substituting V1 = V - V2 we have

L2*V - L2*V2 - M*V2 = L2*V - (L2 + M)*V2 = ...

Solving for (L2 + M)*V2 gives

(L2 + M)*V2 = L2*V - [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'

Solving for V2 gives

V2 = L2/(L2 + M)*V - [(1 - k^2)*L1*L2 + (Z + 1)*M*L2)]*I1'/(L2 + M)

V2 = Tau*V - Gamma*I1'

Note if Gamma is 0 or practically 0 then We have a voltage divider effect.
I.e., V2 ~= Tau*V. Tau depending on L1 and L2 and there ratio sets tau(it is
not linear necessarily but we'll investigate that later).

The main thing is to determine Gamma:

There are 4 cases

(Ideal transformer => k = 1, High load => Z = 0)

For an ideal transformer k = 1 so we have

Gamma = M*L2/(L2 + M)

M = sqrt(L1*L2)

or

Gamma = 1/(1/L2 + 1/sqrt(L1*L2))

Note if L2 or L1 is small then Gamma ~= 0. Only for large inductance do we
get a gamma that is large. (note this doesn't necessarily mean the voltage
divider effect out of play as we will see shortly)

Even for 1H inductances we get Gamma = 1/2

But then

V2 = Tau*V - I1/2

But unless Tau*V is compariable to the current being drawn it doesn't have
much of an effect. Take V = 110V and tau = 0.25 and drawing 3A. Which is
approximately 5% off the ideal case.

For your run of the mill transformers obviously L1 and L2 are much much
smaller and obviously the effect will be insiginificant. (although for very
low voltages and high current's the effect is much stronger)

(Ideal transformer => k = 1, Low load => Z = 1)

Obviously with Z = 1 we at most double the value of I1.

For the Ideal transformer and practical configurations, while theoretically
you are correct, for all practical purposes the mutual inductance has no
effect.

(Completely Non-ideal transformer => k = 0, High load => Z = 0)

In this case we have and additional factor of L1*L2 to boost I1. But again,
for all practical cases it is very small. (M = 0 in this case and the other
term gets knocked out)

(Completely Non-ideal transformer => k = 0, Low load => Z = 1)

Again, this just doubles the effect which is still quite small.

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

Conclusion

While you are theoretically correct about the mutual inductance, it just
doesn't have any effect on the voltage dividing effect in practical terms(at
least in all but the most extreme cases).

So we are left with simply only with V2 = Tau*V

Tau is the voltage dividing effect. It is set by choosing the ratio of L1
and L2.

Tau = L2/(L2 + M)

If L2 = 0 then Tau = 0 and this is exactly what we expect

If M = 0 ==> Either no mutual inductance or L1 = 0 then Tau = 1 and this is
exactly what we expect.

The curve is nonlinear

BUT NOTE

What does L2/(L2 + M) look like??

How bout the resistive divider law!!!!!!!!!

Note this is with a load. Without a load the analysis is much easier... This
is exactly what I said in the beginning.

Any more bright ideas or will you finally give in?

19. ### Jon SlaughterGuest

Slaughter murders yet another analysis!

Here's a puzzle for you:

Apply your inductive voltage-divider "theory" to an autotransformer
set anywhere else except at the halfway point - say, at the 1/3 point.
Then the inductance of the short side is 1/4 of the long side (recall
that L is proportional to the number of turns SQUARED). Therefore the
output voltage should be 1/5 not 1/3. Where did you go wrong?

--

Actually you went wrong wtih your red herring.

Where exactly did I talk about the contactor position? This is something you

Just as a voltage divider we get

V2 = R1/(R1 + R2)*V

It is not linear.

You assume way to much to make yourself look intelligent but in the eyes of
god you are just another imbecile.

20. ### JosephKKGuest

That is the most unbelievably way wrong answer i have seen in a long
time. Are you trying to out stupid AlwaysWrong?