Transformer Load Question

[Guy Paddock]

I have a 110v to 25v step-down transformer that I'm trying to have drive an
electromagnetic coil. The original coil this transformer drove was rated for
4 amps and 25 volts. My digital multimeter tells me that the transformer
itself puts out around 27-28 volts, actually. Anyways, I am building the new
coil myself. I have already wound it and measured its impedance. According
to the original rating, the original electromagnetic coil should have had an
impedance of 6.25 ohms. My coil, at the moment, has an impedance of about 8
ohms. My coil will be applied differently than the original coil however; I
need mine to repell something, not just attract it, so I got a 4 amp, 50PIV
Full-Wave Bridge Rectifier at RadioShack and connected it between the
transformer and the coil. Currently, the transformer powers the coil as
required, with specific magnetic poles, and it is capable of repelling other
magnets, but it only functions for a short time (5-10 minutes). During this
time, the transformer gets very hot, and the coil gets a bit warm as well.

My question is this: shouldn't the fact that my coil has a higher impedance
mean that there should be LESS of a load on the transformer? If the first
coil had a 4 amp draw, and had an impedance of 6.25 ohms, then mine, with an
8 ohm impedance, should only have a draw of 3.125 amps (at rated voltage).
If this is so, why am I experiencing a heat issue?

TIA
--Guy

 

[Ian Bell]

I have a 110v to 25v step-down transformer that I'm trying to have drive
an electromagnetic coil. The original coil this transformer drove was
rated for 4 amps and 25 volts. My digital multimeter tells me that the
transformer itself puts out around 27-28 volts, actually.

If this is the unloaded output volts then this would be about right if the
transformer has the typical 10% regulation at full load - 10% of 28 is just
under 3 and 28 minus 3 equals 25.


Ian

 

[Henry Kolesnik]

You should be more specific becasue you use impedance and resistance and it
is confusing. If the original coil was AC and your homebrew is DC you're
talking about apples and oranges., still fruit but big difference.
hank wd5jfr

 

[Guy Paddock]

I do not mention resistance. Besides, impedance and resistance produce the
same effect, do they not? Are they not the same? Resistance slows the
current flowing through a given point in the circuit, and I was under the
impression that impedance measures the AMOUNT of resistance.

More so, does the fact that the circuit is converted from AC to DC impact it
much? The rectifier simply uses a diode bridge internally, forcing the
current to flow in one direction, so I don't see how forcing the electrons
to flow in one direction for each cycle impacts it. Is there much of a
voltage or amperage loss associated with it?

--Guy

 

[John Popelish]

I do not mention resistance. Besides, impedance and resistance produce the
same effect, do they not? Are they not the same? Resistance slows the
current flowing through a given point in the circuit, and I was under the
impression that impedance measures the AMOUNT of resistance.

More so, does the fact that the circuit is converted from AC to DC impact it
much? The rectifier simply uses a diode bridge internally, forcing the
current to flow in one direction, so I don't see how forcing the electrons
to flow in one direction for each cycle impacts it. Is there much of a
voltage or amperage loss associated with it?

Resistance is an effect that is independent of time, so it works the
same for DC and any frequency of AC. But coils have an inductive
impedance component in addition to the resistive component. And the
effect of inductance is certainly a function of time. Inductance adds
no impedance for DC, while it adds to the resistance (vectorially at
right angles) a reactance that is proportional to frequency. For
example, if your 8 ohm (resistive) coil had an inductance of 100
millihenries, it would have a total impedance at 60 Hz or square root
(100^2 + (2*pi*60*.1)^2) = 107 ohms. But if the inductance were 1
henry, then the total AC impedance would be square root(100^2 +
(2*pi*60*1)^2)= 390 ohms. Both coils would have only 100 ohms
impedance for DC. So DC coils generally have more turns of finer wire
to draw the same current as coils driven by AC. And the higher
resistance DC coil will produce more heat with that same current, due
to this higher resistance. Holding back current with inductance does
not produce heat, but doing so with resistance does.

 

[Guy Paddock]

So I would need to remove some turns from my coil in order to create less
heat?

--Guy

 

[Guy Paddock]

Also, how do I know how many henries my coil is?

--Guy

 

[Guy Paddock]

Perhaps my problem is in the way my coil is wound? Is there a place I could
get a coil that would meet my specs?

Or, perhaps there is some other way to keep the transformer from
overheating?

--Guy

 

[John Popelish]

So I would need to remove some turns from my coil in order to create less
heat?

No. You need a coil with higher resistance. The current the coil
will pass is V/R. The power the coil will produce is V^2/R or I^2*R.
More turns of finer wire will pass less current. The field strength
is proportional to the current and to the number of turns.

So if you use wire with half the cross sectional area, about twice the
turns will fit in the same space, while the resistance will be about 4
times higher (a factor of 2 increase for the smaller area, and another
factor of 2 for twice the number of turns, and so the length). So the
field strength will be about half (1/4 the current but twice the
turns). The heat, in both the transformer and the coil will both
drop. The transformer winding losses will drop by a factor of 16
(because the windings still have the same resistance, but the current
will be about 1/4, by I^2*R. But the coil heat will drop to only 1/4,
because the current is 1/4 but the resistance is times 4.

 

[John Popelish]

Also, how do I know how many henries my coil is?

If you can measure its resistance and also its current draw with a
measured applied voltage, you can calculate the inductance from the
impedance formula.

I=V/(sqrt(R^2 + (2*pi*F*L)^2), where F is the AC frequency, and L is
the inductance in henries.