Coil EMF volt vs current

Discussion in 'Electronic Design' started by Mike Lennis, May 3, 2007.

1. Mike LennisGuest

Given the same solenoid, and assuming a fixed available wattage, what
proportion of 1) voltage vs. 2) current produces the highest possible
field strength?

How is this determined?

Thank you,

Mike Lennis

2. Joe G \(Home\)Guest

Sound like a homework question!

3. Guest

Field strength is entirely determined by the current through the coil.
The voltage across the coil reflects both the rate of change of
current, which generates a voltage across the inductance of the coil,
and the instanteous current through the coil which generates a voltage
drop across the resistance of the coil.

At very high frequencies, one has to take into account the current
flowing throuhg the capacitance between the turns of the coil, which
by-passes both the resistance and the inductance of the coil, and has
no effect on the filed strength.

This is a rather ill-posed question, which suggests that it isn't
homework.

4. Greg NeillGuest

What determines the steady-state current in the
solenoid? (What are the electrical properties of
the solenoid?)

5. Phil AllisonGuest

"Mike Lennis

** The highest current level that is within the available wattage is it.

So, with coil resistance R and wattage W

W is given by I squared R

So I = sq rt ( W/R )

And V = I.R

........ Phil

6. MooseFETGuest

This is not really correct. The current that passes through a
capacitance between the windings still makes a magnetic field. It is
just a lot smaller because it travels a shorter path.

The rules are straight forward. When a current flows, it makes a
magnetic field around its path. The effects of multiple current flows

7. John LarkinGuest

The coil has a resistance, and that resistance fixes the relationship
between the current and the voltage. So you can't vary them
independently, so your question is sort of meaningless.

As the voltage goes up, the current goes up, and there is a single,
unique voltage:current that dissipates the available watts.

John

8. Phil AllisonGuest

"John Larkin"
Mike Lennis

** Looks like a cute, trap question to bugger gullible students.

** As given by the most basic equations for DC power and ohms law.

........ Phil

9. Mike LennisGuest

Yes, very sorry. The fixed coil resistance was an error on my part.

What I am driving at is this. Given the same signal source, consumed
wattage can be increased by 1) lowering the coil resistance OR 2)
applying higher voltage.

Which of the two is preferrable for maximum field strength?

For example, in the extreme case, one could have microvolts (low
resistance coil) and tons of current, or heaps of voltage (high
resistance coil) and negligible current. Yet the wattage is the same.

What is the trade-off point for these two variables, ie. practical
solution?

Thanks again.

Mike Lennis

Thanks again.

Mike Lennis

10. John LarkinGuest

Ampere-turns make magnetic field. For a given available wire winding
volume, the external surface area determines the cooling, so the watts
you can dump into a given coil geometry, for a given temperature rise,
is fixed.

So, imagine you replace a N=100 turn coil with a 200 turn coil. The
cross-section area of the wire must drop by 2:1 to fit into the same
space. Resistance per foot of wire doubles, and you have twice the
wire length, so net resistance goes up by 4:1. Since power is I^2 * R,
we need to drop the current in half to get back to the same power
dissipated. So N*I is unchanged!

So, it doesn't matter. In real life, very fine wire will lose more
winding area to the insulation, so you may start to lose pull with
lots of very fine wire.

Square wire helps, both to pack more copper into the available space
and because the heat flow will be better. Anything to improve cooling
will allow you to dump more watts into the coil, too. High temperature
insulation will allow more watts, too.

Then there's the old Austin-Healey overdrive solenoid trick: apply
lots of power to pull in the solenoid, and then once it's seated, back
off to some lower holding current.

John

11. MooseFETGuest

Make the coil resistance zero by making it super conducting. Infinite
voiltages are much harder to do.

12. Rich GriseGuest

Unless, of course, you've got some superconductor wire. ;-)

Cheers!
Rich

13. BFoelschGuest

Y'all miss the point.

A magnetic field is potential energy. Once the field is created it
exists, that is why a permanent magnet works. It takes no energy to
maintain the field. Hence, "consuming wattage" in this case means
nothing.

Assuming we are talking about DC fields here, all that matters from
the solenoid point of view is ampere-turns. For a fixed voltage, given
a wire of uniform resistance, the MMF is independent of the number of
turns, but the consumed power is inversely proportional to the number
of turns. Hence, (for a fixed field voltage)a solenoid with an
infinite number of turns will develop the same field as a solenoid
with one turn, but with infinitesimally low power consumption. It
will, however, also have infinite inductance, so the field will take
an infinite amount of time to build up.

14. John LarkinGuest

If the winding area is constant, increasing the number of turns by a
factor of N increases the resistance by N^2. Since power is E^2/R,
then for constant E, power falls as N^2, not N. Field strength goes as
1/N.
That assumes you don't change the wire size, so it will also be
infinitely large.

John

15. BFoelschGuest

True enough. I was not concerned with constant area. I presupposed a fixed
wire size, too.
Yup.

I was simply trying to show that steady-state power consumption is no
predictor of field strength. It is all wasted, once the field is created.

16. jasenGuest

no, power consumption is about the same (unless the solenoid is larger)
It'll also have infinite resistance, beacause the wire's thinness and length.

More turns won't make solenoids more efficient, more copper can but that
has limits.

Bye.
Jasen

17. MooseFETGuest

Ignoring build up:

If you hold the pounds of copper constant, changing the number of
turns doesn't change the power. The resistance of the windings runs
as the square of the number of turns and the required current
decreases proportionally so I^2*R remains the same. It is better to
think in terms of how much conducting material you can afford.

The outter most turn can be quite a bit further from the coil center
than the inner most. Adding turns to the outside of a coil increases
resistance more rapidly than N because they are bigger turns. For a
very thick coil, the resistance grows as N^2. The field from each of
the new turns is decreasing by a 1/N curve.

Holding current constant, we can find the rate that power and the
field increases with N.

<=> is proportional to.

P = I^2 * R <=> N^2

Field <=> integral(1/N) <=> ln(N)

Therefor for a very thick coil:

Field/P <=> ln(N)/ N^2

For some large amount of wire, adding more turns stops helping. This
is another good reason to go with super conducting wire.

18. John LarkinGuest

Good points.
Big electromagnets used to be water cooled and now mostly
superconductive. The NMR magnets I work around have 20T fields, are
stable to a ppm field per day, and have zero power consumption.

John

19. Phil HobbsGuest

Scaling arguments--fun.
This assumes that the core is a small perturbation on the inductance of
one turn at the outer edge, which is true for open cores but not for
closed ones. For closed-loop cores, the field is determined by the
total current threading the loop. Position doesn't matter, at least not
until order (iirc) 1/mu**2. If "solenoid" here has the physicsy meaning
of "a coil wound like a Miniductor", this is true, but if it means "a
push-pull magnetic actuator whose magnetic circuit is nearly closed at
all times" then it isn't.

Cheers,

Phil Hobbs

20. MooseFETGuest

To add more turns, once the core is full you need a bigger core. You
get into the same bind with a core once you start going to larger and
larger cores. All the turns get longer when you go to a bigger core.