"DaveM" wrote in message
. ..
wrote in message
ups.com...
I have an AC circuit problem I need help with. At my mother's
boyfriend's house he has a furnace with air conditioner/heat pump
attached. When he bought the house natural gas prices were lower then
electricity and he had the heat pump disconnected but there still is a
flow shutoff valve attached to the air conditioner/heat pump that is
operated by a solenoid. The solenoid is normally energized but the
voltage suppied to it exceeds the normal recommended operating voltage
and gets very hot. The unit is an ALCO Controls, Coil Type DMG, 24
volts, 7 watts, 50 - 60 Hz. The computer in the furnace is supplying 26
to 28 volts.

What I would like to do is attach a power resistor to reduce the
voltage. The assumption I would like to make is the highest voltage to
the solenoid would be 30 volts. The solenoid will activate at as low of
a voltage as 17 volts. I would like to add a resistor that will take
the 30 volts and reduce the voltage to 24 volts.

How do I go about computing the value for the resistor? Is there a
better way of going about this than using a resistor?

FYI: The reason the voltage is so high is the line voltage is 133 to
138 volts. It is then stepped down by a transformer. The secondary
output of the transformer is the passed through a relay, control by a
computer, to the solenoid.

Thanks In Adavnce,
Derek


You find the value of the resistor by first calculating the current drawn
by the
solenoid.
I = P/V
I = 7W/24V = .3A

You can also calculate the impedance of the solenoid's coil.
Z = V/I
Z = 24V/.3A = 80 ohms

Now, calculate the voltage that the resistor needs to drop.
V = Vsupply - Vsolenoid
V = 28 - 24 = 4V

Now, using these two values, calculate the resistor value.
R = V/I
R = 4V/.3A = 13.33 Ohms (a close standard value would be 12 ohms)

Calculate the power that the resistor will dissipate.
P = VI
P = 4V * .3A = 1.2 Watts
Always add a safety factor of at least 2 on the power rating, giving us a
resistor power rating of at least 2.4 Watts. A standard wattage rating
for
power resistors is 3W or 5W.

Putting all this together results in a 12 Ohm/5 Watt resistor.

You can deviate from these values a bit if you need... i.e., rework the
figures
to further reduce the solenoid voltage.


If this were DC, that would be the right way to calculate this, however, he
is dealing with AC. The problem here is that the impedance of the coil is
largely resistive *and* inductive. To make things worse, the inductance will
change, likely a lot when the plunger is in vs. out. Since when the plunger
is out, the magnetic path would have a much larger air gap, and thus making
the resistance of the coil more dominant, and when the plunger is in, the
air path for the magnetic field is much less, and therefore it's inductance
is much higher. The problem comes when the plunger needs to pull in, the
impedance of the coil is at a minimum, causing a higher dissipation in the
resistor, and a much weaker then expected magnetic field in the coil. When
the plunger is in, the impedance increases, dropping the current, and
increasing the voltage. Simply said, adding a series resistor can cause the
plunger to not pull in, and/or keep the power dissipation in the coil high.
I quickly figured out this a long time ago when I wired two contactors with
120V coils in series on a 240V supply. Only 1 contactor would reliably pull
in!