How many steps in a variac

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

Because it's stacking the output voltage on top of the input supply.

A 120-to-150 volt, 150 VA transformer has to size the windings for 150 VA. You can do this...

||--------150v out || ||

120 in ----------|| || || || || || || || || com------------------------com

using a much smaller 120-to-30 volt transformer, rated 30 VA. You give up isolation.

John

So much theorizing, so few facts!

The cheap Calrad variac on my bench goes below 1 mV fully CCW, 146 volts max CW. It has a lot of mechanical stickiness, but usable resolution is around 50 mV. That's about 1 part in 3000, and I'm guessing it has maybe 200 turns.

The windings are flat in the brush area, and the brush is almost exactly 3 windings wide.

John

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

Cheers! Rich

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?

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 
created in your mind.

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.

I'm getting closer to a diagnosis of your personality disorder.

(BTW where are the videos of the home-made 4 layer board? Remember: I'll pay for everything!)

Narcissistic personality disorder sounds about right.

Narcissistic Personality Disorder is characterized by: (At least 5 to diagnose)

1. Grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements.)
2. Is preoccupied with fantasies of unlimited success, power, brilliance, beauty, or ideal love.
3. Believes that he or she is "special" and unique and can only be understood by or should associate with other special or high-status people.
4. Requires excessive admiration.

1. Has a sense of entitlement,i.e., unreasonable expectations of especially favorable treatment or automatic compliance with his or her expectations.

2. Is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own ends.

1. Lacks empathy

2. Is often envious of others or believes that others are envious of him or her.

1. Shows arrogant, haughty behaviors and attitudes.

Has someone just proved that a linear pot is nonlinear?

John

--
The voltages aren\'t proportional to the inductances, the impedances are.

The voltages are proportional to the turns ratio:

    Ep     Np
   ---- = ----
    Es     Ns

Oh yeah, flattened windings, I remember that. If they were all point contact, my above analysis would be correct. However, they are flattened, so the contact area is spread out and, as the brush moves off one and onto another, voltage change can be very smooth. The out- of-phase situation I spoke of would be advantageous when there is some gap between turns, improving the brush's interpolation.

Tim

The Narcissist as Know-it-all

For a more detailed view of pathological narcissism and the Narcissistic Personality Disorder (NPD) - click on these links:

Other Personality Disorders

Pathological Narcissism diagnosed with Other Mental Health Disorders

Cyber Narcissist